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 Question from Brian: I am trying to determine the amount of a liquid remaining in a 55 gallon drum when it is tilted at 45 degrees and the liquid level is low enough so that the liquid does not completely cover the bottom of the drum. Your help is greatly appreciated.

Hi Brian,

We received a somewhat similar question a few years ago from Richard but the tilt was at an arbitrary angle. I developed an expression for the volume and it appears at the bottom of my response to Richard. Since your tilt is 45 degrees and $\tan(45) = \cot(45) = 1$ the expression for the volume $V$ simplifies to

$V = \frac23 \left( 2rh - h^2\right)^{3/2} + 2(h-r) \left(\frac{\pi r^2}{4} - \frac{r^2}{2}\sin^{-1}\left(\frac{r-h}{r}\right) - \frac{r-h}{2}\sqrt{2rh-r^2}\right)\mbox{ cubic units.}$

When using this expression make sure your calculator is in radian mode not degree mode.

Still a somewhat complicated expression that requires $r$ and $h$ where $r$ is the radius of the drum and $h$ is the slant height of the liquid in the drum. Can you measure $h?$ If not what can you measure to determine the amount of liquid in the tank?

Harley

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