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.

Source: Joe’sIt 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.

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}. *When isolating the temperature*, 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 insists on making the light fall on the hemisphere. Sometimes the disc isn’t *real* enough, like when, with the tip of Archimedes’ hat, he averages flux over a hemisphere, which increases solar input and intriguingly gets him 15.5C. More often the sphere is the target of his ire: his hemispherical equilibrium equation supports diagrams that display 30C in contrast to the usual -18C {5}. How does he pull his trick? Probably misspecification {6}. Perhaps fiddling too, for he conceals the -273C on the unlit side. 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, I see no computational to prefer divide by two than by four {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 {8}. So I don’t buy Joe’s appeal to the naturalness of a hemisphere over a disc or a sphere. These toy models are no GCMs!

It’s as if Joe wanted to have the zero-dimensional cake and eat it in one dimension {9}. 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.

{5} Notice how Joe’s diagram omits the Disc and how the 0.5 appears on the input side, as if the hemisphere was the receptor. In our EBM, Power would be *Sun* (in W/m^2) x *Disc* (in m^2), which gives us *Watts*, or *Joule per second*. Only when Joe divides back by the hemisphere (from the output side of the equation) that he gets back flux in W/m^2.

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

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

{8} Paragraph revised thanks to Rajinder’s feedback. I owe him a copy of The Spanish Prisoner.

{9} 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: 2021-05-13]

## § 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

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

> A Poe, surely?

I doubt it:

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

Note the date.

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.)

Fair. Perhaps you’d prefer:

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

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

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.

Since you mention the Moon, you might like:

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.

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.

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.

Link (and text) to above animated GIF …

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

“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?

“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

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

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…

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:

I also said:

Dave,

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

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.

Hello MP,

Seems that you left a note at Joe’s:

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

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.

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!

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.

> 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.

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.

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

> 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.:

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.

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

as expected. If you want to do the full area of the sphere, simply integrate from to .

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 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

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

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

and we’ve integrated longitude from to , rather than from to .

If you carry out this integral, you get

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.

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.

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.

I took my leave at Joe’s:

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?

> 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:

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.

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)

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∕ ⁴Α – 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∕ ⁴global 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∕ ⁴The 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

Drive-by done, Christos.

Thanks.

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.

Joe responds to fan mail:

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.

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.

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

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.

“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).

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.

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.

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:

.

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

where is the effective radiative temperature of the Earth.

If we know the albedo and then we could equate these two to estimate .

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

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 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:

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?

Willard,

OK I have had my say.

“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.

4 W/m^2

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.

aljo1816,

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

> 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.

> 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.

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.

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.

.

MP,

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

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. is the

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

No. Here’s the breakdown:

Earth’s shadow =

Discin my model =Earth’s surface =

Areain my model = .Disc/Area= 1/4Hence why we divide by four.

@ 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

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.

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).

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:

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.

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.

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:

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

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.

> 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:

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

I certainly don’t.

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

Thanks, Russell:

What a great gift!

Bob,

Indeed, I did that in this comment.

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

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?

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.

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.““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

> increasing atmospheric CO2 does not cause undesirable warming now,

Of course not:

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

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/

I tried to plea with Joe.

He continues to gloat.

Oh, well. I tried.

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.

BREAKING

Philip for the win:

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 ; the area of a hemisphere (without the base) is , and the area of a disc is . 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/

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.

[

No thanks, Doc. – W]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

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.

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?

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.

TrueSceptic,

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

is the effective radiative temperature of the Earth given a solar flux at 1 AU of and an albedo of . 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).

ATTP 1:38 pm

I’ve always like Professor Bohren’s definition:

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.

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

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!

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.

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

There’s clearly a on one side and a 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

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:

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).

“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.

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.

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.

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.

> 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:

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

Do you now

seehow 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.

“

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?

> 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!

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.

“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.

> 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.

> 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:

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

“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.

> 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?“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.

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

Joe might disagree with you:

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

You should take that one up with Joe.

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?

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.

Willard:

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

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

showhow 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.

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.

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?

> [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 Ishowedyou 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.

