# Calulating length of cable running along exterior of an axle

This question may seem simple, but I have no idea where to start.

I'm in design phase of a hobby electronics project and need your help. Basically, I'm going to have two platforms with electronics on both of them. I'm planning to have a stepper motor on the bottom platform which will be connected to top platform by an axle. I plan to use a microcontroller to precisely control the intensity of angular velocity of the motor and the angle by which it will rotate. I will also need to use wires to connect the bottom platform to the top platform. My idea is to have wires run on the outside of the axle and be long enough that the axle will be able to rotate by some angle of $±θ$. In normal operation the angle shouldn't be grater than 2π, so I want to avoid constructing some complex mechanism which will transfer current irrelevant of the axle's rotation.

Image (I apologize for my bad drawing skills) shows the two states which are interesting (at least in my opinion) here. The top is when the axle is in neutral position and angle by which it is rotated is 0. While the image does not show it, the wire will be loose. The bottom image shows what I expect will happen after the axle has been rotated by a certain angle. The two black rectangles are soldering joints by which the cable will be connected to the bottom and the top platforms. The red line is wire.
Here's direct link to the image. It is more clear there.

Here's my question: Is I have an axle with length L and radius R, how would I calculate length l of a cable running form the top of the axle to the bottom after axle has rotated by a certain angle θ?

Also, I have no idea which tags would be appropriate, so do feel free to edit them.

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I just want to point out that the answers given by REDace0 and by J. M. are the same, up to a change of notation. – Michael Lugo Sep 20 '10 at 4:21

This is a total and utter guess, but suppose you fix one endpoint of the wire and unroll the axle to make a rectangle? This is the same trick used to justify the rectangle in the surface area of a cylinder. Then the wire is the diagonal of a rectangle of width $L$ and height $r\theta$. So by the Pythagorean theorem, $l = \sqrt{L^2+r^2\theta^2}$.

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It took me a bit of time to understand what you are suggesting. If I understand you correctly, then this would be a simple way to calculate the length. However, to me it looks like it wouldn't work if for some reason I need more than one turn of the cable. I guess I could then "cut" the unrolled axle into smaller rectangles and have each small rectangle's hight be distance between turns of cable. – AndrejaKo Sep 19 '10 at 13:45
@AndrejaKo: Actually, it still works if you have more than one turn. Say your cable is twisted twice, so $\theta = 4\pi$. Imagine drawing a diagonal on a transparent $L\times4\pi r$ rectangle, and wrapping the rectangle twice around the axle. Where does the diagonal go? – Rahul Sep 19 '10 at 16:25
@Rahul I understand now. – AndrejaKo Sep 19 '10 at 17:03

Looks like you can approximate the cable's equation with a helix, $(x\quad y\quad z)=(r\cos(m t)\quad r\sin(m t)\quad \frac{h t}{2\pi})$ where $m$ is the number of turns, $h$ is the height of the cylinder, $r$ is the radius of the cylinder, and the parameter $t$ runs from $0$ to $2\pi$.

If we compute the arclength of this one, we get

$$\sqrt{4\pi^2 r^2 m^2+h^2}$$

This should be a nice starting point, unless there's a better way to approximate the behavior of the cable.

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Ah, yes. Helix was the word I was looking for when I made my post. I believe that cable can be approximated by a helix and two straight lines, so I'll do some research in that direction. – AndrejaKo Sep 19 '10 at 13:32