Parametric Surface Equations for Orthogonal Projection of Torus Knot Tube onto Torus

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What are the parametric equations for the orthogonal projection of the torus knot tube onto the torus surface?

For instance, if we have the equations for the torus knot

$$
vec r(t)= (R+rcos pt)cos qt mathbf i+(R+rcos pt)sin qt mathbf j+(r sin pt) mathbf k
$$

and the tube has diameter $d ll r$ and $d ll R$, can we project the wire diameter orthogonally onto the torus surface and express this as a parametric surface? (by orthogonally, I mean cutting the tube normal to the direction of the knot and projecting the line segment that is the tube diameter in that cross section onto the surface of the torus in the direction normal to the tube diameter line segement).

Would this be a portion of the torus surface spanned between two parallel torus knots?

I've started by looking at the Darboux frame for the torus knot and the answer here Building a Tube Around the Torus Knot, but am not sure how to proceed further. I tried to ask this question a few days ago here Toroidal Knot Patch/Ribbon, but didn't get much of a response so I'm trying to better clarify my question here.

Forgive my imprecise use of mathematics terminology. Any help would be much appreciated. Thank you.

surfaces parametric knot-theory

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asked Jul 28 at 0:24

A. Hendry

165

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up vote
2
down vote

favorite

What are the parametric equations for the orthogonal projection of the torus knot tube onto the torus surface?

For instance, if we have the equations for the torus knot

$$
vec r(t)= (R+rcos pt)cos qt mathbf i+(R+rcos pt)sin qt mathbf j+(r sin pt) mathbf k
$$

and the tube has diameter $d ll r$ and $d ll R$, can we project the wire diameter orthogonally onto the torus surface and express this as a parametric surface? (by orthogonally, I mean cutting the tube normal to the direction of the knot and projecting the line segment that is the tube diameter in that cross section onto the surface of the torus in the direction normal to the tube diameter line segement).

Would this be a portion of the torus surface spanned between two parallel torus knots?

I've started by looking at the Darboux frame for the torus knot and the answer here Building a Tube Around the Torus Knot, but am not sure how to proceed further. I tried to ask this question a few days ago here Toroidal Knot Patch/Ribbon, but didn't get much of a response so I'm trying to better clarify my question here.

Forgive my imprecise use of mathematics terminology. Any help would be much appreciated. Thank you.

surfaces parametric knot-theory

share|cite|improve this question

asked Jul 28 at 0:24

A. Hendry

165

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

What are the parametric equations for the orthogonal projection of the torus knot tube onto the torus surface?

For instance, if we have the equations for the torus knot

$$
vec r(t)= (R+rcos pt)cos qt mathbf i+(R+rcos pt)sin qt mathbf j+(r sin pt) mathbf k
$$

and the tube has diameter $d ll r$ and $d ll R$, can we project the wire diameter orthogonally onto the torus surface and express this as a parametric surface? (by orthogonally, I mean cutting the tube normal to the direction of the knot and projecting the line segment that is the tube diameter in that cross section onto the surface of the torus in the direction normal to the tube diameter line segement).

Would this be a portion of the torus surface spanned between two parallel torus knots?

I've started by looking at the Darboux frame for the torus knot and the answer here Building a Tube Around the Torus Knot, but am not sure how to proceed further. I tried to ask this question a few days ago here Toroidal Knot Patch/Ribbon, but didn't get much of a response so I'm trying to better clarify my question here.

Forgive my imprecise use of mathematics terminology. Any help would be much appreciated. Thank you.

surfaces parametric knot-theory

share|cite|improve this question

asked Jul 28 at 0:24

A. Hendry

165

What are the parametric equations for the orthogonal projection of the torus knot tube onto the torus surface?

