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“Tanks” are cylinders with circular cross-section and axis horizontal. These cylinders are variable in size with radius and length different for each tank.

We need to determine the amount of liquid in a buried tank. We do this by “sticking”, that is, we insert a dipstick through an opening over the deepest part of the buried container until the dipstick touches the bottom, then we pull the stick out and read the liquid level showing on the stick.

I need a formula for converting the height shown on the stick to the volume of liquid in the tank.

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Area found by integration of circle segment.

$$ y= h/R\,;\; Volume = (Length/\pi) ( \cos^{-1}(1-y)-(1-y) \sqrt { y( 2-y)}) $$

Plotted y vs Volume.

VolFraction

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We integrate the equation of a circle of diameter $1$, with center at $x=1/2$, with a correction factor such that the total area equals $1$: $$\frac{4}{\pi}\int_0^h\sqrt{x-x^2}dx$$ and get $$v=\frac{\arcsin(2h-1)+2(2h-1)\sqrt{h-h^2}}\pi+\frac12.$$

This formula gives the fraction of the total volume ($v$ in $[0,1]$) as a function of the fraction of the total height ($h$ in $[0,1]$).

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  • $\begingroup$ This isn't and equation that you can plug in radius, length, and height of liquid in the tank to find the volume of the liquid. $\endgroup$ Oct 13 '14 at 21:27
  • $\begingroup$ No, it is much better than this: it's a universal equation, that will work for any tank. I made it for this very purpose. To match it to an existing tank, you input $h=H/2R$ (liquid height over full height - the diameter), and you multiply the output v by the actual volume, $\pi R^2L$ or the known tank capacity. Actually it depends on how you plan to use the formula: in a program, in an Excel sheet, on a printed table, an abacus... $\endgroup$
    – user65203
    Oct 14 '14 at 7:25
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If the length of the tank is $L$, the radius of the tank is $R$, and the height of the liquid is $h$, then if $h < R$ the volume is $$(R^2 \cos^{-1} \frac{R-h}{R} - (R - h)\sqrt{2Rh - h^2})L$$ This follows from the formula for the area of the segment of circle bounded by a chord and an arc. See, for example, the Wikipedia entry for circular segments.

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  • $\begingroup$ What if the height of the liquid is greater than the radius? $\endgroup$ Oct 13 '14 at 21:26
  • $\begingroup$ Let $S(R,h)$ be the above parenthesized expression. If $h < R$, the volume is $S(R,h)L$. If $h > R$, the volume is $(\pi R^2 - S(R, 2R - h))L$. $\endgroup$
    – Derek
    Oct 13 '14 at 22:26
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I had the same problem. Calculus is nice, but there's a much simpler way.

For a given horizontal cylinder:

V = pi/4 * D^2 * L * h/D

where,

V is the volume of the cylinder

D is the diameter of the cylinder

L is the length of the cylinder

h is the measured height of the liquid in the cylinder, and 0 <= h <= D

It's a simple ratio correction factor. If h = D/2, then the correction factor is 1/2 -- if the height measured was half the diameter, then you would have half the volume of a horizontal cylinder.

Simple.

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