Use of Steradians
Date: 7/26/96 at 13:41:42 From: Anonymous Subject: Use of Steradians Sir, I read your answer pertaining to solid angle measurements. I would like an example or two of how to apply steradians in real life. For instance: Let's say an antenna radiates a beamwidth whose -3db points are 30 degrees x 40 degrees. What is the steradian angle subtended and how does it relate to the square degrees? This leads into my real question: how many square degrees are represented by a shere (4pi sr)? Intuitively, I thought it would be 360 x 360, but think that is wrong. Thank you, Mike
Date: 7/26/96 at 18:24:28 From: Doctor Tom Subject: Re: Use of Steradians Hi Mike, The way I think of steradians is that they are the surface area on the unit sphere (sphere of radius 1) that the solid angle would cut out of the surface. Thus, there are 4*pi steradians in the entire surface of the sphere. The number of "square degrees" in a sphere is not 360x360, since even around the equator 360 degrees don't quite fit (the tops and bottoms overlap a tiny bit). Each layer as you go toward the poles overlaps more and more. Think of the lines of latitude and longitude on the earth. For evenly spaced lines, the regions near the equator are (nearly) square, but they get skinnier and skinnier as you approach the poles, right? So let me tell you a very interesting fact. If you draw figures on the surface of a sphere using parts of great circles (a great circle is the intersection of a plane that passes through the center of the sphere with the surface of a sphere), you get what are called "spherical triangles," "spherical quadrilaterals," and so on. Let's start with triangles. A very, very tiny triangle on the surface will be almost flat (imagine a 1 centimeter triangle on the surface of the earth - it would take amazingly accurate measurements to show that it's not really flat). If you add the angles of a tiny triangle like that, it will be nearly 180 degrees, since it's almost flat. The tinier it is, the closer the number will be to 180 degrees. Now consider a giant triangle that goes to the north pole, and to two points on the equator 90 degrees apart. Its angles are all 90 degrees, so this spherical triangle's angles sum to 3*90 = 270 degrees. The amazing fact is this: the difference between the sum of the angles of a spherical triangle and 180 degrees is the number of steradians that the triangle subtends on the surface. I should have been using radian measure - the pole-to-equator triangle has 3*pi/2 as the sum of its angles. This, minus pi (180 degrees) is the number of steradians: pi/2 in this case. You can check by noticing that 8 of these triangles cover the sphere, and 8 times pi/2 is the 4*pi steradians! So, to calculate the number of steradians in a "square degree," I'd find the equations of 4 planes that mark out a region which, when looked at from the center of the sphere, is one degree on a side. Take the dot products of the normal vectors to these planes to find the cosines of the angles between them, and then work out the angles. Add the four angles, and it will be a bit more than 2*pi degrees. (Remember that in a simple quadrilateral, the interior angles add to 2*pi). The difference between the sum of the four angles and 2*pi is the solid angle represented by a "square degree". Note that your "30 degree by 40 degree" antenna problem will not make a 120 square degree solid angle. You've got to work it out from the equations of the planes that bound the region. I did this a long time ago and have lost the calculations, but if you work it out in detail, you can go directly to the solid angle from the plane equations without having to mess around with an arc-cosine, but the obvious, direct method using the arc-cosine will get you there. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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