Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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/   
    
Associated Topics:
College Non-Euclidean Geometry
High School Non-Euclidean Geometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/