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Test for arbitrary-speed curve lying on a sphere
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My textbook, ONeill's Elementary Differential Geometry, gives as an exercise the following test function to determine if a unit-speed curve lies on a sphere. If the test function is constant then the curve lies on a sphere. For a later exercise, we are invited to extend the test function so that it applies to arbitrary-speed curves. The prescription given is to multiply the derivative in the original formula by the speed of the curve. I tried that with a test curve that I am pretty confident lies on a sphere. But the results do not check out; the test function is not constant. I would like to know the above-stated prescription for taking into account arbitrary speed is as simple as I have suggested.
The test function for unit-speed curves is $rho^2+(rho^primesigma)^2$, where $rho$ is the inverse of the curvature, $sigma$ is the inverse of the torsion, $N$ is the principal normal and $B$ is the binormal.
For the speed, I use $nu=||fracdalphadt||$. So for the modified test function I get $rho^2+(nurho^primesigma)^2$
My curve that I believe lies on a sphere is $alpha(t)=t/sqrtt^2+1, cos(t)/sqrtt^2+1, sin(t)/sqrtt^2+1$
differential-geometry
asked Jul 27 at 4:41
Gene Naden
203
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My textbook, ONeill's Elementary Differential Geometry, gives as an exercise the following test function to determine if a unit-speed curve lies on a sphere. If the test function is constant then the curve lies on a sphere. For a later exercise, we are invited to extend the test function so that it applies to arbitrary-speed curves. The prescription given is to multiply the derivative in the original formula by the speed of the curve. I tried that with a test curve that I am pretty confident lies on a sphere. But the results do not check out; the test function is not constant. I would like to know the above-stated prescription for taking into account arbitrary speed is as simple as I have suggested.
The test function for unit-speed curves is $rho^2+(rho^primesigma)^2$, where $rho$ is the inverse of the curvature, $sigma$ is the inverse of the torsion, $N$ is the principal normal and $B$ is the binormal.
For the speed, I use $nu=||fracdalphadt||$. So for the modified test function I get $rho^2+(nurho^primesigma)^2$
My curve that I believe lies on a sphere is $alpha(t)=t/sqrtt^2+1, cos(t)/sqrtt^2+1, sin(t)/sqrtt^2+1$
differential-geometry
asked Jul 27 at 4:41
Gene Naden
203
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My textbook, ONeill's Elementary Differential Geometry, gives as an exercise the following test function to determine if a unit-speed curve lies on a sphere. If the test function is constant then the curve lies on a sphere. For a later exercise, we are invited to extend the test function so that it applies to arbitrary-speed curves. The prescription given is to multiply the derivative in the original formula by the speed of the curve. I tried that with a test curve that I am pretty confident lies on a sphere. But the results do not check out; the test function is not constant. I would like to know the above-stated prescription for taking into account arbitrary speed is as simple as I have suggested.
The test function for unit-speed curves is $rho^2+(rho^primesigma)^2$, where $rho$ is the inverse of the curvature, $sigma$ is the inverse of the torsion, $N$ is the principal normal and $B$ is the binormal.
For the speed, I use $nu=||fracdalphadt||$. So for the modified test function I get $rho^2+(nurho^primesigma)^2$
My curve that I believe lies on a sphere is $alpha(t)=t/sqrtt^2+1, cos(t)/sqrtt^2+1, sin(t)/sqrtt^2+1$
differential-geometry
asked Jul 27 at 4:41
Gene Naden
203
My textbook, ONeill's Elementary Differential Geometry, gives as an exercise the following test function to determine if a unit-speed curve lies on a sphere. If the test function is constant then the curve lies on a sphere. For a later exercise, we are invited to extend the test function so that it applies to arbitrary-speed curves. The prescription given is to multiply the derivative in the original formula by the speed of the curve. I tried that with a test curve that I am pretty confident lies on a sphere. But the results do not check out; the test function is not constant. I would like to know the above-stated prescription for taking into account arbitrary speed is as simple as I have suggested.
