Mind Your Units

With both sadness and joy I must report that the Sky Dragons {1} invaded Roy’s.

Joy, because I’m having fun. As an editor friend observed (pers. corr.): this place looks like the perfect Thunderdome for you. She’s not wrong. To follow the comments I reinstalled an RSS reader, like the good ol’ times at Judy’s.

Sadness, because the intensity of denial is too damn high! Higher than Tony’s, where guests can sound like voices of reason nowadays. Climateball veterans might recognize MikeF, who took puppet names circa August 2019. A circus of Dark Triad clowns enable him. In short, Roy forfeited. He can’t even be contacted.

The following gem cited by a Sky Dragon made me look into energy balance models:

Out of the mathematical convenience of not having to treat the system in real-time, and with the real power of sunshine, climate scientists average the real-time power of sunshine over the entire surface of the Earth at once, so that they can get rid of day and night, and also so that they can treat the Earth as flat, which makes things easier for them in the math. By spreading the power of sunshine over the entire Earth at once, so that they don’t have to worry about the difference between day and night, the mathematical number required to do this works out to a division of the real incoming power P by the number 4. It is a result of a geometric math problem of transforming a sphere into a flat plane, which is how climate scientists make the simplifications of the real system to something which is not real but is a convenient approximation.

Source: Joe’s

Joe’s story stinks: “real-time” and “at once” stretch incredulity. Zero-dimensional models express with a single equation the balance between the energy in and out of the Earth {3}:

[EMB] (Disc) x (Sun) x (1 – Albedo) = (Area) x (Emissivity) x (SB) x (Temp + Conv)4

Disc is the Earth’s shadow, Sun the Solar constant, Area the Earth’s area, SB the Stefan-Boltzmann constant, Temp the Earth’s temperature, and Conv the conversion constant from Kelvin to Celsius. The notation is adapted from (Kleeman); (ACS), (Lindsay) and (UCAR) provide good intros; (Kiehl & Trenberth) remains the Climateball battleground.

The only parameters we need to discuss here are Disc and Area. The reason why we “divide by 4” for the Sun’s input is simple. The Earth receives light over its shadow {2}:

But the outgoing energy leaves from the whole Earth area. As AT puts it (pers. comm.), the energy we receive per unit time depends on the cross-sectional area (pi r^2), but the energy radiated per unit time depends on the surface area of the sphere (4 pi r^2).

Note how AT lays out the problem in time and space. We’re looking for the energy flux, a specific rate of energy transfered through a surface. Climate scientists speak of Watt per square meter. In SI units, that’d be W⋅m−2, or (equivalently) in J⋅m−2⋅s−1.

In other words, the energy balance model states an equality between two quantities, the left one in Watt per square meter, the right one in Celcius. How can’t it work in real time? A Watt is a rate of work per second, for Newton sake!

Joe’s trouble may be geometric: he wants the light to fall on a hemisphere. The disc doesn’t seem not real enough for him {4}. This is corroborated by a post in which he, with the tip of Archimedes’ hat, increases solar input to 15.5C. His diagrams display 30C, in contrast to the usual -18C {5}. How does he pull his trick? Probably misspecification {6}. Joe’s model for the whole Earth remains elusive. It needs to balance the same way as the other ones or it’s humbug. Meanwhile, his divide-by-two trick fares no better than the divide-by-four one {7}.

A more piecemeal way to account for the flux over a hemisphere would be to correct each part of the surface according to the angle from which the Sun hits it by applying Lambert’s Law. Using a disc saves that integration since the whole of it faces the Sun directly. As AT calculates (pers. comm.), this correction gives the same flux as when taking a disc. So I don’t buy Joe’s appeal to the naturalness of a hemisphere over a disc. These toy models are no GCMs!

It’s as if Joe wanted to have the zero-dimensional cake and eat it in one dimension {8}. Instead of modeling the Earth like a single point, climate scientists can split the Earth into regions, each one with e.g. different temperatures or albedos. See (Huber) for such model.

The argument I offered appeals to basic algebraic and geometric intuition and is supported by my references. I also contacted AT’s, who’s solely responsible for any mistake I made in the post. Kidding. Comments and corrections welcome.

§ Notes

{1} By Sky Dragon I am referring to someone who denies the Tyndall Gas Effect.

{2} The expression “Earth shadow” can refer to another astronomical phenomenon than the geometrical fact outlined in the video, i.e. the cross-section of the Earth.

{3} No idea why the new WP editor fails to grok LaTeX. Nevermind. Let’s do it the programmer’s way. Perhaps one day scientists will borrow it: it solves many notation problems.

{4} See AT’s proof that illustrates how the light already falls on the equivalent of a hemisphere.

{5} How Joe found that 365.5 x 0.5 = 303 is left as an exercise to readers. So is the question: what temperature is on the unlit hemisphere that Joe does not show? Solutions in comments.

{6} Also called Mathurbation. A friend swayed me to shy away from denigrating masturbation.

{7} This paragraph has been heavily edited. Thanks to Rajinder for the feedback. I owe him a copy of The Spanish Prisoner.

{8} Joe is not even wrong about his “flat Earth” jab: the model he attacks has zero dimension! I guess climate scientists call their models so because it takes the Earth as a point. They still measure areas, which means their usage of “zero dimension” differs from geometry. In any event, a sphere is not flat.

[Last revision: 2021-05-05]

§ References on Energy Balance Models

ACS; Energy from the Sun

Huber; 1997-07; One-Dimensional Energy Balance Model

Lindsey; 2009-01; Climate and Earth’s Energy Budget

Kiehl & Trenberth; 1997-02; Earth’s Annual Global Mean Energy Budget

Kleeman; Zero-Dimensional Energy Balance Model

UCAR; 2021; Calculating Planetary Energy Balance & Temperature

About Willard

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519 Responses to Mind Your Units

  1. TrueSceptic says:

    So, someone with an MSc in Astrophysics doesn’t understand simple geometry? A Poe, surely?

  2. Willard says:

    > A Poe, surely?

    I doubt it:

    @ 1) How does the greenhouse effect work in general and why is the additional CO2 enhancing it to exactly the degree the science say (how is this calculated)?

    Well we have the flat Earth diagrams which demonstrate the idea. Although to pin them on how it works is tricky: they start with backradiation, but then when you point out that heat cannot recycle, they switch to “slowed cooling/reduced emission” and what have you, etc. There is no greenhouse effect at all, other than in real greenhouses which stop convection – the open atmosphere can’t do that.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71732

    Note the date.

  3. TrueSceptic says:

    I’m sorry, but I can *not* see how anyone with this level of education can be so obtuse. It truly beggars belief. (I’m also convinced that “Flat Earthers” are just having fun, seeing how far they can stretch credulity.)

  4. Willard says:

    Fair. Perhaps you’d prefer:

    [W]e categorically assert that the result of equation {7} (and {20}) cannot be interpreted so as to be physically equivalent in temperature to the actual average solar heating input. What the Stefan-Boltzmann analysis states is, specifically, the instantaneous average effective spherical radiative output of the system, with the system-ensemble as defined above. It does not state anything further than this. There exists no logical or physical justification for reversing the interpretation of the result of equation {7}, and arbitrarily equating the effective instantaneous [9] spherical output radiative flux with the instantaneous average radiative heating input over the same system geometry. The obvious physical justification for this reality is that, in actuality, only half of the Earth’s surface physically accumulates radiative heating energy from the Sun in any moment. This is the actual and physically real average boundary condition that exists. The true, and physically accurate average of the system, is that half of the surface of the Earth absorbs twice as much energy as the entire surface of the Earth radiates. The incoming solar radiation is not equal, in energy flux density, and thus temperature, to the outgoing terrestrial radiation. Claiming otherwise forgets the reason for the difference in illumination between day and night, and is completely irrational within the frame of physics. Dividing the solar flux by a factor of four and thus spreading it instantaneously over the entire surface of the Earth as an input flux amounts to the denial of the existence of day-time and night-time, and violates the application on the Stefan-Boltzmann Law which deals only with instantaneous radiative flux.

    My emphasis. That’s on p. 8-9 of Joe’s magnum opus:

    Equation {7} is similar to the model EBM I presented.

  5. For the Moon use this Diviner dataset …
    https://pds-geosciences.wustl.edu/lro/lro-l-dlre-4-rdr-v1/lrodlr_1001/catalog/gcpds.cat
    https://pds-geosciences.wustl.edu/lro/lro-l-dlre-4-rdr-v1/lrodlr_1001/data/gcp/

    You would find that the linear weighted mean temperature does not agree with the forth order weighted (SB) mean temperature (as one would expect). The forth order mean temperature is the correct effective mean temperature based on the SB energy relationship. Some infilling of that GCP dataset is required (it has been awhile, but the somewhat newer Polar Cumulative Products (PCP) might help there, might try that one at a date TBD).

    For Mercury (tidally locked to the Sun with virtually no atmosphere) I think we are still left with theory and/or MESSINGER data.

    For the Earth, we would need a dataset for the surface proper (unobscured by clouds) similar in construction to the Diviner dataset. That might be an interesting calculation, for the mean over time, if possible. But useless for current global climate change as the spatial-temporal error bars would swamp any such fourth order absolute temperature calculation.

  6. Willard says:

    Since you mention the Moon, you might like:

    The moon does not rotate about its own center of mass.

    A ball on a string being spun around your head is not rotating about its own center of mass. It is rotating about your head. It is “orbiting” your head, and not “rotating on its own axis”.

    https://www.drroyspencer.com/2021/03/uah-global-temperature-update-for-february-2021-0-20-deg-c/#comment-627765

    Sometimes, all you got to do is to open yourself to the many facets of the human mind.

  7. Mercury completes one rotation every orbital period. An object without rotation (e. g. a rocket in space flying in a proverbial straight line (a simplification) with no roll (might be yaw (or pitch) for orbital mechanics, but I am using naval architecture inertial frame of reference) will always see the same non-moving fixed points in the far field for all times …
    https://en.wikipedia.org/wiki/International_Celestial_Reference_Frame
    “The modeling incorporates the effect of the galactocentric acceleration of the solar system, a new feature over and above ICRF2. ICRF3 contains positions for 4536 extragalactic sources. Of these 303 have been identified as defining sources.”

    Sci-fi series like Lost in Space and Star Trek: Voyager immediately come to mind. The ICRF3 should be good for substantial extragalactic travel.

  8. Actually I got Mercury wrong, it is 3:2 resonance …
    https://en.wikipedia.org/wiki/Mercury_(planet)#Spin-orbit_resonance

    I learn something new each day …

    Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth. Except for libration, this results in the Moon keeping the same face turned toward Earth, as seen in the left figure. (The Moon is shown in polar view, and is not drawn to scale.) If the Moon were not rotating at all, it would alternately show its near and far sides to Earth, while moving around Earth in orbit, as shown in the right figure.

  9. TrueSceptic says:

    “The true, and physically accurate average of the system, is that half of the surface of the Earth absorbs twice as much energy as the entire surface of the Earth radiates”.
    So, the average temperature would keep rising with no upper limit?

    The problem here is that it’s all wrong on a level so basic that it’s hard to get a grip on it.

    I look forward to Joe’s Magic Balls (TM) solving the water crisis in the world’s driest areas. Each Ball will, by simple geometry, double any rainfall compared with what would fall on a circle of ground of the same diameter (the Ball’s “shadow”). How could he miss this chance to get a Nobel?

  10. an_older_code says:

    “I’m also convinced that “Flat Earthers” are just having fun, seeing how far they can stretch credulity.”

    that maybe true for the “celebrity” grifters like Mark Sargent or Bob Nodel, but not their army of followers / supporters who genuinely believe

  11. Entropic Man says:

    Don’t forget such gems as
    “The backradiation model within the atmosphere has been refuted by experiment.”
    and
    “The oceans warm from the bottom and the sides”

    A few of us try to keep it rational, but after a while you lose the will to live

  12. MP says:

    Roasting a rotating chicken alongside a red hot 1000 watt element creates the same climate to get a crispy outside and well cooked inside as using 2 x 500 watt not red hot elements on both sides?

    The geometry of watts falling on the surface area is the same…

  13. Dave_Geologist says:

    Shades of attempts to explain away the greenhouse effect by appealing to the ideal gas law, on the basis that if you apply the ideal gas law to the atmosphere of Earth and other planets where the atmosphere is thin enough to approximate an ideal gas you get the right temperature, because it’s a truism (or perhaps a circular argument). As I said at the time:

    The Earth’s surface temperature absent greenhouse gases would be 255K. The atmosphere is close to an ideal gas, so if there were no GHGs, and you calculated the surface temperature from the surface pressure and density, you’d get 255K. The Earth’s surface temperature with greenhouse gases is 288K. The atmosphere is close to an ideal gas, so if you calculate the surface temperature from the surface pressure and density, you get 288K. It’s true because it’s a truism. One that doesn’t tell you why it works for both 255K and 288K.

    I also said:

    Try a bit of dimensional analysis, then explain how it works when the units of ECS are [K or °C] not J/K/mol. It’s as meaningless as using weight divided by colour (wavelength or wavenumber) to derive speed.

  14. Dave,
    Yes, I wrote a post about the issue with using the ideal gas law to refute the planetary greenhouse effect.

  15. mtntim says:

    If you really wanted to steelman his argument, you could say that energy conservation tells you about the average value of T^4 (denoted {T^4} ), which doesn’t necessarily tell you about the average temperature {T}. So in this sense, converting solar irradiance into an average temperature is not obvious.

    But then it’s fairly easy to prove that {T}^4 <= {T^4}. So really the energy balance argument says the earth’s average temperature would have to be {T}<= 255 K without greenhouse gases, compared to the observed value of 288 K.

    Submitted twice to fix formatting issues.

  16. Willard says:

    Hello MP,

    Seems that you left a note at Joe’s:

    @ Joseph

    I posted this rhetorical question on andthentheresphysics.wordpress.com

    It was deleted within a minute lol. Cognitive dissonance of climate change brainwashed from childhood people is strong.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71742

    I left a note under it:

    https://ibb.co/6DrcXCM

  17. Dave_Geologist says:

    Yes ATTP, I copied and pasted after searching my posts (I tend to be long-winded so type and save them in a text editor in case they get lost in posting or I accidentally delete the tab or whatever 😉

    Couldn’t find it at first 😦 . Searched inside the text files for PV then nRT and found nothing. Then I found a file called … PVnRT.txt 🙂

    Speaking of units, my mind tends to get blown when petroleum articles appear in journals which insist on S.I. Units. What’s with the m² crap, give me millidarcies!

    Wiki tells me why it’s almost-but-not-quite powers-of-ten.

    The odd combination of units comes from Darcy’s original studies of water flow through columns of sand. Water has a viscosity of 1.0019 cP at about room temperature.

    Dang! Near miss, like 9.81 m/s².

    A thoughtful Creator would have rounded the numbers.

    OTOH 9.87 is spookily close to 9.81. Maybe She’s playing with us?

    Wiki also tells me the plural is millidarcys, not millidarcies. A quick check in Recoll (which I recommend for indexing and full-text search if you have a lot of papers on your hard drive) tells me the literature uses millidarcies about ten times more often than millidarcys. Now don’t get me started on fracking vs. fraccing!

  18. Dave_Geologist says:

    MP obviously believes in a Counter-Sun model rather than the more common Counter-Earth.

    Instead of a planet hiding behind the Sun, diametrically opposed to the Earth, a star tracking Earth round its orbit 93,000,000 miles farther out. That must be why the nights are so bright.

    Ironically, the title of the ATTP post I pinched my previous comment from was “A little knowledge”. Followed of course by “is a dangerous thing”. From “An Essay on Criticism” by Alexander Pope. I presume he meant literary criticism but the sentiment can be more widely applied.

  19. Willard says:

    > millidarcys

    Nice one.

    Speaking of units, it seems that heat flux can be expressed in terms of kg⋅s−3 (basic units) and Btu/(h⋅ft2).

    Btu is good for slow cooking.

  20. mtntim says:

    I was the one who originally challenged Joe to solve the easy problem in this article https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/ . The point was to disentangle two lessons from the “layer atmosphere” model:

    1) By using realistic values of average insolation and albedo, one can calculate the “blackbody temperature” for the earth
    2) The model shows the basic premise of the greenhouse effect

    Because Joe has been struggling for years to understand the averaging involved in point 1), I posed the problem differently: forget about the earth, and calculate the steady state temperature for an actual two layer plate in space which constantly faces the sun. The idea was that this would show him how the greenhouse effect works and obeys the laws of thermodynamics in a simple setting, and from there we could apply it to the earth.

    Of course, he was unable to make any headway on solving the problem correctly.

  21. TrueSceptic says:

    So, Joe’s calling you Squirt now? Does that mean something special in his parallel universe?

  22. Willard says:

    > Does that mean something special in his parallel universe?

    You may not want to know.

    Over the years I might have been called worse. I don’t mind much, as it removes communication constraints. One day Climateball players will realize that Manners Maketh the Climateball Player, e.g.:

    Joe,

    I asked rhetorical questions to make a point.

    Speaking of communication, you still fail to own your miscalculation. You are still holding my comments in the pending bin. You are not calling me by my name. And you are still hiding behind your flying monkey.

    Funny that he thought of a chicken:

    > But the same surface

    Have I ever told you the Kentucky-Kiev Paradox, Clint?

    If you give me one chicken, I will bring you two.

    https://www.drroyspencer.com/2021/04/uah-global-temperature-update-for-march-2021-0-01-deg-c/#comment-664189

    Do you happen to know Clint, by any chance?

    To answer your and his silly question, a planet can receive light and still be really cold.

    You know where to find me.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71766

    I should have clarified that I asked rhetorical questions to make a point about rhetorical questions, and predict that Joe won’t get it.

  23. Below is the integral I did to show that if you properly integrate the incoming solar flux over the hemisphere that faces the Sun, you get the same answer as simply considering the cross-sectional area.

    Essentially, if you want to determine the surface area of one hemisphere of a sphere, you can do

    A = \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} R^2 \cos \theta d \theta d\phi = R^2 \pi \int_{-\pi/2}^{\pi/2} \cos \theta d \theta = 2 \pi R^2,

    as expected. If you want to do the full area of the sphere, simply integrate \phi from 0 to 2 \pi.

    If you want to consider how much energy that hemisphere receives per unit time you can carry out the same integration, but multiply it by the incoming flux and \cos of the solar zenith angle (which takes into account the angle at which the incoming solar flux hits each surface element).

    The solar zenith angle is given by

    \cos \Phi = \sin \theta \sin \delta + \cos \theta \cos \delta \cos h,

    where \theta is the latitude, \delta is the declination of the Sun, and h is the hour angle of the Sun. If, for simplicity, we assume that Sun is currently overhead at midday at the Equator, then \delta = 0 and we can assume that h is essentially the longitude. This means the first term above is 0.

    So, our integral becomes (where E is the energy intercepted per unit time, and F is the incoming solar flux):

    E = F \int_{-\pi/2}^{\pi/2} \int_{-\pi/2}^{\pi/2} R^2 \cos \theta (0 + \cos \theta \cos \phi) d \theta d \phi,

    and we’ve integrated longitude from -\pi/2 to \pi/2, rather than from 0 to \pi.

    If you carry out this integral, you get

    E = F 2 R^2 \int_{-\pi/2}^{\pi/2} \cos^2 \theta d \theta = F 2 R^2 \left[ \frac{\theta}{2} + \frac{1}{4} \sin (2 \theta) \right]_{-\pi/2}^{\pi/2} = F 2 R^2 \left[\frac{\pi}{4} - (-\frac{\pi}{4}) \right] = F \pi R^2.

    So, you can see that if you do the integral over the hemisphere and take the solar zenith angle into account, you recover that the energy intercepted per unit time is the incoming flux times the cross-sectional area of the Earth.

    I will admit that I got slightly confused about the angles, so I hope I’ve done this correctly. Happy to be corrected if I have made some error.

  24. Physicks for cranks … where there are no right answers … everyone gets an A+ … all wheels are flat sided (caveperson creates round plate wheel. crankperson attaches perpendicular to normal application and sez go really fast)… or four sided tetrahedrons … how else to explain the Borg.

  25. ATTP,

    Of course you are right. Why are the poles so cold and the equator so hot and what about that seasonal axial tilt thingie anyways.

  26. Willard says:

    I took my leave at Joe’s:

    > I’m not “Clint” either.

    It’s possible. Just as it’s possible you read a forest of comments where I used “kiddo” for a guy you do not know. It’s also POSSIBLE that you’re using caps lock just like Clint for the fun of it. That your prosody matches his could be pure coincidence.

    Everything is possible, including doubling the flux by doubling the area it covers.

    Stay safe.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71775

    So Joe has gone for the food fight instead of owning his division error.

    He might have missed your last comment, AT. I linked to it, but Joe got carried away.

    I hope Mosh will notice the point about prosody. Where’s Mosh?

  27. Willard says:

    > I should have clarified that I asked rhetorical questions to make a point about rhetorical questions, and predict that Joe won’t get it.

    Of course Joe didn’t:

    You asked rhetorical questions after admonishing people about rhetorical questions to make the points that you’re a psycho – got it, Squirt!

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71771

    In Joe’s world, either you’re a Kantian or a psycho.

  28. Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law

    Planet Energy Budget:
    Solar energy absorbed by a Hemisphere with radius “r” after reflection and dispersion:
    Jabs = Φ*πr²S (1-a) (W)

    What we have now is the following:
    Jsw.incoming – Jsw.reflected = Jsw.absorbed
    Φ = (1 – 0,53) = 0,47

    Φ = 0,47
    Φ is the planet’s spherical surface solar irradiation accepting factor.
    Jsw.reflected = (0,53 + Φ*a) * Jsw.incoming
    And
    Jsw.absorbed = Φ* (1-a) * Jsw.incoming
    Where
    (0,53 + Φ*a) + Φ* (1-a) = 0,53 + Φ*a + Φ – Φ*a =
    = 0,53 + Φ = 0,53 + 0,47 = 1

    The solar irradiation reflection, when integrated over a planet sunlit hemisphere is:
    Jsw.reflected = (0,53 + Φ*a) * Jsw.incoming
    Jsw.reflected = (0,53 + Φ*a) *S *π r²

    For a planet with albedo a = 0
    we shall have
    Jsw.reflected = (0,53 + Φ*0) *S *π r² =
    = Jsw.reflected = 0,53 *S *π r²

    The fraction left for hemisphere to absorb is:
    Φ = 1 – 0,53 = 0,47
    and
    Jabs = Φ (1 – a ) S π r²

    The factor Φ = 0,47 “translates” the absorption of a disk into the absorption of a hemisphere with the same radius. When covering a disk with a hemisphere of the same radius the hemisphere’s surface area is 2π r². The incident Solar energy on the hemisphere’s area is the same as on disk:

    Jdirect = π r² S
    The absorbed Solar energy by the hemisphere’s area of 2π r² is:
    Jabs = 0,47*( 1 – a) π r² S

    It happens because a hemisphere of the same radius “r” absorbs only the 0,47 part of the directly incident on the disk of the same radius Solar irradiation.
    In spite of hemisphere having twice the area of the disk, it absorbs only the 0,47 part of the directly incident on the disk Solar irradiation.

    Jabs = Φ (1 – a ) S π r² , where Φ = 0,47 for smooth without atmosphere planets.
    and
    Φ = 1 for gaseous planets, as Jupiter, Saturn, Neptune, Uranus, Venus, Titan. Gaseous planets do not have a surface to reflect radiation. The solar irradiation is captured in the thousands of kilometers gaseous abyss. The gaseous planets have only the albedo “a”.

    And Φ = 1 for heavy cratered planets, as Calisto and Rhea ( not smooth surface planets, without atmosphere ). The heavy cratered planets have the ability to capture the incoming light in their multiple craters and canyons. The heavy cratered planets have only the albedo “a”.

    Another thing that I should explain is that planet’s albedo actually doesn’t represent a primer reflection. It is a kind of a secondary reflection ( a homogenous dispersion of light also out into space ).

    That light is visible and measurable and is called albedo.

    The primer reflection from a spherical hemisphere cannot be seen from some distance from the planet. It can only be seen by an observer being on the planet’s surface.
    It is the blinding surface reflection right in the observer’s eye.
    That is why the albedo “a” and the factor “Φ” we consider as different values.

    Both of them, the albedo “a” and the factor “Φ” cooperate in the Planet Rotating Surface Solar Irradiation Absorbing-Emitting Universal Law:

    Jsw.incoming – Jsw.reflected = Jsw.absorbed
    Jsw.absorbed = Φ * (1-a) * Jsw.incoming

    Total energy emitted to space from entire planet:

    Jemit = A*σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W)

    Α – is the planet’s surface (m²)
    (β*N*cp)¹∕ ⁴ – dimensionless, is a Rotating Planet Surface Solar Irradiation Warming Ability
    A = 4πr² (m²), where r – is the planet’s radius

    Jemit = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W)

    global Jabs = global Jemit
    Φ*πr²S (1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴
    Or after eliminating πr²
    Φ*S*(1-a) = 4σTmean⁴ /(β*N*cp)¹∕ ⁴
    The planet average Jabs = Jemit per m² planet surface:
    Jabs = Jemit
    Φ*S*(1-a) /4 = σTmean⁴ /(β*N*cp)¹∕ ⁴ (W/m²)

    Solving for Tmean we obtain the Planet Mean Surface Temperature Equation:

    Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)

    β = 150 days*gr*oC/rotation*cal – is a Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant
    N rotations/day, is the planet’s axial spin
    cp – is the planet surface specific heat
    cp.earth = 1 cal/gr*oC, it is because Earth has a vast ocean. Generally speaking almost the whole Earth’s surface is wet. We can call Earth a Planet Ocean.

    Here (β*N*cp)¹∕ ⁴ – is a dimensionless Rotating Planet Surface Solar Irradiation Warming Ability
    σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant

    Rotating Planet Spherical Surface Solar Irradiation Absorbing-Emitting Universal Law:

    Jemit = 4πr²σΤmean⁴/(β*N*cp)¹∕ ⁴ (W)

    The year-round averaged energy flux at the top of the Earth’s atmosphere is Sο = 1.361 W/m².
    With an albedo of a = 0,306 and a factor Φ = 0,47 we have
    Tmean.earth = 287,74 K or 15°C.

    This temperature is confirmed by the satellites measured
    Tmean.earth = 288 K.

    https://www.cristos-vournas.com

  29. Willard says:

    Drive-by done, Christos.

    Thanks.

  30. Willard says:

    And so Joe sticks to his guns:

    “Everything is possible, including doubling the flux by doubling the area it covers.”

    That is the most psychotic inversion of what I demonstrate which I have ever come across, Squirt.

    Climate science dilutes the power of sunshine by spreading it over four times the surface area which intercepts it.

    No one, ever anywhere, has doubled the flux by doubling the surface area.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71776

    Poor geometric intuition may not explain how one can fail to grasp that division and multiplication are mutual inverses.

  31. TrueSceptic says:

    My mind is still boggling. Of course, it’s amusing how angry and insulting the terminally delusional can get, otherwise I might feel sympathy for those who just “know” the “establishment” is so obviously dishonest and incompetent and yet has managed to convince the majority of sane, educated people.

  32. Willard says:

    Joe responds to fan mail:

    Yes I did see the calculation…not sure what the point of it is as it just shows that the intercepted energy is that from a disk cross-section of the sphere, and it also shows the area of hemisphere. Everything in it demonstrates my model and the point of my model…so not sure what they’re trying to imply in that comment. Often times, at the very end I’ve noticed, they’ll present something truthful and they’ll even present what you yourself present, but pretend that it is theirs and that it doesn’t show what it factually actually directly shows – it is an insane level of sophistry when they do that! They’ll present your argument as if it is theirs, but then ignore what the result of the argument is! It’s just amazing, really. It’s amazing to see it go that far.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71785

    Not sure where Joe’s model shows the equivalence between taking the shadow of the Earth and integrating a hemisphere, but there you go.

    The fan in question (whom I call “kiddo” because he chose a confusing nickname at Roy’s) is the one who motivated me to write this post, btw.

  33. Christos Vournas ,

    You are, of course, wrong.

    The only situation where the SB relationship would not hold true is if an object always presented its same non-rotating face to any fixed point is space. Which means any point in space, the object has no roll, no pitch and no yaw, no rotations whatsoever, ever.

    The SB (black-body) temperature for the Moon is ~270.4K …
    https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

    The Diviner dataset I mentioned above gives 270.46K for T^4 (SB) relationship and 199.44K for a T^1 (linear) relationship. Mean long term temperatures over many rotations.

    For Mercury, we will find a better SB estimate (than MESSINGER) sometime later this decade …
    https://en.wikipedia.org/wiki/BepiColombo
    Estimation of surface temperatures on Mercury in preparation of the MERTIS experiment onboard BepiColombo
    https://www.sciencedirect.com/science/article/abs/pii/S0019103520304310
    (paywalled but you know the drill)
    … over several Mercury years.

  34. Willard says:

    This might reinforce the idea that the conceptual error isn’t merely geometric:

    I have spent a considerable amount of time trying to explain the “divide by 2” principle to Willard but he just always comes back to “you are doubling the flux by doubling the area it covers, which is impossible”. I point out that the area is in fact being halved, and it just seems to be ignored.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71787

    Back in my days, we were taught early enough that 1/2 was twice 1/4.

