Maximum tilt for a cylindrical glass without spilling Something I've been wondering about for a while after drinking water one day:
Suppose you have a right cylindrical cup with interior radius $R$ and interior height $H$.  The cup is filled with a liquid to height $h$.  Assuming that surface tension is not a factor, what is the maximum angle from vertical that the cup can be tilted without the liquid spilling out?
I presume that there are two cases:


*

*When the cup is at maximum tilt, the bottom of the cup is completely covered by the liquid.

*When the cup is at maximum tilt, the bottom of the cup is only partially covered by the liquid.
For case 1, I intuitively feel like the center of the liquid's top surface always should have distance $h$ from the bottom of the cup: if I take infinitesimally thin cross-sections of the cup, for each cross-section, it seems clear that tilting the cup should not affect the distance from the bottom of the cup to the midpoint of the liquid surface.
This leads me to calculate that the maximum tilt from vertical is 
$$\frac{\pi}{2}-\tan^{-1}\left(\frac{R}{H-h}\right) \text{if $h\ge\frac{H}{2}$}$$
I'm not sure what to do for case 2, however.
(Note that this question is not the same as Liquid levels from the base of a right cylindrical drinking glass.)
 A: 
An alternative way to solve case 1 is to say,
"When the cup is tilted, I can divide it into two parts (see right hand pic above): 


*

*The part from the lip of the cup to the other end of the liquid surface,

*The cylindrical part from the end of the liquid surface to the bottom of the cup." (shown shaded greenish). 
For the second case, the figure at the left applies. (Only it doesn't .... because the stuff in the pink triangle above the horizontal line doesn't stay to the left of the blue/pink dividing line when we tilt, so the rest of my earlier analysis was wrong). 
A: For case 2 one has to compute the volume in the cylinder $\{(x,y,z)\>|\>x^2+y^2\leq R^2, \ 0\leq z\leq H\}$
lying below the plane through the line $x=a, z=0$ and the point $(R,0,H)$. Here $a\in [{-R},R]$ is related to the tilting angle $\alpha$ via $\tan\alpha={H\over R-a}$. Given $h<{H\over2}$ one obtains in the end a transcendental equation for $\alpha$ (or a suitable related quantity) involving no interesting geometric insight, as in case 1. 