For instance, if we have the equations for the torus knot

$$
vec r(t)= (R+rcos pt)cos qt mathbf i+(R+rcos pt)sin qt mathbf j+(r sin pt) mathbf k
$$

and the tube has diameter $d ll r$ and $d ll R$, can we project the wire diameter orthogonally onto the torus surface and express this as a parametric surface? (by orthogonally, I mean cutting the tube normal to the direction of the knot and projecting the line segment that is the tube diameter in that cross section onto the surface of the torus in the direction normal to the tube diameter line segement).

Would this be a portion of the torus surface spanned between two parallel torus knots?

I've started by looking at the Darboux frame for the torus knot and the answer here Building a Tube Around the Torus Knot, but am not sure how to proceed further. I tried to ask this question a few days ago here Toroidal Knot Patch/Ribbon, but didn't get much of a response so I'm trying to better clarify my question here.

Forgive my imprecise use of mathematics terminology. Any help would be much appreciated. Thank you.

surfaces parametric knot-theory

share|cite|improve this question

asked Jul 28 at 0:24

A. Hendry

165

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 0:24

A. Hendry

165

asked Jul 28 at 0:24

A. Hendry

165

asked Jul 28 at 0:24

A. Hendry

165

165

add a comment | 

add a comment | 

1 Answer
1

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up vote
0
down vote



accepted

Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.

A "toroidal knot patch" can be made simply by using one of the following equations:

$$
vec r_1 (t, u) = (R+rcos (p(t+u)))cos qt mathbf i + (R+rcos(p(t+u)))sin qt mathbf j + rsin (p(t+u)) mathbf k
$$

or

$$
vec r_2 (t, u) = (R+rcos (pt))cos(t+u) mathbf i + (R+rcos(pt))sin(t+u)mathbf j + rsin(pt)mathbf k
$$

As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.

To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.

This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve

$$
((vec r(t,u)-vec r(t,0))-((vec r(t,u)-vec r(t,0))cdot hat N(t))hat N(t)) cdot hat B(t) = d/2
$$

for $u$. In the above, $vec r$ can be either $vec r_1$ or $vec r_2$, and $vec r(t,u)-vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $hat N(t)$ and $hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.

Below are graphs plotting using WinPlot

First Equation (Poloidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10(t+u)))cos t mathbf i + (5+2cos (10(t+u)))sin t mathbf j + 2sin (10(t+u)) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

Second Equation (Toroidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10t))cos (t+u) mathbf i + (5+2cos (10t))sin (t+u) mathbf j + 2sin (10t) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

share|cite|improve this answer

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

add a comment | 

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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

Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.

A "toroidal knot patch" can be made simply by using one of the following equations:

$$
vec r_1 (t, u) = (R+rcos (p(t+u)))cos qt mathbf i + (R+rcos(p(t+u)))sin qt mathbf j + rsin (p(t+u)) mathbf k
$$

or

$$
vec r_2 (t, u) = (R+rcos (pt))cos(t+u) mathbf i + (R+rcos(pt))sin(t+u)mathbf j + rsin(pt)mathbf k
$$

As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.

To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.

This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve

$$
((vec r(t,u)-vec r(t,0))-((vec r(t,u)-vec r(t,0))cdot hat N(t))hat N(t)) cdot hat B(t) = d/2
$$

for $u$. In the above, $vec r$ can be either $vec r_1$ or $vec r_2$, and $vec r(t,u)-vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $hat N(t)$ and $hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.

Below are graphs plotting using WinPlot

First Equation (Poloidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10(t+u)))cos t mathbf i + (5+2cos (10(t+u)))sin t mathbf j + 2sin (10(t+u)) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

Second Equation (Toroidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10t))cos (t+u) mathbf i + (5+2cos (10t))sin (t+u) mathbf j + 2sin (10t) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

share|cite|improve this answer

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

add a comment | 

up vote
0
down vote



accepted

Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.

A "toroidal knot patch" can be made simply by using one of the following equations:

$$
vec r_1 (t, u) = (R+rcos (p(t+u)))cos qt mathbf i + (R+rcos(p(t+u)))sin qt mathbf j + rsin (p(t+u)) mathbf k
$$

or

$$
vec r_2 (t, u) = (R+rcos (pt))cos(t+u) mathbf i + (R+rcos(pt))sin(t+u)mathbf j + rsin(pt)mathbf k
$$

As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.

