Given two points laying on the surface of a cylinder, is there a simple equation for the arclength of the geodesic that connects those two points?

In my use case, the cylinder is oriented axially coincident with the x axis. I have two points for which I know their (x,y,z) locations, and I understand that I can convert these coordinates to cylindrical coordinates by the transformation x=x, y=rcos(theta), z=rsin(theta). Beyond that, I am not sure is there is a simple equation for calculating the geodesic length of between these two points without "unrolling" the cylinder into a plane and using the distance formula.

Can someone confirm for me if it is simply: L=SQRT(r^2θ^2+x^2), where x ix the axial distance which separates the points in my example? Is it this easy?

Thank you.

  • $\begingroup$ Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances. $\endgroup$ – Doug M Jan 11 at 21:38

Geodesic of cylinders are known to be

  • 1) either helixical arcs (the shortest helixical arc connecting the two points).

  • 2) or vertical segments

Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :

  • the ordinates of the points stay the same, whereas

  • abscissas are measured by the unrolling of arc lengthes $r \theta$.

The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r \theta_1,y_1)$ and $(x_2=r \theta_2,y_2)$.

Its arc length is thus :

$$\sqrt{(r(\theta_2-\theta_1))^2+(y_2-y_1)^2} \tag{1}$$

(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).

In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $\theta_2=\theta_1$ under the simplified form $$|y_1-y_2|$$

  • $\begingroup$ Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis? $\endgroup$ – kreeser1 Jan 11 at 21:30
  • $\begingroup$ Question 1) Yes, the r term has to be squared (I correct it at once). $\endgroup$ – Jean Marie Jan 11 at 21:33
  • $\begingroup$ Question 2) : what do you mean exactly by "oriented along the x-axis" ? $\endgroup$ – Jean Marie Jan 11 at 21:34
  • $\begingroup$ Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $\theta=\pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $\theta=0$ ? $\endgroup$ – Jean Marie Jan 11 at 21:39
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    $\begingroup$ @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates. $\endgroup$ – user1998586 Jan 12 at 7:39

You can "hide" the unrolling by using trig functions to convert chord-length to arc length.

A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r \arcsin \frac{l}{2r}$

Given points on the cylinder with $\Delta x = x_2 - x_1$, $\Delta y = y_2 - y_1$, $\Delta z = z_2 - z_1$ we then get the geodesic length: $$ \sqrt{(\Delta x)^2 + 4r^2 \arcsin^2 \left(\frac{1}{2r}\sqrt{(\Delta y)^2 + (\Delta z)^2}\right)} $$

  • $\begingroup$ Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable? $\endgroup$ – kreeser1 Jan 11 at 20:15

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