Does Surface Area Determine Temperature?

It’s commonly accepted today that our spherical planet absorbs only ¼ of the radiant energy that the Sun imparts because a sphere has four times the surface area as a flat disk facing the Sun.

This radically reduced irradiance on our planet is the reason that its calculated temperature is very low: 255 Kelvin, or minus 18° Celsius. Since it’s obvious, though, that the Sun only illuminates half a sphere at a time, we ought to examine how light actually adapts to a hemisphere before we accept a ¼ irradiance average for a whole sphere.

At the outset, it’s immediately apparent that a hemisphere simultaneously presents many angles of incidence to sunlight, unlike a flat surface.

Assigning a maximum intensity of 1 to light that falls vertically, i.e., at a 90° angle of incidence, intensities fall gradually off to zero from there. As you see, however, the intensity at 45° – halfway between 90° and 0° – isn’t 50{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} but nearly 71{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of maximum intensity. It’s only natural to wonder, then, what the average of all these intensities might be – because this would tell us the average intensity of sunlight on a hemisphere.

Keep in mind that this is just a generic sphere we’re depicting, not a literal Earth. This means that these radiant angles of incidence pertain to any orientation a sunlit sphere may take.

The green line here, for instance, sees the same average irradiance as the red line. So would any number of lines. Indeed, millions of similar lines would fill out a whole hemisphere. The average irradiance on one line, then, stands for the average irradiance on a hemisphere.

So here’s the thing: The average irradiance on a hemisphere is not 50{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} at all, which means that the average of light on a sphere is not ¼ of the radiant energy on a sun-facing disk.

Read the full article at:

https://principia-scientific.com/wp-content/uploads/2018/04/Does_Surface_Area_determine_Temperature.pdf

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Comments (18)

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    Cosmos

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    Link for the full article goes to a corrupted PDF

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      John O'Sullivan

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      Cosmos, it works fine for us. Try clearing your browser and loading again.

      Reply

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    Ed Bo

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    Good grief! How badly can you screw up simple math?

    You completely neglect the fact that there is more surface area at the lower incidence angles than at higher angles.

    For example, the portion of the hemisphere with angles of incidence less than 30 degrees has a full half of the surface area of the hemisphere. Your analysis implicitly and incorrectly assumes it only has one third.

    If you account for this properly, the discrepancy you claim goes away. But that would require you to do a surface integral in polar and spherical coordinates. It appears you have never done one.

    And more basically, if you understood what the average value of a continuous field actually was, you would know that it is the total flux divided by the area BY DEFINITION. In this case, the total flux is 1368 * Pi * R^2, and the total area of the hemisphere is 2 * Pi * R^2, so the average is 0.5 * 1368 = 684.

    Your claim that the average flux density on the hemisphere is 0.6366 * 1368 = 871 invents a lot of power out of nothing. By the definition of the average, you could multiply this value by the area of 2 * Pi * R^2 to get a total power received of 1742 * Pi * R^2. But the sun can only provide 1368 * Pi * R^2. Where did the extra 374 * Pi * R^2 come from?

    All you have demonstrated here is that you are completely out of your depth, and that you desperately need to review your basic mathematical concepts.

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      lifeisthermal

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      I agree with this. Except your value of TSI. It is 1360.8W/m². Now add diffusion in a double shell volume.
      (1368 * π * r²)/(2πr²)/(4πr³/3)/(4πr³/3)
      This simplifies to
      1/2
      (1360.8/(4/3)²)=σ286.6⁴
      Problem solved, no gh-effect included.

      Reply

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    Alan Siddons

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    Thank you, Ed Bo, but you’re presenting an argument that assumes the very premise it’s purporting to prove. Do you agree, though, that at 90° incidence on a hemisphere, light intensity will be the same as on a sun-facing disk? If you do, then can you describe how intensity on a round surface falls to zero from there WITHOUT falling in a straight line? Remember: you’re constrained to a 50% irradiance relative to a disk. As I showed, you CAN make irradiance conform to a round surface, but within a 50% constraint, you can’t reach equal irradiance at 90° that way — so you end up contradicting what you just agreed to.

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    Peter Champness

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    I think EdBo is correct.

    For a cylinder your analysis works well because all the slices are parallel and can be summed.

    For a sphere the slices which are integrated (summed) must either be parallels of latitiude, becoming smaller at the pole or slices parallel to a meridian through the pole and becoming smaller at the edges.

    I have forgotten how to do that but I will look it up.

    Reply

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    Al Shelton

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    Why can’t you experts agree on the math?
    Please continue until agreement is reached,
    We[me] are waiting for the correct answer.

    Reply

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    Ed Bo

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    Alan, we do agree. Peter pointed out that while your analysis might be correct for a cylinder (where the area does not increase with decreasing incidence), it is not correct for a sphere (where it does).

    And seriously, you need to ponder carefully the second point I made above. You are making a simple problem complicated, then getting confused because you do not have the skills to do the complex analysis required (a surface integral over a hemisphere).

