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I've found an answer to the question I had here.

But in that answer, we assume that the curve at hand has unit speed. In working with the cross-sectional curve of a circular helix, I do not know my curve has unit speed. How can I still show that the cross-sectional curve is a circle? I have demonstrated that it has constant curvature, lies in a plane and has zero torsion.

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    $\begingroup$ Why not reparameterize it so that it has unit speed? $\endgroup$ – MSobak Jul 21 '18 at 7:47
  • $\begingroup$ A circular helix is a curve, and its cross section is a point !? $\endgroup$ – Yves Daoust Jul 21 '18 at 7:51
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The shape of the trajectory does not depend on the speed, so it needn't be unit. (Think that tough you can drive at different speeds, the road remains unchanged. :)

The curvature and torsion formulas are established by computing the curvilinear abscissa, which "normalizes the speed away".

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  • $\begingroup$ I see that the curvature does not depend on the parameterization, but I notice that the proofs at the link below seem to leave out terms. For example, it says Let γ=α+(1/κ)N so γ′=T+1/κ∗(−κT+τB) =T−T=0 This seems to leave out a term (1/kappa)*tau B. The link is math.stackexchange.com/questions/1477797/… $\endgroup$ – Gene Naden Jul 21 '18 at 14:54
  • $\begingroup$ Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh... $\endgroup$ – Gene Naden Jul 21 '18 at 15:04
  • $\begingroup$ @GeneNaden: we are safe now. For a while I have been believing that your car was deforming the road. May be in a relativistic world... $\endgroup$ – Yves Daoust Jul 21 '18 at 15:15

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