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
_____________________________________________

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/ 
Associated Topics:
College Geometry
College Higher-Dimensional Geometry
High School Geometry
High School Higher-Dimensional Geometry
High School Practical 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/