The test function for unit-speed curves is $rho^2+(rho^primesigma)^2$, where $rho$ is the inverse of the curvature, $sigma$ is the inverse of the torsion, $N$ is the principal normal and $B$ is the binormal.
For the speed, I use $nu=||fracdalphadt||$. So for the modified test function I get $rho^2+(nurho^primesigma)^2$
My curve that I believe lies on a sphere is $alpha(t)=t/sqrtt^2+1, cos(t)/sqrtt^2+1, sin(t)/sqrtt^2+1$
differential-geometry
asked Jul 27 at 4:41
Gene Naden
203
asked Jul 27 at 4:41
Gene Naden
203
asked Jul 27 at 4:41
Gene Naden
203
asked Jul 27 at 4:41
Gene Naden
203
203
add a comment |Â
add a comment |Â
3 Answers
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Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $fracdrhods=fracfracdrhodtfracdsdt$
answered Jul 27 at 6:43
Gene Naden
203
add a comment |Â
up vote
0
down vote
There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $tau$ and curvature $kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $|c-p|^2=r^2$, iff
$$
left(frac1kapparight)^2+ left(left(frac1kapparight)'right)^2cdotleft(frac1tauright)^2=r^2.$$
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
add a comment |Â
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Assuming it lies on a sphere, $||alpha-C||^2$ is constant. $rho=1/kappa$, $sigma=1/tau$.
$nu$ is the speed.
$(alpha-C)cdot T=0$
$alpha^primecdot T+(alpha-c)cdot T^prime=0$
$nu=||T||^2=-(alpha-C)cdot T^prime=-(alpha-c)cdotkappanu N$
$kappa(alpha-C)cdot N=-1$
So the component of $alpha-C$ along N is $-frac1kappa=-rho$
$fracddt||alpha-C||^2=2(alpha-C)cdot T=0$
$alpha$ is a linear combination of $B$ and $N$.
$kappa^prime(alpha-C)cdot N=-kappa(alpha-C)N^prime=-kappa(alpha-C)cdot(-kappa nu T+tau B)=-kappa(alpha-C)cdot tau nu B$
$(alpha-C)cdot B=-frackappa^prime(alpha-C)cdot Nkappanutau=frackappa^primekappa^2nutau$
$(alpha-C)cdot B=-frackappa^primekappa^2nutau=-fracrho^primenutau=-fracrho^primesigmanu$
This is the component of $alpha-C$ along $B$
$alpha-c=-rho N-fracrho^primesigmanu B$
So for the curve to lie on a sphere, $||alpha-c||^2=rho^2+(fracrho^primesigmanu)^2$ is constant and vice versa.
answered Jul 28 at 12:15
Gene Naden
203
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $fracdrhods=fracfracdrhodtfracdsdt$
answered Jul 27 at 6:43
Gene Naden
203
add a comment |Â
up vote
0
down vote
Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $fracdrhods=fracfracdrhodtfracdsdt$
answered Jul 27 at 6:43
Gene Naden
203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $fracdrhods=fracfracdrhodtfracdsdt$
answered Jul 27 at 6:43
Gene Naden
203
Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $fracdrhods=fracfracdrhodtfracdsdt$
answered Jul 27 at 6:43
Gene Naden
203
answered Jul 27 at 6:43
Gene Naden
203
answered Jul 27 at 6:43
Gene Naden
203
answered Jul 27 at 6:43
Gene Naden
203
203
add a comment |Â
add a comment |Â
up vote
0
down vote
There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $tau$ and curvature $kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $|c-p|^2=r^2$, iff
$$
left(frac1kapparight)^2+ left(left(frac1kapparight)'right)^2cdotleft(frac1tauright)^2=r^2.$$
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
add a comment |Â
up vote
0
down vote
There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $tau$ and curvature $kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $|c-p|^2=r^2$, iff
$$
left(frac1kapparight)^2+ left(left(frac1kapparight)'right)^2cdotleft(frac1tauright)^2=r^2.$$
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $tau$ and curvature $kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $|c-p|^2=r^2$, iff
$$
left(frac1kapparight)^2+ left(left(frac1kapparight)'right)^2cdotleft(frac1tauright)^2=r^2.$$
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $tau$ and curvature $kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $|c-p|^2=r^2$, iff
$$
left(frac1kapparight)^2+ left(left(frac1kapparight)'right)^2cdotleft(frac1tauright)^2=r^2.$$
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
answered Jul 27 at 10:16
Michael Hoppe
9,55631432
9,55631432
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
add a comment |Â
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
According to my calculations, this formula needs to be modified to take into account the speed. A curve lies on a sphere if $(frac1kappa)^2+((frac1kappa)^prime)^2(frac1nu)^2cdot(frac1tau)^2$ constant.
