# How to Calculate the Average Projection Factor onto a Hemisphere

Written by Joseph E Postma

There has been an ongoing discussion in the comments on previous posts about this problem, and it is also a problem often discussed in climate debate circles due to its importance in understanding how solar radiation reaches the Earth’s surface.

The problem in question is how one calculates the average projection factor of the incident solar radiation onto the hemisphere of the Earth which sunlight falls upon. To begin it might be helpful to look at the schematic of the cross-section of sunlight which is intercepted by the Earth:

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

However, the flux at any given location on the hemisphere is actually a function of the cosine of the solar zenith angle (with zero degrees pointing toward the sun, and 90 degrees toward the terminator at any azimuth), and, there is more surface area at larger zenith angles. Now, the cosine function spends more angular sweep between zero and ninety degrees above 0.5, while, there is increasing surface area above the angle where the cosine function equals 0.5. So we can only use calculus to see how these influences balance out. That is, to get the average projection, we must weight the projection with the surface area any particular projection value falls upon. So we need to do a weighted integrated average. We can exploit existing symmetry in this situation to make the integral more visual.

First, consider that for any given zenith angle the projection factor is the same, and hence, there will be bands about the hemisphere centered on the zenith axis which have the same projection factor. Consider that any given infinitesimal annulus on the input cross-section disk has a constant projection factor and falls onto a circular-section band of the hemisphere:

Each band of width dr falls onto a surface area of a hemisphere, and we should like to know how much surface area on a hemisphere that each dr occupies as dr goes to zero. For this, we exploit **Archimedes’ Hat-Box Theorem**:

-For any two parallel planes which cut off a band on a sphere, the surface area of the band on the sphere is the same as another band on a cylinder enclosing the sphere which is perpendicular to and cut off by the planes.

~or~

-For any two parallel planes with distance h between them which cut off a band on a sphere, the surface area of the band on the sphere is the same as the surface area of a cylinder band of height h with identical radius of that of the sphere.

A diagram with all necessary factors we’ll need is shown next:

So what we need to do is integrate the projection factor multiplied by the weight for each projection factor “P”, and divide by the integral of the weights. The weights for each projection factor are of course the areas dA which they fall upon.

<P> = ∫P(φ)·dA(φ)dφ / ∫dA(φ)dφ

We should know what the integral of all band-area elements dA is already (the denominator above), given they all form a hemisphere, but we can write it out since the function is needed in the top of the equation multiplied by P(φ). Of course, P(φ) = cos(φ) where phi is the zenith angle.

The surface area of each infinitesimal spherical band annulus, dA, via Archimedes’ Hat-Box Theorem is

dA = 2πR·dh.

We want dA as a function of phi and hence dh as a function of phi, where dh is simply the difference in height along the zenith axis between h(φ) and h(φ + dφ), i.e.

dh(φ) = h(φ) – h(φ + dφ)

where h(φ) = R·cos(φ). Using the definition of an derivative, then

dh(φ) = -R·sin(φ)dφ.

That makes sense because dh is negative (h becomes smaller) when phi increases, and it changes only very slowly at first which is what the sin gives you. However, we only need the absolute value of dh since we are calculating area on the cylinder, which is positive. Hence

dA(φ) = 2πR²·sin(φ)dφ

Thus

A = ∫dA(φ)dφ = 2πR²·∫sin(φ)dφ where 0 <= φ <= π/2.

A = -2πR²·[cos(φ)] @φ = π/2 – @φ = 0

A = -2πR²·[0 – 1] = 2πR²

Of course we knew what that area was supposed to be, but we needed the differential form to use in the weighted integral. Thus

<P(φ)> = ∫P(φ)·dA(φ)dφ / 2πR²

= ∫cos(φ)·(2πR²)·sin(φ)dφ / 2πR²

= ∫cos(φ)sin(φ)dφ

= (1/2)·[sin²(φ)] @φ = π/2 – @φ = 0

= (1/2)·[1 – 0] = 1/2.

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

Now, the reason why we’re interested in the integrated average projection factor is because we can then multiply that by the top-of-atmosphere solar flux in order to get the integrated average flux on the input hemisphere. So that gives us the 1370 W/m² divided by 2, as we have in the top figure, and we can convert that to an equivalent forcing temperature as shown there too.

However, the relationship between flux and temperature is not linear but has a fourth-power exponential dependence between them, and so, the integrated average flux converted to temperature will not actually be the same as the integrated average flux when that flux is first converted to an equivalent temperature. The integrated average flux is certainly useful and interesting, but we are probably still more interested in the integrated average flux as converted to a temperature forcing. And so to calculate this we follow the same routine as above, but this time we first convert the flux “F” at each symmetric annulus band to a temperature via the Stefan-Boltzmann Law (T = (F/5.67e-8)^{1/4}) and multiply that value with the weighting (as area) it falls upon.

<T> = ∫T(φ)·dA(φ)dφ / ∫dA(φ)dφ

= ∫ (F(φ)/5.67e-8)^{1/4}·2πR²·sin(φ)dφ / 2πR²

= ∫sin(φ)·(1370·cos(φ)/5.67e-8)^{1/4}dφ

= (1370/5.67e-8)^{1/4}·∫sin(φ)·cos(φ)^{1/4}dφ

= (1370/5.67e-8)^{1/4}·(-4/5)·[cos(φ)^{5/4}] @φ = π/2 – @φ = 0

= (-4/5)·(1370/5.67e-8)^{1/4}·[0 – 1]

= (4/5)·(1370/5.67e-8)^{1/4}

= (4/5)·394K = 315.2°K = 42.2°C.

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.

This still isn’t really that meaningful of a number, because the rotating Earth spends quite a bit of angular sweep underneath input much closer to unity with the top-of-atmosphere flux, and it is this high-powered flux which creates cumulonimbus clouds, and the climate, etc.

The major point here of course being that it is 100% irrational, illogical, and incredibly stupid to average the incoming solar flux evenly over surface area it never exists upon, i.e. the entire surface area of the Earth at once, literally as if the Earth is a flat plane, etc.

New video up next soon, and it will be a doozy.

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