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.

> Just as with Diviner (Moon)

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

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”.

> 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.”

> 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:

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

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

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

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?

> 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) inoneEarth. 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 four, 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 two, 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 rascalsSky 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?

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!

Willard,

Who was that last message meant for?

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

I have yet to see

anybit of discourse Sky Dragons can’t misrepresent, True One.I’m not sure I understand your numbers.

Thanks for the quote!

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.

> 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.

> 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:

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

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?

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.)

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…

> 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.

> 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.

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

> bizzare

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

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?

“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.

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).

“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/

Looks like I made it at Joe’s.

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

Nice find, Keith!

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?

> 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.

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

[

Playing the ref. Let’s play it out over there. -W]This paragraph from an new comment in an old thread at SoD’s might be relevant here:

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.

Sometimes it’s worth it to deal with contrarians:

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.

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.

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).

I think we’re nearing the end:

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.

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.

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.

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.

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

https://www.khaaan.com/

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.

I should have looked back at SoD’s.

First, the graphic:

Second, the quote:

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

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

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

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.

“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.

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.

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

P.s.:https://i.ibb.co/p2YsqMb/Solar-Flux-Petty06-p53.jpg

Nice graph, TYSON!

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

You can hotlink it too.

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.

All our bases belong to Arthur:

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

Beautifully written!

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.

Thanks.

Something like you suggest emerged this evening;

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.”

> 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:

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

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?

Holger,

Sky Dragons do not believe in GHG theory..

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).

@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.

> What errors have I made

Since you insist:

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.

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

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:

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.

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.

> 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.

Willard,

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

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.

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.

@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.

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.

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).

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.

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.

“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.

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.

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

> 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:

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

MP,

This is why I’ve been trying to point out to you that the equality

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.

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.

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?

“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.

> 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:

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

ifthey 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

directya thousand times.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).

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:

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.

>

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^2But it is not happening on average… you do know that, don’t you?

> which we are not justified to do…

I already responded to that point, Christos:

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

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

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

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∕ ⁴So we shall have:

Planet Energy Budget:

Jabs = Jemit

πr²Φ*S*(1-a) = 4πr²σTmean⁴ /(β*N*cp)¹∕ ⁴Solving 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

Christos,

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

Where’s your 0.47 coming from?

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

0,47https://en.wikipedia.org/wiki/Drag_coefficient

I welcomed the

0,47as 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 ≤ 1Thus, 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.

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:

No specular reflection from the planet surface then?

That’s all accounted for in A.

Just to clarify, (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 () and the fraction of the incoming energy that is reflected back into space () then the amount of energy the Earth absorbs per unit time is simply:

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.”[Source:] https://ntrs.nasa.gov/api/citations/19940020024/downloads/19940020024.pdf

Sorry, I have to stop now. It is getting too late.

Shall we continue tomorrow?

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

“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

> Lit surface equals half the entire surface.

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

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.

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.

Or …

1 = sphere

2 = hemispheres

4= discs

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

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

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

Sorry.

“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

nottrying to get “twice the Sun’s power out of twice the area”. Thefluxin is double (480 W/m^2 as opposed to 240 W/m^2) because the total power in is divided byhalfthe surface area (the hemisphere as opposed to the sphere).[

Fixed. Thanks. – W]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.

Seldom is the question asked: are Sky Dragons learning?

No, because they are cognating:

Lyndon LaRouche …

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

Ouch!

It gets better:

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.“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?

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?).

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

outputfrom 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:

where is the solar flux at the radius of the Earth’s orbit and is the albedo. If you carry out the calculation you get that

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:

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:

Therefore, if there is a change in solar flux of 1 W/m^2 (i.e., ) then the change in solar forcing is:

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:

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.

@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).

Sorry, missed this bit

Holger says:

May 1, 2021 at 6:57 pm

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^2Planet 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[

Most obliged. -W]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.

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 portionof the incident solar energy.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.

“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.”> 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

Areain 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.

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:

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

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.

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

> 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.