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    Alan Siddons

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    That’s a fair comment, Peter. But please bear in mind that the orientation of a shape toward a light beam determines the pattern of light upon it. For a cylinder whose butt-end is facing the beam, for example, 1/pi certainly doesn’t apply. That single edge will be exposed and the rest of its length will see virtually nothing. The unique thing about a sphere, however, is that WHATEVER its orientation, a beam of light will form a 1/pi pattern of rings. Believe me, thinking in terms of latitudes and longitudes will just confuse you. The orientation of a shape toward a light beam determines the pattern of light upon it.

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    jerry krause

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    Hi Fellows,

    Am I correct if I note that the system of your focus is a static one. But the real (actual) system to which you pretend your analysis applies is a dynamic one. I was alerted to this fact because of NOVA presentation on PBS last night in which they directed a flashlight beam on a stationary globe instead of the real globe which should actually rotate.

    Have a good day, Jerry

    Reply

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    Alan Siddons

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    No, Ed, what you need to ponder is what you haven’t addressed yet. To repeat what I wrote on the two 50% hemispherical irradiance profiles,
    “If irradiance on a hemisphere at 90° incidence matches the irradiance on a disk, yet its total irradiance is ½, then all other irradiance values along this curved surface must fall off in a straight line in order to conform to a 50% total. This is impossible.”
    On the other hand,
    “If irradiance on a hemisphere peaks at 90° incidence and gradually falls to zero at the terminator, yet its total irradiance is ½ that of a disk’s, then the peak irradiance cannot match disk irradiance at the same angle. This is impossible.”

    By contrast, the 63.66% irradiance profile I presented matches irradiance versus angle of incidence at every fraction of a degree along a hemisphere. For instance, the value at 30° of incidence is 0.5, because at 30° the intensity of a light ray is cut in half. Actually, it’s you who’s making this more complicated than it has to be.

    Reply

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    Alan Siddons

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    That’s cute, Jerry, but I think neither party here is “pretending.” Still, your point about dynamism deserves some attention. You may note, for example that the ESA lists 1/6 as the average irradiance on a cube. But you’d have to continuously rotate that cube all which ways to obtain such an average. A rotating sphere is different because its shape is so uniform. If Ed et al are correct, then the average irradiance on a sphere — whether rotating or not — is 25%. If I am correct, the average is 6.83% more than that.

    I would imagine that long ago someone noticed that light reflecting off a hemisphere looked about half as bright as that from a flat disk. A quick check seemed to verify that estimate, for the difference in surface area is 2 times. Thus was born the surface area myth.

    Reply

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    jerry krause

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    Hi Alan,

    I am sorry the word–pretend–seemed to offend you. I used the word because I consider it far simpler to actually study the actual temperatures that have been measured by the various projects which measure radiations and temperatures. Data from which can be seen the diurnal oscillations (changing incident angles at one location) during a single day and from one day to the next and from one season to the next and from one years. For if you do observe what I know has been observed you will find the system which you claim to be studying is continually changing in some random pattern which can never be reduced to a system which changes ‘dramatically’ from one day to the next day to etc. etc. That is the dynamic system in which I have lived.

    So I do say you are pretending if you believe you can learn how this system behaves from hour to hour without observing the random changes that actually do occur from hour to hour and from day to day etc.

    Have a good day, Jerry.

    Reply

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    Ed Bo

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    Alan:

    You’re not even trying to come to terms with the issue I brought up, and I think you don’t even understand what that issue is. It’s very clear to me now that you have never learned how to perform a surface integral, which is what you are trying to do here.

    When you integrate over two dimensions (i.e. perform a surface integral), when you integrate over the first dimension, you still must use two-dimensional infinitesimal “patches” so your patches have “width”. That is what you are failing to do.

    Let’s get specific to this case. Let’s call the angle of incidence Theta, which varies from 90 degrees (Pi radians) to 0 degrees over our hemisphere, and the angle around the hemisphere Phi, which varies from 0 to 360 degrees (2*Pi radians). The radius of the hemisphere is R.

    First, we want to integrate along Theta from Pi to 0. The length of the infinitesimal patch along theta is RdTheta. But the width of this patch (which you completely failed to consider) is Rcos(Theta)*dPhi. Note that the width increases as Theta decreases.

    The intensity of the flux on a patch is sin(Theta), as you correctly note (with unit intensity at 90 degrees). So when we integrate the intensity over Theta we are integrating the function

    R^2 * sin(Theta) * cos(Theta) *dTheta

    R^2 is constant, so comes out of the integral. The resulting integral is

    (1/2) * sin^2(Theta) [evaluated at Pi and 0, taking the difference]

    Now, sin^2(Pi) is 1 and sin^2(0) is 0, so our result is (1/2) * [1 – 0] = 0.5, and NOT 0.6366! Your failure to use the cos(Theta) “width” in your implicit attempt at integration. led to your error.

    Integrating the 0.5 *R^2 result over Phi from 0 to 2 * Pi, we get 2 * Pi * 0.5 * R^2 = Pi * R^2, as expected. (But we knew that already, so there was really no point in doing this whole process, as I keep pointing out.

    This is standard first-year calculus, and can be found in any calculus textbook. You would do well to get one and study it.