  35. Dr Roys Emergency Moderation Team says:

    “No need to check: minding our units leads to a reductio. Flux is a rate of energy by unit of surface. The more surface for the same energy {6}, the less flux there is; if the Earth was infinite, flux would be nil. One can’t get twice the Sun’s power out of twice the area. That’s absurd.”

    It’s not twice the Sun’s power out of twice the area, it is double the flux out of half the surface area. When you divide by 4 you spread the incoming sunlight over the entire Earth’s surface area. 960 W/m^2 (after factoring in albedo) becomes 240 W/m^2. This equates to a blackbody temperature of 255 K. If you think of the Earth in real time, it is actually only ever receiving sunlight over the lit hemisphere at any given moment. In that same moment, energy leaves from the entire surface area of the globe.

    So what is actually being received, on a second by second basis, is 960 W/m^2 divided by 2 = 480 W/m^2. That’s the real time flux impinging on the lit hemisphere. This balances the approx. 240 W/m^2 leaving from the entire Earth’s surface area at any given moment. Energy is conserved, because in real time the surface area the energy is received over (the lit hemisphere) is only half that over which the energy leaves (the entire sphere).

  36. I should have said “might not” instead of “would not” as there needs to be a heat source other than the CMB. Probably a few other caveats as well.

  37. Willard says:

    Francis,

    I don’t know if you know Christos, but I think it’s best to leave it at one drive-by and one response.

    He has a blog, and he comments at Roy’s.

  38. We seem to be having one of those discussions that end up spread over more than one blog. Maybe those who think it should only be a factor of 2 should consider the following.

    The amount of energy we intercept from the Sun per unit time is:

    E_{in} = F_{\odot} \pi R_E^2.

    The amount of energy we radiate back into space per unit time is:

    E_{out} = 4 \pi R_E^2 \sigma T_E^4,

    where T_E is the effective radiative temperature of the Earth.

    If we know the albedo and F_{\odot} then we could equate these two to estimate T_E.

  39. Francis E Sargent,
    > You are, of course, wrong.

    Why?
    And why you say Moon’s Te = 270 K?
    I presented a Universal Law, and a Planet Universal mean surface temperature theoretical calculation formula…

    What is wrong with that?
    Please, explain, you should have some reasons to say so.

    https://www.cristos-vournas.com

  40. Another thought experiment. Imagine we could increase the atmospheric GHG concentration so as to produce a change in forcing of 1 W/m^2. By definition this applies across the full surface of the Earth. In other words, if we could do this instantaneously, the planetary energy imbalance at that instant would be N = 1 W m^{-2} and the difference between the amount of energy the Earth would be receiving per unit time and the amount radiated back into space per unit time would be:

    \Delta E = 4 \pi R_E^2 N = 4 \pi R_E^2 J s^{-1}.

    If the same effect were to occur through an increase in solar flux, ignoring albedo would the solar flux at the Earth’s orbit have to increase by 1 W/m^2, or 4 W/m^2?

  41. Willard,

    OK I have had my say.

  42. “Back in my days, we were taught early enough that 1/2 was twice 1/4.”

    Yes, but when you divide by 4 you are spreading the incoming solar radiation over the entire Earth’s surface. When you divide by 2 you are spreading the incoming solar radiation over only the lit hemisphere. The lit hemisphere has only half the surface area of the entire sphere (obviously). That’s why I’m saying the surface area is being halved. The flux is higher when spread over only a hemisphere than over the entire sphere because the surface area of the hemisphere is half that of the entire sphere.

  43. aljo1816 says:

    Perhaps it’s because I’m simple-minded and am missing something, but intuitively the reason for dividing by 4 seems quite obvious to me. The quantity we are trying to calculate is the surface flux in units of Watts per square meter. There is an implicit “of earth’s surface area” after the “square meter” in this dimensional analysis. If instead we’re presenting the surface flux implicitly as “Watts per square meter of earth’s sunlit hemisphere” then we cannot equate E_In=E_out at equilibrium without dividing the outgoing energy by 2.

  44. aljo1816,

    then we cannot equate E_In=E_out at equilibrium without dividing the outgoing energy by 2.

    Actually, it’s 4 because the incoming energy depends on the cross-sectional area (\pi R^2) while outgoing energy depends on the whole surface area (4 \pi R^2).

  45. Willard says:

    > Yes, but when you divide by 4 you are spreading the incoming solar radiation over the entire Earth’s surface.

    No, the entire Earth’s surface is equal to 1. If you divide the Earth’s surface by 2, you get a hemisphere. If you divide by 4, you get a disc. That’s the geometric point.

    But my point here is algebraic (or arithmetic if we abstract away the variable): 1/2 = 1/4 + 1/4 = 2 x 1/4.

    So to divide by 2 the way you do is in effect doubling the surface of the Earth’s surface over which is spread incoming solar radiation. If that gets you twice the flux, that means something is wrong with your calculation.

    Joe’s dismissal of AT’s point was pure armwaving.

  46. Willard says:

    > So to divide by 2 the way you do is in effect doubling the surface of the Earth’s surface over which are spread incoming solar variation

    Perhaps I should clarify: compared to the “divide by 4” shortcut.

    Just in case this point gets lost.

  47. Willard, the “Earth’s shadow” disk with a radius the same as the Earth = 1. That has four times less surface area than the entire Earth’s surface and receives approx. 960 W/m^2 (after factoring in albedo). To take that input and spread over a hemisphere you divide by 2 (480 W/m^2) and to spread it over the entire Earth’s surface area you divide by 4 (240 W/m^2). Hope that clears it up.

  48. MP says:

    When the not realistic divide by 4 shortcut algebra for an equilibrium state of energy in and out on earth works, then it doesn’t mean that the average also can be used to determin if the sun can create the climate.

    90% of the earth climate is created around the equator during day time (zenith zone)

    Besides raising the temperature of the surface also subsurface heating, sensible and latent heat creation, gravitational potential energy creation by raising the atmosphere, etc.
    .

  49. MP,
    I think you misunderstand what the relationship below is actually representing.

    S (1 - A) \pi R^2 = 4 \pi R^2 \sigma T_E^4.

    The left hand term is the amount of energy the Earth is receiving from the Sun per unit time. The right is side is the amount of energy the Earth radiates back into space per unit time, in equilibrium. T_E is the effective radiative temperature of the Earth. It is the temperature of a blackbody that would radiate as much energy per second as the Earth radiates into space. In other words, if we were able to measure the energy being radiated into space by the Earth and then determined the equivalent blackbody temperature, we would get T_E.

  50. Willard says:

    > the “Earth’s shadow” disk with a radius the same as the Earth = 1

    No. Here’s the breakdown:

    Earth’s shadow = Disc in my model = \pi R^2
    Earth’s surface = Area in my model = 4\pi R^2 .
    Disc / Area = 1/4

    Hence why we divide by four.

  51. MP says:

    @ ATP

    A virtual blackbody in space doesn’t have a subsurface, doesn’t have an atmosphere, doesn’t get most flux strenght around the equator, and doesn’t have a day and night cycle

  52. MP,
    I’ll try one more time (I’m not talking about a virtual blackbody in space). When we use the equation I mentioned in my previous comment, what we’re determining is the effective radiative temperature of the Earth. This is the temperature of a blackbody that would radiate the same amount of energy per square metre per second as the Earth does when in equilibrium.

  53. Yes, of course, everyone knows that both the Moon and Earth are hollow …

    PS Back to my preferred name, did not notice that one for awhile. Also thinking about Lagrange Points and out of the Solar System ecliptic plane radiation measurements (in general, where these measurements might be biased due to observation location).

  54. Willard says:

    Joe will not dare to come here to tell me what he thinks of me to my face. More importantly, he won’t come here and address AT’s direct refutation of his “divide by two” silly trick:

    [Y]ou haven’t provided a single argument, explanation, or position on anything here yet clearly. We’re waiting.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71803

    The tears of the world are a constant quantity. For each one who begins to weep somewhere else another stops. The same is true of the laugh.

  55. Entropic Man says:

    There’s an unspoken difference in approach here.

    DREMT and his fellows at Roy Spencer are talking about measuring the instantaneous surface energy input of the dayside. Hence his division by 2. This is fine if you want to know the flux at 56N 7W at 4pm local time. It is useful for weather forecasting, which plugs the local flux into short term planetary model grid squares hour by hour.

    The rest of us are thinking about climate and are more interested in the annual average flux and the planetary energy budget.. Hence our division by 4 to cover the whole planetary surface; and averaging of daily, seasonal and longer term variations.

  56. Willard says:

    EM,

    > DREMT and his fellows at Roy Spencer are talking about measuring the instantaneous surface energy input of the dayside. Hence his division by 2.

    That still does not explain why Joe doesn’t get the same result as the division by 4:

    [I]f you do the integral over the hemisphere and take the solar zenith angle into account, you recover that the energy intercepted per unit time is the incoming flux times the cross-sectional area of the Earth.

    Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190372

  57. MP says:

    [Enough peddling, MP. – W]

  58. Bob Loblaw says:

    Oh, I’ll probably regret joining this, as it could become an incredible time sink, but it certainly is amusing to read.

    In the land of the clueless, who can’t see the 1/4 ratio, it appears that they have not yet clued in that angle of incidence is fundamental here. The solar constant, at 1368 W/m^2, is always defined for a surface perpendicular to the sun’s rays. There is only one spot on earth where that happens: where the sun is directly overhead. Every other spot on earth receives sunlight at an angle. Every other spot receives <1368 W/m^2 when you think in terms of the earth's surface rather than the 1m^2 that is pointing directly at the sun.

    The geometry says that the sunlight passing through 1 m^2 measured perpendicular to the sun's rays gets spread over a larger horizontal area on the earth's surface of 1/cos(z) m^2, where z is the zenith angle (the angle between where the sun is in the sky and the vertical line perpendicular to the earth's surface). So, the flux is 1368*cos(z) W/m^2 on 1m^2 of the earth's surface. (Well, at the top of the atmosphere.)

    In the land of the clueless, they have apparently not clued in that the edges of the earth's sphere along the border between night and day are getting essentially zero solar input. Have they not noticed how much less light there is looking straight up just before the sun sets? Have they not noticed that when trying to read a book in a dimly lit room, you're best off turning the page so that it points directly at the light instead of having the page aligned parallel to the light rays? Angle of incidence is fundamental.

    Taking 1/cos(z) into account, and doing the integration over the enter earth and entire day, you get, well, you get what people have already presented here who are not from the land of the clueless. Hint: the answer is 4.

  59. Willard says:

    > Have they not noticed how much less light there is looking straight up just before the sun sets? Have they not noticed that when trying to read a book in a dimly lit room, you’re best off turning the page so that it points directly at the light instead of having the page aligned parallel to the light rays?

    Nicely put.

    Since Sky Dragons play Climateball at dusk or at dawn, they seldom notice wrenches:

  60. Willard says:

    Let’s hope MP does not mind much if Joe pulled the plug:

    I certainly don’t.

  61. Russell says:

    It’s a shame the Shy Dragoons haven’t applied their dimensionless dimensional analysis skills to the surface temperature of the sunlit underside of the un-dimensional staircases in the works of M.C. Escher,

    If terrestrial maps can incorporate inner Asia, Upper Volta, and Lower Silesia,there may be more planetary geometries than are dreamt of in Joe’s philosophy:

    https://vvattsupwiththat.blogspot.com/2020/07/schellenberger-sequel-will-connect-dots.html

  62. Willard says:

    Thanks, Russell:

    What a great gift!

  63. Bob,

    The geometry says that the sunlight passing through 1 m^2 measured perpendicular to the sun’s rays gets spread over a larger horizontal area on the earth’s surface of 1/cos(z) m^2, where z is the zenith angle (the angle between where the sun is in the sky and the vertical line perpendicular to the earth’s surface).

    Indeed, I did that in this comment.

  64. Dave_Geologist says:

    Alexander Pope would have a ball if he was alive and had Internet!

  65. TrueSceptic says:

    I don’t believe it. Joe and his mates still don’t get it! What a bunch of f***wits!

    BTW did no one get my analogy of rain falling on balls?

  66. Reactance (psychology)
    https://en.wikipedia.org/wiki/Reactance_(psychology)

    “Reactance is an unpleasant motivational arousal (reaction) to offers, persons, rules, or regulations that threaten or eliminate specific behavioral freedoms. Reactance occurs when a person feels that someone or something is taking away their choices or limiting the range of alternatives.

    Reactance can occur when someone is heavily pressured to accept a certain view or attitude. Reactance can cause the person to adopt or strengthen a view or attitude that is contrary to what was intended, and also increases resistance to persuasion. People using reverse psychology are playing on reactance, attempting to influence someone to choose the opposite of what they request.”

    And no, reverse psychology will not work with these people as thy tend to be conspiracy theorists and paranoid to boot. Oh and anything that even hints at theory of mind or psychology in any manner or manifestation is bound to get the most vociferous of reactance rejections.

    This is also why they can not come up with a more compelling narrative then the one that climate science offers.

    One wanker at Roy’s even suggested increasing GHG emissions (over and above current emissions) to delay the certain Ice Age that they believe we are now entering.

  67. Bob Loblaw says:

    ATTP: “Indeed, I did that in this comment.

    Indeed, although I hoped that my explanation made it a bit clearer why the 1/cos(z) is needed.

    Your comment was one of the ones that I was referring to when I said “Taking 1/cos(z) into account, and doing the integration over the enter earth and entire day, you get, well, you get what people have already presented here who are not from the land of the clueless.

  68. Bob Loblaw says:

    “One wanker at Roy’s even suggested increasing GHG emissions (over and above current emissions) to delay the certain Ice Age that they believe we are now entering.

    So, increasing atmospheric CO2 does not cause undesirable warming now, but it will cause desirable warming/prevention of cooling later?

    Skeptical Science has a web page for that sort of stuff:
    https://skepticalscience.com/contradictions.php

  69. Willard says:

    > increasing atmospheric CO2 does not cause undesirable warming now,

    Of course not:

    [A]nother way to show them wrong is to accept their bogus solution. Then, as with the “steel greenhouse”, if we use Earth values their nonsense blows up in their faces.

    So, in the 2-plate scenario, insulate one side of the blue plate. Now, using their own calculations, for the green plate to emit 240 W/m^2, the blue plate must be at 303 K, emitting 480 W/m^2.

    But Earth is only at 288 K. They have proved CO2 cools the planet!

    https://www.drroyspencer.com/2021/03/uah-global-temperature-update-for-february-2021-0-20-deg-c/#comment-629896

  70. Willard says:

    An important unit to mind is the nut. It often comes in twos:

    (E) https://climateball.net/but-predictions/
    (V) You could be intersted in my CO2-Chicken-O.Grill!
    (E) Only if it’s a BOFA.
    (V) What is a BOFA?
    (E) The best for deez nutz!

    Source: https://www.reddit.com/r/climateskeptics/comments/mz75gs/fear_keeps_the_masses_moving/gvz6wqk/

  71. Willard says:

    I tried to plea with Joe.

    He continues to gloat.

    Oh, well. I tried.

  72. Willard,
    I must admit to having completely lost track of quite what argument Joe is actually making. That may, of course, be a feature, rather than a bug.

  73. Willard says:

    BREAKING

    Philip for the win:

    “if the area of a sphere is 1, then a hemisphere is found by dividing by 2, and a disk is by 4.”
    Well that would seem to be the end of geometry and calculus..

    Euclid and Newton felled with a single blow.

    https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71871

    Here’s a reminder: the area of a sphere is \ 4 piR^2 ; the area of a hemisphere (without the base) is \ 2 piR^2 , and the area of a disc is \ piR^2 . Check the video, dammit!

    So 1, 1/2, and 1/4. Even I can grok that. It’s not that complex.

    Philip’s the one whose work has been reviewed in the post I cited from Tony’s:

    https://wattsupwiththat.com/2021/04/17/atmospheric-energy-recycling/

  74. Joe is getting lost in the details, things like adiabatic lapse rate, leading to comments like this …

    “If GHG’s increase the emission of the atmospheric body, then they cool that body.”

    The conventional answer of 33K delta (288K – 255K) is to 1st order correct afaik as the surface temperature distribution for the Earth is rather narrow in absolute terms relative to say the Moon. I played around with a normal distribution set to 288K and 10K < sigma < 25K, 1st and 4th power weightings, for that one T^4 weighting goes up to 289-291K. while T^1 remains unchanged. Radiation to space remains unchanged at ~255K as that is its far field mean effective temperature. GHG theory is correct, planet heats up, reaches thermal equilibrium, … planet has more GHG, planet heats up further, …

    That is my current understanding. I could be wrong about things in general though. Corrections welcomed.

  75. angech says:

    [No thanks, Doc. – W]

  76. mtntim says:

    Joe’s whole MO is to over complicate a well-reasoned argument by replying with mountains of nonsense. I don’t know if it’s a bug or a feature, but the guy is certainly unhinged. You can see the depths of his tantrums here, but it’s NSFW: https://www.reddit.com/r/RealClimateSkeptics/comments/mf6abw/comment/gu9x48b

  77. TrueSceptic says:

    To be fair, you could have put that better: if the area of a sphere is 1, then the area of a hemisphere is ½ and that of a disc is ¼.

    It’s obvious that you meant the same thing, but they will find any excuse to attack.

  78. TrueSceptic says:

    ATTP, you said
    “MP,
    I’ll try one more time (I’m not talking about a virtual blackbody in space). When we use the equation I mentioned in my previous comment, what we’re determining is the effective radiative temperature of the Earth. This is the temperature of a blackbody that would radiate the same amount of energy per square metre per second as the Earth does when in equilibrium.”

    Please help me here. You seem to say that you are not talking about a virtual blackbody, then you do!

    PS stupid question: how do I reply to messages so they indent under those messages?

  79. Dave_Geologist says:

    We already stopped the next Ice Age, about ten years ago (unless we use clear-air capture to get CO2 levels down to pre-industrial within the next 1000 years).

    Next one is more than 100,000 years away, when the Milankovitch forcing will be stronger. Plenty of time to fine-tune CO2 content before then.

  80. TrueSceptic,

    Please help me here. You seem to say that you are not talking about a virtual blackbody, then you do!

    Okay, I guess I wasn’t thinking in terms of some virtual blackbody. The temperature, T_E, we get from the following equality.

    pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    is the effective radiative temperature of the Earth given a solar flux at 1 AU of F_{\odot} and an albedo of A. It is, by definition, the temperature of a blackbody that would radiate the same amount of energy per square metre per second as the Earth does into space. It’s just a way of defining an effective temperature. It’s not the temperature of the surface (since that is enhanced by the greenhouse effect) but is close to the temperature that would be estimated if an outside observer were to measure the outgoing spectrum of the Earth. I say “close to” because the spectrum is not a perfect blackbody spectrum and so the temperature that you would estimate would depend on what method was used (total flux, peak of the spectrum, best fit to a blackbody spectrum, etc).

  81. TYSON MCGUFFIN says:

    ATTP 1:38 pm

    I’ve always like Professor Bohren’s definition:

    An average (or better yet, effective) temperature that does have an unambiguous physical meaning is the effective radiative equilibrium temperature Teff of Earth defined as the temperature of a blackbody with a total emission equal to the net solar radiation received by Earth averaged over its entire surface…

    … This is the equivalent blackbody temperature an observer on the moon would infer for Earth looked upon as an infrared sun. Just as we on Earth say that the sun is equivalent to a 6000 K blackbody (based on the solar irradiance), an observer on the moon would say that Earth is equivalent to a 255 K blackbody (based on the terrestrial irradiance). Note that the effective temperature defined by Eq. (1.76) in no (direct) way depends on the emissive properties of Earth’s atmosphere.

    Fundamentals of Atmospheric Radiation: An Introduction with 400 Problems. Craig F. Bohren and Eugene E. Clothiaux

  82. TrueSceptic says:

    Just found this at SkepticalScience It identifies the errors and ends with
    “This work makes extraordinary claims and yet no effort was made to put it in a real climate science journal, since it was never intended to educate climate scientists or improve the field; it is a sham, intended only to confuse casual readers and provide a citation on blogs. The author should be ashamed.’
    Apologies if it’s already been mentioned here.

  83. TrueSceptic says:

    Try again with link, which was removed. SkepticalScience https://skepticalscience.com/postma-disproved-the-greenhouse-effect.htm

  84. Dr Roys Emergency Moderation Team says:

    From the Skeptical Science link:

    “In essence, he would prefer we had one sun delivering 1370 W/m2 of energy to the planet, with a day side and a night side, noon and twilight, etc. instead of the simple model where we average 1370/4=342.5 W/m2 over the planet (so that the whole Earth is receiving the appropriate “average” solar radiation). The number becomes ~240 W/m2 when you account for the planetary albedo (or reflectivity).

    The factor of 4 is the ratio of the surface area to the cross section of the planet, and is the shadow cast by a spherical Earth. It is therefore a geometrical re-distribution factor; it remains “4” if all the starlight is distributed evenly over the sphere; it is “2” if the light is uniformly distributed over the starlit hemisphere alone; with no re-distribution, the denominator would be 1/cosine(zenith angle) for the local solar flux.”

    There you go, Willard – what I said is correct. You can correct your article now!

  85. See also works by P Mulholland SPR Wilde …
    https://scholar.google.com/scholar?hl=en&as_sdt=0%2C25&q=P+Mulholland+SPR+Wilde&btnG=

    Currently under discussion at Where The Hell Am I? Willard has posted several links, the latest is … Trouble in Noonworld, Take 2

    There be Skydragons amongst us.

    PS AFAIK Bob Wentworth (at WTHAI?) is now doing a pretty good takedown of this nonsense.

  86. Dr Roys,
    Maybe the problem (and maybe I’m wildly optimistic here) is that people don’t appreciate the basic point here.

    If all you want to know is the effective radiative temperature of the Earth, you can simply use

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4.

    There’s clearly a \pi R^2 on one side and a 4 \pi R^2 on the other, but you don’t need to think in terms of spreading the incoming solar energy over the whole sphere – it hits one side of the sphere, but the whole sphere radiates back into space.

    However, if you want to compare the impact of increasing solar flux with, for example, increasing CO2 concentrations, this it typically done using what is called changes in forcing. If, for example, we instantaneously increased atmospheric CO2 concentrations so that there was a change in forcing of 1 W/m^2, the impact this would have on the energy balance of the Earth (instantaneously) would be

    \Delta E = 4 \pi R^2 \times 1 = 4 \pi R^2 J s^{-1}.

    This would be the difference between the amount of energy the Earth is receiving per unit time (from the Sun) and the amount being radiated back into space per unit time. By definition, a change in forcing is averaged over the whole sphere.

    If we then ask, how much would the solar flux need to increase by to have the same effect, then we can go back to the first equation in the comment and use that:

    \pi R^2 (1 - A) dF_{\odot} = 4 \pi R^2 \Rightarrow dF_{\odot} = \frac{4}{1 - A} = 5.71 W m^{-2}.

    In other words, for an increase in solar flux to have the same effect on the planetary energy balance as a change in forcing of 1 W/m^2, the solar flux needs to increase by 5.7 W/m^2 (taking an albedo of 0.3 into account). Therefore if we want to convert from a change in solar flux to a forcing, you need to divide by 4 and then multiply by (1 – A).

  87. Dr Roys Emergency Moderation Team says:

    “No need to check: minding our units leads to a reductio. Flux is a rate of energy by unit of surface. The more surface for the same energy {6}, the less flux there is; if the Earth was infinite, flux would be nil. One can’t get twice the Sun’s power out of twice the area. That’s absurd.”

    The problem is, the above is a huge straw man. Postma is *not* arguing that you can get twice the Sun’s power out of twice the area, as I have already explained. This should be corrected.

  88. Willard says:

    Kiddo,

    You’re not at Roy’s here. You can’t wave your arms over and over again. I quoted Joe. You did not.

    A hemisphere is twice the size of a disc. Deal with it.

  89. Dr Roys Emergency Moderation Team says:

    Yes, a hemisphere is twice the size of the disk.

    But back to my point…a hemisphere has half the surface area of the entire sphere.

  90. sciencepublishinggroup is a (well) known pay-to-play predatory journal and will publish anything for the price of admission …
    Google …
    Stephen Paul Rathbone Wilde site:sciencepublishinggroup.com
    Philip Mulholland site:sciencepublishinggroup.com

    Twelve hits for either. Just a warning for those who do not want to be called names or some such.

  91. Willard says:

    > But back to my point

    Your point is that I misrepresented Joe. You don’t quote Joe. I do. Here’s one version of Joe’s figure I alluded to already in my note 5:

    Here’s how he explains it:

    The diagram above simply uses a factor of 0.5, which is what you get when you take a direct linear average by spreading the intercept-disk evenly over the hemisphere, and of course a hemisphere of the same radius has twice the area of a disk, hence the factor of 1/2 = 0.5.

    https://climateofsophistry.com/2019/07/08/how-to-calculate-the-average-projection-factor-onto-a-hemisphere/

    Do you now see how Joe’s division by 2 gives him twice the Sun’s power?

    Also, if you can explain how Joe justifies his “>90% zenith flux,” that’d be great. I asked him directly. He forgot to respond amidst his lulzing.

  92. Bob Loblaw says:

    But back to my point…a hemisphere has half the surface area of the entire sphere.

    Yes, and that hemisphere is lit by the sun – but not equally. Only the one location where the sun is directly overhead does it see the full solar flux at right angles to the sun’s rays. By the time you adjust for the weaker angled sun everywhere else, the average amount of solar radiation received is cut by half again. What the earth’s surface sees is is the amount spread over 1m^2 of the earth’s surface, not 1^m2 of a surface pointing directly at the sun.

    You have noticed how much less light there on a horizontal surface near sunrise and sunset, haven’t you?

  93. Willard says:

    > Only the one location where the sun is directly overhead does it see the full solar flux at right angles to the sun’s rays.

    Check Joe’s figure, Bob. For Joe, more than 90% of the regions or points over a hemisphere at at zenith angle!

  94. Dr Roys Emergency Moderation Team says:

    Willard, it doesn’t double the power. The flux (W/m^2) is doubled, but not because it is spread over a larger surface area. The power is spread over a *smaller* surface area. It is spread over a hemisphere (resulting in 480 W/m^2) as opposed to being spread over the entire sphere (resulting in 240 W/m^2). Try reading the Skeptical Science quote again.

  95. Dr Roys Emergency Moderation Team says:

    “Check Joe’s figure, Bob. For Joe, more than 90% of the regions or points over a hemisphere at at zenith angle!”

    No…the small circle he has drawn is the area of the Earth’s surface that he is saying (at any one moment) receives 90% or more of the solar zenith flux, 1370 W/m^2 minus that which is reflected due to albedo.

  96. Willard says:

    > The flux (W/m^2) is doubled, but not because it is spread over a larger surface area. The power is spread over a *smaller* surface area.

    Let’s finish these two sentences of yours, Kiddo:

    The flux (W/m^2) is doubled when we take a hemisphere instead of a disc.

    The power is spread over a *smaller* surface area than the whole Earth.

    Your verbal trick consists in hiding two different ratios. Anyone who can read the EBM I stated above should see that you’re conflating the left side to the right side of the equation.

    In a way, this Climateball episode is the perfect illustration that technicalities aren’t at the bottom of contrarian obduracy.

  97. Willard says:

    > No…the small circle he has drawn is the area of the Earth’s surface that he is saying (at any one moment) receives 90% or more of the solar zenith flux,

    Good!

    Do you know where he said what was the size of that small circle?

    According to his calculations, it must not be such a small one:

    And so it turns out that the weighted integrated average projection factor on a hemisphere is the same as the simple linear average, even though the weighted projection function is not linear.

    https://climateofsophistry.com/2019/07/08/how-to-calculate-the-average-projection-factor-onto-a-hemisphere/

  98. Dr Roys Emergency Moderation Team says:

    “The flux (W/m^2) is doubled when we take a hemisphere instead of a disc.”

    No, the flux is *halved* when you take a hemisphere instead of the disk. The disk receives 960 W/m^2, after factoring in albedo. The hemisphere receives 480 W/m^2. The full sphere receives 240 W/m^2.

  99. Willard says:

    > No, the flux is *halved* when you take a hemisphere instead of the disk.

    Again with the same trick, kiddo:

    The flux is *halved* when you take a hemisphere instead of the disk, and when you take the disc you *half* that half.

    So when you take a disc instead of a hemisphere, you don’t half the flux: you divide it by four.

    You already agreed that 1/2 was twice 1/4. Do you really dispute that if we *half* some quantity Q instead of dividing it by four, we double what results?

  100. Dr Roys Emergency Moderation Team says:

    “So when you take a disc instead of a hemisphere, you don’t half the flux: you divide it by four.”

    No, when you take a disc instead of a hemisphere, you double the flux. 480 W/m^2 over the hemisphere becomes 960 W/m^2 over the disk, because the disk has half the surface area of the hemisphere.

  101. Willard says:

    > No, when you take a disc instead of a hemisphere, you double the flux.

    Joe might disagree with you:

    Climate scientists take the real power of sunshine, of P = 960 W/m2, equal to +88o Celscius, but divide the power by the number 4 so that they can make the Earth flat and get rid of day and night – for convenience. When they do this, they artificially (it is artificial because it is no longer real, and only a mathematical simplification to make the Earth flat) decrease the power of sunshine to 960/4 = 240 W/m2 which is equal to -18oC.

    https://climateofsophistry.com/2019/07/08/how-to-calculate-the-average-projection-factor-onto-a-hemisphere/

    You should take that one up with Joe.

