If a non-unit-speed has constant curvature and zero torsion, is it a circle necessarily?

Question

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.

A circular helix is a curve, and its cross section is a point !? — , Jul 21 '18 at 07:51

score 1 · Accepted Answer · answered Jul 21 '18 at 07:49

1

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".

answered Jul 21 '18 at 07:49

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 https://math.stackexchange.com/questions/1477797/show-that-curve-alpha-s-is-a-circle-if-curvature-is-constant-and-torsion-is – Gene Naden Jul 21 '18 at 14:54
Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh... – Gene Naden Jul 21 '18 at 15:04
@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... – Jul 21 '18 at 15:15

If a non-unit-speed has constant curvature and zero torsion, is it a circle necessarily?

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