Variable Volumes in an Oblate Spheroid
Date: 12/21/2002 at 10:00:45 From: Greg Subject: Variable volumes in an oblate spheroid To whom it may concern, I have a question and hope you may be able to supply me with an answer. This is a practical question that has application in my field. I am an environmental engineer, and in my chosen profession I manage and am the engineer for a fairly large water utility. In our system we have several elevated water tanks. Their shape is that of an oblate ellipsoid. The problem simply stated is that we need to know how much water is in the tank at any given time. I can get exact and continuous data as to the height of water in the tank, which varies throughout the day. If the tank is 100% full the volume is easy, as this is just the volume equation for the elipsoid V = (4pi/3)abc, where a is the z dimension diameter (height of the water) and b and c are the diameters in the x and y planes. I would like to know a simplified expression to calculate the volume of the water in the tank given any height of water (a). Can you help me ? Sincerely Greg A. Shellito Manager of Treatment and Production
Date: 12/21/2002 at 22:46:23 From: Doctor Peterson Subject: Re: Variable volumes in an oblate spheroid Hi, Greg. You can transform the ellipsoid into a sphere by stretching it vertically, and then use the formula for the volume of a spherical cap: Sphere Formulas - Dr. Math FAQ http://mathforum.org/dr.math/faq/formulas/faq.sphere.html V = (Pi/6)(3r1^2+h^2)h r = (h^2+r1^2)/(2h) where r is the radius of the sphere, r1 is the radius of the "cut surface," and h is the height of the cap. Solving for r1, r1^2 = 2rh - h^2 So in terms of depth h, the volume of a spherical cap is V = pi/6 (6rh - 3h^2 + h^2)h = pi/6 (6rh - 2h^2)h = pi/3 (3r - h)h^2 For an ellipsoid with semiaxes a (vertical) and b (horizontal), the volume will be a/b times that of a sphere of radius r=b, and the cap corresponding to your water volume with depth d will have height h=bd/a. So the volume of our ellipsoidal cap is V = a/b * pi/3 (3b - bd/a)(b^2d^2/a^2) = pi/3 (3a - d)b^2d^2/a^2 As a check, if the depth d is 2a (a full ellipsoidal tank), this gives 4/3 pi a b^2, which is the volume of the whole ellipsoid, and if d is a we get half that. Note that your b and c are equal (if it's truly oblate), and are semiaxes, or "radii," not diameters. If you have different b and c, just replace "b^2" in my formula with "bc". If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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