Volume by Inch of a Cone-Shaped Tank

Date: 03/03/2003 at 21:33:18
From: Matt
Subject: Volume by inch of a cone-shaped tank

I have a chemical tank that has a cone bottom. I can get the total 
volume of the cone part. I used your formula under "Frustum of a 
Right Circular Cone."

 http://mathforum.org/dr.math/faq/formulas/faq.cone.html#conefrustum 

I need to know the volume of each inch of the cone. The top radius is 
16.25", the bottom radius is 3.25, and the total cone height is 22.5". 
I can get the total volume but I need to know the volume as the level 
changes. I cannot calculate the top radius as the level changes.

Date: 03/04/2003 at 01:50:04
From: Doctor Jeremiah
Subject: Re: Volume by inch of a cone-shaped tank

Hi Matt,

What we need is an equation to calculate the top radius at any point.  
What we have is a straight line graph like this:

                       height
                          |
                          |                    /
                     22.5 - - - - - - - - - - + (16.25,22.5)
                          |                  /
                          |                 /
                          |                /
                          |               /
                          |              /
                          |             /
                          |            /
                          |           /
                          |          /
    ----------------------+---------+---------------> radius
                          |        / (3.25,0)

Because it is a straight line graph we can use the equation for a 
straight line: y=mx+b

For us the "y" is the height and the "x" is the radius so the equation 
is really   height = m x radius + b

And we have these two points on the graph:

                  height     |     radius
          -----------------------------------------
                     0       |      3.25
                   22.5      |     16.25

If we put the first row into height = m x radius + b we get:

                  height = m x radius + b
                     0   = 3.25 m + b

If we put the second row into height = m x radius + b we get:

                  height = m x radius + b
                   22.5  = 16.25 m + b

If we solve these two equations we get:

                       m = 22.5/(16.25-3.25)
                       b = -3.25 x 22.5/(16.25-3.25)

Which gives us this equation for a straight line:

  height = 22.5/(16.25-3.25) x radius - 3.25 x 22.5/(16.25-3.25)

But it's all backward. We need an equation with radius = ...

  radius = 3.25 + height x (16.25-3.25)/22.5

Now, this kind of makes sense because with a height of 0 we get radius 
= 3.25 and with a height of 22.5 we get radius = 3.25 + (16.25-3.25)

So that all means that the radius of the liquid at any height is:

  radius = 3.25 + height x (16.25-3.25)/22.5

Which you can use as the top radius of your frustum of a cone.

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