– Gene Naden
Jul 28 at 11:42
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
This formula does assume arclength parametrization (because of the differentiation).
– Ted Shifrin
Jul 29 at 18:05
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
No, it doesn’t. If $c$ is a Frenet-curve satisfying $|c-p|=r^2$, then $c=p_0-frac1kappaN-left(frac1kapparight)'cdotfrac1taucdot B,$ where $N$ and $B$ denote the normal and binormal vector, resp. From here the formula follows.
– Michael Hoppe
Jul 29 at 18:16
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
@MichaelHoppe: No, there's a factor of $|c'|$ in the denominator of the coefficient of $B$.
– Ted Shifrin
Jul 30 at 22:38
add a comment |Â
up vote
0
down vote
Assuming it lies on a sphere, $||alpha-C||^2$ is constant. $rho=1/kappa$, $sigma=1/tau$.
$nu$ is the speed.
$(alpha-C)cdot T=0$
$alpha^primecdot T+(alpha-c)cdot T^prime=0$
$nu=||T||^2=-(alpha-C)cdot T^prime=-(alpha-c)cdotkappanu N$
$kappa(alpha-C)cdot N=-1$
So the component of $alpha-C$ along N is $-frac1kappa=-rho$
$fracddt||alpha-C||^2=2(alpha-C)cdot T=0$
$alpha$ is a linear combination of $B$ and $N$.
$kappa^prime(alpha-C)cdot N=-kappa(alpha-C)N^prime=-kappa(alpha-C)cdot(-kappa nu T+tau B)=-kappa(alpha-C)cdot tau nu B$
$(alpha-C)cdot B=-frackappa^prime(alpha-C)cdot Nkappanutau=frackappa^primekappa^2nutau$
$(alpha-C)cdot B=-frackappa^primekappa^2nutau=-fracrho^primenutau=-fracrho^primesigmanu$
This is the component of $alpha-C$ along $B$
$alpha-c=-rho N-fracrho^primesigmanu B$
So for the curve to lie on a sphere, $||alpha-c||^2=rho^2+(fracrho^primesigmanu)^2$ is constant and vice versa.
answered Jul 28 at 12:15
Gene Naden
203
add a comment |Â
up vote
0
down vote
Assuming it lies on a sphere, $||alpha-C||^2$ is constant. $rho=1/kappa$, $sigma=1/tau$.
$nu$ is the speed.
$(alpha-C)cdot T=0$
$alpha^primecdot T+(alpha-c)cdot T^prime=0$
$nu=||T||^2=-(alpha-C)cdot T^prime=-(alpha-c)cdotkappanu N$
$kappa(alpha-C)cdot N=-1$
So the component of $alpha-C$ along N is $-frac1kappa=-rho$
$fracddt||alpha-C||^2=2(alpha-C)cdot T=0$
$alpha$ is a linear combination of $B$ and $N$.