  102. Bob Loblaw says:

    DREMT:

    You do know that a hemisphere is roughly, approximately, half a sphere, don’t you? There is another factor of two you are ignoring?

    And that disks have two sides? They are like The Force: they have a Light Side, and a Dark Side, and it holds the earth together?

  103. Dr Roys Emergency Moderation Team says:

    The “divide by 4” spreads the 960 W/m^2 received by the disk over the entire Earth’s surface area, resulting in the 240 W/m^2 figure I already mentioned (and was mentioned in the Skeptical Science link). There is no disagreement there between what Joe is saying and what I am saying and what Skeptical Science is saying.

  104. Bob Loblaw says:

    Willard:
    The diagram you posted from Joe only appeared to me after I posted my comment. Bogosty index appears to be high.

  105. Willard says:

    You’re simply waiving your arms right now, kiddo.

    But let’s mind our units one last time:

    The “divide by 4” trick divides the flux by four and 1/4 corresponds to the ratio Area/Disc.

    The “divide by 2” trick divides the flux by two and 1/2 corresponds to the ratio Area/Hemisphere.

    So taking the Hemisphere instead of the Disc gets you a number that is 2:1.

    There’s no way you can get out of this, and you can’t show how that misrepresents Joe’s argument. You can keep repeating the same thing over and over again, but that will have to be at Roy’s.

    Therefore I suggest you stop waving your arms right now.

  106. Dr Roys Emergency Moderation Team says:

    The disk receives 960 W/m^2, after factoring in albedo.
    The hemisphere 480 W/m^2.
    The entire sphere 240 W/m^2.

    You misrepresent Joe’s argument because he is *not* arguing that you can get twice the Sun’s power out of twice the area, as I have already explained. He is arguing that double the 240 W/m^2 flux is impinging on the lit hemisphere at any given moment. That is double the flux (so, 480 W/m^2), impinging on *half* the surface area that the 240 W/m^2 is spread over. A hemisphere as opposed to a sphere.

  107. TrueSceptic says:

    I agree with Dr Roys. The total energy intercepted is that covering the earth’s shadow (the “disc”). It actually hits the “day” hemisphere, which of course is 2x the area and therefore average energy per unit area is halved. Over the whole planet, we double area and halve average energy per unit area again. What’s crucial is the total energy for all 3 is identical. Clearly we must divide by 4 for the whole earth.

    Now, what is Joe’s actual claim? Isn’t he claiming that we should divide by 2?

  108. Willard says:

    > [Joe] is *not* arguing that you can get twice the Sun’s power out of twice the area

    I’m not saying that this is what Joe says, kiddo. I’m saying that this is what he does. And I showed you that he does.

    I quoted him. I cited his graph. You have yet to quote him. All you got left is to spin the windmills of your mind. Mind your units instead.

  109. Joe sez …

    “When they do this, they artificially (it is artificial because it is no longer real, and only a mathematical simplification to make the Earth flat)”

    Unfortunately Joe is wrong. Just as with Diviner (Moon) or MESSENGER (Mercury (spelled it right this time)) or even a perfect 1:! tidally locked body we would always consider the entire body of revolution from a polar orbit wrt the body in question.

    Only later might we consider biased sampling, for instance taken from the Lagrange Points or out-of-plane of the orbital pair.

  110. Willard says:

    > Just as with Diviner (Moon)

    Please don’t mention the Moon, Everett. Kiddo’s using the same semantic tricks regarding her.

  111. TrueSceptic says:

    Willard,
    “The flux is *halved* when you take a hemisphere instead of the disk, and when you take the disc you *half* that half.”

    That is careless! words matter as much as units, do they not? The 2nd “disc” should be “sphere”.

  112. Willard says:

    > The 2nd “disc” should be “sphere”.

    No, it should not. The sphere is in all our divisions: it’s the 1 in the numerator. The 1 refers to the sphere (Area), the 2 to the hemisphere, and the 4 to the disc.

    And just to make sure, it’s the first “disc.” Kiddo’s using the American “disk.”

  113. Willard says:

    > Now, what is Joe’s actual claim? Isn’t he claiming that we should divide by 2?

    Since nobody here quotes Joe, I guess I will once again:

    If we wish to determine the physically instantaneous solar input energy density (Wattage per square meter) and corresponding heating temperature, via the Stefan-Boltzmann equation, we must use the correct actually-physical geometry. Thus, with a day-light hemisphere of half the surface area of an entire sphere, we must write the hemispherical equilibrium equation as: […]

    Source: https://principia-scientific.com/publications/The_Model_Atmosphere.pdf

  114. My bad, MESSENGER was not in a polar orbit, however BepiColombo will be in a polar orbit. Sorry about that one.
    https://en.wikipedia.org/wiki/MESSENGER
    https://en.wikipedia.org/wiki/BepiColombo

  115. “Please don’t mention the Moon, Everett. Kiddo’s using the same semantic tricks regarding her.”

  116. TrueSceptic says:

    Willard,

    Your sentence was confusing without that context. No wonder Joe’s lot find it easy to misrepresent you.
    Sphere = 1 (flux = ¼ = 240)
    Hemisphere = ½ (flux = ½ = 480)
    Disc/k = ¼ (flux = 1 = 960)
    Yes?

  117. Willard says:

    > The 1 refers to the sphere (Area), the 2 to the hemisphere, and the 4 to the disc.

    Let’s rewrite this as if I was speaking to a real kid while watching this video:

    The Earth looks like a sphere. It’s not exactly a sphere, but it’s close enough to one.

    A sphere has two hemispheres. Like the Earth. For our problem, instead of dividing the Earth at the Equator, we divide it according to how it faces the Earth. There is a small tilt, but you get the idea. One side of the Earth looks at the Sun, and the Sun shines on one side at a time. The last geometry bit we need to solve our problem is this: there are four “shadows” (or discs, or disks) in one Earth. Look at the video for the demonstration.

    (I skip the bits about zenith, Solar constants, albedo, etc.)

    Our problem is to determine how much Sun rays hit the Earth. There are two ways to do that. We can divide by for the amount of Sun rays, or divide by two. If we divide by four, we can take each point on the disc as facing the Sun at right angle. We can also divide by four, but then we have to adjust each point according to the angle from which it receives the Sun rays.

    (Cue to Bob’s examples.)

    Now, get this. The two methods lead to the same result. One is more complex than the other. So scientists simply divide by 4 and don’t have to do integrals. That is simpler, but then they did not think that some rascals could try to fool people on the Internet by trying to suggest that their trick hid something nefarious. We call these rascals Sky Dragons. When you look at it from a geometrical point of view, the trick is a bit silly, but not everyone knows that the two methods give the same results.

    Is that better?

  118. TrueSceptic says:

    Willard,

    Another quote from https://principia-scientific.com/publications/The_Model_Atmosphere.pdf
    “The true, and physically accurate average of the system, is that half of the surface of the Earth absorbs twice as much energy as the entire surface of the Earth radiates.”

    Let’s see Dr Roys explain that one away!

  119. TrueSceptic says:

    Willard,

    Who was that last message meant for?

  120. Willard says:

    > No wonder Joe’s lot find it easy to misrepresent you.

    I have yet to see any bit of discourse Sky Dragons can’t misrepresent, True One.

    I’m not sure I understand your numbers.

    Thanks for the quote!

  121. Dr Roys Emergency Moderation Team says:

    Yes, TrueSceptic, that is wrong as written. It should say, “The true, and physically accurate average of the system, is that half of the surface of the Earth absorbs twice as much *flux* as the entire surface of the Earth radiates.”

    480 W/m^2 flux received over half the surface area at every moment vs. 240 W/m^2 leaving from the entire Earth’s surface area. Energy is conserved then, because the area the flux is received over is only half that from which it leaves. Bad choice of words on Postma’s part.

  122. Willard says:

    > Who was that last message meant for?

    A kid. Myself. The audience.

    It’s my post as an elevator speech. That’s how I’d pitch it to my friends and family. That’s how I did it a few times already.

  123. Willard says:

    > The true, and physically accurate average of the system, is that half of the surface of the Earth absorbs twice as much *flux* as the entire surface of the Earth radiates.

    And that true, physically accurate average of the system, doubles the power of the Sun compared to scientists who divide by four instead.

    Hence why Joe claims that the Tyndall gas effect does not exist, or as he says in the abstract of his opus:

    We develop and describe the standard model of the atmospheric radiative greenhouse effect. […] A new starting-point model is introduced with physically accurate boundary conditions, and this will be understood to physically negate the requirement for a postulation of a radiative atmospheric greenhouse effect.

    Source: https://principia-scientific.com/publications/The_Model_Atmosphere.pdf

  124. Dr Roys Emergency Moderation Team says:

    Round and round we go…

    [Snip. You were warned, kiddo. You’re not Roy’s moderator. I’m AT’s moderator. – W]

  125. TrueSceptic says:

    Dr Roys,

    Thanks. How is the correct version a revelation? The following sentences, however, grossly misrepresent accepted science. Joe loves his straw men, doesn’t he?

  126. TrueSceptic says:

    Willard,

    Average flux has be doubled if you halve the area and want the same total.
    240*4 = 480*2 = 960
    (It’s ridiculous that I’m having to say this.)

  127. Dr Roys Emergency Moderation Team says:

    I tried posting a link but I think anything I post with links is going straight into the spam filter. Anyway, I think I might be outstaying my welcome already…

  128. Willard says:

    > I think I might be outstaying my welcome already…

    You can’t keep repeating the same points refuted a thousand times here, kiddo. You’re not at Roy’s. See you there.

    Meanwhile, take some time to think about the small circle.

  129. Willard says:

    > It’s ridiculous that I’m having to say this.

    It’s always better to say what one means, but as TS Eliot observed, it is impossible to say just what I mean!

    Language is a social art. Some use it to communicate. Otters use it to prevent communication.

    I’ve been more than a week at Roy’s, almost a month in fact, and you can be sure that kiddo is on the second camp.

  130. The latest argument on Joe’s blog seems bizarre. He seems to be suggesting that you can’t add energy fluxes.

  131. Willard says:

    > bizzare

    That’s because you don’t read Roy’s, AT, e.g.:

    Did TF choke on his own nonsense?

    We have to remember, TF believes fluxes add. So your example of a “red hot object”, say an iron pulled from a fire at 800C, could be made hotter with ice cubes, TF would claim!

    It might take several ice cubes, but TF could do the calculation.

    (Yeah, he’s an idiot.)

    https://www.drroyspencer.com/2021/04/uah-global-temperature-update-for-march-2021-0-01-deg-c/#comment-657475

    Clint and Joe share many beliefs. They have a similar prosody. I bet they use the same examples over and over again. Clint’s Climateball toolkit includes

    – the hammer
    – the cone
    – the two ice cubes
    – the steel plate

    Perhaps I should check at Joe’s to see if he likes these examples too?

  132. aljo1816 says:

    “The latest argument on Joe’s blog seems bizarre. He seems to be suggesting that you can’t add energy fluxes.”

    There seems to be an enormous amount of mental gymnastics going by Postma and his followers on to dodge the fact that Postma’s model requires some mechanism by which radiant energy emitted by a cooler object magically(?) veers away from any warmer objects or that the energy simply vanishes from existence upon being absorbed by the warmer object.

    The reality is (as I understand it, and please correct me if I’m wrong) that in an object of any temperature there will be some molecules in excited energy states and some in lower energy states, with only the average energy of the molecules being higher than the average energy of a cooler object. The low-energy state molecules will be ready to absorb incoming radiation in the right wavelengths, regardless of the temperature of the radiation’s source. The reason that the net flow will be from warmer > colder is not because of any magical property of photons to read temperatures, but because the warmer object will be shedding energy via radiation faster than the cooler one could ever supply it.

  133. aljo1816,
    Yes, the net transfer of energy is always from the hot object to the cold energy. That doesn’t mean that energy from the cold object can’t be absorbed by the hot object, just that in a situation where there is a transfer of energy between a hot object and a cold object, the net effect is that the cold object will gain energy and the hot object will lose energy (assuming there is no other energy generation, or transfers, taking place).

  134. Keith McClary says:

    “You should really quit while you can Joe, you are making a fool of yourself when you make such claims that are so easily disproven. – Anthony”
    https://wattsupwiththat.com/2013/05/27/new-wuwt-tv-segment-slaying-the-slayers-with-watts/

  135. Willard says:

    Looks like I made it at Joe’s.

    Perhaps not in the comments anymore, but still. Great success!

  136. Willard says:

    Nice find, Keith!

  137. TrueSceptic says:

    Keith McClary says:
    April 28, 2021 at 8:41 pm
    “You should really quit while you can Joe, you are making a fool of yourself when you make such claims that are so easily disproven. – Anthony”
    https://wattsupwiththat.com/2013/05/27/new-wuwt-tv-segment-slaying-the-slayers-with-watts/

    That’s great. I know this was 8 years ago, but the real Willard has gone up a few notches (from a low point) in my estimation.

    But why does Roy Spencer allow this nonsense to be peddled on his blog?

  138. Willard says:

    > the real Willard

    I resemble that remark!

    To answer your question, here’s how I see it. Curating a blog takes time. Roy is retired. His WP install is deprecated: his encoding sucks, his spam filter is more capricious than ours, etc. He tried to ban Mike Flynn and other Sky Dragons a few times. They keep coming back; now there’s even one who plays the Ref over there.

    The only way would be to moderate his blog and to zamboni threads after threads. But each thread gets more than a thousand of comments. That takes time. What’s the upside? So my guess is that Roy uses noise to cancel noise. Nobody can read the threads without an RSS reader, and nobody but a few ninjas use them nowadays. Half of the comments are made by the same guys. Those who are used to read Roy’s must skip them.

    Even I can’t read them all. I skim, and I comment according to my communication objectives. But I do that to improve my Climateball Bingo and Manual. Once these are done I’ll probly turn to Reddit. Who knows?

    So there you go.

  139. Nick Stokes has a post rebutting Roy Spencer’s post.

  140. Dr Roys Emergency Moderation Team says:

    [Playing the ref. Let’s play it out over there. -W]

  141. Willard says:

    This paragraph from an new comment in an old thread at SoD’s might be relevant here:

    Planck’s Law is for the intensity of radiation emitted in all directions – over a sphere of 4Pi steradians, which is why “sr” [is] present in W/m2/sr/um. This is appropriate for emission by GHG molecules in the atmosphere. For radiation emitted by the surface of the earth, radiative cooling to space (OLR), or DLR or SWR arriving at the surface of our planet, we usually think in terms of the “flux” of radiation perpendicular to a surface in units of W/m2 or possibly the “spectral flux” (W/m2/um). We talk about fluxes perpendicular to a surface because “viewing angle” is critical to radiative transfer of power (Lambert’s cosine law). When you convert the intensity (W/m2/sr) emitted from a point on a surface over a hemisphere (2Pi steradians) into the flux emitted perpendicular to the surface, you pick up factor of Pi. 1 W/m2/sr becomes 3.14159 W/m2 radiated perpendicular to the surface. When considering GHGs emitting in all directions in the atmosphere, 2 W/m2/sr becomes 3.14159 W/m2 traveling upward (OLR), 3.14159 W/m2 traveling downward (OLR) and horizontal fluxes that cancel. The is called the “two-stream” approximation. Climate science usually deals with incoming or outgoing flux (W/m2) perpendicular to the surface, not intensity (W/m2/sr).

    https://scienceofdoom.com/2010/06/01/the-sun-and-max-planck-agree/#comment-163359

    Nothing can prevent Sky Dragons to pretend that Joe is not getting a number that is twice the usual one by dividing the sphere by an area that is twice the size of a disc, misapplying Lambert’s law along the way.

  142. Willard says:

    Sometimes it’s worth it to deal with contrarians:

    You’re still evading a very basic point, kiddo:

    Unless and until Joe can prove that every point on the hemisphere are [sic] at zenith, he can’t divide by two {1}.

    Yet Joe already conceded that only a small disc was at zenith.

    Seldom do we see such a pure reductio ad absurdum in Climateball.

    Thank you for helping me find it!

    {1}: We can indeed divide by four because we take all the points of the disc as being at zenith.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-678017

    The small disc I’m referring to is on Joe’s diagram.

    ADD. Let’s hope that the concept of zenith will allow less wiggle room than the concept of power.

  143. Russell says:

    Willard
    The Zenith concept is indeed an important one, as the angle of the sun above the horizon has a massive effect on solar heating of the Earth’s surface, because most of that surface is covered with water.

    The reflectivity of water at normal incidence, i.e., with the sun at the zenith, is only 7%. That’s a very low figure- asphalt is only slighty brighter , by as the sun angle falls the , reflectivity rises, for at angles below 45.8 degrees, calm water reflects sun and sky.

    A good thing it does– were water’s refractive index any lower , less solar energy would bounce off the Arctic ocean, and more would chew on sea ice.

  144. Bob Loblaw says:

    With respect to the two-stream approximation, you need to also think in terms of the radiation incident on the surface. That also comes in from all directions. In the case of IR, it is generally fairly close to isotropic – equal from all directions in the sky. Why? Because in all directions there is similar atmosphere emitting in all directions… (Obvious exceptions – low scattered cloud where the IR from clouds will be different from the IR making its way through the gaps from the clear sky above).

    Radiation from any other direction besides directly overhead (zenith) needs to be adjusted according to the zenith angle. So, to get the radiation incident on a surface, you need to integrate over all solid angles of incidence (the hemisphere that the sky represents, in spherical coordinates)). So, without the two-stream approximation, you need to emit in all directions, and also receive from all directions. When you do all the integrating, you’ll end up with the W/m^2 result.

    For solar radiation, the diffuse sky radiation is a little less isotropic – especially with scattered cloud – which complicates things. For direct solar radiation, things are far from isotropic and angle of incidence is extremely important. Any decent solar radiation measurement system includes three instruments: global (unshaded, full hemisphere view), diffuse (shaded, full hemisphere view), and direct (narrow view of sun disk only, tracking the sun).

  145. Willard says:

    I think we’re nearing the end:

    > The lit hemisphere absorbs 480 W/m^2 from the sun, all things considered, including angles, yes.

    And to do so Joe divides by 2 only once, both to spread the flux on the hemisphere, and to correct for the angles.

    Don’t you see the problem here?

    If Joe divides by two for each, he gets the divide by 4 trick everyone but him understands and uses!

    AT and Bob already provided the geometrical proof, btw.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-678124

    This episode might be the first time where I see an advantage in the Socratic procedure. That it’s a geometry discussion might not be accidental. Nevertheless, I now can appreciate that the first virtue of going step by step to verify at each point if the interlocutor is well understood parries artful dodging.

    We shall see if that’s enough. I would not bet the farm on a positive outcome.

  146. gator says:

    This is supposed to be about units, right? Maybe it is better to think in terms of power, J/s, instead of flux, J/s/m2. Physically, the earth absorbs a certain power, and radiates the same power in equilibrium. The point of flux vs. disk area vs. hemisphere etc is really just to make calculating *power* easier. Whatever combination of flux&area you pick, the power input from the sun should be the same.

    BTW, all this nonsense about heat flow, cold objects affecting temperature of a warmer object etc. How do people think a thermos works? Having designed a refrigerator that operates at mili-Kelvin, I guarantee that Eli’s plate model is real, and that operating at mili-Kelvin requires multiple plates to cut heat flow between room temperature and the cold plate.

  147. Willard says:

    Thanks gator.

    I think it’s better to think in all the terms possible. The more the merrier {1}. Your usage of power seems to be compatible with mine. Perhaps I err. I don’t need it: all I need is another concept than flux to help describe Joe’s trick.

    {1} Perhaps not at the same time, as we’re having problems with only a few concepts.

  148. Bob Loblaw says:

    The down side of going to power instead of flux density is that the geometry is inherently spherical and four-dimensional (including time). Forget the fourth dimension and Marty doesn’t understand time travel. Forget the third dimension, and Khan loses to Kirk.

    It’s difficult to explain spherical trigonometry to someone that can’t understand Cartesian trigonometry,

    Solar radiation received by earth is essentially a set of parallel rays, and the strength decreases by the inverse square law as distance from the sun increases. W/m^2 is the most reasonable representation.

    With power, you still need to multiply by areas, so you are just moving where area is used in the calculations. Six of one; half dozen of the other.

  149. Willard says:

    > Forget the third dimension, and Khan loses to Kirk.

    https://www.khaaan.com/

  150. Willard says:

    Come to think of it, the idea of converting to Joule per second square meter might be fruitful at the argumentative level, since you make the time dimension explicit. The “second by second” argument becomes harder to deploy. Also more difficult to forget that the overall input can be measured as a yearly average. Which then begs the question: over what area has our measurement been made?

    Then all we need to understand is that irradiance is measured perpendicular to the incoming sunlight and along with some algebraic geometry we should be good to go.

  151. Willard says:

    I should have looked back at SoD’s.

    First, the graphic:

    Second, the quote:

    Look back at the graphic – is the sun shining equally on every part of the earth every second, for all 24 hours of the day? It’s not. It’s shining onto one side of the earth. It’s night time for half the world at any given moment.

    So think of it like this – the absolute maximum area receiving the sun’s energy on average can only be half of the surface area of the earth – 2πr2 (=4πr2/2)

    But that’s not the end of the story. Picture someone where the sun is right down near the horizon. It’s still daytime but obviously that part of the earth is not receiving 1367W/m2 – they are receiving a lot less. In fact, the only spot on earth where someone receives 1367W/m2 is where the sun is directly overhead. So the effective area receiving the solar constant of 1367 W/m2 can’t even be as high as 2πr2.

    Linking Incoming Solar Radiation to the Earth’s Outgoing Radiation

    The earth radiates out energy in a way that is linked to the surface temperature. In fact it is proportional to the fourth power of absolute temperature.

    As we think about the earth radiating out energy, it might be clearer why we labored the point earlier about the area that the sun’s energy was received over.

    Take a look at that graphic again. The energy from the sun hits an effective 2d disc with area = πr2.

    https://scienceofdoom.com/2010/02/06/the-earths-energy-budget-part-one/

    Ah, well. A week of Climateball can save ten minutes reading.

  152. Bob Loblaw says:

    The usual term is “extraterrestrial radiation” for the amount arriving on a horizontal surface at the top of the atmosphere (i.e., parallel to the earth’s surface, not perpendicular to the sun’s rays).

    And even Wikipedia has a decent write-up, with equations and diagrams. This is undergraduate climate stuff. I learned it it third year.

    https://en.wikipedia.org/wiki/Solar_irradiance

  153. It is even simpler than that Bob …
    Stefan–Boltzmann law
    https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Effective_temperature_of_the_Earth

    No one, and I do mean no one, disagrees with that derivation except for you know who, those who follow the Sky-Dragon law. Which you will only ever see in dorky (pay-to play-predatory) journals or knuckle dragging blog posts at contrarian websites.

  154. “Because of the greenhouse effect, the Earth’s actual average surface temperature is about 288 K (15 °C), which is higher than the 255 K effective temperature, and even higher than the 279 K temperature that a black body would have.”

    Glad to see that we are all in agreement then. 288K – 255K = 33K due to GHG’s. Now we can move on to more important stuff.

  155. Then he should publish his stuff in any well respected climate science journal. Nature and/or Science would be more than glad to publish such a Nobel Prize winning article. Or not, for all to obvious reasons.

    There is a new contrarian journal, he should publish there, the name of the journal is Manure In Nature.

  156. TYSON MCGUFFIN says:

    Spending a few hours in Physical Meteorology Lab doing this exercise is sometimes worthwhile.

  157. Willard says:

    Nice graph, TYSON!

    If you add the jpg on a line alone, it shows up:

    You can hotlink it too.

  158. Bob Loblaw says:

    I’ve seen graphs like that dozens of times over the last 45 years, and given them to students. I even know how to make them myself. Nothing new here.

  159. Willard says:

    All our bases belong to Arthur:

    Source: https://arxiv.org/pdf/0802.4324.pdf

    Beautifully written!

  160. gator says:

    My suggestion to think in terms of power is simply that power in = power out. Flux can be different (will be different) because the area “catching” the power in is the disc, the area radiating power out is the whole sphere. Flux in and out will never match unless the sphere is surrounded by a uniform temperature. But like you say there are many ways to get to the right answer. And it’s a good sign when you can use multiple lines of reasoning to come up with the same answer.

  161. Willard says:

    Thanks.

    Something like you suggest emerged this evening;

    Allow me to rephrase “What happens when we double the Disc variable”: suppose some Joker J, instead of using

    (1) Power = Disc x Sun

    wanted to switch to

    (2) Power = Hemisphere x Sun

    Wouldn’t you say that J is doubling the Sun’s power?

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-679484

    As you can see, no “divide by four” can get in the way. So thank you for that.

    In the above, Sun refers to irradiance, and Disc or Hemisphere refers to “the area the planet subtends in the plane perpendicular to the radiant propagation direction,” to borrow Arthur’s wording.

    One difficulty I have with minding units the normal way is: what happens when contrarians won’t calculate?

    The one big breakthrough that came out of the exchange is that kiddo at last conceded that he is “having a bit of fun.”

  162. Willard says:

    > The one big breakthrough

    Well, after each breakthrough some regression to the mean is to be expected. Kiddo will soon turn to copypasting random snippets from Joe’s. The first one ends thus:

    With an energy input of 1.22 x 10^17 Joules over a hemisphere in one second from the Sun, and an energy output from the Earth of 1.22 x 10^17 Joules from the entire globe, i.e. both hemispheres, it is not physically correct to equate these energy values in terms of flux. These values are true and totally correct in terms of energy. They can not be made to be equal in terms of flux.

    For example, if we say that the Earth is in numerical flux equilibrium with the Sun, and mistake this for conserving energy, then that would mean that the Earth must emit the same flux of energy as it receives the Sun. Therefore the Earth must emit 480 W/m^2 on average since that is what it receives from the Sun on an instantaneous basis.

    Well, the Earth does not emit this flux of energy. That is way too high of a value. If you converted that value into total energy emitted per second over the entire globe, it would be more energy than actually comes in. The known and measured value for the flux output from the Earth is 240 W/m^2.

    https://climateofsophistry.com/2013/09/25/fraud-aghe-18-conserving-wattage-not-physics-rant-free/

  163. Holger says:

    Amazing, some here (Roy…, True Skeptic) seem to make the same errors many of our students do in undergraduate course. They forget that you usually have scalar products with the surface normal when doing the flux integrals. Instead they know of simple examples one does for illustration purpose (usually flat surfaces) leading to the direct change in flux with the surface area. It seems they are falling into this trap here, too, by talking of a factor 2 in flux when taking one or both hemispheres, but forget, that the incoming flux is _not_ approaching the hemisphere radially, that is true only for the emitting part.
    If they would talk about light coming in radially at each point of a hemisphere, their argument would make sense.
    Ho can one, after so many discussions or even after ATP’s integral calculation/demonstration still not understand this?

  164. Holger,

    Sky Dragons do not believe in GHG theory..

  165. TrueSceptic says:

    Holger says:
    May 1, 2021 at 9:24 am
    Amazing, some here (Roy…, True Skeptic) seem to make the same errors many of our students do in undergraduate course.

    What errors have I made (quote me, do not paraphrase).

  166. Holger says:

    @TrueS You are not reading what others wrote and say. ATP and Willard are referring to the flux coming in from the sun, which is _not_ going in radially towards the hemisphere.
    You are talking about something completely different, which has no relevance to the problem at hand:
    “Average flux has be doubled if you halve the area and want the same total.240*4 = 480*2 = 960”
    If you talk about a sphere radiating out (along the surface normal) and then half the surface area, while keeping the resulting integral value over the surface constant, then surely the flux has to double. But that is a completely different problem, something ATP and Willard try to explain to some here for some time now.

  167. Willard says:

    > What errors have I made

    Since you insist:

    Please acknowledge that both the Sun’s constant and the disc give W/m^2 values that are at zenith whereas the hemisphere does not, kiddo.

    By Sun’s constant, I am referring to this:

    [T]he net amount of solar radiation arriving on a 1 m2 area (perpendicular to sun) on the earth’s surface […]

    https://math.nyu.edu/faculty/kleeman/zero_dim_ebm.html

    When I write “at zenith” I am using a shortcut to refer to the situation where the Sun hits the surface at perpendicular angle.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-679206

    The problem is that when we divide by 2 we have not corrected the angles. We need to divide by 2 twice. Then we get a division by four.

    Joe’s trick is a mere double accounting trick.

  168. Willard says:

    In fairness to True One, Chris’ wording could have been better:

    The factor of 4 is the ratio of the surface area to the cross section of the planet, and is the shadow cast by a spherical Earth. It is therefore a geometrical re-distribution factor; it remains “4” if all the starlight is distributed evenly over the sphere; it is “2” if the light is uniformly distributed over the starlit hemisphere alone; with no re-distribution, the denominator would be 1/cosine(zenith angle) for the local solar flux.

    https://skepticalscience.com/postma-disproved-the-greenhouse-effect.htm

    That said, I’m the last Climateball player to be in a position to complain about wording. Moreover, he cited Arthur’s paper. I’m thankful he did. If you check back the screenshot I provided in a recent comment of mine, you’ll notice that it ends thus:

    Integrating this [s(x, t) = cos(θ(x, t))S(t)] on a spherical planet (including only the lit side) gives the πR^2 factor in equation 1.

