Segment of an Ellipse

Date: 09/06/2001 at 03:37:34
From: Martin Braaksma
Subject: Segment ellipse

Dear Dr. Math,

I am a Dutch technical aid worker (volunteer) in Africa looking for 
the formulas to calculate the segment of an ellipse. The reason is 
that we often use horizontal oval tanks for storing drinking water 
and fuel, and we would like to be able to calculate the contents.

I found your formulas for calculating the segment of a circle, so 
that solved the same question with regard to round tanks. I got stuck 
with calculating the segment of an ellipse, though, because in this 
case I only seem to be able to know d and not c/2 or the hypotenuse,
and I need to know 2 sides of the triangle in order to calculate 
theta, don't I? In the case of a circle, the hypotenuse is known as 
well because it equals r, so there I did not have a problem. Could you 
help us out?

Thank you in advance,

Martin Braaksma
Doctors without Borders

Date: 09/06/2001 at 13:23:37
From: Doctor Peterson
Subject: Re: segment ellipse

Hi, Martin.

I'm not sure what dimensions you know. Assuming you are using the 
diagram in our FAQ:

  http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment   

I would expect h and c (depth and width) to be the easiest numbers to 
get, and d one of the hardest. In any case, the ellipse is somewhat 
different. You will need three numbers to determine the exact shape, 
since it takes two numbers to define the ellipse itself. I will use 
these pictures, showing your ellipse, together with a circle obtained 
by compressing the ellipse:

             *************                  *****
        *****      |      *****          ***  |  ***
      **           |b          **       *     |b    *
     *             |       a     *     *      |  b   *
    *              +--------------*    *      +------*
     *             |d   c'/2     *     *      |d c/2 *
      **-----------+-----------**       *-----+-----*
        *****      |h     *****          ***  |h ***
             *************                  *****

The area of the segment of the ellipse will be a/b times the area of 
the corresponding segment of the circle, since the ellipse is 
stretched by a factor of a/b horizontally. The hard part is to 
determine the value of b, which will be used in place of the radius r 
in the formula. I have written c' for the chord of the ellipse, to 
distinguish it from c in the circle; c'/c = a/b. If you use an angle 
in the formula, it will be the angle in the circle, not in the 
ellipse.

If you can tell me what you can measure (and, in fact, how you know 
the shape is an ellipse), I can help work out the details.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 09/06/2001 at 19:26:09
From: Martin Braaksma
Subject: Re: Segment ellipse

Hi, Doctor Peterson,

First of all, thank you for your quick response.

Since this is about a real-life case, I guess that we would only be 
able to measure the height, width, and length of the tank and the 
height of the fluid. We would measure the height of the fluid with a 
measuring stick from the top of the tank, probably using a cm-scale, 
or through a transparent parallel hose. So d, for example, could be 
derived. An oval tank (ellipse) could have any shape, I guess, as long 
as b < a.

I hope this information is sufficient for you to provide us with some 
more calculation details. Looking forward to your response again.

Thank you,

Martin Braaksma

Date: 09/06/2001 at 22:06:06
From: Doctor Peterson
Subject: Re: Segment ellipse

Hi, Martin.

I just realized that I was picturing a slightly harder problem, in 
which you only have the segment of the ellipse, so you have to deduce 
the axes from the height, width, and something else about what you 
have. Since you presumably have the whole ellipse, or at least the 
bottom half of it, it's not so hard.

Referring to my diagram, and my earlier comments, I'll use the 
following formula from the FAQ for a segment of a circle:

    K[circle] = r^2 arccos(d/r) - d sqrt(r^2-d^2)

Applying this to the circle I drew, r is actually b, the vertical 
semiaxis of the ellipse. The area of the segment of the ellipse, as I 
pointed out, is a/b times this area, or

    K[ellipse] = (a/b)[b^2 arccos(d/b) - d sqrt(b^2-d^2)]
               = ab arccos(d/b) - ad sqrt(1-(d/b)^2)

If you know the semiaxes a and b and the depth, you just need to find 
d = b - h and plug that, together with a and b, into this formula to 
find the area, then multiply by the length of the tank to get its 
volume. Note also that none of this really depends on b being less 
than a.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/