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Volume of a Partially Filled Cylindrical Tank


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,

Thanks for the quick reply.

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 
differentiation already done.

           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:
College Calculus

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