    Joe’s trick works better on diagrams, hence why he’s always using them to convey his point. When he works with equations he needs to break the equality between the inputs and the outputs. That leads him to question SB. But then that turns the discussion into physics, whereas the double accounting trick rests on algebra and geometry.

  169. TrueSceptic says:

    Holger says:
    May 1, 2021 at 2:28 pm
    @TrueS You are not reading what others wrote and say. ATP and Willard are referring to the flux coming in from the sun, which is _not_ going in radially towards the hemisphere.

    No error. I am talking about the *total* energy, which is all that matters here. It doesn’t matter what the flux is at any point, and calculating and integrating that just adds pointless complication. It is this very simple point that has caused all the stupid to-and-fro between Joe’s… and Willard.

  170. Willard says:

    > I am talking about the *total* energy

    Talking is a big word, True One.

    The only time you did you felt silly.

    One problem with playing Socrates is that people may not bite.

  171. TrueSceptic says:

    Willard,

    I don’t know what your problem with simple English is. Whose side do you think I’m on anyway?

  172. Willard says:

    True One,

    You’re not listening. Everything you say can be used against you on a Climateball court. That can extend to people that contrarians think are on “my” side. I spent my day yesterday trying to correct your [wording] at Roy’s.

    If you want to make a case, make it. Don’t ask anyone else to make it for you. If you think you can do better with Sky Dragons by using energy instead of flux, be my guest.

    So far, it has been the opposite of helping.

  173. TrueSceptic says:

    This is getting strange. Why would you be “trying to correct [my] wording” at Roy’s?

    I’m getting the idea that you’d rather play the smart alec than engage in discussion.

  174. Holger says:

    @Skeptic
    You still don’t get it. To get the total incoming energy you need _first_ to do an integration to account for the difference in angle of the incoming radiation to the surface normal. For a given location x on the planetary surface, the normal to the plane of the local surface makes an angle θ(x, t) with the radiation propagation direction of this stellar source (the radiation is practically coming in parallel towards the earth, it is not coming in radially), giving you a effectively F_in(t)*cos(theta(x,t)). This is what ATP did similarly in his comment on April 26, 2021 at 4:56 pm. This gives you the F_in*pi*r^2.
    THIS is the total incoming energy. The outgoing radiation is along the surface normal and involves the full 4 pi r^2. The ratio gives you the factor 4.

  175. ATTP,

    Is there any way to determine if the Sun has an asymmetry in its black body radiation? Meaning an asymmetry from equator/ecliptic to its poles (e. g. the Sun rotates close to the ecliptic as does much of the Solar System, Earth’s orbit being the ecliptic reference afaik).

    Or do we have to wait for Solar Orbiter …
    https://en.wikipedia.org/wiki/Solar_Orbiter
    “During the mission the orbital inclination will be raised to about 24°.”

    I would think that a polar orbit about the Sun at say one AU (or less) might shed some light 😉 on this. More, same, less or too small to measure or matter. Or perhaps solar/stellar theories.

  176. ATTP,

    I think I have answered my own question …

    A journey of exploration to the polar regions of a star: probing the solar poles and the heliosphere from high helio-latitude (posted 22 Apr 2021)
    Source: https://arxiv.org/pdf/2104.10876.pdf

    See … 2.3 Science goal 3: To determine the solar irradiance at all latitudes (starting on p. 13)

    You may know the lead author (or other authors) of that manuscript (it looks like it has been submitted (or will be submitted) to a journal called Experimental Astronomy).

  177. MP says:

    Complication #1:

    The 4th root problem Say that two spots on your blackbody sphere are being exposed to 50 and 100 watts per square meter. (Due to curvature, remember, a single light source gets spread out and becomes weaker.) Using the Stefan-Boltzmann equation, thetwo temperatures will be about 172 and 205 Kelvin respectively, i.e., an average of 188.5K. But the average irradiance is 75 W/m², which corresponds to 191K. That’s 2.5 degrees off the mark. In other words, average temperaturedoes not agree with average irradiance, and vice versa. Take three spots at 100, 200, and 300 W/m². The average of course is 200 W/m². The temperatures are 205, 244, and 270 respectively, averaging about 240K. But 200 W/m², the average, equals 244K. Now you’re 4 degrees off the mark. And so on, as you proceed to compare irradiance with temperature on each and every angle of a half-lit sphere. It’s a huge problem to tackle. Throw in rotation (i.e., the irradiance is constantly changing) and the heat-retention of various three-dimensional substances, and the problem runs out of control.

  178. 190.63K (SB 4th order weighting) versus 188.5K (linear 1st order weighting).
    243.95K (ditto) vs 239.67K (ditto)

    I am quite happy with both sets of numbers.

    You do realize what a narrow banded distribution is? Like the Earth’s oceans ~2/3 open water, no matter what time of day or year and all of that at the same ~geoid (surface) elevation. Range in Kelvins 271.15 to 303.15, so delta max of 32K at, I don’t know, but say 287.15K simple mean … to 1st order both calculations 288K linear estimate and whatever the SB gives will very likely be within 3K of each other.

    You should download the Diviner dataset (my 1st post here) and do the math or take any GCM/ESM from CMIP5/CMIP6 and … do the math. Suggest you use PI and something on the order of one to three hours for your time step.

    I also played around with a very simple distro for the Earth and that is all you need, a distro for the body in question (preferably from a polar orbit or numerical model data). I am too busy right now (building an i9-10980XE computer or two), otherwise I would do most of the heavy lifting.

  179. “to 1st order both calculations 288K linear estimate and whatever the SB gives will very likely be within 3K of each other.” is for the entire Earth. The oceans will be very close to its simple linear average and a SB (4th order) weighting, very likely significantly less then 1K delta. The oceans are the easier part to illustrate here.

  180. Bob Loblaw says:

    MP:
    So, you create three-dimensional climate models. Are you trying to point out that zero-dimensional climate models are not three-dimensional?

    You can do one-dimensional (either zonal or radiative-convective) and two dimensional (rare, but I’ve heard they exist) on your way…. Climate scientists did.

  181. MP says:

    It has been demonstrated that the widely-accepted divide-by-4rule cannot reliably predict the actual temperature conditions on a globe due to the deviations inherent in a 4th power law, which is also a 4th root law.

    To explain further, 16 times more energy brings about a doubling of temperature because temperature conforms to the fourth root of the radiant energy.

    Thus,

    1 unit of radiance = 4√1, i.e., one unit of temperature
    2 units of radiance = 4√2, or 1.189207 units of temperature
    4 units of radiance = 4√4, or 1.414214 units of temperature
    8 units of radiance = 4√8, or 1.681793 units of temperature
    16 units of radiance = 4√16, or 2 units of temperature

    In detail, then, the divide-by-4 practice consists of mistakenly dividing a uniform disk temperature by the fourth root of 4

  182. Willard says:

    > In detail, then, the divide-by-4 practice consists of mistakenly dividing a uniform disk temperature by the fourth root of 4

    If only climate scientists knew algebra.

    A simple workaround is to abstract all irrelevant details with a variable, say T:

    If you accept that

    [EMB] Disc x Sun = Area x T

    Joe can’t divide by 2 unless he wishes to do an energy-balance model for half of the Earth, like Russell suggests here:

    https://vvattsupwiththat.blogspot.com/2020/07/schellenberger-sequel-will-connect-dots.html

    But even then that’s a generous interpretation, for that kind of hemisphere is 3\pi^2.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-680048

  183. MP,
    This is why I’ve been trying to point out to you that the equality

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    gives us the effective radiative temperature of the Earth. In a sense, it’s a truism. Given that the system will tend to energy balance, the effective radiative temperature of the Earth will be equivalent to that of a blackbody that radiates the same of energy per square metre per second.

    Also, this is not meant to represent the temperature of the surface, because the greenhouse effect amplies the surface temperarture, relative to this effective radiative temperature. At best, it’s an approximation for the temperature at an altitude of 5km in the atmosphere. Noone is claiming that this effective radiative temperature is the same as you would get if you were to somehow measure the temperature and average.

  184. Dave_Geologist says:

    ISTM it’s really simple. Childishly simple, since I’m sure I learned about the area of a circle and of a sphere at school, not as an adult.

    The Sun is so far from the Earth we can treat its rays as arriving in parallel. From the POV of those rays the Earth looks like a disc, and intercepts a disc’s worth of outgoing solar radiation. Area intercepted: πR². The Earth radiates infrared from its entire surface, in all directions. Radiating area: 4πR². Ratio: 4.

    If your thermodynamics is stuck in the mid-19th century and you think cooler objects can’t radiate towards hotter ones, subtract a tiny amount for the area of the Sun’s disc as seen from Earth. But ask yourself a couple of questions first. How does that photon choosing not to head towards the Sun know it’s heading towards the Sun when it won’t get there for another eight minutes? What happens to that heat that you think can’t escape? Bear in mind that the Earth is rotating and the Sun is only momentarily at the zenith.

  185. Bob Loblaw says:

    MP: “It has been demonstrated that the widely-accepted divide-by-4rule cannot reliably predict the actual temperature conditions on a globe due to the deviations inherent in a 4th power law, which is also a 4th root law.

    To explain further…

    Right. To explain further, we know that a zero-dimensional climate model, which does not resolve the horizontal geography of the globe, and does not resolve the complexities of the vertical variations in the atmosphere, is not perfect.

    Other climate models do resolve horizontal and vertical variations in the earth-atmosphere system.

    The “divide by four” rule is a simple matter of geometry, not climatology. Are you trying to claim that our understanding of geometry is incorrect? Are you trying to claim that you can “fix” the lack of details in the climate model portion of the zero-dimensional model by changing the fundamentals of geometry?

  186. Dr Roys Emergency Moderation Team says:

    “My opponent has nothing else on his side…he has no model.”

    Willard, I explained this to you yesterday and the day before. The only “model” we need is:

    Total power absorbed = Total power emitted = 1.22 x 10^17 Watts

    The 1.22 x 10^17 Watts comes from taking the solar constant (Joe typically uses 1,370 W/m^2) and multiplying it by the surface area of the disk (pi x r^2) intercepting the Sun’s energy, then multiplying the result by 0.7 to factor in albedo.

    So, it is 1,370 W/m^2 x pi x 6,371,000 meters x 6,371,000 meters x 0.7 = 1.22 x 10^17 Watts.

    That is the total power that the Earth absorbs, so it must be the total power that the Earth emits. In any one second, the Earth absorbs over the lit hemisphere. The area of the hemisphere is 2.55 x 10^14 square meters.

    1.22 x 10^17 Watts divided by 2.55 x 10^14 square meters equals approx. 480 W/m^2.

    In any one second, the Earth emits over the entire sphere. The area of the sphere is 5.1 x 10^14 square meters.

    1.22 x 10^17 Watts divided by 5.1 x 10^14 square meters equals approx. 240 W/m^2.

    At any given moment, the Earth absorbs 480 W/m^2 and emits 240 W/m^2. Flux is not conserved, but energy is, because the area the Earth absorbs the energy over is half that of the area that the energy leaves from.

  187. Willard says:

    > The “divide by four” rule is a simple matter of geometry, not climatology.

    To clarify, it’s also a matter of algebra. Energy balance models balance the energy emitted with the energy received by comparing the flux received, in Watt (per meter second) to the temperature that the Earth emits (in Kelvin). For some philosophical reason, Joe rejects that balancing act for fluxes:

    With an energy input of 1.22 x 10^17 Joules over a hemisphere in one second from the Sun, and an energy output from the Earth of 1.22 x 10^17 Joules from the entire globe, i.e. both hemisphere’s, it is not physically correct to equate these energy values in terms of flux. These values are true and totally correct in terms of energy. They can not be made to be equal in terms of flux.

    https://climateofsophistry.com/2013/09/25/fraud-aghe-18-conserving-wattage-not-physics-rant-free/

    Hence why I don’t believe that reducing everything to energy may not work.

    Sky Dragons interpret the divide-by-4 trick as some kind of magic that “spreads” the flux over the “entire globe,” as Joe says. Instead they prefer to divide by 2, to represent the hemisphere that receives light. One problem with that approach, as I underlined in my post, is that if they seriously take the hemisphere as some kind of True Receptor of the divine light from the Sun, they need to deal with the geometrical identity between a disc and a hemisphere corrected for its angular reception. So by that logic, we would not need to divide at all.

    Sky Dragons seem to fail to appreciate that the sphere has been introduced on the right side of the equation. It represents the emitter part of our model, not the receptor part. If the emitter part of our model does not receive the flux from the receiving part, from where does it get its energy?

    Thus Joe’s shenanigans about divorcing energy and fluxes.

    ***

    Philosophy is littered with mistaken ideas. One could argue that once a philosophical idea works, it becomes science. (It’s not a correct idea, but it evokes a credible tendency in idea formation.) When we go back and review these historical documents, we need to develop a mind flexible enough to be able to understand ideas that don’t work. That’s the only way to explain to ourselves why they don’t. In my experience, scientists have less patience with that kind of endeavor, at least outside their academic obligations.

    In a sense it is true to say that geometry isn’t involved into our energy considerations. But for Sky Dragons it’s critical. If we do not refute that mistaken idea directly, it will persist. It might still persist if we refute it, but at least it will become a point refuted directy a thousand times.

  188. I’ll make the point I made a few days ago. In most circumstances you don’t need to divide by 4 (i.e., just equate the energy in and the energy out). However, if you do want to – for example – compare a change in GHG forcing with a change in solar forcing, then you do need to divide by 4 (and take the albedo into account) because a change in forcing is defined as being an average across the whole sphere (or, more correctly, it’s how some change influences the energy balance which can then be described in units of W/m^2).

  189. Willard says:

    I just released kiddo’s last comment from spam, and thank him for taking his best shot so far.

    By serendipity, it confirms everything I said in my last comment.

    In Sky Dragon world, energy is Joule, flux is Watt per meter per second, but Energy is not flux.

    Perhaps gator was onto something when he said we should Joule all the things:

    Maybe it is better to think in terms of power, J/s, instead of flux, J/s/m2. Physically, the earth absorbs a certain power, and radiates the same power in equilibrium. The point of flux vs. disk area vs. hemisphere etc is really just to make calculating *power* easier. Whatever combination of flux&area you pick, the power input from the sun should be the same.

    Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190565

    But again, how could that solve the difficulty of shadowboxing arguments such as “the Earth absorbs the energy over is half that of the area that the energy leaves from”?

    Hence the post.

  190. > At any given moment, the Earth absorbs 480 W/m^2 and emits 240 W/m^2. Flux is not conserved, but energy is, because the area the Earth absorbs the energy over is half that of the area that the energy leaves from.

    Well, what I think, if estimate the incoming and outgoing radiative energy on the average, which we are not justified to do…
    But let’s say we look at the matter on the average, which does not work the average way, since the vast majority of the IR outgoing emission occurs during the surface solar irradiated hours…
    And only a tiny portion is IR outgoing emission during the night-hours…

    Let’s see then:
    Energy in = energy out
    Energy in = Φ(1-a)So πr^2 = 0,47*0,7*1361 πr^2 = 444 πr^2 ( W )
    or, on average on the cross-section disk
    Energy in = 444 πr^2 ( W ) / πr^2 ( m^2 ) = 444 W/m^2
    and
    the Energy out on average from the entire Earth’s surface
    Energy out = 444 * πr^2 ( W ) /4 πr^2 = 111 W/m^2
    But it is not happening on average… you do know that, don’t you?

  191. Willard says:

    > which we are not justified to do…

    I already responded to that point, Christos:

    > because planet surface does not emit on the average VALUES

    The planet surface only receives the values you take to be granted on average, Christos.

    So by your logic you can’t do *any* algebraic operation on your model.

    None whatsoever.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-680852

  192. Please explain… On some example… I do not understand…

  193. Christos,
    I don’t follow your equation. Where does this come from?

    Energy in = Φ(1-a)So πr^2 = 0,47*0,7*1361 πr^2

  194. It is the Solar Flux on the TOA – Reflected portion (specular + diffuse) = Energy in
    [1 – Φ(1-a) ]So = ( 1 – Φ + Φa )So…… is the reflected portion
    Φ = 0,47
    a = 0,3
    (1 – 0,47 + 0,47*0,3)So = (0,53 + 0,141)So = 0,671So
    0,671*So is the reflected portion (specular + diffuse) of the So
    when Solar flux is perpendicular to planet cross-section disk.
    And, what is left to “absorb” (the not reflected portion) = Φ(1-a)S πr^2 = Jemit
    Also, according to the Planet Solar Irradiation Absorbing-Emitting Universal Law
    the total surface Jemit =
    Jemit = 4πr²σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W)
    So we shall have:
    Planet Energy Budget:
    Jabs = Jemit
    πr²Φ*S*(1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴ (W)
    Solving for Tmean we obtain the PLANET MEAN SURFACE TEMPERATURE EQUATION:
    Tmean.planet = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (K)
    Also, please visit

    SOLAR ENERGY BUDGET
    https://www.cristos-vournas.com/448704125

  195. Christos,
    Except, the amount of energy we receive from the Sun per unit time is simply:

    E = \pi R^2 (1 - A) F_{\odot}.

    Where’s your 0.47 coming from?

  196. When I realized that for the smooth planets surface the parallel solar rays irradiation should be not only diffusely reflected, but also specularly reflected, when I realized that I started searching for papers describing the solar parallel beams reflection from the spheres.

    No matter how hard I searched for, there were none. So, I concluded there are not papers about the sphere’s the parallel solar beams specular reflection.
    Since I could not let it there unsolved, I decided to look somewhere else.

    I knew about the different shape bodies having different resistance to the liquid parallel flow when the Reynold’s Number is Re < 10000.
    So I looked for a Table of Measured Drag Coefficients.
    And, for a smooth sphere the Measured Drag Coefficient is
    0,47

    https://en.wikipedia.org/wiki/Drag_coefficient
    I welcomed the 0,47 as the planet spherical surface solar irradiation accepting factor for the smooth surface planets without-atmosphere (an analogue of parallel liquid flow resistance).

    Φ = 0,47 (Φ is from the Greek word Φως [phos], the light. Like in the Phosphorus.
    And Φ varies from 0,47 for smooth surface without-atmosphere planets to 1 for heavy cratered planets, and 1 for gases planets.
    0,47 ≤ Φ ≤ 1
    Thus, solar system planets surfaces have been divided in three major categories – the smooth without atmosphere ones, the heavy cratered without atmosphere and the gaseous.

    The heavy cratered planets and the gases planets do not reflect specularly, they reflect only diffusely.
    During their multibillion years History, every planet has developed either a smooth surface Φ = 0,47 , or a heavy cratered Φ = 1.
    Only Triton (Neptune’s satellite) does not get in either categories. In the case of the Triton Φ is neither 0,47 , no 1, but somewhere in between.

  197. Christos,
    You don’t need to do any of that. The amount of energy the Earth intercepts from the Sun per unit time is simply:

    E = \pi R^2 (1 - A) F_{\odot}.

  198. No specular reflection from the planet surface then?

  199. That’s all accounted for in A.

  200. Just to clarify, A (the albedo) is the fraction of incoming radiative energy that is reflected back into space. So, if we know the solar flux at the radius of the Earth’s orbit (F_{\odot}) and the fraction of the incoming energy that is reflected back into space (A) then the amount of energy the Earth absorbs per unit time is simply:

    E = \pi R^2 (1 - A) F_{\odot}.

  201. Here is a paper about the satellite measured planet albedo being a diffuse reflection only.

    19940020024.pdf (nasa.gov)

    NASA Technical Memorandum 104596

    An Earth Albedo Model

    A Mathematical Model for the Radiant Energy Input to an Orbiting Spacecraft Due to the Diffuse Reflectance of Solar Radiation From the Earth Below

    Thomas W. Flatley
    Wendy A. Moore
    Goddard Space Flight Center
    Greenbelt, Maryland

    National Aeronautics and
    Space Administration
    Goddard Space Flight Center
    Greenbelt, Maryland

    1994

    (NASA-TM-IO459&) AN EARTH ALBEDO MODEL: A MATHEMATICAL MODEL FOR THE RADIANT ENERGY INPUT TO AN ORBITING SPACECRAFT DUE TO THE DIFFUSE REFLECTANCE OF SOLAR RADIATION FROM THE EARTH BELOW (NASA) 33 p

    Page 1

    “With specular reflection (as commonly occurs with mirrored surfaces) some or all of the incoming solar rays are reflected with the angle of reflection equal to the angle of incidence. Since a spacecraft would receive very little energy from even an entire Earth which was specularly reflecting this type of reflection is ignored here.

    Here, we consider the sunlit potion of the Earth to be a uniform, diffuse reflector and will use the word “albedo” in a limited sense, i.e. the albedo constant will be taken to be the ratio of the energy diffusely radiated from a surface to the total energy incident on the surface.”

    Page 2

    “According to Wertz, Fsun, the solar constant in the vicinity of the Earth, is approximately 1358 wad/m2. The sunlight strikes the Earth with this intensity at point B. At locations away from this point, the intensity of the incoming sunlight decreases proportional to cos_, so that the solar flux reaching any given incremental area is:

    Fin = Fsun(ne’,S) wa_/m2

    This incoming solar flux is partially absorbed and partially reflected. The amount of light reflected is proportional to the incident light by an albedo constant, ALB, which depends on the Earth’s surface characteristics. (See Appendix II.) This model assumes that the albedo constant does not vary over the Earth’s surface, neglecting the variation of diffuse reflectance with geographical features. A good estimate of the Earth’s annual average albedo constant is 0.3″

    Page 15

    “Conclusions

    This simplified albedo model was developed for use in spacecraft control system simulations, specifically, for modeling Coarse Sun Sensors. It is based on several approximations. Only diffuse reflectance is included; specular reflectance is neglected. For an elliptical orbit, the unit vectors associated with the incremental areas should change direction with altitude; instead, this algorithm assumes a circular orbit. The albedo constant is set to the annual global average for the entire Earth; Appendix II illustrates how the percentage of light reflected truly varies with geographical features. The Earth is considered a perfect sphere which does not rotate; it was unnecessary to model rotation since the albedo constant was not varied.”

  202. Christos Vournas says:

    Sorry, I have to stop now. It is getting too late.
    Shall we continue tomorrow?

  203. Willard says:

    Fine. Try not to post too many comments, and make sure to be responsive to what AT says.

  204. “At any given moment, the Earth absorbs 480 W/m^2 and emits 240 W/m^2.”

    Should be …

    “At any given moment, the Earth absorbs 480 W/m^2 on its lit surface and emits 240 W/m^2 over its entire surface.” Lit surface equals half the entire surface. Energy is conserved and energy flux is conserved. I actually do not see a difference in using either, see …
    Energy flux
    https://en.wikipedia.org/wiki/Energy_flux
    Radiative flux
    https://en.wikipedia.org/wiki/Radiative_flux

    :/

  205. Willard says:

    > Lit surface equals half the entire surface.

    The lit surface at zenith equals the fourth of the entire surface. It’s the disc:

    Below is the integral I did to show that if you properly integrate the incoming solar flux over the hemisphere that faces the Sun, you get the same answer as simply considering the cross-sectional area.

    Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190372

    Division happens to isolate the Temperature on the output side. That means that the divide by four trick expresses the ratio between the receiving and the emitting surfaces.

    Joe has no rationale to divide by 2 in his model except verbiage. In his magnum opus, he adds two temps (for nights and days) to cover his tracks, but fails to recoup the geometry of the Earth (e.g. eq 24-27). And so he ends up breaking the relationship between flux and energy (while rejecting SB!) with more verbiage.

    My “probly miscalculation” was too generous. He misspecified big time.

  206. Lit surface equals hemisphere which is twice the disc area and half the total surface of the sphere.

    1 = disc
    2 = hemisphere
    4 = sphere

    Energy/radiation thru the disc =1 hits the hemisphere = 2 = incoming energy/radiation
    Energy/radiation through the sphere = 4 = outgoing energy/radiation
    Energy/radiation is conserved

    I have not, and will not, read Joe’s stuff. I am of an opinion that this is really simple stuff, so much so, that if there were some error in the current conventions that they, whomever they may be, should be able to make a precise statement, using math, that leaves no ambiguities in their derivation whatsoever. To date, all that I have seen is the same error(s) in their derivation(s), ad infinitum, ad nauseam. Welcome to the internet.

  207. Or …

    1 = sphere
    2 = hemispheres
    4= discs

    To me, multiply by or divide by, same difference.

  208. Willard says:

    > I have not, and will not, read Joe’s stuff.

    You are commenting in a thread on Joe’s stuff.

    Sorry.

  209. Dr Roys Emergency Moderation Team says:

    “No need to check: minding our units leads to a reductio. Flux is a rate of energy by unit of surface. The more surface for the same energy {6}, the less flux there is; if the Earth was infinite, flux would be nil. One can’t get twice the Sun’s power out of twice the area. That’s absurd.”

    One last time…Postma is not trying to get “twice the Sun’s power out of twice the area”. The flux in is double (480 W/m^2 as opposed to 240 W/m^2) because the total power in is divided by half the surface area (the hemisphere as opposed to the sphere).

    [Fixed. Thanks. – W]

  210. Notice anything missing in this list of publications …
    https://www.researchgate.net/profile/Joseph-Postma
    … versus this list of publications …
    https://scholar.google.com/scholar?hl=en&as_sdt=0%2C25&q=greenhouse+author%3AJE+Postma&btnG=

    Like no P-S papers and only a note circa 2015 wrt GHG’s. Not even a single citation according to Google Scholar for anything related to those GHG papers either. Funny that.

  211. Willard says:

    Seldom is the question asked: are Sky Dragons learning?

    No, because they are cognating:

    Experimentally, the postulate of a radiative greenhouse effect is simple to test. Such experimental methods will be discussed here, but first, we must understand what it is exactly we are testing. In his book, “Now, Are You Ready To Learn Economics”, American Patriot and polymath Lyndon LaRouche (www.larouchepac.com) describes the act of cognition as something qualitatively unique and superior to the simple act of learning. We read on pp 81-82:

    What is Cognition?

    The discoveries of what are later experimentally validated as universal physical principles, are prompted by the demonstration of those qualities of paradoxes, the which are not susceptible of formal solution by means of the deductive and other methods of the philosophical reductionists. Such paradoxes are typified by the ontological paradox of Plato’s Parmenides dialogue; the impossibility of solving such by deductive methods, is typified by the case of that historical Parmenides, whose method Plato referenced in that dialogue. […] After Plato, this became the age-old Classical method of cognitive education in globally extended European civilization.

  212. Willard says:

    It gets better:

    In physics the units we use to describe the EM energy emitted by a Blackbody are Joules per second per square meter, or Watts (J/s) per square meter. Abbreviating, we then have W/m2. The Watts part is what we call “power” in physics, and the dependency on the area (the per square meter term) gives the density of this power.

    So, a soul is an instance of the EM spectrum, and an EM spectrum has units of power per square meter. When the source of the EM spectrum raises in temperature, then the power of the EM spectrum increases, and the density of power increases as well. The EM spectrum also shifts to higher frequency. Given a fixed area, or given a fixed object such as a monad, then if you increase the temperature of the monad you increase the power it contains, and this necessarily increases the density of the power at the same time. A higher density of power requires higher frequency energy components in order to fit that additional energy into the same space.

    So now we have determined the objective, singular, mathematical, physical, and scientific, definition of merit, in the context of defining merit as that which promotes Becoming, which means increasing power.

    I’ll simplify this definition in a moment, but I would very much like to point out that the only group which has independently objectively defined merit, arriving at this definition discussed here, is Lyndon LaRouche and his “LaRouchePAC” political action committee. If you are an Illuminist or if you wish to support meritocratic and developmental principles, then you can contribute financially to them.

    https://climateofsophistry.com/2015/11/11/how-to-define-merit-in-ontological-mathematics/

    Never forget: the objective definition of Merit is that which increases energy flux density.

  213. Dr Roys Emergency Moderation Team says:

    “How Joe found that 365.5 x 0.5 = 303 is left as an exercise to the reader.”

    The values in question are 360.5 K and 303 K. A blackbody at 360.5 K emits approx. 960 W/m^2. A blackbody at 303 K emits approx. 480 W/m^2. So you could say that a flux of 480 W/m^2 equates to a “blackbody temperature” of 303 K. Either way, it is the 960 W/m^2 (solar constant of 1,370 W/m^2 corrected for albedo) that Joe is dividing by 2, or rather multiplying by 0.5. Does that help shed some light on Joe’s diagram?

  214. Dr Roys,
    That just sounds like he’s calculating something different. The 360.5K seems to be effective radiative temperature of a 1 square metre patch on the surface of a non-rotating planet with no atmosphere with the Sun directly overhead. The 303K seems to be the effective radiative temperature of the sun-facing hemisphere of a non-rotating planet (with no atmosphere). These are similar to the calculatings I do in my class when I introduce how we might estimate planetary temperatures and also as an illustration of the greenhouse effect. (i.e., if you do the calculation for the full sphere you get a lower temperature than we know our surface temperature to be. Why is that?).

  215. Dr Roys Emergency Moderation Team says:

    Postma agrees that the effective temperature of the Earth is 255 K. You will see that figure on his diagram, too. A blackbody at 255 K emits 240 W/m^2. So the beloved “divide by 4” is still present on his diagram (960 W/m^2 divided by 4 = 240 W/m^2), he just notes that this is the output from the Earth, not the input to the Earth.

  216. he just notes that this is the output from the Earth, not the input to the Earth.