    Reply

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    Alan Siddons

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    Here’s what I’ve learned from you, Ed.
    1. That invited twice to address two impossible irradiance profiles for a semicircle or hemisphere whose value, relative to a disk, is 0.5, you will simply ignore the problem and continue to lecture about integrals.
    2. That the math-adept can prove most anything to their own satisfaction, for instance that a two-sided atmospheric layer absorbing one unit of radiance from the surface below will radiate one unit back (thus doubling surface radiation) and one unit to outer space. 2 units of light out of 1. Professors like Lindzen and Archer patiently explain that to us, after all, and they apparently believe it.

    My aim in investigating the angles of incidence along an irradiated semicircle was to find the average angle. The resultant sine with 1° increments was intriguing, but with 1/10th degree increments, and later in 100ths, the result kind of pops out at you — it’s the inverse of pi/2. Which means that the average angle of incidence on a semicircle is 39.54°, not 30°. You’re right, though, some calculus might have helped. Rather than a 99.998% accuracy I could have gotten 100% and obviated getting to the inverse of pi/2 the hard way. No amount of calculus, however, could squeeze a 63.66% average irradiance into a 50% container. Nor can you.

    Surprisingly, you do acknowledge that 1/pi applies for a cylinder, i.e., a circle or a ring edge-on to a light beam, but not for a sphere. So I must repeat: The orientation of a shape toward a light beam determines the pattern of light upon it. The pattern on a cylinder or a sphere is the same, the difference only consisting in the arrangement of irradiated rings. You still fail to grasp that.

    Reply

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    Alan Siddons

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    You’re a funny one, Jerry, but I’m glad I flushed you out: I took offense because — as you admit — you meant offense! Yet for what? My essay’s sole aim was to test an irradiance claim, not to explain various weather conditions or a “climate system.” Is jumping to conclusions the only exercise you get?

    Reply

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      jerry krause

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      Hi Alan,

      “I took offense because — as you admit — you meant offense!” Where did I admit this:?? I reread what I wrote and found I clearly stated the reason I used the word pretend because I believe you must study the real, actual, natural system to learn about it. I believe by studying some artificial model that one cannot learn anything about the natural system.

      My purpose in commenting as I have in this case is to alert a reader there is an alternative which seems not commonly practiced by you and others. NOAA and the Department of Agriculture has spent money to fund projects to actually do what needs to be done if one is to do ‘good’ science. I only recently (relative to their lifetimes) the results of these projects because I had not read of anyone using their data. And I do consider this failure to focus upon what has been observed about the natural system to be BAD science.

      Have a good day, Jerry

      Reply

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    Ed Bo

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    Alan:

    I have addressed your question multiple times, including my original comment before you even asked it. I addressed it without math several times, but since you did not understand any of those arguments, I followed up with the rigorous mathematics to demonstrate my point definitively. You did not understand that either.

    I now realize that not only do you not understand integration, you don’t really understand averaging either. I will make one more attempt.

    First a simple example. You have two subsets of numbers. The first is [4, 4]. The second is [3, 3, 3, 3]. Obviously, the average of the first subset is 4, and the average of the second subset is 3. But what is the average of the full set that includes both subsets?

    Your method would simply take the average of the two subset averages and get 3.5. The correct method would take to sum all of the elements [4, 4, 3, 3, 3, 3] = 20 and divide by 6 to get 3.333. The method you are using for the hemisphere fails to recognize that there are more points on the hemisphere at low angles of incidence than at high angles.

    Now back to the hemisphere. You didn’t follow the exact solution using calculus, so let’s revert to the type of “finite element” analysis you used for a decent approximation. To make it concrete, let’s think of a hemispherical loaf of bread that we take a 1-degree “wedge” slice out of. The hemisphere has 360 of these, all identical.

    You simply summed and averaged zero-D points along one edge of this wedge. Your results for this line were correct, but irrelevant to the solution because of their lack of dimension. You must consider the varying width of the wedge before you can consider the full hemisphere.

    If the intensity at a 90-degree angle of incidence is 1.0, then the intensity at 60 degrees is 0.866, and the intensity at 30 degrees is 0.5. This is as you have calculated.

    But the width of the wedge at 60 degrees is 0.5 times the width of the wedge at 0 degrees, and the width of the wedge at 30 degrees is 0.866 times the width at 0 degrees. So the amount received at 60 degrees is 0.866 * 0.5 = 0.433. The amount received at 30 degrees is 0.5 * 0.866 = 0.433, the same value! Your analysis fails because it does not take account of the increasing width of the wedge (and therefore greater surface area) at lower angles of incidence, and so concludes that a lot less is received at 30 degrees than at 60 degrees.

    Note carefully that this effect is not present for a (right circular) cylinder — and of course we are talking here about a cylinder whose axis of rotation is perpendular to the light direction. In this case we take a slice with parallel edges first, then sum along the length. Because the slice edges are parallel, the cylinder slice does not have the “varying width” issue.

    All of this is immediately obvious on inspection to anyone who has taken and understood first-year university math.

    Oh, and you are completely wrong in your 1st Law of Thermodynamics argument as well. It is obvious that you have never done the simplest energy balance analysis on any defined control mass.

    Reply

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