0
$\begingroup$

I need some help in calculating the wrapping angle of a spiral helix wrapped on a torus with constant angle against all the meridians of the torus. The wrapping angle (or the angle measured around and/or against the torus circular cross-section See here a (13,6)tori knot) always remains constant as the helix curve spirals around the torus. In layman's terms means it spirals with the same angle always. if I were to know the arc length of one such turn, my solution will be trivial: $$Arclen = {2\pi R_2\over cos(wrapping\_angle)} .(1)$$ I know this formula is counterintuitive to be true and valid regardless of R1 (torus major radius), but it is. I have an algorithm to test it out and it is true beyond any shadow of a doubt. It took me a while to figure it out, but indeed it does seem to suggest that the helix is on a cylinder, but it's on a curved cylinder (a torus). It doesn't make sense, but the formula is correct. I have a way to test it out. I'm trying to see that the tangent of any point on the curve is constant (derivative=0?? i don't know) so it projects(unwinds) as a straight line in a 2D plane. All i know is that equation 1 is correct. From here i can find my wrapping angle, where R2 is torus minor radius. This formula is valid only for 1 turn spiral helix and ONLY if I have constant wrapping angle. The thing is I don't know the arc length of one turn, but I do know R1, R2 and the azimuth angle of 1 turn (or 1 turn step-angle or the angle measured around torus central axis see it here). I want to find the arc length of 1 turn of the helix as a function of step-angle and using such formula in combination with the one I already have, I can find my wrapping angle. I don't know how to integrate but I see something in my mind along these lines: $$Arclen={ \int_0^{stp} \pi {{ \int_0^{2\pi}(R_1-R_2cos(\theta) )d\theta }\over 180} }\beta d\beta =something .(2)$$ I don't know what this is. I believe it's some sort of a double integration in polar coordinates(?) of infinitesimal circle-arcs which i know nothing about. stp is my step_angle, $\theta$ is the infinitesimal angle made by R2 in respect to equatorial plane of torus, and $\beta$ is supposed to be the step_angle chopped down into infinitesimal bits. I know equation (2) is wrong, but let's say it's not, then i write: $$equation(1)=equation(2) => wrapping\_angle=arccos({{2\pi R_2} \over something})$$ Since equation(2) is wrong, I also see in my mind something like this: $$Arclen=\int_0^{stp} (\sqrt{(\pi [\int_0^{2\pi}(R_1-R_2cos(\theta)) d\theta]*{\beta \over 180})^2+(\int_0^{2\pi}{{\pi R_2\alpha} \over180}d\alpha)^2})d\beta . (3)$$ I don't know how to integrate. Please don't ask me where i got these equations. Maybe they can provide a place to start for you to help me. I just need a little bit of help to correct this mess. Can you help?

$\endgroup$
  • 1
    $\begingroup$ This is a nice question, but consider using MathJax to make it more easily readable :) $\endgroup$ – YiFan Feb 15 at 20:15
  • 1
    $\begingroup$ Although it's immediately obvious to me that your constant wrapping angle interpretation of constant torsion would be correct on a (flat) cylinder, it's far from obvious to me on a torus sitting in $\Bbb R^3$ (or are you thinking of a flat torus in $\Bbb R^4$?), since its curvature varies as you go around the cross-sectional circles. I guess I'd like to start by asking you what you mean by a helix on the torus (unless it's in the $\Bbb R^4$ interpretation). $\endgroup$ – Ted Shifrin Feb 15 at 21:36
  • 1
    $\begingroup$ I have no idea what you just said to me. I know equation 1 is counterintuitive to be true and valid regardless of R1, but it is. I have an algorithm to test it out and it is true beyond any shadow of a doubt. It took me a while to figure it out, but indeed it does seem to suggest that the helix is on a cylinder, but it's on a curved cylinder (a torus). It doesn't make sense, but the formula is correct. I'm trying to see that the tangent of any point on the curve is constant (derivative=0?? i don't know) so it projects as a straight line in a 2D plane. All i know is that equation 1 is correct. $\endgroup$ – Niculae George Razvan Feb 15 at 21:43
  • 1
    $\begingroup$ I'm trying to work out the parametric equation of a curve that makes constant angle with the (small) circles on the torus. That's what you're calling a wrapping angle. Helices on cylinders are totally different, as cylinders are the same everywhere around; this is far from true on a torus. On the torus, the curve must twist around the torus faster on the inside of the torus than it does on the outside of the torus. I'll write down an explicit parametric equation in a minute. Then computer algebra can compute the torsion ... $\endgroup$ – Ted Shifrin Feb 15 at 22:05
  • 1
    $\begingroup$ "On the torus, the curve must twist around the torus faster on the inside of the torus than it does on the outside of the torus". Not always true. There are exceptions. See the pictures attached. The curve in the pictures (torus knot with constant torsion) can clearly see it doesn't twist faster on the inside of the torus than on the outside. The twist (torsion) remains constant because the wrapping angle is constant. $\endgroup$ – Niculae George Razvan Feb 15 at 22:22
2
$\begingroup$

