Richard: Good post, up until the end, where you used the wrong volume
formula.

Richard Riley wrote:

I'm not engineer, and I'm not currently playing one on TV, but I think
it's straightforward. Just so we're clear, I'm assuming -

Each end is a pentagon, each one of *it's* sides is equal length. The
two ends are parallel to each other. One pentagon is larger than the
other. They are perpendicular to a line drawn from the center of one
to the center of the other.

Good summary of the relevant assumptions!

First, find the area of the large pentagon. The formula is

(the length of one side) squared * 1.7
I used 1.7205 in my calcs, just to get additional significant digits
in the result.

Then multiply by the length of the tank to get the volume if both the
ends were the size of the large one.
So far, so good.

Then do the same thing with the small end.

Now you have 2 volumes. Add them together, divide by 2.
But this is where your formula falls down. You are basically saying
1/2*h*(B1+B2) where B1 and B2 are the areas of the bases. The correct
formula is 1/3*h*(B1+B2+sqrt(B1*B2)). There is no difference if the
two ends are the same size, and the difference grows with the size
differential between the ends to the limiting case of the tank coming
to a point (zero area "base") where your formula overstates the
correct volume by 50%.

As a "gut check", note that the volume of a cone is 1/3*h*pi*r^2.

For the pentagon examples, this breaks down to .5735*h*(s1^2 +s2^2
+s1*s2)

Average 2040

And after that, we got almost identical results. Your use of 1.7 in
the pentagon area fromula rather than more significant digits almost
exactly offset the relatively small error in the formula, with ends
this close to the same size. I calculated: .5735*24*(64+36+48)=2037.
Total 8.83 gallons.
Same
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