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Volume of a Partially Filled Cylindrical Tank

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Date: 04/28/2000 at 04:36:19
From: Mark Pinese
Subject: Inverse of the segment area equation

I am trying to find the inverse for the equation used to find the area
of a circle's segment. I have managed to derive the area formula, but
finding its inverse seems impossible.

One version of the formula follows:

Given a circle of radius 5, and the distance from the center of the
circle to the center of the chord defining the segment, h,

a = 25arccos((5 - h) / 5) - (5 - h)sqrt(10h - h^2)

The formula itself is no trouble to find, being just the difference
between the area of the sector the segment is part of and the triangle
formed by the two endpoints of the chord and the circle's center. I
need to transform it such that it is expressed in terms of h.

If the transformation is too difficult, it may help that this question
is part of a bigger problem. Briefly, a tank, filled from the top and
drained from the bottom, is made from a hemispherical prism, such that
it lies with its axis parallel to the ground and its flat surface
facing upwards (a D with the flat section upwards). The radius of the
hemisphere is 5, and the tank is 10 deep. The problem is to design a
dipstick such that it can be used to find the volume of liquid in the
tank. I had planned to use the inverse to find h for convenient
volumes of liquid, and use that value to specify markings on the
dipstick. I have written a computer program that solves the problem
using successive iterations to reduce error, but it is not really an
acceptable solution.

I have done a little calculus (just differentiation and integration),
but no solving of differential equations (which is what I think may be
needed here, as the derivative of the area would be related to the
derivative of h), so that is little help here unless I hit the books.
Any suggestions?
```

```
Date: 04/28/2000 at 08:09:50
From: Doctor Jerry
Subject: Re: Inverse of the segment area equation

Hi Mark,

Consider a circle of radius a (the end of the tank) and imagine that
the liquid has reached height h, measured from the lowest point on the
circle. Note that 0 <= h <= 2a. The area A of the segment of the
circle covered by the liquid is

A = pi*a^2/2 - a^2*arcsin(1-h/a) - (a-h)*sqrt(h(2a-h))

The volume of liquid is just A*L, where L is the length of the tank.
From this, a dipstick can be calibrated.

This formula is related to the formula you gave, but it is expressed
in terms of the depth of the liquid, which I think is useful.

It is not possible to solve either of our two formulas for h - at
least not in "closed form."

I'm not clear about the shape of the container. I can't understand why
you would need the area of a segment. You need this for finding the
volume of a horizontal cylinder with hemispherical ends, but I don't
think this is what you mean.

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

```
Date: 04/28/2000 at 09:21:18
From: Mark Pinese
Subject: Re: Inverse of the segment area equation

Hi Doctor Jerry,

I might have got a bit carried away with describing the shape; I'll
try again. Imagine a cylinder lying on its curved side, then remove
the top half of the cylinder.  You will end up with something that
*should* look a little like this:

oooooooo|oooooooo
o       |       o
oo     |     oo
ooooooooooo

The line passing through the centre is the dipstick. The problem,
basically, is to design a diptick to use to determine the volume of
liquid in the tank, using a suitable graduation of volume (e.g. m^3,
or 100 L).

Mark
```

```
Date: 04/29/2000 at 07:46:32
From: Doctor Jerry
Subject: Re: Inverse of the segment area equation

Hi Mark,

I interpret your drawing as a cross-section of the horizontal
cylinder. What threw me off, I think, was your use of the term
hemisphere. I think you meant semicircle. It would, of course, be
possible to consider the dipstick problem on a horizontal cylinder
with hemispherical ends; however, I'll act on the assumption that you
have the bottom half of a horizontal cylinder with semicircular ends.
The radius of the semicircular ends is 5 meters and the (horizontal)
length of the tank is 10 meters. Using the formula:

V(h) = 10[pi*a^2/2 - a^2*arcsin(1-h/a) - (a-h)*sqrt(h(2a-h))]

for the volume of such a tank, where a = 5 and h, the depth of the
liquid, lies between 0 and a = 5, our problem is to determine h when
given various convenient values of V(h).

I find V(5) = 392.7 sq meters, approximately.

Suppose we determine h so that

V(h) = 350 sq meters.

I'll use Newton's method on the function

f(h) = V(h)-350.

Letting h_1 be our first guess, which we can take as, say, 4.5, the
next guess comes from

h_2 = h_1-f(h_1)/f'(h_1).

All other guesses follow the same pattern.

I find

h_2 = 4.572538...
h_3 = 4.572487...

So, it looks as if a depth of 4.57 yields 350 sq meters. You could, of
course, specify an amount in liters and, before using the above
formulas, convert the liters to square meters.

By f'(h_1) I mean the derivative (quite messy) of f, evaluated at h_1.

I hope this shows the way in sufficient detail. Don't hesitate to
write if I've gone astray.

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

```
Date: 04/29/2000 at 09:10:42
From: Mark Pinese
Subject: Re: Inverse of the segment area equation

Jerry,

Thanks for that. I did mean semicircle, not hemisphere - I can now see
why you were confused. The method you have shown will work fine, and I
should be able to do the differentiation given enough time (and enough
use of the chain rule to drive myself insane.)

Can you see any easier way of solving this problem? I ask this as my
knowledge of calculus was not taught at school, and so far we have
only covered in class basic area and volume formulae, so I can't help
but think an easier method exists.

Even if not, thanks very much for your help - it's been invaluable.

Mark
```

```
Date: 04/30/2000 at 08:02:24
From: Doctor Jerry
Subject: Re: Inverse of the segment area equation

Hi Mark,

I don't see an easier way of solving the problem. However, I can offer
a simplified version of the iteration formula, with the

0.5*(5 + h_1 - (5*(-7 + 5*ArcCos[1 - h_1/5]))
h_2 = ---------------------------------------------
Sqrt[-(-10 + h_1)*h_1])

As before, h_2 means "h sub 2" and h_1 means "h sub 1". h_{n+1} is
obtained similarly from h_n.

With this I obtained the results I mentioned earlier. I used
"FullSimplify" in Mathematica to obtain this formula.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
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