Calculating the Angular Dependence of Planar Radiation Upon a Sphere

The incidence of planar radiation upon a sphere, such as that from the sun upon the Earth, is an important calculation for climate science, so I am going to demonstrate that calculation.  This post repeats the previous post content, but adds graphs of properties as a function of surface normal angles of the sphere.

Let the amplitude of the planar radiation be P₀, which is commonly expressed as W/m².   The incident radiation on the sphere surface normal to the planar radiation will have an intensity of P₀.  As one examines a unit surface area of the sphere which is non-normal to the planar wave of radiation, the unit area of the sphere intercepts less and less of the planar radiation the greater the surface normal angle is with respect to the direction of the plane wave.  Let us call the angle from the center of the sphere to the point on the surface upon which the planar wave has normal incidence and the surface annulus of points equidistance from the normal incidence point θ.  The surface of the hemisphere facing the planar radiation then consists of a normal point and a series of annuli, one for each Δθ that we choose.  The annuli have increasing radii perpendicular to the planar wave of radiation.  The angle θ to describe these annuli varies from 0 to π/2 radians or 90̊.  We will use the radian units.  Let the radius of the sphere be R.

Each annulus has a circumference of 2πR sin θ and it intercepts an annulus of the planar radiation wave with the same circumference, but with a narrower width than the annulus on the sphere surface has for θ > 0.  The width of the planar wave annulus that projects onto the sphere surface is given by Δθ R cos θ.  Thus, at angle θ the area of the plane wave of radiation intercepted by the sphere at angle θ, is

A_I (θ) = 2πR² sin θ cos θ Δθ

Integrating over the range from 0 to π/2 for θ, we find the area of the planar wave of radiation which is intercepted by the sphere.  The integral is

A_I = 2πR² ∫₀^π/2 sin θ cos θ dθ

The value of the integral is ½ sin²θ, which for these limits of integration yields a value of  ½.  Consequently, we have an obvious result that

A_I = πR²

which helps to confirm that we have A_I (θ) right. The total incident planar wave power on the sphere (all on one hemisphere) is then

I_T = P₀ A_I = πR² P₀

The total area of the sphere, A_S, is 4πR², so we can rewrite this as

I_T = (1/4) A_SP₀

Averaging over the entire area of the sphere, the total incident plane wave intensity average per unit area is then P₀/4.

Let us examine the intensity of radiation on equal areas of the sphere surface now.  The first area is a circle on the sphere whose surface normal is perpendicular to the wave front and whose area is given by 

π ( R sin θ₀ )²

The irradiance of this area is P₀ times this area, so finding the irradiance per unit area means we divide by the area of this circle and arrive at P₀.  To normalize this to one, we further divide by P₀. to make the result more universal for any radiant power.

The subsequent total irradiance values are those of the annuli, which are given by A_I (θ).  To find the irradiance per unit area on the annuli surface of the sphere for comparison with the irradiance per unit area on the normal facing circle, the A_I (θ) values are all divided by the circumference of the annuli times the width of the annuli, so the per unit area relative irradiance values to compare with that of the center circle normal to the radiation wavefront are:

(2πR² sin θ cos θ Δθ) / (2πRsin θ)( R Δθ) = cos θ

If we make the inner circle sized to have an arc of 1̊ or  π/180, then θ₀ = 0.5̊.  Using a step size of 1̊  each annulus will subsume an arc of π/180, then we have the normalized irradiance per unit surface area on the sphere for θ shown in the plot below:

Let us now plot the total irradiance of the sphere as a function of the angle between the wave front and the sphere surface normal.  This is just

A_I (θ) = 2πR² sin θ cos θ Δθ,

where πR² is just the intercepted area of the plane wavefront.  So to keep this a generalized function for any sphere radius R, we will divide A_I (θ) by πR².  Recall that the integral of the product sin θ cos θ over the range of angles from 0̊ to 90̊ is ½, so the integral of A_I (θ) / πR² over this range is 1.  What this means is that A_I (θ) / πR² is the fraction of the total incident radiation on the sphere at the surface normal angles θ.  This fraction of the radiation incident on the sphere for θ is plotted below using a 1̊ step size (The sum of the fractions of the incident radiation is 0.999898506, which is pretty close to 1.):

The total radiation incident upon the sphere maximizes at 45̊ even though the radiation density is greatest at 0̊.  This is because the circular area with a 1̊ arc at 0̊ is very small and the area of the annuli for larger angles gets larger and larger as θ increases, even as the radiation density decreases.

9 July Updated Version: Do IR-absorbing Gases Warm or Cool the Earth's Surface?

Preface:  This is still another updated and improved version of my earlier posts on this issue.  In this version, IR-absorbing gases heated by IR radiation emitted from a warm Earth surface or by IR reflected from the Earth's surface, cool only partially by re-emitting IR radiation, which Alan Siddons was good enough to note was a needed correction to my earlier versions.  Strangely, I had discussed that this was the case, but I had not accounted for it in the calculation!  The net result of the calculation now appears to be a surface cooling effect averaged over the daytime cooling effect and the nighttime warming effect.

This version has been removed since a much improved version is to be published in a book called Slaying the Sky Dragon, Death of the Greenhouse Gas Theory.

Congress May Have Done Something Useful

In May, Congress passed a bill to stop adding oil to the Strategic Petroleum Reserve. This bill takes effect on 1 July. Economist Philip Verleger, as reported in the 16 June issue of Forbes, says that this will result in an immediate drop in the price of sweet (low sulfur), light crude of $20/barrel. How is this possible, given that only 60,000 barrels of oil a day are being added? This is only 0.3% of U. S. oil consumption.

The market for light, sweet crude oil is very tight and this is what is mostly being put into the Strategic Petroleum Reserve. Only 10 million barrels of oil produced each day are light, sweet crude out of the total of 87 million barrels. This is the oil that has the price we hear quoted all the time of $135/barrel. When 130,000 barrels a day of Nigeria's Bonny Light oil was lost to saboteurs earlier this year, prices for sweet, light crude shot up. The light, sweet crude comes mostly from Saudi Arabia, Nigeria, and the North Sea.

In a 29 November 2004 column in Forbes, Steve H. Hanke of The Johns Hopkins University estimated that the filling of the reserve was responsible for about $10 of the then sweet, light crude price of $55.

The government gets 185,000 barrels of oil each day as a royalty payment from its Gulf of Mexico oil field leases. It sells some and trades some for the light, sweet oil it mostly puts into the reserve. After 1 July, the government will sell all of this oil and have more income from it. This will amount to about $1.5 billion extra income in the 2nd half of 2008. We can be sure that Congress will quickly use the fact of that income increase to spend another $2 billion in the 2nd half of the year.

Steve Forbes, in the same 16 June issue, notes that Federal Reserve policy has brought about a reduction in the value of the dollar by adding to the money supply. He believes that the Fed should aim to adjust the money supply so that gold has a price between $550 and $600 an ounce. He notes that the falling value of the dollar has been responsible for a large part of the cost increase for oil.