Consider this figure. The length of the cylinder is '2R', length of the wire is 'L' and it makes an angle of '$\theta$' with the X-axis. The origin is as indicated in the figure.

Now if we wrap this wire over the cylinder to get this, then what is the equation of the resulting helix in the coordinate system shown?


The approach I tried is borrowed from this question. If I define the following parameters:

  • $R$ to be the radius of the cylinder
  • $D$ to be the distance between two screws
  • $t$ to be a parameter between $[0, 1]$ determining the position on the helix
  • $h_1$ to be the end height on the cylinder of the helix

then I get my angular speed coefficient $\alpha$ to be,

$\alpha$ = $\frac{2\pi h_1}{D}$, where

$D$ = $2\pi R \cdot tan(\theta)$. My parametric coordinates then become:

$x$ = $R cos(\alpha t)$

$y$ = $h_1 t$

$z$ = $R sin(\alpha t)$

However upon tracing out this curve although I do get a helix, it's not the same one I get upon wrapping the wire upon the cylinder as seen in this figure. The dull coordinate axis correspond to the point (0, 0) while the brighter coordinate axis correspond to the case where I take $t = 0.3$. Upon varying $t$ I do see that I get a helix but the pitch of the helix is not correct.

It would be helpful if I could get some comments on the correctness of my equations.

Thank you.

  • 1
    $\begingroup$ Welcome to stackexchange. You are more likely to get answers rather than downvotes or votes to close if you edit the question to show what you tried and where you are stuck. $\endgroup$ – Ethan Bolker Feb 24 '18 at 14:33
  • $\begingroup$ Thank you for the feedback. I've edited the question now and included my approach. $\endgroup$ – Chaitanya Prasad Feb 24 '18 at 15:01
  • $\begingroup$ Your equations are correct. You don't give any details about what difference you see between your model and the equations, so I can't help there. $\endgroup$ – Ross Millikan Feb 24 '18 at 15:12
  • $\begingroup$ I've tried to show the difference between the model and the equations through another edit. Since I'm using a CAD based software, this is the closest representation I can provide as of now. $\endgroup$ – Chaitanya Prasad Feb 24 '18 at 15:31

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