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I have a cylindrical tank but on it's side level and flat with a known diameter and volume. I need to solve for the actual physical height of a particular percent of the volume of the tank. For example, half full is half the length of the circumference but it's not linear beyond that point in either direction. If I need to know how high the liquid is if the tank is 28% full, what would be the formula? Thanks in advance.

- Daddio

Daddio wrote:I have a cylindrical tank but on it's side level and flat with a known diameter and volume. I need to solve for the actual physical height of a particular percent of the volume of the tank. For example, half full is half the length of the circumference but it's not linear beyond that point in either direction. If I need to know how high the liquid is if the tank is 28% full, what would be the formula? Thanks in advance.

It is not a pretty formula, and you may have to solve graphically, as it is easier to go from depth to percent full; hard to solve the other way.

If d is depth, and R is radius (half the diameter), introduce computational variable s = 1 - d/R.

(s = 1 when tank is empty, 0 when half full, and -1 when full.)

then V/Vmax = 1/(pi) * [arccos(s) -s*SQRT(1 - sÂ²)] (Your calculator MUST be in radian mode when taking the arccos)

The arccos term varies from 0 to pi radians, the second term is a correction, but zero at s = -1, 0, 1.

I would just at intervals of s = 0.1 or 0.05 and estimate graphically. But I suppose you could differentiate the above equation (messy) and use Newton's method to solve iteratively.

- Guest

2 posts
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