“Albedo (prounounced /ælˈbiːdoʊ/; Latin: albedo, meaning ‘whiteness’) is

the measure of the diffuse reflection of solar radiation out of the total solar radiationand 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

Christos,

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

Ok, I will highlight only the

diffuselyChristos,

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:

“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?

Christos,

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

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?

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.

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

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 KWhat \alpha do you prefer, Christos?

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.

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

Willard, I prefer a, not α…

Christos,

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

> 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?

[

Let’s stick to the albedo issue for the moment. -W]Willard,

>

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

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

“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.

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.

Everett,

>

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

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

Christos,

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

> 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.

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?

Willard,

I only care about Joe’s algebraic justification to divide by 2.I am sure Joe divides by 2 for a good reason.

Christos,

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

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

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.

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.”

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

“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.

“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.> 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.

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”.

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

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.

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.

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.

> 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

poweras 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:

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

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).

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?

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

TS, the lit hemisphere

isreceiving the input, in real time. Hence itdoesmake 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.

> 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.

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?

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?

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.

> He does divide by 4.

Wait.

Whatdoes he divide by 4, again?Here is the 0.637 diagram:

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.

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?

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.

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?

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.

> 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,

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

Dr Roys,

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.

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 sphereThe correct:

.

..it captures the fact thatin real time, the sunlight falls only on a hemisphere whilst energy leaves from the entire sphere.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 . This

clear insightis so obvious few others thought it worth explicitly highlighting.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.

AT,

>

Also, a simple energy balance calculation also takes into account that the energy only falls onto one hemisphere (that’s why it’shttps://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)

Christos,

You can insist all you like, but the correct form for the energy balance for the Earth is:

where .

Eh… AT,

Let’s see what we have today here:

Planet Mars black-body temperature (effective temperature) Te misfortunate coincidenceWe 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.…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).

Christos,

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.

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

onehemisphere, whilst the output of 240 W/m^2 leaves from both hemispheres.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.

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.

@ 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.

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?

I stand by what I have said, because it is correct. Yes, obviously, 240*4 = 480*2.

Dr Roys,

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.

Which article?

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?

MP,

We might be getting somewhere. Why is it at 5km?

> 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:

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?

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 ?

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).”

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.

Dr Roys,

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.

“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.

@ 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

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?

Dr Roys,

Not sure what there is to correct. If Postma now accepts that the correct energy balance equation is:

that’s great. If he always accepted this, then that certainly wasn’t clear.

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?

MP,

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?

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.

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

> 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

stillneed to be powered by 386 to bring Joe’s model on the table, so be it.@ 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

Christos,

No, I think I explained myself very carefully.

Correction.

Should said much less dense over 100km above the median range (not colder per se, temperature rises somewhat in certain higher layers),

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?

MP,

You’re still not explaining why this is

Willard, my comment from May 2 @ 4:08 PM already shows you how Postma balances his numbers.

> 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

equivalentto 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?

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.

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.

> That is all there is to it.

We have

notestablished how Joe’s model and everybody else’s areequivalent, and we have notseenJoe’s model yet.If the temperature on the lit hemisphere is 30C in Joe’s model, what is it on the other side?

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,

thatcorresponds to the effective temperature of the Earth. 255 K. But the 240 W/m^2 is anoutput. Not an input.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.”

” 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.

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.

[

I insist that you fulfill my request. Keep your repetitions for Roy’s. – W]“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.

[

Enough peddling for one thread, Christos. Wait for another one. Thanks. -W]> 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?

> 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

sphererepresents flatness.All this is very silly.

Has anyone watched The Spanish Prisoner?

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?

> They can have 303K on the sunny side if they accept 3K on the dark side.

Nicely put.

@ 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.

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.

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?

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).)

“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.

“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.

> 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!

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.

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?

“…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.

> [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.

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.

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.

> 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:

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.

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.

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?

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

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.

“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.

aljo,

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.

“So you admit that the SB temperature has to be 3K on the dark side”

Only if the Earth were not rotating.

EFS,

I think it would be 76K.

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.

MP,

What about the planetary greenhouse effect? What role does that play?

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

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.