$kappa^prime(alpha-C)cdot N=-kappa(alpha-C)N^prime=-kappa(alpha-C)cdot(-kappa nu T+tau B)=-kappa(alpha-C)cdot tau nu B$
$(alpha-C)cdot B=-frackappa^prime(alpha-C)cdot Nkappanutau=frackappa^primekappa^2nutau$
$(alpha-C)cdot B=-frackappa^primekappa^2nutau=-fracrho^primenutau=-fracrho^primesigmanu$
This is the component of $alpha-C$ along $B$
$alpha-c=-rho N-fracrho^primesigmanu B$
So for the curve to lie on a sphere, $||alpha-c||^2=rho^2+(fracrho^primesigmanu)^2$ is constant and vice versa.
answered Jul 28 at 12:15
Gene Naden
203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming it lies on a sphere, $||alpha-C||^2$ is constant. $rho=1/kappa$, $sigma=1/tau$.
$nu$ is the speed.
$(alpha-C)cdot T=0$
$alpha^primecdot T+(alpha-c)cdot T^prime=0$
$nu=||T||^2=-(alpha-C)cdot T^prime=-(alpha-c)cdotkappanu N$
$kappa(alpha-C)cdot N=-1$
So the component of $alpha-C$ along N is $-frac1kappa=-rho$
$fracddt||alpha-C||^2=2(alpha-C)cdot T=0$
$alpha$ is a linear combination of $B$ and $N$.
$kappa^prime(alpha-C)cdot N=-kappa(alpha-C)N^prime=-kappa(alpha-C)cdot(-kappa nu T+tau B)=-kappa(alpha-C)cdot tau nu B$
$(alpha-C)cdot B=-frackappa^prime(alpha-C)cdot Nkappanutau=frackappa^primekappa^2nutau$
$(alpha-C)cdot B=-frackappa^primekappa^2nutau=-fracrho^primenutau=-fracrho^primesigmanu$
This is the component of $alpha-C$ along $B$
$alpha-c=-rho N-fracrho^primesigmanu B$
So for the curve to lie on a sphere, $||alpha-c||^2=rho^2+(fracrho^primesigmanu)^2$ is constant and vice versa.
answered Jul 28 at 12:15
Gene Naden
203
Assuming it lies on a sphere, $||alpha-C||^2$ is constant. $rho=1/kappa$, $sigma=1/tau$.
$nu$ is the speed.
$(alpha-C)cdot T=0$
$alpha^primecdot T+(alpha-c)cdot T^prime=0$
$nu=||T||^2=-(alpha-C)cdot T^prime=-(alpha-c)cdotkappanu N$
$kappa(alpha-C)cdot N=-1$
So the component of $alpha-C$ along N is $-frac1kappa=-rho$
$fracddt||alpha-C||^2=2(alpha-C)cdot T=0$
$alpha$ is a linear combination of $B$ and $N$.
$kappa^prime(alpha-C)cdot N=-kappa(alpha-C)N^prime=-kappa(alpha-C)cdot(-kappa nu T+tau B)=-kappa(alpha-C)cdot tau nu B$
$(alpha-C)cdot B=-frackappa^prime(alpha-C)cdot Nkappanutau=frackappa^primekappa^2nutau$
$(alpha-C)cdot B=-frackappa^primekappa^2nutau=-fracrho^primenutau=-fracrho^primesigmanu$
This is the component of $alpha-C$ along $B$
$alpha-c=-rho N-fracrho^primesigmanu B$
So for the curve to lie on a sphere, $||alpha-c||^2=rho^2+(fracrho^primesigmanu)^2$ is constant and vice versa.
answered Jul 28 at 12:15
Gene Naden
203
edited Jul 28 at 19:58
answered Jul 28 at 12:15
Gene Naden
203
answered Jul 28 at 12:15
Gene Naden
203
answered Jul 28 at 12:15
Gene Naden
203
203
add a comment |Â
add a comment |Â
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