Shortest distance between two points on the surface of a closed cylinder

Question

What is the shortest distance between two points on the surface of a closed cylinder? I understand simple euclidean distance will work if both points are on curved surface, but I am looking for a generic solution where one point can be on the curved while other might be on the top/bottom surface. The parameters we have for point(say A) is its distance from the top of cylinder, and for point B is - its distance from top of the cylinder and angle subtended with the other point along the circumference of cylinder and the centre point of cyclinder.

Answer 1

Let me try to help with expressing your problem. The following is related:

Find minimum distance between two points on a cylinder, one on the top flat surface $A$ (not its circle center) and another point $B$ on its curved surface.

EDIT1:

$$ AO=a\,;\angle AOQ =\alpha; AOP =\theta; \, QB= z ; $$

The problem is best formulated using development of the cylinder curved surface.

Given $(r,\alpha, a,z) $ you want to minimize total length

$$ AP +PB = \sqrt{a^2+r^2-2 a r \cos \theta} +\sqrt{z^2+r^2( \alpha-\theta)^2 }\tag1$$

This should be differentiated w.r.t $\theta$ and set to zero leading to :

$$ \frac {a \sin \theta}{r (\alpha-\theta)} = \sqrt{\frac {a^2+r^2-2 a r \cos \theta}{z^2 +{r^2( \alpha-\theta)^2 }} }\tag2$$

which should be solved. A numerical solution can be attempted as it is a transcendental equation having no closed form solution.

A minimum is seen somewhere in the middle $\theta \approx 0.86$ angle in domain for values assumed $(a,r,\alpha,z)=(.5,1, \pi/2,1.25 ) $ where $ \theta $ is on x-axis and total length of two line segments $ APB$ on y-axis.

EDIT2:

It is instructive here to mention geodesics in 3D. Label the vertex opposite to $Q$ as $C,$ to make $ PQBC $ a rectangle. Now the point $P$ should move such that line $APB$ should be straight for minimum length, or in other words

$$\boxed{ \angle APO= \angle BPC = \psi } \tag3 $$

$$ \sin \psi =\frac{a \sin \theta}{\sqrt{ z^2 +r^2( \alpha-\theta)^2 }} =\frac{r( \alpha-\theta) } {\sqrt {z^2 + r^2( \alpha-\theta)^2 } } \tag4 $$

Minimum distance from axis to line on flat top surface is the Clairaut constant for surfaces of revolution. $$ = a \sin \psi \tag5 $$

It is thus seen that the surface of revolution can have abrupt change in slope needing no smoothness and continuity.

Answer 2

WLOG, let the radius be unit.

The shortest length between two points on the lateral surface is given by

$$\sqrt{\Delta^2\theta+\Delta^2z}.$$

Now, for a trajectory starting from $(1,0,z)$ (on the side) and reaching a point $(x,y,0)$ (on the top) through $(\cos\theta,\sin\theta,0)$ where $\theta$ is unknown, the total path length is

$$\sqrt{\theta^2+z^2}+\sqrt{(x-\cos\theta)^2+(y-\sin\theta)^2}.$$

When $(x,y)=(r,0)$, by symmetry the solution is with $\theta=0$ (the function is even), and is made of two line segments. But in the general case, the minimization yields a transcendental equation that has no closed-form solution.

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For two points on the side, there might as well exist a short path through the top. This case can be handled by introducing two $\theta$ angles corresponding to the entry and exit points. The minimization will be even less tractable.