0
$\begingroup$

I find myself with a cylinder of radius r, positioned along the x-axis, with equation y^2 + z^2 = r^2.

If I have two arbitrary points which lie on the surface of such a cylinder, is there a"two-point formula" which would give me some equation of a curve (which also curves along the surface of the cylinder) connecting the two points?

I envision an equation similar to the two-point equation of a line in 2D or 3D, but I do not know if such an equation is possible or makes sense.

My use case is wanting to render a curve connecting two points laying on such a cylinder, in a program I am writing in Python. I would like to render this curve but also understand the underlying mathematics.

Thank you in advance.

$\endgroup$
  • $\begingroup$ There are infinitely many of such curves. Are you looking for one with the shortest path? $\endgroup$ – Dylan Dec 10 '17 at 5:38
  • $\begingroup$ Hi, thank you for replying. I am looking for the shortest curve that connects the two points along the surface of the cylinder. If an insect were walking from one point on the surface of a cylinder to another point on the surface taking the shortest path, is there an equation that defines that segment? $\endgroup$ – kreeser1 Dec 10 '17 at 5:42
1
$\begingroup$

You can convert the problem into cylindrical coordinates $(x,y,z)\mapsto (x,\theta,r) $

$$ x = x, \ y = r\cos\theta, \ z = r\sin\theta $$

The constant surface $r = r_0$ in this coordinates system is somewhat equivalent to a "plane" in traditional Cartesian, in that any point on the surface is only dependent on two coordinates $(x,\theta)$

Suppose our two points are described by $(x_1,\theta_1)$ and $(x_2,\theta_2)$, then the shortest path between them is the linear parametrization (equivalent to defining a line in Cartesian space)

$$ x(t) = x_1(1-t) + x_2t $$ $$ \theta(t) = \theta_1(1-t) + \theta_2t $$

where the angles are picked so that $|\theta_2-\theta_1|\le \pi$

$\endgroup$
  • $\begingroup$ Thank you for your help Dylan, I appreciate the insight. $\endgroup$ – kreeser1 Dec 10 '17 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.