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Total Solid angle

  1. Aug 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    I am looking at a unit sphere. Two squares are projected onto the sphere on opposite ends, as shown in figure 1 (the figure only shows one square, the other one is at the opposite end).

    There are two more sets of these squares, each set in its own dimension, so there are a total of six squares projected onto the sphere opposite to eachother pairwise. The side of each square is 0.25 long. What I want to find is the total solid angle which is not covered by the squares.

    Can I get a hint to what the easiest way to do this is? I guess I need to do the integral over θ and phi, but it isn't trivial to me what the boundaries are.

    Best,
    Niles.
     

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    Last edited: Aug 17, 2012
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  3. Aug 17, 2012 #2

    tiny-tim

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    Hi Niles! :smile:

    I'd choose a quarter of a square (with a 90° angle at the centre), and multiply by 24. :wink:

    Or maybe an eighth, and multiply by 48.

    (Choosing half a square, cut diagonally, is a bad idea since the angle at the corner isn't 90° !)
     
  4. Sep 27, 2012 #3
    Hi tiny-tim

    Sorry for my late reply. I'm not sure I understand your suggestion, are you implying that I don't need to do any integrals? It just occured to me that the projected squares can even be circles - does that simplify the matter?

    Best,
    Niles.
     
  5. Sep 27, 2012 #4

    tiny-tim

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    Hi Niles! :smile:
    No, you do need to integrate …

    your θ will go from 0 to 90° (or 45°), and your r will go from 0 to a function of θ that you'll need to find. :wink:
    You mean a circular cap?

    Then yes, that's lot easier, since your r will be constant.
     
  6. Sep 27, 2012 #5
    Yes, exactly! I was searching for that word. OK, I'll keep working on it, I'll let you know how it goes. Thanks.

    Best.
     
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