The curves you have described do not have constant torsion (in the usual technical sense of differential geometry). The condition that the curve make a constant angle with the meridian circles on the torus leads to a very complicated parametric equation. For any value of the constant $c$, here is such a curve on the torus obtained by rotating a circle of radius $1$ around a circle of radius $2$:

$$\alpha(t) = \big((2+\cos t)\cos\theta(t),(2+\cos t)\sin\theta(t),\sin t\big),\quad \text{with }\theta(t)= c\arctan\Big(\dfrac{\tan(t/2)}{\sqrt3}\Big).$$

You could replace $2+\cos t$ with $2+\cos at$ for some constant $a$. I haven't had the fortitude to work everything out in that generality.

Now, for the constant value $c$ appearing in the formula above, we find that the speed of $\alpha(t)$ is $\sqrt{1+c^2}$, and the arclength of one trip around the torus will be $$L=\int_0^T \|\alpha'(t)\|dt = T\sqrt{1+c^2},$$ with $T$ given by $\int_0^T \theta'(t)\,dt = 2\pi$. So this gives us $$2\pi = \theta(T) = c\arctan\Big(\dfrac{\tan(T/2)}{\sqrt3}\Big), \quad\text{so ... drumroll ...} \\ T = 2\arctan\Big(\sqrt3\tan\big(\frac{2\pi}c\big)\Big).$$

Finally, the arclength is $$L=(\sqrt{1+c^2})T = 2\sqrt{1+c^2}\arctan\Big(\sqrt3\tan\big(\frac{2\pi}c\big)\Big).$$ I should add that your wrapping angle is given by $$\cos(\text{wrapping angle}) = \frac1{\sqrt{1+c^2}}.$$

$\endgroup$
  • $\begingroup$ "The condition that the curve make a constant angle with [ALL] the meridian circles on the torus..." FINALLY someone understands me and what I'm after. "[such a condition]... leads to a very complicated parametric equation" Yeah...I know. Obviously ...that's why I'm here. $\endgroup$ – Niculae George Razvan Feb 16 at 0:00
  • $\begingroup$ I'm scratching my head like crazy here. You don't know who you're talking to. I'm as dumb as a rock and have a complete nonsensical terminology. That's why nobody seems to understand me. Anyway I'm trying to decipher your equations, and I just can't visualize them. What exactly is c? where did that $\theta (t)$ came from? that sqrt(3) smells strange to me. Where did that came from? What is "speed of α(t)" ? Never heard of such a concept. Is someone chasing my curve and it's on a run? I'm confused and a complete idiot. How do I visualize "speed of α(t)"? $\endgroup$ – Niculae George Razvan Feb 16 at 0:01
  • $\begingroup$ What is this new variable T that came out of nowhere? Is that my step_angle? (see pictures in the original post) I need to picture all these variables in my head and what they represent in 3D space. That's when I can say I finally understand your reply. $\endgroup$ – Niculae George Razvan Feb 16 at 0:01
  • $\begingroup$ You've lost me at defining the parametric equation: α(t)= ( x=(R1+R2*cos(t))cos(u) y=(R1+R2*cos(t))sin(u) z=R2*sin(t) ) where u=c*atan( tan(t/2) / sqrt(3) ) R1 is torus major radius R2 is torus minor radius t - the incremental unit angle forming the curve. It's values are from 0 to k*2*pi ; where k is number of full 360degrees trips around the torus main axis my curve does. (See what I'm doing? I'm explaining all my variables and what they represent in 3D space such that any simpleton like me understands the math behind it) $\endgroup$ – Niculae George Razvan Feb 16 at 0:02
  • $\begingroup$ c - an unknown mysterious constant (unable to visualize in my mind in 3D space what this c is and does) All these I believe are the parametric equations for the Cartesian coordinates(x,y,z) of any given point on my curve in 3D space. I'm just extremely confused why you haven't wrote them like I did from the very beginning? I'm as dumb as a rock. Why confuse me even further? $\endgroup$ – Niculae George Razvan Feb 16 at 0:03
0
$\begingroup$