    I’ll try to explain the point again. Noone (credible) is claiming that the input to the Earth is 4 times smaller than it actually is. If you want to simply consider energy balance, then just use:

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    where F_{\odot} is the solar flux at the radius of the Earth’s orbit and A is the albedo. If you carry out the calculation you get that T_E = 255 K.

    However, a standard quantity in climate science is a change in radiative forcing. This is essentially how some external change (volcanic eruption, change in solar flux, increase in atmospheric greenhouse gas concentration,….) influences the energy balance of the planet. It has units of W/m^2 and is essentially defined in terms of the difference between the incoming and outgoing energy, averaged over the planet. In other words, if the system was initially in energy balance and then this change occurs, what is the change in this energy balance averaged over the planet.

    So, if we think in terms of a change in solar flux, then if the system is initially in energy balance, the change in this energy balance becomes:

    \pi R^2 (1 - A) F_{\odot,1} - \pi R^2 (1 - A) F_{\odot, 2} = \pi R^2 (1 - A) (F_{\odot,1} - F_{\odot,2}).

    However, this is the total change in the energy balance, which has units of J/s. To convert this into a change in radiative forcing, we need to then divide by the surface area of the planet (by definition). Hence, the change in forcing becomes:

    dF = \dfrac{\pi R^2 (1 - A) (F_{\odot,1} - F_{\odot,2})}{4 \pi R^2} = \dfrac{(1 - A)}{4} (F_{\odot,1} - F_{\odot,2}).

    Therefore, if there is a change in solar flux of 1 W/m^2 (i.e., F_{\odot,1} - F_{\odot,2} = 1 W m^{-2}) then the change in solar forcing is:

    dF = \dfrac{(1 - A)}{4} \times 1 = 0.175 W m^{-2}.

  217. Dr Roys Emergency Moderation Team says:

    Referring back to my comment from 4:08 PM, yesterday (sorry, I can’t seem to include links, when I do the comment does not post), all Postma is saying (at least in regards to what I was initially discussing with Willard, which is what inspired him to make this post in the first place) is that in real time the Earth receives 480 W/m^2 over half of its surface area, whilst in the same moment it emits 240 W/m^2 from the entire surface area. The flux values in and out are not equal, but energy is conserved, because the flux in is over half the surface area that the flux out leaves from. That’s it.

    I think what has happened is, people have read all sorts of things into this that aren’t really being argued by Postma…and everything has sort of escalated, and got out of hand.

    Anyway, at least I hope the clarification with regards to:

  218. Dr Roys Emergency Moderation Team says:

    Sorry, don’t know what happened there.

    Anyway, at least I hope the clarification with regards to:

    “How Joe found that 365.5 x 0.5 = 303 is left as an exercise to the reader.”

    has helped.

  219. TrueSceptic says:

    @Skeptic
    “You still don’t get it. To get the total incoming energy you need _first_ to do an integration to account for the difference in angle of the incoming radiation to the surface normal.”

    No, we don’t. We already know the answer: it’s 1361 (at top of atmosphere, ~960 allowing for albedo) W/m^2 totalled over the “disc” (earth’s cross-section).

  220. TrueSceptic says:

    Sorry, missed this bit
    Holger says:
    May 1, 2021 at 6:57 pm

  221. Energy in = energy out
    Energy in = Φ(1-a)So πr^2 = 0,47*0,7*1361 πr^2 = 444 πr^2 ( W )
    or, on average on the cross-section disk
    Energy in = 444 πr^2 ( W ) / πr^2 ( m^2 ) = 444 W/m^2
    and
    the Energy out on average from the entire Earth’s surface
    Energy out = 444 * πr^2 ( W ) /4 πr^2 = 111 W/m^2

    Planet not only reflects diffusely Albedo a = 0,3
    The diffusely reflected portion is 0,3*So

    But Planet reflects specularly, and diffusely, and these two kinds of reflection do not summarize with each other, so the total reflected portion is:

    (1 – 0,47 + 0,47*0,3)So πr^2 = (0,53 + 0,141)So πr^2 = 0,671*So πr^2

    The “absorbed” (not reflected), which is the emitted portion Is:
    (1 – 0,671)*So πr^2 = 0,329*So πr^2 = 444 W/m^2

  222. TrueSceptic says:

    [Most obliged. -W]

  223. Christos,
    I think that is simply wrong. The albedo is simply the fraction of the incoming energy that is radiated back into space, and it is about 0.3.

  224. To continue with the Planet specular reflection issue…
    A smooth surface not visible planet (very dark, not diffusely reflecting surface) has Albedo a = 0.
    Nevertheless, this planet still reflects specularly the 0,53 So πr^2 portion of the incident solar energy.

  225. Did you read the NASA Technical Memorandum which I posted here yesterday? It says clearly the Earth’s Albedo a = 0,3 is a diffuse reflection. The Earth’s specular reflection is neglected.

  226. “Since a spacecraft would receive very little energy from even an entire Earth which was specularly reflecting this type of reflection is ignored here.”
    “Here, we consider the sunlit potion of the Earth to be a uniform, diffuse reflector and will use the word “albedo” in a limited sense, i.e. the albedo constant will be taken to be the ratio of the energy diffusely radiated from a surface to the total energy incident on the surface.”

  227. Willard says:

    > The flux values in and out are not equal, but energy is conserved, because the flux in is over half the surface area that the flux out leaves from. That’s it.

    That does not explain where Joe took the hemisphere that now appears on the left side of the equation. In other models, it comes from the right side of the equation: it’s Area in EMB above. If Joe has a sphere on the right side of his equality, to divide by 2 on the left side implies that there’s a 2 that remains on the right side. At least if there’s an equality to preserve between the equation terms at each transformation.

    Also, there is more than one diagram. In the first link I presented, he divides by 0.637, not by 0.5. In the second link I presented, he divides by 0.5. In the 0.637 version, he gets 322K; in the 0.5 version, he gets 303K.

    It would help if Joe presented or cited a canonical version of his model along his diagram.

  228. Christos,
    And I’m telling you that I’m pretty sure that the global albedo is around 0.3. For example, from this 2015 paper:

    According to these CERES EBAF data, the global, annual mean all‐sky reflected flux is 99.7 W m−2 (equivalent to a global albedo of 0.293)

    Annoyingly, I think Table 2 has an error and has written the global albedo as 0.23, rather than 0.293.

  229. In fact, if you consider Table 2 of the paper, 86.9 W/m^2 of the 99.7 W/m^2 comes from the atmosphere and 12.9 W/m^2 comes from the surface.

  230. I visited Joseph Postma site and I have the final result copy-pasted here:

    “If we want the value with the average absorptivity of 0.7 included, then we need to put the factor of 0.7 in with the flux to get

    (4/5)·(0.7*1370/5.67e-8)1/4

    = 288.5°K = 15.5°C.”

    But Joseph has not considered the smooth surface Planets reflecting not only diffusely, but also reflecting specularly…
    Also Joseph does not consider the Planet rotational spin as major parameter in the planet mean surface temperature – the Planet surface ROTATIONAL WARMING phenomenon..

    The Joseph’s calculations are for a planet Earth without atmosphere…
    We can check then the above equation on the case of the Moon:
    Moon Albedo a = 0,11
    And
    4/5)·(0.89*1370/5.67e-8)1/4

    = 306°K = 33 °C

    And in case of the planet Mars:
    Mars Albedo a = 0,25
    And
    4/5)·(0.75*1370/5.67e-8)1/4

    = 296°K

  231. Willard says:

    > smooth surface Planets reflecting not only diffusely, but also reflecting specularly…

    I think everyone got that point, Christos. It’s your 12th “smooth” so far on this page.

    Thank you for not repeating it.

  232. “Albedo (prounounced /ælˈbiːdoʊ/; Latin: albedo, meaning ‘whiteness’) is the measure of the diffuse reflection of solar radiation out of the total solar radiation and measured on a scale from 0, corresponding to a black body that absorbs all incident radiation, to 1, corresponding to a body that reflects all incident radiation.”

    https://en.wikipedia.org/wiki/Albedo

  233. Christos,
    Why are you higlighting that link? I think we know the definition of albedo.

  234. Ok, I will highlight only the

    diffusely

  235. Christos,
    The albedo is about 0.3. If you want to argue that it is substantively different than 0.3, you’ll need to do better than highlight a document from 1994 that says:

    A good estimate of the Earth’s annual average albedo constant is 0.3.

  236. “Using satellite measurements accumulated since the late 1970s, scientists estimate Earth’s average albedo is about about 0.30.”

    https://earthobservatory.nasa.gov/images/84499/measuring-earths-albedo

    You also do use the Earth’s Albedo a = 0,3 don’t you?

  237. TrueSceptic says:

    Christos,

    Everyone is using 0.3 (or maybe 0.29) as the albedo, are they not? What’s your point?

  238. Christos,
    Maybe I’m missing your point, but I thought you were suggesting that the albedo was greater than 0.3. Is this what you’re claiming?

  239. https://en.wikipedia.org/wiki/Bond_albedo
    https://en.wikipedia.org/wiki/Geometric_albedo

    Bond is the correct one for our current discussion afaik*. Using drag coefficients from aerodynamics/hydrodynamics is a no go for radiation calculations as both have a fluid which is not present in radiation (to space) calculations afaik*.

    *afaik just in case of climate emergency break greenhouse gas, hmm err, glass, pun intended.

  240. “fluid” should be “working fluid” so no free lunches of non-working fluids afaik*, pun intended.

  241. My point is Earth reflects also specularly. The Te = 255 K Earth’s effective temperature is estimated wrongly.
    The not reflected portion on the planet cross-section disk of So is not 970 W/m^2

    The not reflected portion is 444 W/m^2
    Thus the Earth’s corrected effective temperature is Te = 210 K

  242. Willard says:

    What \alpha do you prefer, Christos?

  243. Christos,
    And you point is wrong. As EFS points out, the relevant albedo for determining energy balance is the Bond albedo which is ~0.3.

  244. AT, Earth’s Albedo is the solar lit hemisphere’s diffuse reflection a = 0,3.
    Φ(1 – a)So is the (not reflected) portion

    Earth’s total reflected portion is

    1 – Φ(1 – a) = 1 – 0,47 + 0,47*0,3 = 0,671

    Earth’s specular reflection is

    0,671 – 0,3 = 0,371
    or
    Earth’s specularly reflected portion of the incident on the cross-section disk (solar lit hemisphere) is
    0,371 So W/m^2

  245. Willard, I prefer a, not α…

  246. Christos,
    No, the Bond albedo of the Earth is 0.3. This is the relevant albedo if you want to do energy balance calculations.

  247. Willard says:

    > I prefer a, not α…

    I personally prefer something more explicit, like *Albedo* or Albedo_Corrector.

    What I was asking is: which value do you prefer?

  248. [Let’s stick to the albedo issue for the moment. -W]

  249. Willard,
    > What I was asking is: which value do you prefer?
    I do not understand then… Are there several values of Albedo?

  250. What is the commonly accepted Te for the Moon Christos? Citations required. TIA

  251. Dr Roys Emergency Moderation Team says:

    “That does not explain where Joe took the hemisphere that now appears on the left side of the equation. In other models, it comes from the right side of the equation: it’s Area in EMB above. If Joe has a sphere on the right side of his equality, to divide by 2 on the left side implies that there’s a 2 that remains on the right side. At least if there’s an equality to preserve between the equation terms at each transformation.”

    You are massively over-thinking this. The “divide by 2” (960 W/m^2 becomes 480 W/m^2) is simply a shortcut to doing what I outlined in my 4:08 PM comment from yesterday. There is nothing more to it than that. Nothing untoward is being done. Nobody is doubling the Sun’s power.

    Yes, there is more than one diagram. There is a comment under the Archimedes Hat post where he explains that he no longer goes with the 0.637 version, and why. Sorry, again, I cannot post links for some reason.

  252. Willard says:

    Christos,

    Well, you seem to use a lot of other values along with 0.3.

    For instance, you write:

    > 1 – Φ(1 – a) = 1 – 0,47 + 0,47*0,3 = 0,671

    I have no idea what 1 – Φ(1 – a) does here if not correcting the albedo corrector.

  253. Everett,
    > What is the commonly accepted Te for the Moon Christos? Citations required. TIA

    The SB (black-body) temperature for the Moon is ~270.4K …
    https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

    The Diviner dataset I mentioned above gives 270.46K for T^4 (SB) relationship and 199.44K for a T^1 (linear) relationship. Mean long term temperatures over many rotations.

    My estimation for Corrected Te for the Moon is Te = 224 K

  254. Willard,
    > I have no idea what 1 – Φ(1 – a) does here if not correcting the albedo corrector.

    Φ(1 – a) – is the “absorbed” (not reflected) portion

    Φ – is the planet surface solar irradiation accepting factor (the planet surface spherical shape and roughness coefficient)
    Φ = 0,47 for smooth surface planet Earth
    a = 0,3 is the Earth’s average Albedo
    ” 1 ” is the 100 % Solar flux on the TOA

    1 – Φ(1 – a) – is the reflected portion, or 0,671 *So

  255. Christos,
    What you’re suggesting is just simply wrong. I suspect this has been pointed out to you before.

  256. Willard says:

    > The “divide by 2” (960 W/m^2 becomes 480 W/m^2) is simply a shortcut to doing what I outlined in my 4:08 PM comment from yesterday. Nothing untoward is being done. Nobody is doubling the Sun’s power.

    That would imply that Joe is wrong in saying that ordinary energy balance models “decrease the power of sunshine to 960/4 = 240 W/m2,” as I underlined a few days ago.

    I can change my expression to “increase the power of Sunshine” (or anything else reasonable) if you prefer. There’s nothing deeper to that claim. If that’s just spin from Joe’s part, then Sky Dragons are in no position to criticize spin.

    ***

    > Nothing untoward is being done

    I know why Joe is dividing by 2. What I don’t know is where it comes from in his model. To what model of his are you referring?

    This is a problem. If one transforms Energy_In = Energy_Out to isolate T on the right side, something has moved from the right side. In ordinary EBMs, it’s 4 because that’s what we got on the right side. In other words, the shortcut has an algebraic justification.

    I could not care less if Joe’ strick is untoward or not. I only care about Joe’s algebraic justification to divide by 2. His equations do not make that clear at all.

  257. AT,
    Christos,
    What you’re suggesting is just simply wrong. I suspect this has been pointed out to you before.

    No specular reflection for a smooth planet Earth then?

  258. Willard,
    I only care about Joe’s algebraic justification to divide by 2.

    I am sure Joe divides by 2 for a good reason.

  259. Christos,
    From an energy balance perspective, all that matters is the Bond albedo, which is ~0.3.

  260. Willard says:

    > I am sure Joe divides by 2 for a good reason.

    One reason would be to have twice the temperature we usually have with the same numbers, i.e. by turning x = 4y into x/2 = 2y.

    “Because, flat Earth” isn’t a reason, more so that a sphere is far from being flat.

    Come to think of it, I could simply say that Joe doubles triples the temperature…

  261. [The constructive part of that exchange has run its course, Christos. -W]

  262. Dr Roys Emergency Moderation Team says:

    I have left a comment at CoS with your latest remarks, Willard, under the OID article. Perhaps Joe can explain it in a way that satisfies. I’m sure this can all be resolved.

    Thanks for deleting the offending paragraph. Hopefully the “How Joe found that 365.5 x 0.5 = 303 is left as an exercise to the reader” mystery is now resolved too.

  263. Willard says:

    My pleasure, Rajinder. The paragraph had sense only in one direction the Poe could have taken. It took another.

    It still would be nice if you could help me find a zippy way to describe what Joe does, in effect or not, to get 480 instead of 240 W/m^2.

    ADD. The expression chosen for now is “increases solar input.”

  264. Russell says:

    Dr Roys Emergency Moderation Team says: May 3, 2021 at 4:45 pm

    “That does not explain where Joe took the hemisphere that now appears on the left side of the equation.”

    It fits perfectly inside the hemisphere on the right side , –

    See illustration at Willard says: April 27, 2021 at 3:51 am

  265. Dr Roys Emergency Moderation Team says:

    “Joe’s trouble must be purely geometric: he wants the light to fall on a hemisphere, not a disc. This is corroborated by another post in which he, with the tip of Archimedes’ hat {4}, in effect increases solar input, from 480K instead of 240K {5}. How did he pull his trick? Probably miscalculation {6}.”

    480 W/m^2 instead of 240 W/m^2, not K. Mind your units!

    Also, you now know how he does it, and that your {6} is not a miscalculation.

  266. Dr Roys Emergency Moderation Team says:

    “Joe’s trouble must be purely geometric: he wants the light to fall on a hemisphere, not a disc.”

    …and with this, again…Joe wants the light to fall on a hemisphere, not over the entire sphere. The disk is used as per my 4:08 PM comment from yesterday.

  267. Willard says:

    > 480 W/m^2 instead of 240 W/m^2, not K. Mind your units!

    Edited. Since these figures don’t appear in Joe’s 0.5 diagram, I cited his final results instead.

    > miscalculation.

    Changed to “misspecification.”

    > he wants to fall on the hemisphere

    I already wrote this. That’s not an algebraic justification. Also, we know that what receives a Disc is equivalent to what receives a hemisphere corrected for angles.

  268. Dr Roys Emergency Moderation Team says:

    Yes, you wrote: “Joe’s trouble must be purely geometric: he wants the light to fall on a hemisphere, not a disc”, which implies that Postma does not understand the geometry, which you must surely realize by now, is not the case. He does understand it. Also, as I said, Postma is objecting to the idea that the sunlight falls on the entire sphere. In real time, the sunlight falls only on the lit hemisphere.

    You have written: “he wants the light to fall on a hemisphere, not a disc”.

    Whereas, he actually “wants the light to fall on a hemisphere, not over the entire sphere”.

  269. Does Postma really think that climate scientists think that the sunlight magically falls across the whole sphere, rather than onto only one hemisphere?

  270. Dr Roys Emergency Moderation Team says:

    If you had the input to the Earth as 240 W/m^2 and the output from the Earth as 240 W/m^2 then you would effectively be saying that the sunlight magically falls across the whole sphere. As if the Earth were flat, and thus able to receive the sunlight across the entire surface at once (this is where the “flat Earth” comments come from). Whereas with an input of 480 W/m^2 and an output of 240 W/m^2 it captures the fact that in real time, the sunlight falls only on a hemisphere whilst energy leaves from the entire sphere.

  271. gator says:

    I lost track of what the whole argument was about so I went back to the original link at Joe’s.
    “Let us consider energy budgets. If anyone is familiar with my work, then they know about the so-called “P/4 issue”, which indicates that the standard approach of climate science is to average-out the actual real-time power of sunshine by dividing its real power, P, by the number 4. Now to be sure, the real power of sunshine is this value we call “P”. It has a numerical value of about 1370 Watts per square meter. This is the real power of sunshine and it can be converted into a temperature, which has a value of 121 degrees Celsius – boiling hot! Some of this sunshine power is actually reflected by the Earth though, about 30%, and therefore doesn’t cause any heating; when you factor this in, the real power of sunshine is about 960 W/m2 which is a temperature of about 88oC.”

    Why is no one calling out the complete bullshit here? You can’t equate a flux directly to a temperature like this. According to Joe the sun is “boiling hot”! Oh gee, that sure is hot! This naive conversion of energy flux to a blackbody temperature is at the root of all the confusion. Measured flux will generally fall off as 1/r^2 away from the radiating surface (OK, you’re far enough that the source looks like a ball) – so if you’re equating energy flux and temperature, your estimate of the blackbody changes as your distance from the blackbody changes. Which should tell you that this is not a valid way of measuring temperature. Energy flux is not equivalent to temperature. Not to even get into spectral issues…

    Arguing about flux and whether the area considered is a disk, hemisphere or sphere is only “important” because Joe thinks flux => temperature. Since that is not true in any sense, arguing about flux goes away.

  272. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 3, 2021 at 10:51 pm
    “If you had the input to the Earth as 240 W/m^2 and the output from the Earth as 240 W/m^2 then you would effectively be saying that the sunlight magically falls across the whole sphere.”

    That’s stupid. Everyone knows the 240 input is an *average* over the sphere, just as 480 is the *average* over the day hemisphere.

  273. Willard says:

    > You have written: “he wants the light to fall on a hemisphere, not a disc”. Whereas, he actually “wants the light to fall on a hemisphere, not over the entire sphere”.

    As already mentioned, Arthur defines power as energy per unit time. In the model in the post, that means Sun x Disc; see his equation (1) for a fancier notation.

    In Joe’s diagrams, the light clearly falls on a disc just like other energy balance models. I doubt we can say that Power is Sun x Disc / Hemisphere. So Joe stretches his disc to a hemisphere after Power is established. I have yet to know an algebraic justification for his division-by-2.

    Also, as AT tried to underline yesterday:

    In most circumstances you don’t need to divide by 4 (i.e., just equate the energy in and the energy out).

    Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190698

  274. Dr Roys Emergency Moderation Team says:

    I think Joe’s point is it doesn’t make sense to average the input over surface area which is not receiving it (at least not in real time…obviously over 24 hours the Earth rotates so that all the surface will be exposed to the sunlight, but on a second by second basis light is clearly not falling on the entire Earth’s surface).

  275. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 4, 2021 at 12:07 am
    “I think Joe’s point is it doesn’t make sense to average the input over surface area which is not receiving it”

    In which case he shouldn’t use the average over the day hemisphere either. Does he do that?

  276. Dr Roys Emergency Moderation Team says:

    Why is nobody aware of what Postma’s arguments actually are?

    TS, the lit hemisphere is receiving the input, in real time. Hence it does make sense for him to average the input over the lit hemisphere’s surface area.

    The output is leaving from the entire Earth’s surface area, in real time. Hence it makes sense for him to average the output over the entire sphere’s surface area.

    480 W/m^2 input over the lit hemisphere balances 240 W/m^2 output from the entire sphere. In real time.

  277. Willard says:

    > 480 W/m^2 input over the lit hemisphere balances 240 W/m^2 output from the entire sphere. In real time.

    What’s the unit of that real time, seconds?

    If that’s the case, then there might be an imbalance of 240 W/m^2 each second on Earth since the dawn of time, which is not an unit, just an expression.

    Pending a paradigm shift, arguments don’t replace an energy-balance model.

  278. Dr Roys Emergency Moderation Team says:

    Willard, there is no energy imbalance of 240 W/m^2. The 480 W/m^2 is absorbed over half the surface area that the 240 W/m^2 leaves from. So in terms of energy, it balances. I thought you already understood this?

  279. Willard says:

    I’m being tongue in cheek for that seems to be the only way for you to you volunteer information. Otherwise you mostly ignore what I’m saying.

    So at each second we have a model under equilibrium whereby the Earth gets its energy on one hemisphere, and releases its energy on her whole sphere.

    My more serious query is how does that translate into temperature. I know how to proceed in EBMs: you isolate T. How does Joe succeed in not dividing by four?

  280. Dr Roys Emergency Moderation Team says:

    He does divide by 4. He agrees that the effective temperature of the Earth is 255 K. A blackbody at 255 K emits 240 W/m^2. Check his diagram again, it’s right there…as the output.

  281. Willard says:

    > He does divide by 4.

    Wait. What does he divide by 4, again?

    Here is the 0.637 diagram:

  282. Dr Roys Emergency Moderation Team says:

    Why would you show the 0.637 diagram when you know he has replaced it with the 0.5 diagram?

    He divides the solar constant, corrected for albedo (960 W/m^2) by 4.

  283. Willard says:

    That’s the first link I got. It’s from the post in which I picked up the quote.

    So I guess my question is: where has the hemisphere (represented by 0.5 or 0.637) gone?

  284. Dr Roys Emergency Moderation Team says:

    The lit hemisphere receives 480 W/m^2. That equates to a “blackbody temperature” of 303 K. Hence, if you had the correct diagram on screen, you would see it fine. Kind of hard to discuss a diagram that isn’t on screen.

  285. Willard says:

    No problem. Here it is:

    My question is about the hemisphere that is supposed to be represented by 0.5: where is it when we isolate T?

  286. Dr Roys Emergency Moderation Team says:

    I don’t understand the question. Have you considered asking Postma directly (and with less snark and condescension) what exactly it is you don’t understand? He might respond if you formulate clearly and precisely what your issue is. I think you two got off on the wrong foot.

  287. Willard says:

    > I don’t understand the question.

    Fair. We can postpone to another moment. It’d be cool to do something else today.

    Joe can come here to discuss his model with AT anytime. This is AT’s blog, and he’s the astrophysicist while I’m just a ninja.

    Considering how Joe treated me so far, I hope you don’t mind if I don’t contact him ever again. He should have come here a long time ago.

    What you’re doing right now is commendable. It gives me hope that Roy’s Thunderdome can evolve. You should get your nick back: you deserve better than to be an ironic moderator.

    Enjoy your evening,

  288. Dr Roys Emergency Moderation Team says:

    OK, thanks, you too. I will leave a comment extending your invitation.

  289. Dr Roys,

    If you had the input to the Earth as 240 W/m^2 and the output from the Earth as 240 W/m^2 then you would effectively be saying that the sunlight magically falls across the whole sphere.

    No, you wouldn’t. The input to the Earth is clearly 240 W/m^2. That it all falls on one hemisphere doesn’t change this. Of course, if you wanted to be pedantic, you could say the input is 480 W/m^2 on one hemisphere and 0 W/m^2 on the other, but that just averages to an input of 240 W/m^2.

  290. Dr Roys,

    > If you had the input to the Earth as 240 W/m^2 and the output from the Earth as 240 W/m^2 then you would effectively be saying that the sunlight magically falls across the whole sphere. As if the Earth were flat, and thus able to receive the sunlight across the entire surface at once (this is where the “flat Earth” comments come from). Whereas with an input of 480 W/m^2 and an output of 240 W/m^2 it captures the fact that in real time, the sunlight falls only on a hemisphere whilst energy leaves from the entire sphere.

    It is a very clear insight!

    it captures the fact that <b?in real time, the sunlight falls only on a hemisphere whilst energy leaves from the entire sphere

  291. The correct:
    ...it captures the fact that in real time, the sunlight falls only on a hemisphere whilst energy leaves from the entire sphere.

  292. Christos,
    You do realise that when full 3D climate models are run, they do include that the sunlight only falls on one hemisphere? Also, a simple energy balance calculation also takes into account that the energy only falls onto one hemisphere (that’s why it’s \pi R^2 (1 - A) F_{\odot}. This clear insight is so obvious few others thought it worth explicitly highlighting.

  293. Yeah, climate models, go figure. And that is what Joe Postma is arguing for, a (climate) model that accounts for all those (internal) dynamics. Which they have been doing for several decades now.

  294. AT,
    > Also, a simple energy balance calculation also takes into account that the energy only falls onto one hemisphere (that’s why it’s
    https://s0.wp.com/latex.php?latex=%5Cpi+R%5E2+%281+-+A%29+F_%7B%5Codot%7D&bg=ffffff&fg=333333&s=0&c=20201002&zoom=2

    AT, I insist the correction should be done Φ*https://s0.wp.com/latex.php?latex=%5Cpi+R%5E2+%281+-+A%29+F_%7B%5Codot%7D&bg=ffffff&fg=333333&s=0&c=20201002&zoom=2

    For smooth Earth’s surface Φ = 0,47
    Φ is the planet solar irradiation accepting factor (the planet spherical shape and the planet surface roughness coefficient)

  295. Christos,
    You can insist all you like, but the correct form for the energy balance for the Earth is:

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    where A \simeq 0.3.

  296. Eh… AT,
    Let’s see what we have today here:

    Planet Mars black-body temperature (effective temperature) Te misfortunate coincidence
    We have calculated the Corrected Effective Temperature for Mars Te.correct.mars = 174 K

    But let’s see what happened when the Effective Temperature of Mars was not yet corrected. Te.mars = 209,8 K
    https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

    Tsat. mean.mars = 210 K
    https://en.wikipedia.org/wiki/Mars

    We have here planet Mars mean temperature measured by satellites:
    Tsat.mean.mars = 210 K

    We have the Mars black-body temperature
    Te = 209,8 K

    These temperatures the Tsat.mean.mars = 210 K and the black-body temperature Te.mars = 209,8 K are almost identical.
    These two very important for planet Mars temperatures are almost identical, but it is a coincident.
    It is a coincident, but with very important consequences.

    Let’s explain:
    Tsat.mean.mars = 210 K measured by satellites is almost equal with Te.mars = 209,8 K

    When measuring by satellites the Tsat.mean.mars = 210 K and calculating Mars black-body temperature Te.mars. = 209,8 K scientist were led to mistaken conclusions.
    First they concluded that the planet’s effective and mean temperatures should normally be equal, which is wrong.
    Secondly they concluded that Earth without atmosphere should have an average surface temperature the black-body temperature (effective temperature), Te.earth = 255 K (https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html)

    Then they compared the Te.earth = 254 K with the measured by satellites Tsat.mean.earth = 288 K (https://nssdc.gsfc.nasa.gov/planetary/factsheet/)
    The difference of 288 K – 255 K = Δ33 oC was then attributed to the Earth’s atmosphere greenhouse warming effect.

    Now we have the Mars Corrected Effective Temperature
    Te.correct.mars = 174 K.

    The fact that the Corrected Effective Temperature of Mars is Te.correct.mars = 174 K, which is not even close to the satellite measured Tsat.mean.mars = 210 K debunks the above syllogism that the planet without atmosphere mean surface temperature is the planet black-body temperature (effective temperature).