To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.

This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve

$$
((vec r(t,u)-vec r(t,0))-((vec r(t,u)-vec r(t,0))cdot hat N(t))hat N(t)) cdot hat B(t) = d/2
$$

for $u$. In the above, $vec r$ can be either $vec r_1$ or $vec r_2$, and $vec r(t,u)-vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $hat N(t)$ and $hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.

Below are graphs plotting using WinPlot

First Equation (Poloidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10(t+u)))cos t mathbf i + (5+2cos (10(t+u)))sin t mathbf j + 2sin (10(t+u)) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

Second Equation (Toroidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10t))cos (t+u) mathbf i + (5+2cos (10t))sin (t+u) mathbf j + 2sin (10t) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

share|cite|improve this answer

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.

A "toroidal knot patch" can be made simply by using one of the following equations:

$$
vec r_1 (t, u) = (R+rcos (p(t+u)))cos qt mathbf i + (R+rcos(p(t+u)))sin qt mathbf j + rsin (p(t+u)) mathbf k
$$

or

$$
vec r_2 (t, u) = (R+rcos (pt))cos(t+u) mathbf i + (R+rcos(pt))sin(t+u)mathbf j + rsin(pt)mathbf k
$$

As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.

To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.

This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve

$$
((vec r(t,u)-vec r(t,0))-((vec r(t,u)-vec r(t,0))cdot hat N(t))hat N(t)) cdot hat B(t) = d/2
$$

for $u$. In the above, $vec r$ can be either $vec r_1$ or $vec r_2$, and $vec r(t,u)-vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $hat N(t)$ and $hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.

Below are graphs plotting using WinPlot

First Equation (Poloidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10(t+u)))cos t mathbf i + (5+2cos (10(t+u)))sin t mathbf j + 2sin (10(t+u)) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

Second Equation (Toroidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10t))cos (t+u) mathbf i + (5+2cos (10t))sin (t+u) mathbf j + 2sin (10t) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

share|cite|improve this answer

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.

A "toroidal knot patch" can be made simply by using one of the following equations:

$$
vec r_1 (t, u) = (R+rcos (p(t+u)))cos qt mathbf i + (R+rcos(p(t+u)))sin qt mathbf j + rsin (p(t+u)) mathbf k
$$

or

$$
vec r_2 (t, u) = (R+rcos (pt))cos(t+u) mathbf i + (R+rcos(pt))sin(t+u)mathbf j + rsin(pt)mathbf k
$$

As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.

To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.

This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve

$$
((vec r(t,u)-vec r(t,0))-((vec r(t,u)-vec r(t,0))cdot hat N(t))hat N(t)) cdot hat B(t) = d/2
$$

for $u$. In the above, $vec r$ can be either $vec r_1$ or $vec r_2$, and $vec r(t,u)-vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $hat N(t)$ and $hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.

Below are graphs plotting using WinPlot

First Equation (Poloidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10(t+u)))cos t mathbf i + (5+2cos (10(t+u)))sin t mathbf j + 2sin (10(t+u)) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

Second Equation (Toroidal Direction):

$$
p = 10, R = 5, r = 2, q = 1
$$
such that
$$
vec r(t) = (5+2cos (10t))cos (t+u) mathbf i + (5+2cos (10t))sin (t+u) mathbf j + 2sin (10t) mathbf k
$$

$ 0 le u le 0.1 $

$ 0 le u le 0.3 $

$ 0 le u le 0.5 $

share|cite|improve this answer

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 15:58

edited Jul 30 at 15:58

edited Jul 30 at 15:58

answered Jul 28 at 22:37

A. Hendry

165

answered Jul 28 at 22:37

A. Hendry

165

answered Jul 28 at 22:37

A. Hendry

165

165

add a comment | 

add a comment | 

 

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