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

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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.




differential-geometry




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    Why not reparameterize it so that it has unit speed?
    – Sobi
    Jul 21 at 7:47










  • A circular helix is a curve, and its cross section is a point !?
    – Yves Daoust
    Jul 21 at 7:51














up vote
0
down vote

favorite












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.




differential-geometry




share|cite|improve this question

















  • 1




    Why not reparameterize it so that it has unit speed?
    – Sobi
    Jul 21 at 7:47










  • A circular helix is a curve, and its cross section is a point !?
    – Yves Daoust
    Jul 21 at 7:51












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.




differential-geometry




share|cite|improve this 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.





differential-geometry





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share|cite|improve this question




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edited Jul 21 at 8:41









Chandler Watson

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asked Jul 21 at 7:37









Gene Naden

203




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  • 1




    Why not reparameterize it so that it has unit speed?
    – Sobi
    Jul 21 at 7:47










  • A circular helix is a curve, and its cross section is a point !?
    – Yves Daoust
    Jul 21 at 7:51












  • 1




    Why not reparameterize it so that it has unit speed?
    – Sobi
    Jul 21 at 7:47










  • A circular helix is a curve, and its cross section is a point !?
    – Yves Daoust
    Jul 21 at 7:51







1




1




Why not reparameterize it so that it has unit speed?
– Sobi
Jul 21 at 7:47




Why not reparameterize it so that it has unit speed?
– Sobi
Jul 21 at 7:47












A circular helix is a curve, and its cross section is a point !?
– Yves Daoust
Jul 21 at 7:51




A circular helix is a curve, and its cross section is a point !?
– Yves Daoust
Jul 21 at 7:51










1 Answer
1






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0
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accepted










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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  • 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/…
    – Gene Naden
    Jul 21 at 14:54











  • Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
    – Gene Naden
    Jul 21 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...
    – Yves Daoust
    Jul 21 at 15:15











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










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".






share|cite|improve this answer





















  • 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/…
    – Gene Naden
    Jul 21 at 14:54











  • Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
    – Gene Naden
    Jul 21 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...
    – Yves Daoust
    Jul 21 at 15:15















up vote
0
down vote



accepted










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".






share|cite|improve this answer





















  • 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/…
    – Gene Naden
    Jul 21 at 14:54











  • Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
    – Gene Naden
    Jul 21 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...
    – Yves Daoust
    Jul 21 at 15:15













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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".






share|cite|improve this answer













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".







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 21 at 7:49









Yves Daoust

111k665204




111k665204











  • 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/…
    – Gene Naden
    Jul 21 at 14:54











  • Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
    – Gene Naden
    Jul 21 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...
    – Yves Daoust
    Jul 21 at 15:15

















  • 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/…
    – Gene Naden
    Jul 21 at 14:54











  • Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
    – Gene Naden
    Jul 21 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...
    – Yves Daoust
    Jul 21 at 15:15
















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/…
– Gene Naden
Jul 21 at 14:54





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/…
– Gene Naden
Jul 21 at 14:54













Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
– Gene Naden
Jul 21 at 15:04




Sorry, I goofed. Tau is equal to zero, so the term drops out. Duh...
– Gene Naden
Jul 21 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...
– Yves Daoust
Jul 21 at 15:15





@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...
– Yves Daoust
Jul 21 at 15:15













 

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