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

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

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.


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:

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


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

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 $$ \vec r(t) = (5+2\cos (10(t+u)))\cos t \mathbf i + (5+2\cos (10(t+u)))\sin t \mathbf j + 2\sin (10(t+u)) \mathbf k $$

Poloidal Patch: 0 < u < 0.1 $ 0 \le u \le 0.1 $

Poloidal Patch: 0 < u < 0.3 $ 0 \le u \le 0.3 $

Poloidal Patch: 0 < u < 0.5 $ 0 \le u \le 0.5 $

Second Equation (Toroidal Direction):

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

Toroidal Patch: 0 < u < 0.1 $ 0 \le u \le 0.1 $

Toroidal Patch: 0 < u < 0.3 $ 0 \le u \le 0.3 $

Toroidal Patch: 0 < u < 0.5 $ 0 \le u \le 0.5 $


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