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Say I have 2 astronomical angle pairs defining a confined region on the visible hemisphere:

(minAzimuth, minElevation) & (maxAzimuth, maxElevation)

How can we calculate the solid angle of the projected area of the spherical rectangle on the unit sphere?

I assume that azimuth makes little difference and results in a linear multiple: (maxAzimuth - minAzimuth)/(2*Pi).

My intuition tells me that the elevation factor can be computed as: Sin(maxElevation) - Sin(minElevation)

Thus the answer as a proportion of 1 would be: (Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth)/(2*Pi).

and for the answer in steradians multiply by 2*Pi.

Is this correct?

Many thanks for your efforts.

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  • $\begingroup$ I have edited this question since the original post $\endgroup$ – Trojanian Apr 20 '15 at 16:05
  • $\begingroup$ I attempted an answer with Gauss Bonnet, could n't get it again. $\endgroup$ – Narasimham Apr 20 '15 at 16:07
  • $\begingroup$ I haven't seen that. I'm not familiar with Gauss Bonnet but I'll look into it, thanks. $\endgroup$ – Trojanian Apr 20 '15 at 16:14
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    $\begingroup$ What is your definition of a “spherical rectangle”? The way you write it, I assume you mean all the points whose azimuth is in one interval and whose relevation is in another. Such a rectangle would not be delimited by great circles. Is that what you want? $\endgroup$ – MvG Apr 21 '15 at 19:38
  • $\begingroup$ @MvG This is what I want, I have found the answer by integration, I will add now. $\endgroup$ – Trojanian Apr 22 '15 at 12:21
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"Yes", the formula in your original question appears to be correct.

Let $S$ be a sphere, $\ell$ a line through the center of $S$, and $C$ the circumscribed cylinder with axis $\ell$. A remarkable theorem of Archimedes asserts that axial projection away from $\ell$ is area preserving.

axial projection from a sphere to a circumscribed cylinder

Assume $S$ is a unit sphere and the region in question is defined by $$ \theta_{1} \leq \text{Azimuth} \leq \theta_{2},\qquad \phi_{1} \leq \text{Elevation} \leq \phi_{2}. $$ The area of the "rectangular" patch on the cylinder, and hence of the "rectangular" patch on the sphere, is $$ (\theta_{2} - \theta_{1})(\sin\phi_{2} - \sin\phi_{1}). $$ The fraction of the upper hemisphere covered is, as you initially suspected, $$ (\theta_{2} - \theta_{1})(\sin\phi_{2} - \sin\phi_{1})/(2\pi). $$

When you obtained the extra factor of three, is it possible you were computing a volume?

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  • $\begingroup$ Indeed I was. Thank you - I the volume calculation is referred to in my answer - I have edited it to include the steradians reference. :-) $\endgroup$ – Trojanian Apr 22 '15 at 14:40
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Having done the volume integration I have found that the actual answer to this question is:

(Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth)*(1/3) as a solid angle.

So my intuitive answer was correct. As the volume of a hemisphere with radius 1 comes out as 2*Pi/3, so as a proportion of 1 the answer would be:

(Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth)*(1/3)/(2*Pi/3)) = (Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth)/(2*Pi) as a proportion of 1. Or simply (Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth) in steradians.

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    $\begingroup$ (+1) My post and your edit crossed paths; I'll leave my answer in case the diagram is useful. :) $\endgroup$ – Andrew D. Hwang Apr 22 '15 at 13:56
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    $\begingroup$ I did not know about the axial projection theorem. Amazing insight - thank you! $\endgroup$ – Trojanian Apr 22 '15 at 14:41

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