Ted Shifrin answer to my original question is incomplete and I can't use it the way it is. The almost-complete version is like this:

$$\alpha(t)=\begin{pmatrix} x=\Big(R_1+R_2\cos(t)\Big)\cos\biggr(c\arctan\Big(\dfrac{(R_1-R_2)\tan(t/2)}{\sqrt{R_1^2-R_2^2}}\Big)\biggr) \\y=\Big(R_1+R_2\cos(t)\Big)\cos\biggr(c\arctan\Big(\dfrac{(R_1-R_2)\tan(t/2)}{\sqrt{R_1^2-R_2^2}}\Big)\biggr) \\z=R_2\sin(t) \end{pmatrix}$$

I have injuries for scratching my head trying to more or less do semi-educated guess what extra terms goes where by trail and error (https://en.wikipedia.org/wiki/Trial_and_error) That's because I'm an engineer not a mathematician which should make sense to as why I did it like this.

Using "consecrated" math terminology from here: https://en.wikipedia.org/wiki/Toroidal_and_poloidal (I hope it's ok to use the word "consecrated" in the meaning to "to devote to a purpose with deep dedication" (according to Merriam-Webster dictionary) english is not my native language so I'm asking for leniency)

Let there be the term : "toroidal direction step angle" or the angle done in toroidal direction by only 1 poloidal turn of my curve (a STEP in the right way - literally).

Therefore I can't show mathematically, but have proven undoubtedly in numerous tests and experiments that: $$c=\frac{\text{step angle°}}{180} $$

$$\text{1 poloidal turn wire length}=\frac{2\pi R_2}{\cos(\text{wrapping angle})} $$

$$\text{step angle°}=\frac{360R_2\tan(\text{wrapping angle})}{\sqrt{R_1^2-R_2^2}}$$

$$\text{wrapping angle°}=\arctan\biggr(\frac{\text{step angle}\sqrt{R_1^2-R_2^2}}{360R_2}\biggr)$$ Again, emphasis on how these equations were determined: semi-educated guesses, trial and error a dose of madness and insanity and lots of dedication and perseverance. These formulas never failed me so far for any value of $R_1$ , $R_2$, c, step angle or wrapping angle.

Of course for the answer to be complete, I would have to replace $R_1+R_2\cos(t)$ with $R_1+R_2\cos(at)$ for some constant "a" representing how many turns in the poloidal direction my curve does in the same step angle, the final goal being to generate a special family of closed curves tori knots (with curve tangent(aka velocity vector) always constant against all meridians and parallels of the torus)

I was asked to provide more context to my question: https://www.researchgate.net/publication/311742883_Visualization_in_3D_of_Dynamics_of_Toroidal_Helical_Coils_in_quest_of_optimum_designs_for_a_Concordian_Mandala

I would like to add some sentences (which most likely will be deleted by someone):

  1. Nothing less than a miracle or a fluke for a professor like Ted Shifrin answering my question.

  2. My question was written with sloppy terminology.

  3. It would have been amazing to have seen the math he did behind the scenes for the parametrization of such curves.

  4. Can I hope for yet another miracle?

Why is speaking like this forbidden on this forum? Reading between the lines 1st sentence shows gratitude towards altruism, respect and awe. 2nd- introspection, constructive self-criticism and honesty. 3rd mild regret, call for compassion and desire. and 4th expectation and hope Why speaking (in and with) these terms illegal on this forum? Aren't these aspects extremely important in any form of communication? As a newbie on this forum I'm very confused... Yes... stick to the math and disregard and delete everything else...monstrously strict and rigid.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.