    The above wrong syllogism happened because of the wrongly estimated Mars black-body temperature.
    It was calculated assuming planet absorbing incoming solar energy as a disk. We know now that planet absorbs the incoming solar energy as a sphere, and not as a disk.

  297. TrueSceptic says:

    …and Then There’s Physics says:
    May 4, 2021 at 7:18 am
    “No, you wouldn’t. The input to the Earth is clearly 240 W/m^2. That it all falls on one hemisphere doesn’t change this. Of course, if you wanted to be pedantic, you could say the input is 480 W/m^2 on one hemisphere and 0 W/m^2 on the other, but that just averages to an input of 240 W/m^2.”

    Exactly! If we can’t agree on that, we can’t agree on the most basic arithmetic. Of course all the input is to one hemisphere, but that is constantly changing as the Earth spins, so in a real sense the energy *is* being averaged over the sphere. It would be a different matter entirely if the Earth presented the same physical hemisphere to the Sun at all times (the night hemisphere would never receive any energy directly from the Sun and the climates of the 2 hemispheres would be hugely different).

  298. Christos,

    The difference of 288 K – 255 K = Δ33 oC was then attributed to the Earth’s atmosphere greenhouse warming effect.

    The difference is because of the greenhouse warming effect. Given that we receive 240 W/m^2 from the Sun, the effective radiative temperature of the Earth is 255K. This comes from basic energy balance. The surface temperature, however is more like 288K, 33K higher than this effective radiative temperature. This enhanced surface temperature is a consequence of the planetary greenhouse effect.

  299. Dr Roys Emergency Moderation Team says:

    TS and AT, as I explained, Joe’s point is that you should not average the input over surface area which is not receiving it. In real time, the input is 480 W/m^2 to one hemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres.

  300. Dr Roys,
    And Joe’s point doesn’t make any sense. The input to the Earth is ~960 W/m^2 when considering the cross-sectional area of the Earth, ~480W/m^2 when considering the hemisphere facing the Sun, and ~240W/m^2 when averaging over the whole sphere. There is nothing wrong with doing the latter.

  301. Dr Roys Emergency Moderation Team says:

    In real time, the input is 480 W/m^2 to one hemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres. That is the physical reality of what is happening, on a second by second basis. Of course if you average over longer time periods, you get a different story.

    Will the article be corrected further? It is still wrong as currently written.

  302. MP says:

    @ ATTP May 4, 2021 at 12:54 pm

    Earths surface has depth. Surface, plus Subsurface (water heats around 100 meter of depth, wetland 1 meter, dry land 15 cm), and it has an outersurface in the form of atmospheric gasses, water, ice, and dust particles

    Since the earth surface (including subsurface and outersurface) has depth and a not equal temperature distribution you won’t find the effective temperature of -18 degree at the earth surface layer..You will find it higher up in the atmosphere, at around 5 km hight.

  303. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 4, 2021 at 1:25 pm
    “TS and AT, as I explained, Joe’s point is that you should not average the input over surface area which is not receiving it. In real time, the input is 480 W/m^2 to one hemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres.”

    The result has to be the same, because the total energy is the same and warms the whole planet, not just half of it. We agree that 240*4 = 480*2, don’t we?

  304. Dr Roys Emergency Moderation Team says:

    I stand by what I have said, because it is correct. Yes, obviously, 240*4 = 480*2.

  305. Dr Roys,

    In real time, the input is 480 W/m^2 to one hemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres.

    Why are you having some much trouble getting that 480 W/m^2 averaged over one hemisphere is the same as 240 W/m^2 averaged over the whole sphere? It’s not exactly complicated.

    Will the article be corrected further? It is still wrong as currently written.

    Which article?

  306. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 4, 2021 at 2:30 pm
    “I stand by what I have said, because it is correct. Yes, obviously, 240*4 = 480*2.”

    OK, so why would that make any difference?

  307. MP,

    Since the earth surface (including subsurface and outersurface) has depth and a not equal temperature distribution you won’t find the effective temperature of -18 degree at the earth surface layer..You will find it higher up in the atmosphere, at around 5 km hight.

    We might be getting somewhere. Why is it at 5km?

  308. Willard says:

    > Yes, obviously, 240*4 = 480*2.

    The same should apply to temperatures, DREMT.

    So, how does Joe balance out the various temperatures of the Earth from his model? Here’s how Tim balances his:

    A little physics …

    The sun provides a total power to the earth of P = S*A where S = solar solar constant = 1370 W/m^2 and A = cross-sectional area of earth. Because of albedo, only part is actually absorbed, which we will assume to be P = (960 W/m^2)*A.

    Now, working with that much total power, we could imagine various ways to distribute that much power over the entire 4*A surface area of the globe, and calculate the average temperature of a sphere with emissivity = 1).

    1) 240 W/m^2 uniformly over the whole sphere:
    *** 255 K

    2) 480 W/m^2 uniformly over 1/2 of sphere (one face always to sun:
    *** 153 K average = 1/2 at 303K + 1/2 at 2.7 K e

    3) 960 W/m^2 uniformly over 1/4 of sphere (one face always to sun:
    *** 91 K average = 1/4 at 360K + 3/4 at 2.7 K

    4) Actual distribution over 1/2 of sphere (just for fun).
    *** 145 K average

    We could do many more. We could include rotation. But no matter what we do, the *highest* possible average is 255 K for uniform illumination. *Any* other illumination/rotation/heat capacity/thermal conductivity will have a lower average temperature than 255 K. The more uniform, the higher the average temperature.

    *That* is physics.

    https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-682316

    If two models agree on the amount of energy that comes in and out of the Earth, the two models should translate into one another. Otherwise equality does not mean anything anymore.

    To check where two models diverge, we need to look into them. There’s an equation on his diagrams at the top right, but it’s left unexplained. I checked Joe’s magnum opus, to no avail.

    I need a Sky Dragon expert to find it. An expert like you, DREMT. That’s why I brought you here.

    Where’s Joe’s energy balance model?

  309. AT
    > Given that we receive 240 W/m^2 from the Sun, the effective radiative temperature of the Earth is 255K. This comes from basic energy balance. The surface temperature, however is more like 288K, 33K higher than this effective radiative temperature. This enhanced surface temperature is a consequence of the planetary greenhouse effect.

    Let’s see,
    Given that Moon receives 303 W/m^2 from the Sun, the effective radiative temperature of the Moon (without taking in consideration the Moon spherical smooth surface’s specular reflection)
    is 270 K.

    What is the Moon’s mean surface temperature then?
    270 K ?
    250 K ?
    220 K ?
    193 K ?

  310. Dr Roys Emergency Moderation Team says:

    240 W/m^2 equates to a blackbody temperature of 255 K. 480 W/m^2 equates to a blackbody temperature of 303 K. 960 W/m^2 equates to a blackbody temperature of 361 K. The real time flux from the Sun is enough to melt ice and evaporate water. If you think of the input from the Sun as being 240 W/m^2, that equates to a temperature below freezing, -18 C. As Tallbloke put it:

    “The glib answer is because the Sun really does only heat half the Earth [at any one moment] and so our conceptualization should match that reality.

    But you don’t have to go too much further to realize that whilst glib, it is important.
    Postma gives the example that warmists say that without GHG’s the Earth would be at -18C and that would turn it into a snowball which would cool further due to high albedo.

    With the half lit model in mind and the appreciation that dayside would be well above freezing, which would melt and evaporate water and so reduce albedo (black ocean) and trap night-time heat (fluffy clouds).”

  311. Christos,
    With an albedo of 0.12 and with an incoming solar flux of 1360 W/m^2, the Moon has an effective radiative temperature of 270K. That is what it is. In other words, it radiates as much energy back into space per square metre per second as a 270 K blackbody.

  312. Dr Roys,

    Postma gives the example that warmists say that without GHG’s the Earth would be at -18C and that would turn it into a snowball which would cool further due to high albedo.

    Postma might want to spend a little bit of time understanding what people are saying. All else being equal, the Earth without a greenhouse effect would radiate to space directly from the surface and would still have an effective radiative temperatore of 255K. However, noone who understands the topic actually thinks that if we didn’t have a greenhouse effect, that everything else would remain the same. The albedo would certainly be different, for example.

  313. Dr Roys Emergency Moderation Team says:

    “Postma might want to spend a little bit of time understanding what people are saying.”

    …but AT, the author of this article has not spent enough time understanding what Postma is saying. The article is still wrong. This article. That is the article I was referring to earlier. Comments and corrections welcome, yes? Well…

    “Joe’s trouble must be purely geometric: he wants the light to fall on a hemisphere, not a disc. This is corroborated by another post in which he, with the tip of Archimedes’ hat {4}, in effect increases solar input {5}. His model gives 30C instead of the usual -18C. How did he pull his trick? Probably misspecification {6}.”

    Should be:

    “Joe understands the geometry of the “divide by 4”. His issue is that he wants the light to fall on a hemisphere, rather than being averaged over the entire sphere. This is corroborated by another post in which he, with the tip of Archimedes’ hat {4}, uses 480 W/m^2 rather than 240 W/m^2 as the real time solar input. This is because [insert DREMT’s explanation here].”

    Your note {6} should be either deleted or corrected, for the reasons I explained earlier.

  314. MP says:

    @ ATTP

    Quote “We might be getting somewhere. Why is it at 5km?”

    In the 5 to 6 km zone there is a median in terms of atmospheric patricle density. And there is around the same amount of particle pressure below it as the over 100km atmosphere above it

    So if you try to find an average effective temperature in an (outer) surface layer with depth you should try to find it there. And that average -18c temperature is found there

  315. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 4, 2021 at 2:54 pm
    “240 W/m^2 equates to a blackbody temperature of 255 K. 480 W/m^2 equates to a blackbody temperature of 303 K. 960 W/m^2 equates to a blackbody temperature of 361 K. The real time flux from the Sun is enough to melt ice and evaporate water. If you think of the input from the Sun as being 240 W/m^2, that equates to a temperature below freezing, -18 C. As Tallbloke put it:

    “The glib answer is because the Sun really does only heat half the Earth [at any one moment] and so our conceptualization should match that reality.”

    OK, accepting that premise, what do you do about the night hemisphere? Pretend it doesn’t exist?

  316. Dr Roys,
    Not sure what there is to correct. If Postma now accepts that the correct energy balance equation is:

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    that’s great. If he always accepted this, then that certainly wasn’t clear.

  317. TrueSceptic says:

    MP says:
    May 4, 2021 at 3:09 pm
    “In the 5 to 6 km zone there is a median in terms of atmospheric patricle density. And there is around the same amount of particle pressure below it as the over 100km atmosphere above it

    So if you try to find an average effective temperature in an (outer) surface layer with depth you should try to find it there. And that average -18c temperature is found there”

    What is the temperature of Earth as seen from space?

  318. MP,

    In the 5 to 6 km zone there is a median in terms of atmospheric patricle density. And there is around the same amount of particle pressure below it as the over 100km atmosphere above it

    That still doesn’t explain why it’s at 5 to 6 km. So, again, why does the Earth radiate to space (on average) from an altitude of 5 to 6 km, rather than directly from the surface?

  319. Dr Roys Emergency Moderation Team says:

    Well, TrueSceptic, all that is being brought up now are good questions and challenges to Postma’s arguments. This is because those arguments are now understood a bit better. Look back at your first comment on this thread. The article had led you to believe that it was being suggested Postma does not understand basic geometry. The article still does lead you to think that, with the way it is written. This should be changed. That is why I am here. I am not here to answer every single question on Postma’s behalf that everyone wants to throw at me. What I would like is for the article to accurately represent his arguments and level of understanding.

  320. AT,
    > the Moon has an effective radiative temperature of 270K. That is what it is. In other words, it radiates as much energy back into space per square metre per second as a 270 K blackbody.

    Do you imply by that the Moon has a uniform surface temperature 270 K ?

    Lunar Reconnaissance Orbiter measured the lowest summer temperatures in craters at the southern pole at 35 K (−238 °C; −397 °F)[140] and just 26 K (−247 °C; −413 °F) close to the winter solstice in the north polar crater Hermite. This is the coldest temperature in the Solar System ever measured by a spacecraft, colder even than the surface of Pluto.[
    https://en.wikipedia.org/wiki/Moon#Seasons

  321. Willard says:

    > Your note {6} should be either deleted or corrected, for the reasons I explained earlier.

    Give me Joe’s model, kiddo, and I’ll see what I can do. Unless I see how Joe balances his numbers, no deal. And if the numbers he balances don’t match everybody else’s, no deal.

    If you still need to be powered by 386 to bring Joe’s model on the table, so be it.

  322. MP says:

    @ TrueSceptic

    Quote “What is the temperature of Earth as seen from space?”

    The average effective temperature of outgoing radiation is -18c

    However. Outgoing radiation seen from space is party from the lower surface area, partly from the median range, and partly from the much colder over 100km atmosphere above the median range

  323. Christos,

    Do you imply by that the Moon has a uniform surface temperature 270 K ?

    No, I think I explained myself very carefully.

  324. MP says:

    Correction.

    Should said much less dense over 100km above the median range (not colder per se, temperature rises somewhat in certain higher layers),

  325. Willard says:

    MP,

    I don’t understand why we should be distracted by these details when we can’t even have Joe’s model.

    Do you have Joe’s model?

  326. MP,
    You’re still not explaining why this is

    However. Outgoing radiation seen from space is party from the lower surface area, partly from the median range, and partly from the much colder over 100km atmosphere above the median range

  327. Dr Roys Emergency Moderation Team says:

    Willard, my comment from May 2 @ 4:08 PM already shows you how Postma balances his numbers.

  328. Willard says:

    > already shows you how [Joe] balances his numbers.

    No it does not. I want the temperatures to match the model I have.

    The kind of models we’re talking about is just a glorified function. It takes an input and produces an output. Equivalent inputs and equivalent outputs need to balance.

    Unless Joe’s doing something equivalent to the divide-by-4 trick somewhere, his model won’t balance out. He can go after a new physics all he wants, he needs to follow the same geometry and algebra as everybody else.

    I believe that the Earth’s temperature is function of the energy it receives from the Sun. How about you?

  329. Dr Roys Emergency Moderation Team says:

    Willard, he is not challenging the Earth’s effective temperature. He agrees that it is 255 K. My comment shows you that total power in balances total power out. In real time, the power in is divided by the hemisphere’s surface area, to give a flux of 480 W/m^2. In real time, the power out is divided by the whole sphere’s surface area, to give a flux of 240 W/m^2.

    That is all there is to it.

  330. Dr Roys,
    Well, if he’s getting an effective temperature of 255K, then he’s doing the same calculation as everyone else and his whole complaint about dividing by 4 doesn’t make any sense.

  331. Willard says:

    > That is all there is to it.

    We have not established how Joe’s model and everybody else’s are equivalent, and we have not seen Joe’s model yet.

    If the temperature on the lit hemisphere is 30C in Joe’s model, what is it on the other side?

  332. Dr Roys Emergency Moderation Team says:

    You both seem to want Postma to be arguing something other than he is. He is arguing that in real time, the input is 480 W/m^2 to one hemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres. That is the physical reality of what is happening, on a second by second basis. Now, since the 240 W/m^2 leaves from the entire sphere at once, that corresponds to the effective temperature of the Earth. 255 K. But the 240 W/m^2 is an output. Not an input.

  333. Willard says:

    Look, DREMT.

    You have returned to your ways at Roy’s. Repeating the same irrelevant point over and over again won’t do here. Your next comment needs to answer one of these two questions:

    – where is Joe’s model, chapter and verse;
    – what is the temperature of the other side of the Earth in one of Joe’s diagrams.

    Here’s where we stand. We have two equations:

    [E] xy = 4z

    [J] xy/2 = ?

    We have agreed that they’re equivalent. I have no idea what’s on the right side of J, and I don’t know how the 2 cancels out to give E.

    If both equations are equivalent and Joe indeed divides by 4, he needs to account for the 2 he has decided to put on the left side, “because flat earth.”

  334. aljo1816 says:

    ” All else being equal, the Earth without a greenhouse effect would radiate to space directly from the surface and would still have an effective radiative temperatore of 255K.”

    An ignorant question, but would it actually be true that the Earth without an absorbing atmosphere would radiate to space from an “effective layer” a little above the surface, but much lower than the effective layer with an absorbing atmosphere? I assume that the surface would lose heat directly to space as radiation, and also some heat to the atmosphere via conduction. Any radiation from the atmosphere would then escape directly to space (and some back to the ground), so you’d still set up a scenario with emission taking place on average from a theoretical layer in the atmosphere.

  335. Bob Loblaw says:

    What is all this “in real time” crap?

    In real time, virtually no point on earth (horizontal or vertical) has a radiation balance of zero.

    In real time, virtually no point on earth (horizontal or vertical) has a non-radiative energy balance of zero.

    In real time, virtually no point on earth (horizontal or vertical) has a balance between its radiative inputs/outputs and other energy transfers.

    In real time, virtually no point on earth (horizontal or vertical) has a a non-changing temperature.

    In real time, everywhere goes through minutely, hourly, daily, seasonal and longer-term variations in radiation, thermal energy storage/transfer, latent heat transformations, etc. It’s called weather. Climate is averages of that weather.

    Anyone with a “climate model” that is saying “oh, you can’t average sunlight between day/night sides of the earth” but is also saying “It’s perfectly reasonable to average sunlight across latitude and morning/noon/afternoon values” is very probably quite unaware of what they are actually modelling.

  336. Dr Roys Emergency Moderation Team says:

    [I insist that you fulfill my request. Keep your repetitions for Roy’s. – W]

  337. Bob Loblaw says:

    “the Earth without an absorbing atmosphere would radiate to space from an “effective layer” a little above the surface”

    I think maybe you are not understanding the phrase “effective layer”? That phrase does not mean that the earth is actually radiating from such a layer. It is is saying that the overall radiation as seen from space is essentially equivalent to what would be seen if the earth was radiating from such a layer.

  338. [Enough peddling for one thread, Christos. Wait for another one. Thanks. -W]

  339. Willard says:

    > What is all this “in real time” crap?

    Metaphorical flourish that has been inflated in a philosophical quest for mathematical ontology. It goes along the meme that to divide by 4 is “flat earth.”

    If Joe divides by 4 like everybody else to get his output, does it mean that half of his Earth is flat?

  340. Willard says:

    > If Joe divides by 4 like everybody else to get his output, does it mean that half of his Earth is flat?

    Alternatively, we could say that Joe’s Earth is Schrödinger-flat.

    But even then we’d have to presume that a sphere represents flatness.

    All this is very silly.

    Has anyone watched The Spanish Prisoner?

  341. gator says:

    Dr Roy’s:
    “240 W/m^2 equates to a blackbody temperature of 255 K. 480 W/m^2 equates to a blackbody temperature of 303 K. 960 W/m^2 equates to a blackbody temperature of 361 K. The real time flux from the Sun is enough to melt ice and evaporate water. If you think of the input from the Sun as being 240 W/m^2, that equates to a temperature below freezing, -18 C.”

    This is the whole of the argument. This is bullshit. Correct this and the whole argument about divide by 2 or divide by 4 goes away. The incoming energy has no “blackbody temperature”. Think of some power flux J/s/m^2 falling on an object – you can’t calculate an equivalent blackbody temperature of that flux. That energy is not in equilibrium with anything. You could increase the flux by concentrating the beam, without changing the total energy content. Does the object absorbing the energy suddenly get hotter? No – because the object’s temperature depends on how it gets rid of the energy, in the case of the earth by radiating away like a blackbody.

    The whole schtick is that Joe and apparently Dr. Roy’s are trying a fast one – “the flux is higher, therefore ‘hotter’ and climate scientists divide by 2 to make it ‘colder’!” That’s nonsense. If you want to fight them on that then obviously on the dark side of the earth the temperature should be 3k right since there’s no sunlight on that side of the earth. They can have 303K on the sunny side if they accept 3K on the dark side. See how stupid that is?

  342. Willard says:

    > They can have 303K on the sunny side if they accept 3K on the dark side.

    Nicely put.

  343. MP says:

    @ ATTP May 4, 2021 at 3:30 pm

    Quote “You’re still not explaining why this is”

    The sun provides at daytime enough heat at mainly the zenith zone to maintain the atmosphere hight and average pressure equilibrium , by evaporation of water and release of NO and O2 from the water there and then.

    The distribution to higher layers is pressure and density related and follow what is described in the gas laws. Convection is the bulk carrier.

  344. Dr Roys Emergency Moderation Team says:

    Willard and Gator, you want me to say the “dark side” is 3 K and the “light side” is 303 K. That might make some sense if the Earth were not rotating, but it is rotating. All the while, the lit hemisphere is receiving 480 W/m^2 in real time.

  345. Willard says:

    Yes, Rajinder. The Earth is rotating. If only we had a model of the Earth that took that into account. Perhaps you know one that does. How about Joe’s?

    WHERE IS JOE’S MODEL?

    If my online calculator is correct, 3K is -273,15C. That figure does not appear in Joe’s diagrams.

    So here’s a challenge: try to convince Joe to put on his diagram that behind that 30C hemisphere, there’s another one that is at -273C, or whatever figure you deem fit. Specific numbers are not the object right now.

    If he wants to get to ground zero of mathematical ontology, that’s the minimum he can do for you, don’t you think?

  346. ATTP or anyone else for that matter …

    So if the lit side of the Earth has a SB temperature of 303K (taken directly from JP diagram above) what would the dark side of the Earth temperature have to be in order to satisfy the SB law?

    (Ignoring the CMB, starshine, earthshine, etceteras for the moment (a pedantic, such as myself, has confirmed that these are negligible to at least nine significant figures).)

  347. aljo1816 says:

    “I think maybe you are not understanding the phrase “effective layer”? That phrase does not mean that the earth is actually radiating from such a layer. It is is saying that the overall radiation as seen from space is essentially equivalent to what would be seen if the earth was radiating from such a layer.”

    Phrased another way, I’m asking if we wouldn’t still “see” the radiation as if it were coming from some point above the surface even without an absorbing atmosphere, just a point much lower than we do now. Removing GHGs would, as I understand, keep lowering the effective height of emission, but it doesn’t seem obvious to me that it would lower it right onto the ground. I could simply be confused.

  348. Bob Loblaw says:

    “That might make some sense if the Earth were not rotating, but it is rotating.”

    Yes, and last time I looked, that rotation kept changing which part of the earth is lit, and which is unlit. And it is doing this “in real time”.

    The thing that is not making sense is ignoring the rotation at one point, and arguing that it is important at another.

  349. Willard says:

    > The thing that is not making sense is ignoring the rotation at one point, and arguing that it is important at another.

    As a way to get to the bottom of the final ontology of mathematics, it does not.

    As a way to run a con, however, it makes a lot of sense!

  350. Dr Roys Emergency Moderation Team says:

    The point I am trying to get across to you is, there is no “dark side” or “light side”. There is a rotating Earth, receiving 480 W/m^2 across one hemisphere, whilst the whole surface emits 240 W/m^2. All parts of the Earth’s surface receive sunlight…but how warm each part of the surface remains on average is going to depend on how warm it gets during the daylit hours and how efficiently it cools during the night. The problem becomes so complex that I do not think it is able to be solved. So my answer is one big “I don’t know”, if that’s what you want to see.

    Postma does not, to my knowledge, have one definitive model which will give you one temperature value for the whole Earth’s surface that is higher than 255 K, if that is what you think he has.

  351. Bob Loblaw says:

    but it doesn’t seem obvious to me that it would lower it right onto the ground.

    If the atmosphere is completely transparent to IR radiation, then yes, what you would see from space would be what is emitted from the surface.

    From a visible light analogy, what happens when you are seeing an object, and you place a completely transparent sheet of glass in front of it? Are you now looking at something coming from the glass, or are you still seeing that original object, unaltered?

  352. Bob Loblaw says:

    “…but how warm each part of the surface remains on average is going to depend on how warm it gets during the daylit hours and how efficiently it cools during the night. The problem becomes so complex that I do not think it is able to be solved.”

    That is an argument from incredulity. Three-dimensional climate models do this all the time. So do weather models.

  353. Willard says:

    > [Joe] does not, to my knowledge, have one definitive model which will give you one temperature value for the whole Earth’s surface that is higher than 255 K, if that is what you think he has.

    I’d settle for any model. I don’t think he has one that solves his 2 problem.

    Sometimes, DREMT, I think you think I’m some kind of dummy. I admit to have made many mistakes. I also admit that some of them were mischievous. If you need to correct me to help me find Joe’s model and make you cooperate, to look like a fool has been worth it to me.

    You know, Climateball isn’t about winning. It’s about winning something. You want to be right and to reach truth. I want Joe’s model, but I also want to feel humanity and a way out of our silly predicament. This piece has made our interests converge.

    If we continue to play well, we can both win that Climateball exchange.

  354. aljo1816 says:

    From a visible light analogy, what happens when you are seeing an object, and you place a completely transparent sheet of glass in front of it? Are you now looking at something coming from the glass, or are you still seeing that original object, unaltered?

    That makes sense to me, but in your example the glass pane is held above the surface? The ground is in contact with the atmosphere, so at least a little energy is exchanged via that contact, and some of that energy must be radiated from the atmosphere out to space. If I’m correct, that would still put the effective height of emission at least a little way above the surface. I think (think being the operative word) that you’d only “see” the radiation as coming directly and only from the surface if there were no atmosphere at all.

  355. Dr Roys Emergency Moderation Team says:

    I should have written “will give you one temperature value for the whole Earth that is higher than 255 K”…He does not think the effective temperature applies to the Earth’s surface itself, as MP explained.

    Willard, he has no “2 problem”. I have explained why he divides by 2 for the input, and 4 for the output. There is no predicament.

  356. Willard says:

    > I have explained why he divides by 2 for the input, and 4 for the output.

    No you have not. Every time you can near to approach the problem you desisted. Yesterday you offered that I talk to Joe to settle that matter. To solve it, you need replace the “?” with something in my earlier equations. You are free to replace them with anything you want.

    The joy of model is it frees you to build your own. In return, once you add marks on the paper, you’re bound to the inner logic of the signs you put down. These signs have meanings, they are of certain types, they have units.

    I don’t see how Joe’s equations preserve any of that. But if they do, they’re equivalent to the ones everybody else offer. For the same inputs, the same outputs obtain.

    If in the end every single sentence of my post needs to be rewritten, I won’t mind. That’s how humans float:

    We are like sailors who on the open sea must reconstruct their ship but are never able to start afresh from the bottom. Where a beam is taken away a new one must at once be put there, and for this the rest of the ship is used as support. In this way, by using the old beams and driftwood the ship can be shaped entirely anew, but only by gradual reconstruction.

    https://en.wikipedia.org/wiki/Neurath%27s_boat

    Did I tell you I want Joe’s model?

    If you give me Joe’s model, I will send you a copy of The Spanish Prisoner.

  357. Dr Roys Emergency Moderation Team says:

    It’s just a shortcut to what I explained in my May 2 @ 4:08 PM comment. You divide the solar constant, corrected for albedo (so 960 W/m^2) by 2, and it gets you the same result as the long way round. Same with dividing it by 4. If you look through what I do in that comment and just think about it for a moment, you should see why.

    I am multiplying the solar constant by the “Earth’s shadow” disk surface area. Then correcting the result for albedo. That gives you the total power in watts that the Earth absorbs. You then divide that figure by the total surface area of the hemisphere to get 480 W/m^2. Since the hemisphere’s surface area is double the disk surface area, all you actually have to do is divide the solar constant, corrected by albedo (so 960 W/m^2) by 2, to get the same result.

  358. Willard says:

    Repeating yourself won’t solve the equations, Rajinder.

    Solve these equations and win:

    [E] xy = 4z

    [J] xy/2 = ?

    I believe in you more than I believe in me. You know me. I’m either dishonest or delusional. How could I ever solve this very complex algebraic problem that preserves the geometry of the Earth?

    I’m sure Joe’s model solves that. Where is it?

  359. Dr Roys Emergency Moderation Team says:

    Flux in (W/m^2) = xy(1-a)/z

    Where x = the solar constant, y = disk surface area, a = albedo and z = the area of the hemisphere.

    z = 2y

    So:

    Flux in (W/m^2) = xy(1-a)/2y = x(1-a)/2

  360. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 4, 2021 at 3:18 pm
    “Well, TrueSceptic, all that is being brought up now are good questions and challenges to Postma’s arguments.”

    Well, that is promising. Given that we don’t disagree over the basic energy balance, just whether it matters if the input is considered to spread over the hemi- or whole sphere, it’s hard to see how Joe can reject the GHG theory, unless that really is a crucial distinction and results in a x2 error somewhere.

  361. “Willard and Gator, you want me to say the “dark side” is 3 K and the “light side” is 303 K. That might make some sense if the Earth were not rotating, but it is rotating. All the while, the lit hemisphere is receiving 480 W/m^2 in real time.”

    So you admit that the SB temperature has to be 3K on the dark side (or 0K as it is such a small amount taken to the fourth power wrt 303K taken to the fourth power) in order to satisfy the SB law for the entire sphere.

    You absolutely cannot slip in 303K on the lit side without slipping in 0K-3K on the dark side. The Earth has a temperature distribution over its surface and it is not uniform. I have done the math for said surface distribution (linear T and fourth power (SB) T). Anyone else can do this math correctly. JP has abjectly decided not to do so …

    Turns out that T (linear) = 286.7K and T (SB 4th power) = 287.8K (or about a degree higher and in good agreement with the conical 33K delta.