> Flux out (W/m^2) = xy(1-a)/w

To be an energy

balancemodel, you need something that looks like energy in = energy out, not justenergy 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.

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.

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.

EFS,

That’s what I thought I was doing, but have clearly made some silly error.

@ 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.

MP,

Can I just clarify. Are you describing something other than the greenhouse effect, or are you trying to describe the greenhouse effect?

@ 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

MP,

Seems like you’re partly describing the planetary 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.

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.

“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.

> That’s the “model”.

Thanks. The only geometrical entity in your model is a disc.

Thatis 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!

@ 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

“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?

> 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!

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.

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).

> 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?

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.

You can think whatever you please, kiddo.

Tell me where to get Joe’s model, or we’re done.

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.

> 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!

[

I asked you to confim if Joe’s model was in his Magnum Opus. I want a yes or a no. -W]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:

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:

OK. Part 2 has only a diagram:

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.

More pictures for part 3, first the famous Trenberth diagram, then his 0.637 diagram. Follows an intriguing question:

https://climateofsophistry.com/2012/11/07/the-fraud-of-the-atmospheric-greenhouse-effect-part-3-in-pictures/

A question I have for

youis: if Joe clearly put on his diagram that the unlit side is -273C, would his readers still pick his?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:

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.

The Copernicus paper is a short one and is mostly a salespitch. We notice Joe’s 0.5 diagram, prefaced by beautiful poetry:

The July paper is what I called Joe’s Magnum Opus.

Alright. I’m fed up. Until tomorrow.

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.

“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.

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.

[

Answer AT’s question, please. -W]Lol, OK.

Oneof 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 theinputfrom the Sun. He describes this as “flat Earth”, since the only way the Earth could actually absorb 240 W/m^2 from the Sunat oncewould 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.

“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

notabout 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.> 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.

Thisis 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

~~Disc~~Albedo. 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 Albedo is not the same unit as Power x Albedo / 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.NowhereI have seen Joe admit that the unlit hemisphere is -270C.> He wants the light to fall on a hemisphere.

The light

already fallson a hemisphere.A hemisphere, when corrected at zenith,

isa disc! Check AT’s proof. How many times must I repeat that geometrical fact?There’s a reason why I’m citing Joe’s Mad Hat post, dammit!

Joe,

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.

[

Repond to what is being said, keep repetition for Roy’s, and please consult Joe’s Magnum Opus, for Newton’s sake! -W]> 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.

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?> 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!

“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.

> 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?

I don’t disagree with AT’s integration, if that’s what you are asking.

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.

> 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?

“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

notreceive 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.

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.

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)?

> 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.

…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.

> I think the issue is conceptual, not mathematical.

If the issue was mathematical, it would not be me who would have written this post!

No, you will have to ask Joe on that one.

That will be hard.

Send him this for me:

https://www.gfdl.noaa.gov/blog_held/2-linearity-of-the-forced-response/

Dr Roys,

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.

So, Willard, dare I ask…are you going to correct the article?

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).

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.

> 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.

> 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.

I explained at 1:30 PM what I disagreed with. I still do. Nothing has changed.

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

willfind a way to put the -270C in the text, and not only allude to it in a note.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.

“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

doesbalance energy.> 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

~~Disc~~Albedo. 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?

“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 ™> What’s really happening?

Suppose that what’s really happening is really happening. Does it settle if it is

reallyreally happening?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.

> 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 modeldoes 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.

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

doesbalance energy.Dr Roys,

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.

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!”

🙂

> 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 ahemisphericalmodel. He simply replaced a 4 by a 2.Check equations (21-22) in Joe’s Magnum Opus. Report.

> I’m not saying it can’t be done.

I’ll raise that with a quote from a comment on which nobody commented:

Source: https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/#comment-190878

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

doesbalance energy. I have done that, several times.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.

“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

isbeing 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:

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).

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 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.

> but

That’s a big but there, Rajinder.

One does not simply sell a

hemisphericalmodel 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.

And the equality

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?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.

“Except nobody (that I am aware of) is saying that. You are arguing a straw man.”

