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Spherical tank proportion question

  1. May 29, 2005 #1
    I need to know how to figure out "r" in this picture from "R" and "h". It is some kind of proportion or integral i'm guessing, but I can;t think of it. "R" is the radius of a spherical tank. "r" is the radius of the water level (not the same as "R") "h" is the height of the water level. Here is the pic. The water level will be draining out, thats why I need to figure out a GENERAL proportion.
    Can someone help me!

    here is the pic
    http://img.photobucket.com/albums/v217/sk3499/mathQ1.bmp
     
  2. jcsd
  3. May 29, 2005 #2

    arildno

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    Think of Pythagoras' theorem.
     
  4. May 29, 2005 #3
    it's changing at a non constant rate if the water goes down. I can't see how pythagorian thm applies. I know how to do a conic tank, but it seems liek I would need to use the circles equation here for one of the sides, mabye the hypotenues, is that right?
     
  5. May 29, 2005 #4

    arildno

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    Regard the triangle with one vertex in the centre, and sidelengths R-h,r and R.
    How can you relate these sidelengths by Pythagoras' theorem?
     
  6. May 29, 2005 #5
    r=Sqrt((R^2) - ((r-h)^2) )

    is that right? Even then the curvature is greater near the bottom?
     
  7. May 29, 2005 #6

    arildno

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    It should be:
    [tex]r=\sqrt{R^{2}-(R-h)^{2}}=\sqrt{2Rh-h^{2}}[/tex]
    What do you mean by curvature??
     
  8. May 29, 2005 #7
    ok, thanks, thats making sense now.

    so if I go (pi)*(r^2) with r equal to what we just came up with, I get the cross sectional area of the water at any time , right?
     
  9. May 29, 2005 #8

    arildno

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    What do YOU think?
    Don't be too unsure of yourself.

    Welcome to PF.
     
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