# Volume of Cylindrical Wedge (not Segment)

What is the volume of liquid in a cylindrical tanker (a fuel truck say) on an incline, when the liquid is touching the back circular face, but not the front?

The tank has clear tubes at the front and back to indicate the height of the liquid at those points (in plane with the cross section) - however as the liquid is not touching the front face, that tube is measuring zero so can be ignored.

In addition, how is this affected if the tank is 1. elliptical 2. rolled due to road camber

Edit for clarification: the incline is known, volume in terms of the incline and height of liquid in the tube would be nice, but is not required. The cylinder is lying down, as in a petrol truck say.

Edit 2: The reason I ask is that we have a pump out sewage system at my workplace. They calculate the volume by measuring the height of liquid at the font and back of the tank (which are scaled to read the volume of the tank) and averaging them.

We investigated as we suspected that we were being overcharged when they sent a bill for a volume greater than that of our tank. It turns out that if they only have a reading at one end of the tank, then they just halve it, and as our workplace is at the start of their run, they usually start with an empty tank - clearly this is not accurate.

-
This will require some clarifications. What do you mean by the liquid "touching" a face? Is the answer to be given in dependence on the measurement of the tube at the back? Which cross section are you referring to? It seems that even if we have some measurement at one end, the surface of the liquid could end anywhere over the length of the cylinder without changing the measurement at the one end? Please try to put yourself into the shoes of someone trying to understand your question without having access to the images you have in mind while writing it. – joriki Oct 26 '11 at 3:31
See edits - The cylinder is lying down, but on an angle such that the liquid is in contact with the circular face at the back, but not the front of the tank (think of a mostly empty fuel truck). The tubes are vertical and bisect the circular ends of the tank. – Peter Gibson Oct 26 '11 at 4:31
I don't understand this sentence: "They calculate the volume by measuring the height of liquid at the font and back of the tank (which are scaled to read the volume of the tank) and averaging them." To determine the volume of liquid at zero inclination, the height measurement would have to be converted nonlinearly (i.e. not just scaled) into circular segment areas and then multiplied by the length of the tank. Then for non-zero inclination there would be two different ways of averaging: averaging the heights before converting or averaging the areas after converting. Which do they do? – joriki Oct 26 '11 at 7:18
Yes sorry, converted nonlinearly. They get a reading for volume from the front and back (the conversion does assume zero inclination as you say) and average those readings. – Peter Gibson Oct 26 '11 at 22:27

There's still some room for interpretation in the question, as the relative orientation of the tubes and the inclination hasn't been specified. I'll assume that the cylinder is inclined by rotating it about a horizontal axis that is perpendicular both to the central axis and to the tubes bisecting the circular ends, which are originally vertical (but cannot remain so upon inclination).

Then the liquid takes the form of a cylindrical wedge. By equations (5) and (9) of the linked page, its volume is given by

$$V=R^3\frac hb\left(\sin\phi-\phi\cos\phi-\frac13\sin^3\phi\right)\;,$$

where $R$ is the radius of the circular end, $b$ is the height indicated on the tube, $h$ is the height of the wedge along the axis of the cylinder and $\phi=\arccos\left(1-b/R\right)$. The ratio $h/b$ is the cotangent of the angle of inclination $\alpha$, so

$$V=R^3\cot\alpha\left(\sin\phi-\phi\cos\phi-\frac13\sin^3\phi\right)\;.$$

Substituting $\phi=\arccos\left(1-b/R\right)$ leads to an unwieldy expression that doesn't seem to allow any useful simplifications.

-
Thanks for the help joriki :) – Peter Gibson Oct 26 '11 at 22:28