Nobody

saysthat…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.“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.

> is also not intended to model the Earth.

OMG. I am wrong again! I think I will die.

*Faints*.

Are you starting to

getwhat 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/

“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.

No, it would be 2.725K!

> 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

reallythink that we’re running GCMs out of a single equation? That’s justsanity check. If you want a better model, go check Isaac’s two-box model. It’s still simple, and Isaac is cool.Willard, before you write your next article, are you going to correct this one?

> 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?

“Nobody says that…but it is what the average of 240 W/m^2 implies.”No, it does not.> Nobody says that…but

When I had a Twitter account, I had a series called *Erisology*. It was based on the Darth 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.

[

I asked for one sentence, Rajinder. One thing at a time. You’re not here to spam Sky Dragon propaganda. -W]“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”.

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?

[

Playing the ref. -W]> 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?

Maybe Joe should correct this

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.

Read Joe’s Mad Hat post, AT.

Tell me what you think of it.

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!

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.> He does not

Proof by assertion, 0-2.

Changing the subject was 0-1.

> 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.

“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”.

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.

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.

[

Playing the ref once more. This time it counts: 0-3. -W]“

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?

[

Please answer my questions. They are related to your request. If you don’t answer them, we can’t proceed. -W]“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.

> 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

beforewe 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:

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, why is Joe dividing 1370 by 2 here?

I’ll let you work on that one, Rajinder.

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

> 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.

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.

> True, looks like a mistake in the post.

Perhaps not, for here’s how he continues:

We’re almost done. After some computation Joe gets a number.

I’ll let you the joy of rediscovering it.

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.

Here’s the number Joe gets:

So yeah. 15C. Surprising, isn’t it?

Now, we

couldtry 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.

I personally never read too much into that number. I can see how it could be misleading though.

> 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 bypasses 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 first in the post.

[NARRATOR: This sequence (I’d call it the

Mad Hat scene if it was a play)established how Joe’s Mad Hat post supports the first version of the post. Since Rajinder does not grok that post, his misunderstanding is natural. As of this morning, he’s still powered by 386 and believes I’m wrong (2021-05-07).]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?

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.

Well, multiplying by 0.5 is the same as dividing by 2.

> Well, multiplying by 0.5 is the same as dividing by 2.

Right. Sorry, my mind was elsewhere.

Then the questions become

whatdoes 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?

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.”

> 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.

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.

Haven’t we had vastly more than enough on this trivial point?

Audits never end, David.

Come back next week for a new Climateball episode!

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

outgoingflux, since that does indeed leave from the entire sphere at any given moment. However, at any given moment, theincomingflux is received over only the lit hemisphere. So…480 W/m^2.David B Benson says: May 5, 2021 at 7:20 pmHaven’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!).

> 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.

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.

“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

averagethat 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

meaningfulthan averaging over the entire sphere.And “climate” isn’t instantaneous.

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.

> “climate” isn’t instantaneous.

As Kermit would say, that’s a very good point,

“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.

Dr Roys,

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?

Bob,

Except, I’d argue that that isn’t really an example since it is simply energy in = energy out.

…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 Hansenet al (1981).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]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.

Dr Roys,

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.

As I see it, flatness is about how we model the atmosphere:

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 littleto do with dividing by 4.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.

> 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

computationallysuperior to divide-by-4?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.

It would be fun if we could install this on a phone:

SZA stands for Solar Zenith Angle.

I got that cite from the last comment to Joe’s Mad Hat post.

“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

I don’t see any logic in Dr Roys equation xy(1-a)/z since x is the solar constant and already an average, the _mean_ solar irradiance per unit area, measured on a surface _perpendicular_ to the ray .

The additional division by 2y is arbitrary.

The energy_in = energy_out is independent of arbitrary surfaces used in defining the averages, but it doesn’t fit some people’s narrative to arbitrarily define an averaged flux (now over a hemisphere by dividing by z) based on using an already averaged flux (solar constant).