  362. aljo,

    An ignorant question, but would it actually be true that the Earth without an absorbing atmosphere would radiate to space from an “effective layer” a little above the surface, but much lower than the effective layer with an absorbing atmosphere?

    Technically, the hypothetical isn’t realistic. However, if there is no planetary greenhouse effect then the atmosphere would be radiatively inactive, which means it would neither absorb, nor emit, radiation. Hence, even if the surface did transfer energy to the atmosphere, this energy could not then be emitted to space from the atmosphere and all the energy would be radiated from the surface. However, this is really an unrealistic “all else being equal” hypothetical.

  363. Dr Roys Emergency Moderation Team says:

    “So you admit that the SB temperature has to be 3K on the dark side”

    Only if the Earth were not rotating.

  364. EFS,

    So if the lit side of the Earth has a SB temperature of 303K (taken directly from JP diagram above) what would the dark side of the Earth temperature have to be in order to satisfy the SB law?

    I think it would be 76K.

    303^4 + T^4 = 2 \times 255^4 \Rightarrow T = 76 K.

  365. Willard says:

    You’ve just rewritten the left part, Rajinder.

    It’s easy to see because you left the 4 out.

    An energy balance model is a model that balances the energy that comes in and the energy that comes out.

  366. MP,

    The sun provides at daytime enough heat at mainly the zenith zone to maintain the atmosphere hight and average pressure equilibrium , by evaporation of water and release of NO and O2 from the water there and then.

    What about the planetary greenhouse effect? What role does that play?

  367. Dr Roys Emergency Moderation Team says:

    Flux out (W/m^2) = xy(1-a)/w

    Where x = the solar constant, y = disk surface area, a = albedo and w = the area of the sphere.

    w = 4y

    So:

    Flux out (W/m^2) = xy(1-a)/4y = x(1-a)/4

  368. ATTP,

    It is a weighted average, so ((302K)^4/2 + (2.7K)^4/2)^0.25 = 254K or the fourth power of each half summed then take the fourth root (I get 254K instead of 255K due to slightly different constants).

    From my Excel spreadsheet =(0.5*B7^4+0.5*B12^4)^0.25, B7 = 302K and B12 = 2.7K

    That is how I did the Diviner data set but at much higher temporal and spatial resolutions.

  369. Willard says:

    > Flux out (W/m^2) = xy(1-a)/w

    To be an energy balance model, you need something that looks like energy in = energy out, not just

    energy in = abc
    energy out = def

    At some point you need to state that abc = def.

    I need to get going. Will check back later if what you wrote does that.

  370. Dr Roys Emergency Moderation Team says:

    Total power in = Total power out = xy(1-a)

    Where x = the solar constant, y = disk surface area, a = albedo

    That’s the “model”. The other calculations which result in “flux in” or “flux out” just convert the total power (either in or out) involved to flux by dividing by the surface area involved.

  371. TrueSceptic says:

    Bob Loblaw says:
    May 4, 2021 at 4:44 pm
    “What is all this “in real time” crap?

    Anyone with a “climate model” that is saying “oh, you can’t average sunlight between day/night sides of the earth” but is also saying “It’s perfectly reasonable to average sunlight across latitude and morning/noon/afternoon values” is very probably quite unaware of what they are actually modelling.”

    That’s how I see it, but did not know how to put it. Thanks.

  372. EFS,
    That’s what I thought I was doing, but have clearly made some silly error.

  373. MP says:

    @ ATTP

    Quote “What about the planetary greenhouse effect? What role does that play?”

    Heat flows down the gradiënt from hot to cold by bouncing molecules, pressure difference related convection, latent and potential heat transfer, and also by radiation.

    Ground level atmosphere is thanks to gravity the most pressurized zone where the heat received at the surface flows through, hence the higher temperature.

  374. MP,
    Can I just clarify. Are you describing something other than the greenhouse effect, or are you trying to describe the greenhouse effect?

  375. MP says:

    @ ATTP

    Quote “Are you describing something other than the greenhouse effect, or are you trying to describe the greenhouse effect?’

    Depends on what you mean with greenhouse effect.

    1 – if it means direct exrta surface heating by colder backradiation, then i describe something other than the greenhouse effect,

    2 if it means there is slowing of heat going out compared to a fictional black/grey body, resulting in an higher equilibrium temperature/state. Then there are many other slowing down causing factors at play, as described above

  376. MP,
    Seems like you’re partly describing the planetary greenhouse effect.

    if it means direct exrta surface heating by colder backradiation, then i describe something other than the greenhouse effect,

    Well, there has to be back-radiation. If there wasn’t, the surface would be far from energy balance and would be cooling.

    if it means there is slowing of heat going out compared to a fictional black/grey body, resulting in an higher equilibrium temperature/state. Then there are many other slowing down causing factors at play, as described above

    What other slowing factors can there be? There surface radiates as much energy per square metre per second as a 288K blackbody. There has to be something that prevent this all from being radiated directly into space, otherwise we’d be cooling. This is essentially the basics of the greenhouse effect.

  377. Bob Loblaw says:

    “The ground is in contact with the atmosphere, so at least a little energy is exchanged via that contact, and some of that energy must be radiated from the atmosphere out to space. “

    …but in order to radiate as IR, the emissivity has to be greater than zero somewhere in the IR wavelength range. And if the emissivity is greater than zero, then the absorptivity is also greater than zero, because they are equal (at a specific wavelength).

    So, now you have a gas/atmosphere that absorbs in the IR, and now you have the greenhouse effect in action.

  378. Willard says:

    > That’s the “model”.

    Thanks. The only geometrical entity in your model is a disc. That is a flat earth model!

    More seriously, you only stated that xyz = xyz, where z is (1-a ); your model has no means to convert power into temperature; and you have no division by 4.

    You can add all the letters you want, but I want my division by 4!

  379. MP says:

    @ ATTP

    Quote “What other slowing factors can there be?”

    Radiative transfer through the atmosphere is faster than the internet. Radiation travels with the speed of light and co2 absorbtion+radiation is in 1 10.000 th second. So even with many bouncing very fast

    Oceans have besides shorter term cycles very long term cycles, around 35 year net heating and 35 years net cooling

    Wetland subsurface can be heated the first 1 to 1.5 meter depth, Resulting in months surface heating later in the year when it is colder outside

    The slow convection bulk carrier takes several minutes to go high up

    Hadley cell pushes convected energy back down, causing a delay

  380. Bob Loblaw says:

    “Well, there has to be back-radiation. “

    Do I win a ClimateBall Bingo square if I am correct in guessing that MP is a person that thinks that back-radiation does not exist?

  381. Willard says:

    > Do I win a ClimateBall Bingo square

    The Bingo does not give away its squares! They are eternally there for every player to enjoy foreva!

  382. Dr Roys Emergency Moderation Team says:

    The other calculations which result in “flux in” or “flux out” just convert the total power (either in or out) involved to flux by dividing by the surface area involved. I showed you the divide by 4 with the “flux out” calculations, and to convert from flux to temperature you use the SB Law. You have everything there that you require.

  383. MP,
    It seems to me that you’re on the verge of simply describing the planetary greenhouse effect. You do realise that the greenhouse effect is essentially that the greenhouse gases in the atmosphere prevent the radiation in some bands from being radiated directly into space from the surface. A consequence of this is that some of the energy radiated into space comes from within the atmosphere (as you’ve already highlighted). That there is a lapse rate (atmospheric temperature profile) which is set by convection, means that the atmospheric temperature increases as you go down towards the surface (within the troposphere, at least). Given that we will tend towards a state in which the amount of energy being radiated into space (~240 W/m^2) is the same as the amount of energy we receive from the Sun (also ~240 W/m^2) means that the surface temperature will end up warmer than the effective radiative temperature (~255 K). In the case of the Earth, this ends up with the surface temperature being about 33K warmer than this effective radiative temperature (i.e., ~288K rather than ~255K).

  384. Willard says:

    > You have everything there that you require.

    I certainly don’t.

    You obviously have great interactional expertise on Joe’s model. But you can’t reconstruct it. Tell me where to look. Titles will be enough. Do you think I can find Joe’s canonical model in his Magnum Opus?

  385. Dr Roys Emergency Moderation Team says:

    You act like you don’t have everything that you require because that enables you to keep up this nonsense rather than correcting your article.

  386. Willard says:

    You can think whatever you please, kiddo.

    Tell me where to get Joe’s model, or we’re done.

  387. gator says:

    The incoming energy flux has no “temperature”. Connecting energy flux with a blackbody temperature only makes sense for an object in thermal equilibrium *radiating* that power. And even then, that flux has no temperature. The flux may be 100 W/m^2 1 meter from the object, but then it will be 25 W/m^2 2 meters from the object. (Assuming 1/r^2 type radiation.) Is the object suddenly cooler?? Is the flux “cooler”?? Nope, it’s just not how an energy flux connects to a temperature.

    What’s the solar flux hitting Mars? Less than hitting earth certainly. Does that meant the sun is now cooler? Nope it just means the energy radiating from the sun is spread over a bigger area.

    So the assertion that the earth “should be” 303 K because 480 W/m^2 is hitting it is nonsense. Who cares what the flux *in* is. The only thing setting the blackbody temperature of the earth is how much energy it is radiating out in equilibrium. So total energy in, total energy out – radiating out over the total surface area of the earth.

    Once you understand that, the whole thing of Joe saying the earth *should be* 303 K because “that’s the temperature of the flux” is seen as just wrong. It comes from a naive application of an equation without understanding what the equation means or where it is correct to use it.

  388. Willard says:

    > So the assertion that the earth “should be” 303 K because 480 W/m^2 is hitting it is nonsense. Who cares what the flux *in* is. The only thing setting the blackbody temperature of the earth is how much energy it is radiating out in equilibrium. So total energy in, total energy out – radiating out over the total surface area of the earth.

    That’s the best justification of the divide-by-4 trick if I ever seen one. Thanks!

  389. Dr Roys Emergency Moderation Team says:

    [I asked you to confim if Joe’s model was in his Magnum Opus. I want a yes or a no. -W]

  390. Willard says:

    Let’s corroborate if there are mentions of Joe’s model in his corpus.

    The first one is from the post I quoted in the post:

    I will end this post with a diagram of a reality-based model I have created which represents the Earth and power of sunshine as it actually exists in reality. […] If you want to read more about why this model was created, before I write Part 2, you can see this paper here: Copernicus Meets the Greenhouse Effect.

    https://climateofsophistry.com/2012/11/06/on-the-absence-of-a-measurable-greenhouse-effect-part-1-the-failure-of-ipcc-energy-budgets/

    Follows the 0.637 diagram. Part 2 looks promising:

    In Part 2 I will explain this model and show that, because it is based on actual reality, it is far superior to the fictional flat earth models of climate science and the AGHE.

  391. Willard says:

    OK. Part 2 has only a diagram:

    So let us have a look at the reality-based model once again and briefly develop an understanding of how reality actually works with the real power of sunshine. An updated version (it is a work in progress) of the global energy model is shown below.

    https://climateofsophistry.com/2012/11/06/the-fraud-of-the-atmospheric-greenhouse-effect-part-2-moving-to-reality/

    The 0.637 diagram again.

    There’s an equation at the top right of the diagram. It is left unspecified, and is also only for real-time heat flow.

  392. Willard says:

    More pictures for part 3, first the famous Trenberth diagram, then his 0.637 diagram. Follows an intriguing question:

    A question I have for you is: If you were to compare my model and the climate science model side-by-side, which one would you pick? Even forget about paying attention to the numbers and what they mean, and just based visually on the aesthetic and the shapes within the model, which would you pick as the model which looks like reality? Does the picture below help make the decision?

    https://climateofsophistry.com/2012/11/07/the-fraud-of-the-atmospheric-greenhouse-effect-part-3-in-pictures/

    A question I have for you is: if Joe clearly put on his diagram that the unlit side is -273C, would his readers still pick his?

  393. Willard says:

    In the first post of the series, Joe cites four papers:

    A Discussion on the Absence of a Measurable Greenhouse Effect
    Understanding the Thermodynamic Atmosphere Effect
    The Model Atmospheric Greenhouse Effect
    Copernicus Meets the Greenhouse Effect

    These are the ones I checked. I’ll check them once more to make sure.

    First, the quote from the first should show why I picked my title:

    If you average the input power of sunlight over the entire globe it makes it appear as though sunlight can only provide -180C worth of heating at the surface. But you have to look at the units of the metric you are dealing with here: W/m2or J/s/m2. Units mean something…you need to pay attention to what they mean. The units pose the question: what is the energy, in one second, over a square meter? These all have to occur together, simultaneously. The time and space in which incoming solar energy also impinges the night-side of the planet simply does not physically exist. It is only an imaginary mathematical trick that does not actually occur. What does occur in one second, and in the square meters where sunlight actually impinges, is illumination of a hemisphere with an intensity projection factor that goes as the function of the cosine from the zenith. If you integrate to the average projection factor and combine this with the Stefan-Boltzmann Law and terrestrial albedo, then the real-time instantaneous heat input is constantly +490C. At the zenith it has a maximum of +1210C, constantly, when the albedo is zero.

    A question I have for you is: where is the Earth?

    Follows a paragraph as to why power density can’t be averaged, another on mathematical ontology, then another on mathematical ontology, followed by another on mathematical ontology.

  394. Willard says:

    The Copernicus paper is a short one and is mostly a salespitch. We notice Joe’s 0.5 diagram, prefaced by beautiful poetry:

    The standard model greenhouse effect makes a simple but critical mistake in not differentiating between the concepts of energy flux density and total energy, in the context of the Law of Conservation of Energy. Along with this comes the un-realistic model of a dim, cool Sun, simultaneously shining on both sides of the Earth at once. This establishes a cognitive boundary condition, or paradigm, which does not conform to reality. In reality, an actual per-second, per-square meter snapshot has the sunlight shining upon one side of the Earth in any moment, and the Earth returning energy to space over both the light and dark sides. This was all formally explained and proven mathematically in the July paper. A model which incorporates the actually physical boundary conditions as they really exist was presented in that paper, and is reproduced below in Figure 3.

    The July paper is what I called Joe’s Magnum Opus.

    Alright. I’m fed up. Until tomorrow.

  395. TrueSceptic says:

    Bob Loblaw says:
    May 4, 2021 at 6:15 pm
    “…but how warm each part of the surface remains on average is going to depend on how warm it gets during the daylit hours and how efficiently it cools during the night. The problem becomes so complex that I do not think it is able to be solved.”

    That is an argument from incredulity. Three-dimensional climate models do this all the time. So do weather models.”

    Yeah, whoda thunkit? Anyone would think that some fake sceptics don’t have a f***ing clue what they are attacking.

  396. Dr Roys Emergency Moderation Team says:

    “Once you understand that, the whole thing of Joe saying the earth *should be* 303 K because “that’s the temperature of the flux” is seen as just wrong”

    gator, you are attacking a straw man. Joe is not saying the Earth should be 303 K. Joe has the effective temperature of Earth at 255 K, same as everybody else.

    “What is all this “in real time” crap?

    In real time, virtually no point on earth (horizontal or vertical) has a radiation balance of zero.

    In real time, virtually no point on earth (horizontal or vertical) has a non-radiative energy balance of zero.”

    What is all this “point on Earth”, crap, Bob? We were talking about the entire lit hemisphere, in real time, receiving 480 W/m^2, whilst the entire sphere emits 240 W/m^2.

  397. Dr Roys,
    What is Joe’s actual argument then? What’s he actually criticising? If all it amounts to is “the Sun only falls on one hemisphere” then that’s just silly. Everyone knows that. As has been pointed out numerous times already, energy balance just means energy in = energy out. It makes no difference if you consider the energy falling on the planet as being intercepted onto a circle of radius R (cross-sectional area), averaged across only one hemisphere, or averaged across the whole sphere.

  398. Dr Roys Emergency Moderation Team says:

    [Answer AT’s question, please. -W]

  399. Dr Roys Emergency Moderation Team says:

    Lol, OK. One of Joe’s many “arguments”, and the one which kicked off the discussion that started this post, is simply that, in real time, the Sun shines on only one hemisphere, with a flux of approximately 480 W/m^2 averaged over that hemisphere, whilst at the same moment 240 W/m^2 leaves from the entire sphere. Joe thinks it is incorrect to use 240 W/m^2 as the input from the Sun. He describes this as “flat Earth”, since the only way the Earth could actually absorb 240 W/m^2 from the Sun at once would be if the Earth were a flat plane in space.

    I put “arguments” in quotes because all he is really doing is stating a fact.

  400. Dr Roys Emergency Moderation Team says:

    “Joe’s trouble may be geometric: he wants the light to fall on a hemisphere. The disc doesn’t seem not real enough for him {4}. This is corroborated by a post in which he, with the tip of Archimedes’ hat, increases solar input. His diagrams display 30C, in contrast to the usual -18C {5}. How did he pull his trick? Probably misspecification {6}. Joe’s model for the whole Earth remains elusive. It needs to balance the same way as the other ones or it’s humbug. Meanwhile, his divide-by-two trick fares no better {7}.”

    Ultimate, eternal sigh. Joe has no problem understanding geometry. It is not about the hemisphere vs. the disk, for crying out loud! He wants the light to fall on a hemisphere, as it does, in real time, as opposed to being averaged over the whole sphere. Do you need to see one of his diagrams that display the 480 W/m^2 input, and the 240 W/m^2 output? Because he has those, where it is actually spelled out for you. Joe’s model is not elusive. I have explained how he gets the 480 W/m^2 input dozens of times. This whole paragraph is an insult to our discussion, Willard, and for you to claim I have helped you come up with it is outrageous.

  401. Willard says:

    > If all it amounts to is “the Sun only falls on one hemisphere” then that’s just silly.

    Here’s what I think is Joe’s conceptual problem:

    When the light falls on Earth, it falls on hemisphere. A hemisphere, when corrected at zenith, is a disc. That’s what every EBM I know does.

    This is the point of the post. It rests on a theorem.

    The only state of the equation that refers to what the Earth absorbs is when the left part only contains Power x DiscAlbedo. That’s by definition. As soon as we start moving terms around, the left part stops being the part about what the Earth receives. Should we think that what the Earth emits has been reduced to [SB x] T because we move the 4 at the left? That would be absurd.

    Hence why AT’s argument according to which we don’t always need to divide by four matters. As you yourself said elsewhere (as some kind of gotcha, no less) Power x Disc is not the same unit as Power x Disc / 4. They’re not of the same type. So as I understand it, Joe’s argument commits a type error.

    Joe is stuck with the 2 he introduces. It needs to come from right side. He also needs to be able to put it back there. If he does that, where is the geometry of the Earth? To accept that the Earth emits 240 is not enough: he needs to model the whole Earth too!

    I asked so many times now (I lost count at 10), it’s obvious you haven’t checked Joe’s Magnum Opus. All he got is a hemispherical equilibrium model. It’s equation 21.

    Nowhere I have seen Joe admit that the unlit hemisphere is -270C.

  402. Willard says:

    > He wants the light to fall on a hemisphere.

    The light already falls on a hemisphere.

    A hemisphere, when corrected at zenith, is a disc! Check AT’s proof. How many times must I repeat that geometrical fact?

    There’s a reason why I’m citing Joe’s Hat theorem post, dammit!

  403. Joe,

    Joe thinks it is incorrect to use 240 W/m^2 as the input from the Sun. He describes this as “flat Earth”, since the only way the Earth could actually absorb 240 W/m^2 from the Sun at once would be if the Earth were a flat plane in space.

    So, basically Joe struggles to divide by 2? That’s a bit embarassing. Seriously, 480 W/m^2 times the area of one hemisphere is the same as 240 W/m^2 times the area of a sphere.

  404. Dr Roys Emergency Moderation Team says:

    [Repond to what is being said, keep repetition for Roy’s, and please consult Joe’s Magnum Opus, for Newton’s sake! -W]

  405. Willard says:

    > basically Joe struggles to divide by 2

    As I see it, he’s refusing to accept that the energy that the Earth receives goes to the other side instantaneously. It looks like he’s building a model ab initio, in which every parcel of energy is communicated to the next for the first time. Hence why he spends so much time talking about how the heat moves.

    It’s as if Joe never saw a GCM at work. Heck, it’s not even clear he knows anything about one-dimensional models.

  406. Dr Roys Emergency Moderation Team says:

    Yes, indeed, AT. But the globe is not receiving 240 W/m^2 from the sun at once, is it? It is receiving 480 W/m^2, over only the lit hemisphere, in real time. Yes?

  407. Willard says:

    > It is receiving 480 W/m^2, over only the lit hemisphere, in real time

    What real time?

    Do you or Joe really think that every single second is the same?

    Do you really think that the unlit hemisphere is -270?

    Do you really think that you can get a GCM out of an EBM as is?

    Why do you still dodge the geometrical fact that the hemisphere, when corrected for angles, is a disc?

    I can argue by leading questions too!

  408. Dr Roys Emergency Moderation Team says:

    “The only state of the equation that refers to what the Earth absorbs is when the left part only contains Power x Disc. That’s by definition…”

    And Power x Disc gives you a result in watts that the Earth absorbs, if you want to get a value for the flux (W/m^2), you have to divide by the surface area absorbing it. Which, in real time, is the hemisphere’s surface area.

  409. Willard says:

    > if you want to get a value for the flux (W/m^2), you have to divide by the surface area absorbing it.

    Cope.

    I want you to agree or disagree on the main point of this post: a hemisphere, when corrected (correctly, I might add) for angles, is a disc.

    Yes or no?

  410. Dr Roys Emergency Moderation Team says:

    I don’t disagree with AT’s integration, if that’s what you are asking.

  411. TrueSceptic says:

    Dr Roys Emergency Moderation Team says:
    May 5, 2021 at 1:49 pm
    “Yes, indeed, AT. But the globe is not receiving 240 W/m^2 from the sun at once, is it? It is receiving 480 W/m^2, over only the lit hemisphere, in real time. Yes?”

    By definition, the total input is identical, as stated and agreed many times.

  412. Willard says:

    > I don’t disagree with AT’s integration, if that’s what you are asking.

    I’m asking you to do the next step.

    The light that falls on a hemisphere, when we correct for the angles, equals the light that falls on a disc.

    Yes or no?

  413. Bob Loblaw says:

    “What is all this “point on Earth”, crap, Bob? We were talking about the entire lit hemisphere, in real time, receiving 480 W/m^2, whilst the entire sphere emits 240 W/m^2.”

    The point of “point on earth” is that every location (AKA point) on earth is different. Geometry tells us this, as every single location sees the sun at a different location in the sky. And that location in the sky changes every second – in real time – as the earth rotates.

    I happen to know this because I have programmed computer-controlled tracking systems designed to point solar radiation instruments directly at the sun throughout the day. In real time.(Well, I actually knew it may years before that, because I paid attention in climatology class as an undergrad.)

    And just about every one of those locations does not receive 480 W/m^2.

    I know this because I used to make a living measuring solar radiation. In real time. (Well, I actually knew it may years before that, because I paid attention in climatology class as an undergrad.)

    The only way you get 480 W/m^2 is by averaging over a large number of locations across an entire hemisphere. You know. Averaging. The kind of thing that some claim is wrong to do for the entire sphere?

    The internal inconsistency in your position is absolutely amazing. In real time.

  414. Dr Roys Emergency Moderation Team says:

    TrueSceptic, if you wish to divide the incoming power over surface area that is not receiving said power, then by all means use 240 W/m^2 as the real time solar input. Just know that the reality of the situation is, the incoming power is actually only received by the lit hemisphere at any given moment, so the incoming flux is really 480 W/m^2.

  415. Dr Roys Emergency Moderation Team says:

    Sure, Willard, of course…but that is the total power, in watts. You need to then decide what measure of surface area you are going to divide that total power by. The hemisphere (480 W/m^2) or average it over the whole sphere (240 W/m^2)?

  416. Willard says:

    > You need to then decide what measure of surface area you are going to divide that total power by.

    If you want to model the Earth, the choice is rather limited.

    Quick question: in Joe’s diagrams, there is an equation at the top right. Do you know what it means? I only found it at one place in his work.

  417. TrueSceptic says:

    …and Then There’s Physics says:
    May 5, 2021 at 1:38 pm
    “So, basically Joe struggles to divide by 2? That’s a bit embarassing. Seriously, 480 W/m^2 times the area of one hemisphere is the same as 240 W/m^2 times the area of a sphere.”

    Isn’t this the crux of the matter? Joe insists that the 2 cases give different results, enough to make the greenhouse effect unnecessary. I think the issue is conceptual, not mathematical.

  418. Willard says:

    > I think the issue is conceptual, not mathematical.

    If the issue was mathematical, it would not be me who would have written this post!

  419. Dr Roys Emergency Moderation Team says:

    No, you will have to ask Joe on that one.

  420. Dr Roys,

    Yes, indeed, AT. But the globe is not receiving 240 W/m^2 from the sun at once, is it? It is receiving 480 W/m^2, over only the lit hemisphere, in real time. Yes?

    But it’s also the case that not every square metre on the lit hemisphere is receiving 480 W/m^2. A one square metre patch with the Sun directly overhead receives ~960 W/m^2 while a 1 square metre patch with the Sun on the horizon receives almost nothing. So, it’s still an average.

  421. Dr Roys Emergency Moderation Team says:

    So, Willard, dare I ask…are you going to correct the article?

  422. Willard says:

    Read back. Tell me if you still disagree with anything.

    And in return to your cooperation, Joe explains his equation in A Discussion.

    Start with equation (7).

  423. Dr Roys Emergency Moderation Team says:

    That’s right, AT. The 480 W/m^2 is still an average. But, it’s at least an average that reflects what is really happening. 240 W/m^2 implies the whole Earth receiving the Sun’s energy at once, which is of course impossible.

  424. Willard says:

    > I happen to know this because I have programmed computer-controlled tracking systems designed to point solar radiation instruments directly at the sun throughout the day.

    Nice.

  425. Willard says:

    > 240 W/m^2 implies the whole Earth receiving the Sun’s energy at once

    I don’t think it does. We’ll have to agree to disagree on that one.

    A model is a model is a model. But all energy balance models need to balance energy. And Joe’s stuck with his 2.

  426. Dr Roys Emergency Moderation Team says:

    I explained at 1:30 PM what I disagreed with. I still do. Nothing has changed.

  427. Willard says:

    You are free to disagree, of course.

    I explained well enough why I think Joe’s geometrical intuition sucks.

    Every time you ask me to correct the text, you help me improve my case. If you insist, I will find a way to put the -270C in the text, and not only allude to it in a note.

  428. Dr Roys Emergency Moderation Team says:

    Willard, you may believe that your 1:33 PM comment is coherent. Perhaps, inside your head, with whatever it is that is going on in there, the words you are saying make some sort of sense. I challenge any other commenter here to try to translate what point you believe Willard is making, in that comment, into something comprehensible.

  429. Dr Roys Emergency Moderation Team says:

    “A model is a model is a model. But all energy balance models need to balance energy. And Joe’s stuck with his 2.”

    480 W/m^2 input over the hemisphere and 240 W/m^2 output over the whole sphere does balance energy.

  430. Willard says:

    > I challenge any other commenter here to try to translate what point you believe Willard is making,

    I make four points in that comment, kiddo. You just agreed on the first two:

    The light already falls on a hemisphere. When corrected for angles, it’s a disc. You agree on that.

    The energy that gets in is defined by Power x DiscAlbedo. You agree on that.

    Joe is stuck with his division by 2.

    Joe is a scoundrel who hides that his unlit hemisphere is -270C.

    So we have two points to discuss. Which one will you pick?

  431. Bob Loblaw says:

    “That’s right, AT. The 480 W/m^2 is still an average. But, it’s at least an average that reflects what is really happening.”

    What’s really happening? Really? It’s no more “really happening” than any other average. That 480 W/m^2 does not represent what is happening at any point on the hemisphere – expect for a few isolated points that happen to receive that value for a brief second.

    After all, a recent statistic survey has show that what is really happening is that the average person has one testicle and one breast. In Real Time ™

  432. Willard says:

    > What’s really happening?

    Suppose that what’s really happening is really happening. Does it settle if it is really really happening?

  433. Dr Roys Emergency Moderation Team says:

    Yes, Bob, really. On a second by second basis, the Earth really is absorbing 480 W/m^2 across the lit hemisphere, whilst it emits 240 W/m^2 from the entire sphere. That is the reality of the situation. Or, at least, it is far closer to the reality of the situation than saying that, on a second by second basis, sunlight falls across the entire sphere, so that the input is 240 W/m^2, whilst 240 W/m^2 is emitted from the entire sphere.

  434. Willard says:

    > 480 W/m^2 input over the hemisphere and 240 W/m^2 output over the whole sphere does balance energy.

    I have yet to see a model where Joe does that. And a hemispherical energy model does not cut it. You did not read Joe’s papers. I did.

    Joe’s philosophical meanderings fizzle when we observe that the light falls on the hemisphere already in ordinary zero-dimensional energy models. The divide-by-four he perceives as a bug is a feature, and when he refuses it he’s shooting himself in the foot. His “flat earth” mockery reveals that he has built a strawman out of a toy model he cannot replicate with his silly division by 2.

    All this to con people like you.

    That’s very sad. But what we accomplished is great. Thank you.

