Formula for Arclength of Geodesic Connecting Two Points in the Surface of a Cylinder 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.
 A: 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:
$$
\sqrt{(\Delta x)^2 + 4r^2 \arcsin^2 \left(\frac{1}{2r}\sqrt{(\Delta y)^2 + (\Delta z)^2}\right)}
$$
A: 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 $$|y_1-y_2|$$
