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### Segment of an Ellipse

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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?

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

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