Using the solar constant is in itself a shortcut. Yes, it is an average, but it does remain fairly…constant. Replacing z with 2y (because the hemisphere has double the surface area of the disk) is necessary to explain why the “y”s cancel out to leave x(1-a)/2. I was just trying to explain the “divide by 2” shortcut to Willard.

Dr Roy’s: “That just gives you the equivalent “blackbody temperature” associated with the 480 W/m^2 input…”

This is nonsense. (Specifically the INPUT part there.) There was also something above about converting an energy flux to an equivalent temperature forcing or somesuch. Nonsense. This comes from treating the problem like algebra, not like physics.

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

“The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature.” Note that although this is an *equation* the physics is one-way. If I know I have a blackbody at some temperature I can tell you what energy flux that body will radiate. The equation does NOT say every energy flux has a temperature. The equation only works when each element of the equation matches the physical situation being described.

This is why Joe thinks it is so important to insist that 480 W/m^2 on the lit side vs. 240 W/m^2 over the whole sphere. 480 W/m^2 looks “hotter” which means the evil (or just stupid?) cabal of climatologists are forcing the greenhouse farce on us. But when you realize that 480 W/m^2 input has NO TEMPERTURE, and the energy balance is EXACTLY the same either way, you realize Joe’s argument is simply wrong because he doesn’t understand physics.

“Using the solar constant is in itself a shortcut. Yes, it is an average, but it does remain fairly…constant.”It is defined at the mean earth-sun distance. That distance varies by +/-3% over the course of a year. That makes for a +/-6% variation in the intensity reaching earth.

So, if you are doing things In Real Time ™ then for sure you are taking this into account, too? After all, if you consider it inappropriate to average over time scales long enough for the dark side and lit side of the earth to average out, then surely you’re not averaging solar irradiance over an entire year? Are you?

The solar constant does not vary very much on a second by second basis.

> The solar constant does not vary very much on a second by second basis.

How do you know?

I don’t. But according to Wikipedia:

“The actual direct solar irradiance at the top of the atmosphere fluctuates by about 6.9% during a year (from 1.412 kW/m2 in early January to 1.321 kW/m2 in early July) due to the Earth’s varying distance from the Sun, and typically

by much less than 0.1% from day to day”So I assume it doesn’t start varying wildly as you go down from day to day to second by second. Maybe it does.

> The additional division by 2y is arbitrary.

It’s based on the idea that using the hemisphere instead of a sphere is [insert your favorite success word].

If you want to know how Joe goes further in his magnum opus, check where he shows that even if you add two temps, one for both hemispheres, you’ll get -255 anyway as soon as the two temperatures are averaged:

That might explain his fight against averages and translating energy and flux.

All this not to divide by 4.

> The actual direct solar irradiance at the top of the atmosphere fluctuates by about 6.9% during a year

Fair.

The overall points are still that (a) it’s really hard not to average stuff in models, and (b) we’re trying to balance the energy of a planet from a climate perspective. We’re not building a holodeck.

“It’s based on the idea that using the hemisphere instead of a sphere is…”

You said it! Yes! Finally! Thank you. A hemisphere instead of a sphere. Exactly.

Nota hemisphere instead of a disk. I think you’ve got it.It is about having the sunlight received over the lit hemisphere, instead of spreading it over the entire sphere. So this sentence:

“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” can now be changed.

> “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” can now be changed.

I already explained where I got this, kiddo.

Joe can want two things, you know.

Let’s support that claim furthermore. Here’s what follows the last excerpt:

Where’s the hemisphere now?

“The solar constant does not vary very much on a second by second basis.”The measurable output from the sun will vary during the rotation of the sun as sun spots move across the face. (They are darker- output drops when there are more of them/greater are visible.) Not sure how large the output variation is, but the rotation rate averages about 28 days.

…but on a second by second basis – which second during the year is your “In Real TIme ™ ” happening at? Surely you are not ignoring a >6% variation in one of the key drivers of the system, are you? If it is soooo important to do these calculations “In Real TIme ™ ” then surely it is important to get the right solar input value?

