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.
Best Answer
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.