Does the velocity vector have to have unit speed parametrisation in calculations of the frenet frame?
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Consider the curvature of a curve $beta$ at a point s.
This is given by $kappa(s):=|T'(s)|$, where $T(s)=beta '(s) $.
similarly we define the fields in the frenet frame ${T,N,B}$ by
$$T(s)=beta'(s)$$
$$N(s):=tfrac{T'(s)}{kappa(s)}$$
$$B(s):=T(s)times N(s)$$
My question is does $beta'(s)$ have to be parametrised to unit speed in all of these calculations ?
differential-geometry frenet-frame
asked Nov 14 at 23:07
exodius
924417
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up vote
1
down vote
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Consider the curvature of a curve $beta$ at a point s.
This is given by $kappa(s):=|T'(s)|$, where $T(s)=beta '(s) $.
similarly we define the fields in the frenet frame ${T,N,B}$ by
$$T(s)=beta'(s)$$
$$N(s):=tfrac{T'(s)}{kappa(s)}$$
$$B(s):=T(s)times N(s)$$
My question is does $beta'(s)$ have to be parametrised to unit speed in all of these calculations ?
differential-geometry frenet-frame
asked Nov 14 at 23:07
exodius
924417
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the curvature of a curve $beta$ at a point s.
This is given by $kappa(s):=|T'(s)|$, where $T(s)=beta '(s) $.
similarly we define the fields in the frenet frame ${T,N,B}$ by
$$T(s)=beta'(s)$$
$$N(s):=tfrac{T'(s)}{kappa(s)}$$
$$B(s):=T(s)times N(s)$$
My question is does $beta'(s)$ have to be parametrised to unit speed in all of these calculations ?
differential-geometry frenet-frame
asked Nov 14 at 23:07
exodius
924417
Consider the curvature of a curve $beta$ at a point s.
This is given by $kappa(s):=|T'(s)|$, where $T(s)=beta '(s) $.
similarly we define the fields in the frenet frame ${T,N,B}$ by
$$T(s)=beta'(s)$$
$$N(s):=tfrac{T'(s)}{kappa(s)}$$
$$B(s):=T(s)times N(s)$$
My question is does $beta'(s)$ have to be parametrised to unit speed in all of these calculations ?
differential-geometry frenet-frame
differential-geometry frenet-frame
asked Nov 14 at 23:07
exodius
924417
asked Nov 14 at 23:07
exodius
924417
asked Nov 14 at 23:07
exodius
924417
asked Nov 14 at 23:07
exodius
924417
asked Nov 14 at 23:07
exodius
924417
924417
add a comment |
add a comment |
1 Answer
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Yes. More precisely, these formulas only work for curves parametrirized by the arclength. Otherwise, you would not have, for instance, that $bigllVert T(s)bigrrVert=1$.
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes. More precisely, these formulas only work for curves parametrirized by the arclength. Otherwise, you would not have, for instance, that $bigllVert T(s)bigrrVert=1$.
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
add a comment |
up vote
1
down vote
accepted
Yes. More precisely, these formulas only work for curves parametrirized by the arclength. Otherwise, you would not have, for instance, that $bigllVert T(s)bigrrVert=1$.
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes. More precisely, these formulas only work for curves parametrirized by the arclength. Otherwise, you would not have, for instance, that $bigllVert T(s)bigrrVert=1$.
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
Yes. More precisely, these formulas only work for curves parametrirized by the arclength. Otherwise, you would not have, for instance, that $bigllVert T(s)bigrrVert=1$.
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
answered Nov 14 at 23:19
José Carlos Santos
141k19111207
141k19111207
add a comment |
add a comment |
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