Or, are the calculations you are presenting only applicable to a few seconds at some unspecified time of the year? Perhaps during those few seconds where the earth is actually at the mean earth-sun distance? How useful is a model that only applies to a few seconds each year? Doesn’t sound much like a climate model to me.

[

Editorializing, 0-4. -W]Willard, all the highlighted sentence means is that the local surface area underneath the solar zenith is small enough that it can be approximated as a disk, in other words the curvature of the Earth can be ignored.

> all the highlighted sentence means […]

No it does not, and that’s irrelevant to the point made, which is that Joe sometimes divides by 1.

Here’s the new version of the paragraph:

Does that work for you?

Try to be constructive and to understand what I’m trying to say.

Also, chill.

[

You already know the drill, kiddo. One sentence, one objective reason why I should correct it. This is not your text. As soon as you don’t follow through, we’re done, the text stands as it is, and I thank you for having tried a new way to play Climateball. -W]> If I know I have a blackbody at some temperature I can tell you what energy flux that body will radiate. The equation does NOT say every energy flux has a temperature. The equation only works when each element of the equation matches the physical situation being described.

Does that make any sense to you, gator:

https://archive.ph/4Bgvb#selection-359.492-363.69

You can also check the last screenshot I provided, where Joe establishes the temperature of a disk underneath the solar zenith by using \piR^2 on the SB side.

“It needs to balance the same way as the other ones or it’s humbug”

As you agree, the 480 W/m^2 input over the lit hemisphere and 240 W/m^2 output from the entire sphere

doesbalance energy in and energy out.Show me the model where Joe does that.

May 2, 2021 at 4:08 pm

Are you Joe?

I’m asking for Joe’s model.

So you accept my model, which comes from my understanding of Joe’s arguments, but won’t change your article because you haven’t seen him write out in full what I explained?

What you present is not a model. It’s a series of calculations to hide that you’re stuck with a division by 2.

If even you, Joe’s biggest fan, can’t find the model I’m asking for, do you agree it’s elusive?

[

No citation to Joe’s model or an admission you don’t have it, no more comment from you in this thread. Is that clearer? -W]I don’t understand what your problem is.

(ESTR) Pozzo’s shoes needs to balance the same way as the other ones or it’s time to stop waiting for Godot.

(VLAD) But look here! I can balance my shoes!

(ESTR) Are they Pozzo’s shoes?

(VLAD) Look at me!

(ESTR) I’m asking for Pozzo’s shoes.

(VLAD) What?

(ESTR) Do you have Pozzo’s shoes?

(VLAD) I don’t understand what you mean.

(ESTR) OK. Let’s wait.

They wait.[

This was your 36th try since yesterday night, Rajinder. Too late. Goodbye. -W]Tim Folkerts, a physicist from an American university that needs not be named, soldiers on:

https://www.drroyspencer.com/2021/04/an-earth-day-reminder-global-warming-is-only-50-of-what-models-predict/#comment-684950

arguments with flatearthers tend to go this way

until you understand that they simply don’t accept the fundamental principle that there is no up or down in space

*Vladimir prepares himself to wash his umbrella. The rain filled up the shower tank. (They live outside.) He needs Estragon to help him with opening the shower faucet while he holds the umbrella.*

(VLAD) The water needs to fall on all the spots of my umbrella.

(ESTR) We must be well coordinated. There is the same amount of water drops in the tank as there are spots on your umbrella.

(VLAD) I want to see it in real time.

(ESTR) It might be hard. Know that the tank is the same size as your umbrella.

(VLAD) But second by second it does not.

(ESTR) If we’re perfectly synchronized, it’ll work. Now, concentrate.

*They proceed. They succeed. Vladimir is happy. Estragon is surprised. Then they wait.*

One more comment?

Tim Folkerts conveniently ignores the fact that the sunlit hemisphere is not lit uniformly, so it is not going to have a uniform temperature or IR emission flux.

Baby steps, Bob. Baby steps. Sky Dragons must learn to do proper accounting first. I like Tim’s style at Roy’s. You are welcome to join in!

Time to bring this thread to a close.

I extend my infinite gratitude to those who played a part in it, and my most profund admiration to those who read it.