  435. Dr Roys Emergency Moderation Team says:

    Willard, this is where you disagree with seemingly every other commenter here. Most commenters here have already understood the point that 480 W/m^2 input over the hemisphere and 240 W/m^2 output over the whole sphere does balance energy.

  436. Dr Roys,

    But, it’s at least an average that reflects what is really happening. 240 W/m^2 implies the whole Earth receiving the Sun’s energy at once, which is of course impossible.

    Except, this is the energy input to the climate system. And, no, it doesn’t imply the whole planet receives the Sun’s energy at once, it just implies that the Earth is absorbing an average of 240 J of Solar energy every second per square metre.

  437. TrueSceptic says:

    Willard says:
    May 5, 2021 at 2:23 pm
    “If the issue was mathematical, it would not be me who would have written this post!”
    🙂

  438. Willard says:

    > Most commenters here have already understood the point that 480 W/m^2 input over the hemisphere and 240 W/m^2 output over the whole sphere does balance energy

    You still don’t get it, kiddo, do you?

    I’m not saying it can’t be done. I’m saying that Joe does not do it. The 2 comes from a hemispherical model. He simply replaced a 4 by a 2.

    Check equations (21-22) in Joe’s Magnum Opus. Report.

  439. Willard says:

    > I’m not saying it can’t be done.

    I’ll raise that with a quote from a comment on which nobody commented:

    So, how does Joe balance out the various temperatures of the Earth from his model? Here’s how Tim balances his […]

    Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190878

  440. Dr Roys Emergency Moderation Team says:

    So your argument is that I am doing something Joe is not doing? Because I have demonstrated that 480 W/m^2 input over the hemisphere and 240 W/m^2 output over the whole sphere does balance energy. I have done that, several times.

  441. Willard says:

    Alright. You will never read Joe’s Magnum Opus. Here’s where the 30C figure comes from:

    How does that model the Earth?

    It does not.

    ***

    Now, do you know two-box models?

    AT, tell our guest about two-box models.

  442. Bob Loblaw says:

    “Or, at least, it is far closer to the reality of the situation than saying that, on a second by second basis, sunlight falls across the entire sphere”

    Except nobody (that I am aware of) is saying that. You are arguing a straw man.

    What is being said is that it averages to 240 W/m^2 when you consider the same area that is emitting IR (the entire sphere). And that is just as real as 480 W/m^2 when averaged over a hemisphere (only half the earth).

    When it comes to global climate (and climate models), the unlit side of the planet is just as real as the lit side. It just happens to be a bit darker ( on average).

    You’re like Humpty Dumpty, with averaging instead of words:

    “When I use averaging,” Humpty Dumpty said, in rather a scornful tone, “it means just what I choose it to mean—neither more nor less.”
    “The question is,” said Alice, “whether you can average so many different things.”
    “The question is,” said Humpty Dumpty, “which is to be master—that’s all.”

  443. Dr Roys Emergency Moderation Team says:

    That’s, right, it does not model the Earth. It is not supposed to. That just gives you the equivalent “blackbody temperature” associated with the 480 W/m^2 input, but Postma is aware (and agrees) that the Earth’s effective temperature is 255 K (associated with 240 W/m^2).

  444. AT, tell our guest about two-box models.

    Well, there’s a good description of the simple one-box model in this Isaac Held post and a description of the two-box model in this post.

    The term C \dfrac{dT}{dt} is basically the area averaged heat capacity of the system multipled by the rate of change of temperature, which then gives units of W/m^2 and is essentially the planetary energy imbalance.

  445. Willard says:

    > but

    That’s a big but there, Rajinder.

    One does not simply sell a hemispherical model of the Earth as a more “realistic” alternative to a model of the Earth, more so when one conceals that it’s -270C on the other side.

    And if Joe knows what he’s doing, he’s running a con. This will be the topic of my next post.

  446. That’s, right, it does not model the Earth. It is not supposed to. That just gives you the equivalent “blackbody temperature” associated with the 480 W/m^2 input, but Postma is aware (and agrees) that the Earth’s effective temperature is 255 K (associated with 240 W/m^2).

    And the equality

    \pi R^2 (1 - A) F_{\odot} = 4 \pi R^2 \sigma T_E^4,

    is also not intended to model the Earth. It’s simply an energy balance calculation that gives the effective radiative temperature given the incoming flux, the albedo, and an assumption that the system will tend towards energy balance. Why are Postma’s simplifications okay, and other people’s are not?

  447. Bob Loblaw says:

    And if Joe knows what he’s doing,

    I suspect I first heard this phraseology from someone (Ray L?) over at RC, but here it goes:

    I think that we and Joe share one thing in common: none of us has any idea what Joe is talking about.

  448. Dr Roys Emergency Moderation Team says:

    “Except nobody (that I am aware of) is saying that. You are arguing a straw man.”

    Nobody says that…but it is what the average of 240 W/m^2 implies. That the Earth is continuously receiving this low flux of energy from the Sun, over its whole surface area, in real time. As soon as you say, “the Earth actually receives 480 W/m^2 over the lit hemisphere, at any given moment” you get…well, you get the reactions that we see from the comments here.

  449. Bob Loblaw says:

    “Why are Postma’s simplifications okay, and other people’s are not?

    Oh, I know! Pick me! Pick me!

    …because certain people like Postma’s conclusions, and don’t like other people’s conclusions.

  450. Willard says:

    > is also not intended to model the Earth.

    OMG. I am wrong again! I think I will die.

    *Faints*.

    Are you starting to get what I’m doing here, Rajinder?

    We have one life. It’s meant to be fun. I’d rather be wrong a thousand times a day than not have fun. Try it. It’s worth it. Everybody around you will smile more.

    It does not matter to be wrong. We all make mistakes. Let go of 386. It is ruining your life.

    ***

    Everyone, if you like the Climateball project, please consider donating to Clowns Without Borders. You can find the link below:

    Source: https://climateball.net/colophon/

  451. Dr Roys Emergency Moderation Team says:

    “One does not simply sell a hemispherical model of the Earth as a more “realistic” alternative to a model of the Earth, more so when one conceals that it’s -270C on the other side.”

    One, he’s not doing that…and two, the Earth would only be 3 K on the other side if the Earth were not rotating, and had no atmosphere.

  452. the Earth would only be 3 K on the other side if the Earth were not rotating, and had no atmosphere.

    No, it would be 2.725K!

  453. Willard says:

    > One, he’s not doing that…

    That’s your opinion, kiddo.

    ***

    > and two, the Earth would only be 3 K if

    That’s another cope.

    Do you really think that we’re running GCMs out of a single equation? That’s just sanity check. If you want a better model, go check Isaac’s two-box model. It’s still simple, and Isaac is cool.

  454. Dr Roys Emergency Moderation Team says:

    Willard, before you write your next article, are you going to correct this one?

  455. Willard says:

    > are you going to correct this one?

    Which sentence would you like me to correct next, and where do I send you a copy of The Spanish Prisoner?

  456. Bob Loblaw says:

    “Nobody says that…but it is what the average of 240 W/m^2 implies.”

    No, it does not.

  457. Willard says:

    > Nobody says that…but

    When I had a Twitter account, I had a series called *Erisology*. It was based on the Dark Maul Double Saber meme. It looked like this:

    If nobody says that, nobody says that, and that’s about it.

    If you detest misrepresentation, imagine climate scientists who read Joe’s.

  458. Dr Roys Emergency Moderation Team says:

    [I asked for one sentence, Rajinder. One thing at a time. You’re not here to spam Sky Dragon propaganda. -W]

  459. Dr Roys Emergency Moderation Team says:

    “Joe’s trouble may be geometric: he wants the light to fall on a hemisphere. The disc doesn’t seem not real enough for him”.

    Should be:

    “Joe understands the geometry of the “divide by 4”: his issue is that he wants the incoming solar radiation to be averaged only over the lit hemisphere, rather than being averaged over the entire sphere”.

  460. “Joe understands the geometry of the “divide by 4”: his issue is that he wants the incoming solar radiation to be averaged only over the lit hemisphere, rather than being averaged over the entire sphere”.

    And this is just silly. If you’re using energy balance to estimate an effective radiative temperature, it doesn’t matter if you treat the incoming energy as being intercepted by a circle with radius equal to the radius of the Earth, being average over the lit hemisphere, averaged over the whole sphere, or simply doing it by determining total incoming energy per second and total outgoing energy per second. You. get. the. same. answer. Insisting that people do it in some specific way is nonsensical. Why is this so difficult to get?

  461. Dr Roys Emergency Moderation Team says:

    [Playing the ref. -W]

  462. Willard says:

    > Joe understands the geometry of the “divide by 4”

    You’re changing the subject. That does not work.

    Have you read Joe’s Mad Hat post? I linked to it in the paragraph you help me edit. It’s here:

    https://archive.ph/4Bgvb

    *Puts on his Columbo trenchcoat.*

    Can you help me understand what Joe is doing there?

  463. Maybe Joe should correct this

    Out of the mathematical convenience of not having to treat the system in real-time, and with the real power of sunshine, climate scientists average the real-time power of sunshine over the entire surface of the Earth at once, so that they can get rid of day and night, and also so that they can treat the Earth as flat, which makes things easier for them in the math.

    Having written the above, it’s really hard to not interpret it as Joe objecting to a perfectly reasonable calculation. You do get that noone is actually suggesting that the incoming sunlight is actually spread across the whole sphere? Hence, Joe’s suggestion is simply wrong. All that’s being done is a calculation that determines an effective temperature based on equating incoming and outgoing energy. If anything, Joe should be embarassed that he’s kicked up a fuss about something so trivial.

  464. Willard says:

    Read Joe’s Mad Hat post, AT.

    Tell me what you think of it.

  465. Dr Roys Emergency Moderation Team says:

    Your article implies Joe has trouble with basic geometry. He does not. So that should be corrected. Look at the very first comment your article received!

  466. I should also add, that 3D GCMs do not do what Joe has suggested. They treat the Earth as a rotating sphere with an atmosphere and that is only illuminated on one hemisphere. Therefore, climate scientists do not average the real-time power of sunshine over the entire Earth. They do sometimes do energy balance calculations which simply balance incoming and outgoing energy, but full climate models assume a much more realistic geometry.

  467. Willard says:

    > He does not

    Proof by assertion, 0-2.

    Changing the subject was 0-1.

  468. Willard says:

    > climate models assume a much more realistic geometry

    Exactly what I said in one of my first comments in response to Rajinder regarding this at Roy’s, AT.

    But I wasn’t wrong. So it fell into deaf ears.

  469. Dr Roys Emergency Moderation Team says:

    “he wants the light to fall on a hemisphere. The disc doesn’t seem not real enough for him”.

    Should be:

    “his issue is that he wants the incoming solar radiation to be averaged over only the lit hemisphere, rather than being averaged over the entire sphere”.

  470. Dr Roys,
    You keep repeating that. Why? Noone is actually suggesting that the energy falls equally on all areas of the sphere. All that is being done is a simple energy balance calculation that equates incoming and outgoing energy so as to estimate an effective temperature. As has already been pointed out, it doesn’t matter how you go about doing this as long as you do end up with a calculation that equates incoming and outgoing energy.

  471. Willard says:

    You yourself said that Joe wants the light to fall on a hemisphere. I could produce at least ten quotes from you about “spreading” the light! Have you changed your mind?

    Also, I asked you if you read Joe’s Mad Hat post. Did you? If you have not, I will read it to you.

    Here’s where we’re going. You ask me to correct my perception of Joe’s geometrical intuition. I read you his Mad Hat post. Win-win.

  472. Dr Roys Emergency Moderation Team says:

    [Playing the ref once more. This time it counts: 0-3. -W]

  473. Bob Loblaw says:

    I should also add, that 3D GCMs do not do what Joe has suggested.”

    In addition, Real Time(tm) models do not assume radiative equilibrium at any point in time or space, and they include other important energy fluxes (thermal, latent heat, etc.) in both the atmosphere and the subsurface (land, ocean). They do conserve energy.

    3D GCMs are very similar to weather models. Weather model output varies over time. That’s how they forecast different weather on different days. Whooda thunk?

  474. Dr Roys Emergency Moderation Team says:

    [Please answer my questions. They are related to your request. If you don’t answer them, we can’t proceed. -W]

  475. Dr Roys Emergency Moderation Team says:

    “You yourself said that Joe wants the light to fall on a hemisphere. I could produce at least ten quotes from you about “spreading” the light! Have you changed your mind?”

    No, I have not changed my mind. Joe wants the incoming solar radiation to be averaged over only the lit hemisphere, rather than being averaged over the entire sphere. It is hemisphere vs. sphere, not hemisphere vs. disk. Please correct the article.

    “Also, I asked you if you read Joe’s Mad Hat post. Did you? If you have not, I will read it to you.”

    Yes.

  476. Willard says:

    > No, I have not changed my mind.

    It was a rhetorical question. The way you anwer it, once again, sidesteps the reason why I’m asking it. The very idea of spreading on a hemisphere isn’t related to the left side of the equation before we spread anything. That is the type error I’m talking about. The type error is cause by misguided geometric intuition.

    ***

    Perhaps I should kick things off. Here’s what I think is the central para in Joe’s Mad Hat post:

    And so it turns out that the weighted integrated average projection factor on a hemisphere is the same as the simple linear average, even though the weighted projection function is not linear.

    So as I understand Joe wakes up Archimedes to realize that yes indeed we can divide a hemisphere by 2. And why does he do that? Here’s what he provides:

    Now, the reason why we’re interested in the integrated average projection factor is because we can then multiply that by the top-of-atmosphere solar flux in order to get the integrated average flux on the input hemisphere. So that gives us the 1370 W/m² divided by 2, as we have in the top figure, and we can convert that to an equivalent forcing temperature as shown there too.

    Now, why is Joe dividing 1370 by 2 here?

    I’ll let you work on that one, Rajinder.

  477. Dr Roys Emergency Moderation Team says:

    As already explained:

    Flux in (W/m^2) = xy(1-a)/z

    Where x = the solar constant, y = disk surface area, a = albedo and z = the area of the hemisphere.

    z = 2y

    So:

    Flux in (W/m^2) = xy(1-a)/2y = x(1-a)/2

  478. Willard says:

    > xy(1-a)/z

    You’re trying to dodge my geometric point with algebra.

    And I don’t see how 1370 / 2 = xy(1-a)/z.

  479. Dr Roys Emergency Moderation Team says:

    True, looks like a mistake in the post. He should be dividing the 1370, corrected for albedo (so 960 W/m^2), by 2. To get 480 W/m^2.

  480. Willard says:

    > True, looks like a mistake in the post.

    Perhaps not, for here’s how he continues:

    However, the relationship between flux and temperature is not linear but has a fourth-power exponential dependence between them, and so, the integrated average flux converted to temperature will not actually be the same as the integrated average flux when that flux is first converted to an equivalent temperature. The integrated average flux is certainly useful and interesting, but we are probably still more interested in the integrated average flux as converted to a temperature forcing. And so to calculate this we follow the same routine as above, but this time we first convert the flux “F” at each symmetric annulus band to a temperature via the Stefan-Boltzmann Law (T = (F/5.67e-8)1/4) and multiply that value with the weighting (as area) it falls upon.

    We’re almost done. After some computation Joe gets a number.

    I’ll let you the joy of rediscovering it.

  481. Dr Roys Emergency Moderation Team says:

    No, pretty sure he intended to say:

    “Now, the reason why we’re interested in the integrated average projection factor is because we can then multiply that by the top-of-atmosphere solar flux in order to get the integrated average flux on the input hemisphere. So that gives us the 1370 W/m² [corrected for albedo, 960 W/m^2] divided by 2, as we have in the top figure, and we can convert that to an equivalent forcing temperature as shown there too.”

    If by “top figure” he is referring to his diagram that we have been discussing, then certainly it shows the “equivalent forcing temperature” for 480 W/m^2 of 303 K.

  482. Willard says:

    Here’s the number Joe gets:

    = 288.5°K = 15.5°C.

    This still isn’t really that meaningful of a number, because the rotating Earth spends quite a bit of angular sweep underneath input much closer to unity with the top-of-atmosphere flux, and it is this high-powered flux which creates cumulonimbus clouds, and the climate, etc.

    So yeah. 15C. Surprising, isn’t it?

    Now, we could try to interpret that number. A simple way would be to ask: what has Joe done? To answer that question, we need to track back what Joe did, and put it into a sentence.

    I knew all along where you got your “but rotation.” It’s a cope.

  483. Dr Roys Emergency Moderation Team says:

    I personally never read too much into that number. I can see how it could be misleading though.

  484. Willard says:

    > I personally never read too much into that number.

    Me neither. Here’s what I read:

    Joe’s title is How to Calculate the Average Projection Factor onto a Hemisphere.

    In the post, he says that the question is how one calculates the average projection factor of the incident solar radiation onto the hemisphere of the Earth which sunlight falls upon.

    His computations never take the disc into account, in fact they replace it with a weighted hemisphere.

    His illustration of the Hat theorem should remind you of something we do with one-dimensional models: latitudes.

    Correct me if I’m wrong, but Joe indeed bypass the disc to use the hemisphere instead. And we already know by AT’s derivation that a hemisphere, when weighted correctly, should correspond to the disc. Something’s amiss. But even granting Joe all his computations and final number, we can safely say that what he’s doing is to replace the disc with a hemisphere.

    As written in the post.

  485. Dr Roys Emergency Moderation Team says:

    As I showed, you certainly use the disk in my calculations when coming up with the “divide by 2” factor…but my calculations simply use the fact that the hemisphere is double the surface area of the disk. This is what Postma refers to as the “simple linear average”, I think. What Postma is looking for is an alternative to that, to come up with a better factor than the “divide by 2”…but what he found out is that, for all his calculations, it still came back to “divide by 2”.

    Either way, the number being divided by 2 is the solar constant, corrected for albedo.

    I don’t think there is anything amiss. Maybe AT could weigh in on the post?

  486. Willard says:

    I’m not sure where you got that 2 in Joe’s post, Rajinder.

    As a token of appreciation, my editor friend (the same one I quote in the post) just sent me this:

    I’m serious about sending you a copy of The Spanish Prisoner. Unless you already saw the movie. Have you? If you haven’t, you’re very lucky: it explains almost all of Climateball.

  487. Dr Roys Emergency Moderation Team says:

    Well, multiplying by 0.5 is the same as dividing by 2.

  488. Willard says:

    > Well, multiplying by 0.5 is the same as dividing by 2.

    Right. Sorry, my mind was elsewhere.

    Then the questions become what does he divide by 2, and why does he get half of his trademark 30C.

    If with the same inputs one blackbox B gives 30 and another one D gives 15, then B and D are not functionally equivalent. Do you agree?

  489. Dr Roys Emergency Moderation Team says:

    The number being divided by 2 is the solar constant (1,370 W/m^2), corrected for albedo (so, 960 W/m^2).

    The reason he gets the 15 C instead of 30 C is as he explains:

    “However, the relationship between flux and temperature is not linear but has a fourth-power exponential dependence between them”.

    It is difficult to explain. MP wrote a comment about it:

    “The 4th root problem Say that two spots on your blackbody sphere are being exposed to 50 and 100 watts per square meter. (Due to curvature, remember, a single light source gets spread out and becomes weaker.) Using the Stefan-Boltzmann equation, thetwo temperatures will be about 172 and 205 Kelvin respectively, i.e., an average of 188.5K. But the average irradiance is 75 W/m², which corresponds to 191K. That’s 2.5 degrees off the mark. In other words, average temperaturedoes not agree with average irradiance, and vice versa. Take three spots at 100, 200, and 300 W/m². The average of course is 200 W/m². The temperatures are 205, 244, and 270 respectively, averaging about 240K. But 200 W/m², the average, equals 244K. Now you’re 4 degrees off the mark. And so on, as you proceed to compare irradiance with temperature on each and every angle of a half-lit sphere. It’s a huge problem to tackle.”

  490. Willard says:

    > It is difficult to explain

    I see. Well, that’s one problem with following equations just by their numbers. The important thing is to identify the entities of the model. The EMB I presented in the post can be reduced to

    Power x Albedo = SB x Earth x ET

    (I’m using ET to represent Effective Temperature, because it’s a detail that matters, and because it’s extraterrestrial. Wink wink.)

    That notation is helpful because it helps abstract away all the fussy details under the hood. Power can contain Disc and Sun if you follow your typing properly. Etc. That notation can lead astray, as when I wrote Power x Disc earlier. Notations comes with tradeoffs. Programmers have become quite good with that.

    So, what are the entities that Joe computes in his post?

    I need your help here. There is stuff I need to do around the house, and I think I contributed enough to earn some respect.

    ADD. I just added Earth. I forgot the Earth! I need a break.

  491. Dr Roys,
    Everyone who understands this understands that the effective temperature that you get from an energy balance calcuation is not the same as you would get if you simply did some kind of average of the temperature. If you consider a synchronously-rotating planet with no atmsphere at 1AU from a Sun-like star and that still have an albedo of ~0.3, it would still have an effective radiative temperature of ~255K, but would be very hot on the sun-facing side (~300K) and very cold on the anti-sunward side (much less than 100K). So, if you averaged the temperature of the two hemispheres, it would be much less than 255K.

    Similarly, if you even consider the Earth, it’s not even clear what temperature you would average to get an average temperature that was equivalent to the effective radiative temperature. Some of the energy is being radiated directly from the surface, some from various altitudes in the troposphere, and some even from within the stratosphere. We often suggest that the energy is effectively radiated into space from an altitude of about 5km in the atmosphere where the temperature is close to 255K, but this is a simplification.

  492. David B Benson says:

    Haven’t we had vastly more than enough on this trivial point?

  493. Willard says:

    Audits never end, David.

    Come back next week for a new Climateball episode!

  494. Dr Roys Emergency Moderation Team says:

    The post is just about attempting to come up with a better factor for multiplying the solar constant (corrected for albedo) by than 0.5 (or dividing it by 2) as a shortcut to getting the absorbed incoming flux over the lit hemisphere. In the end, he decided multiplying it by 0.5 (or dividing it by 2) works as the shortcut in any case. That is the super-abridged version of his post. The flux value is 480 W/m^2, either way.

    His issue, ultimately, is with dividing the solar constant (corrected for albedo) by 4, so that the absorbed incoming flux is over the entire sphere. 240 W/m^2. He is fine for 240 W/m^2 to be the outgoing flux, since that does indeed leave from the entire sphere at any given moment. However, at any given moment, the incoming flux is received over only the lit hemisphere. So…480 W/m^2.

  495. TYSON MCGUFFIN says:

    David B Benson says: May 5, 2021 at 7:20 pm
    Haven’t we had vastly more than enough on this trivial point?

    Yes, but unfortunately this is how posts proceed at Dr Roy’s Bodega (Eli’s term!).

  496. Willard says:

    > His issue, ultimately, is with dividing the solar constant (corrected for albedo) by 4

    Why I invoke Joe’s Mad Hat post is not because of that. You worked well, so I’ll allow that repetition. Next repetitions to that effect will be deleted in whatever form you might try. No ifs, no buts.

    Joe should have come here a long time ago. I already knew he was an asshat. Now I know he’s a coward.

  497. Actually, maybe Dr Roys can provide an example of where climate scientists do divide the incoming solar flux by 4. I’ve just realised that even though we’ve discussed this for a good deal of time, I don’t think I’ve seen a good example of what Joe is complaining about.

  498. Bob Loblaw says:

    “However, at any given moment, the incoming flux is received over only the lit hemisphere. So…480 W/m^2.”

    And one more time, at any given moment, this is an average that actually does not happen anywhere, at any given moment (except by miniscule odds).

    You keep treating this averaging as better than averaging over the entire sphere. It is no more meaningful than averaging over the entire sphere.

    And “climate” isn’t instantaneous.

  499. Dr Roys Emergency Moderation Team says:

    Well, look at the input on the K/T energy budget diagram. It has the incoming solar radiation as 340 W/m^2 which is just the solar constant divided by 4.

  500. Willard says:

    > “climate” isn’t instantaneous.

    As Kermit would say, that’s a very good point,

  501. Bob Loblaw says:

    “Actually, maybe Dr Roys can provide an example of where climate scientists do divide the incoming solar flux by 4. “

    https://science.sciencemag.org/content/213/4511/957

    Equation 1.

  502. Dr Roys,

    Well, look at the input on the K/T energy budget diagram. It has the incoming solar radiation as 340 W/m^2 which is just the solar constant divided by 4.

    Okay, so how would you draw the diagram and how would you do it in a way that didn’t make it very complicated to understand?

  503. Bob,

    Equation 1.

    Except, I’d argue that that isn’t really an example since it is simply energy in = energy out.

  504. Bob Loblaw says:

    …and an equation we’ve seen multiple times here…..

    …but at least you can now reference it as a peer-reviewed publication (in Science, no less) as Hansen et al (1981).

  505. Dr Roys Emergency Moderation Team says:

    AT, 480 W/m^2 over the lit hemisphere whilst 240 W/m^2 leaves from the entire sphere, in real time, is also energy in = energy out. In answer to your question, I don’t know. I thought you just wanted an example of dividing the incoming flux by 4?

    I have to go. Sorry that you will have nobody to argue with for a while. [Snip. -W]

  506. Willard says:

    It’s a great illustration, Bob.

    Just look at the PDF. The manuscript has been typewritten. The Science typographer added stuff with glue and scissors.

    I think it’s important to realize that doing science for real requires work. Imagine if instead of writing (1) they started to work out flux with a round atmosphere. Even with computers that’s not cheap.

  507. Dr Roys,

    In answer to your question, I don’t know. I thought you just wanted an example of dividing the incoming flux by 4?

    Yes, and I had forgotten about that graph. However, it’s an extremely informative graphic, so I would be interested in how you would redesign it so that it was still informative but satisfied Joe’s insistence that we must never divide the incoming flux by 4.

  508. Willard says:

    As I see it, flatness is about how we model the atmosphere:

    Early climate and weather models, constrained by computing resources, made numerical approximations on modeling the real world. One process, the radiative transfer of sunlight through the atmosphere, has always been a costly component. As computational ability expanded, these models added resolution, processes, and numerical methods to reduce errors and become the Earth system models that we use today. While many of the original approximations have since been improved, one—that the Earth’s surface and atmosphere are locally flat—remains in current models. Correcting from flat to spherical atmospheres leads to regionally differential solar heating at rates comparable to the climate forcing by greenhouse gases and aerosols. In addition, spherical atmospheres change how we evaluate the aerosol direct radiative forcing.

    https://www.pnas.org/content/116/39/19330

    If flatness bugs Joe, he needs to start with the disc. Which is what I think I underlined with the Mad Hat post. This has very little to do with dividing by 4.

  509. Bob Loblaw says:

    The Science typographer added stuff with glue and scissors.

    Given the date, my guess is that the journal was printed using offset press technology. Typically, “camera-ready copy” was prepared on paper, then a photographic image taken as a full scale negative, and that was used to burn the printing plate (a thin metal sheet with a photo-sensitive coating). Ink was rolled onto the plate, where it would only stick to the coating, and then the plate rolled onto paper where it deposited the ink.

    The physical process of cutting and pasting could have been done at the camera-ready paper stage, or with the negative photo image. When we did this for the school newspaper, the most common cut-and-paste at the negative stage was to cut in photographs, which had to be half-toned as a separate step. We also used opaque liquid (printer’s version of white-out) to mask over any pin-holes in the negative that would have let light through (and created black print). Although we did use scissors to cut paper, we used a special wax gun to coat the back to “paste” it. Much easier to peel back off and reposition than glue.

    Although even in 1981 there would have been better ways that a typewriter to create camera-ready copy, I would not be surprised if figures and equations were done separately from text, and pasted onto the camera-ready copy. Authors would have had to provide figures in camera-ready format as high quality photographic prints.

  510. Willard says:

    > This has very little to do with dividing by 4.

    Perhaps that’s still too strong. So let’s think of it this way. Forget appeals to success words like realness and second-by-second instantaneousness: how is divide-by-2 computationally superior to divide-by-4?

  511. Bob Loblaw says:

    Flat atmosphere? Spherical atmosphere? I haven’t read the full paper you link to, Willard, but I know that solar position and radiative transfer calculations, etc. have included atmospheric refraction for a long, long time. Refraction is probably a more important issue for low sun angles than sphere/planar atmospheric layers. It’s a standard adjustment when calculating apparent solar position.

  512. Willard says:

    It would be fun if we could install this on a phone:

    Solar-J version 7.6c, presented here, offers 4 options: option 0, flat, uses the flat atmosphere optical mass in each layer with a fixed air mass fraction of 1/cos(SZA) and a rigid cutoff at 90° SZA (Fig. 1B); option 1, sphr, adds true straight-line light paths (Fig. 1A) assuming the geopotential height grid, allowing for twilight; option 2, refr, allows these light paths to refract, extending surface sunset to SZAs of ∼90.5°; and option 3, geom, assumes increased optical mass in spherically expanded layers on a geometric height grid with refracted light paths.

    SZA stands for Solar Zenith Angle.

    I got that cite from the last comment to Joe’s Mad Hat post.

  513. Bob Loblaw says:

    “flat atmosphere optical mass in each layer with a fixed air mass fraction of 1/cos(SZA)”.

    That’s the case for no refraction. When I was a poorly grad student, Kasten’s 1965 table was often used. His paper references early work in the 1904-1907 range. (Yes, over a century ago.) This is old stuff.

    https://link.springer.com/article/10.1007/BF02248840

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