Find wrapping angle of helix on a torus with constnt torsion

Find wrapping angle of helix on a torus with constnt torsion

I need some help in calculating the wrapping angle of a spiral helix wrapped on a torus with constant torsion.
Please correct me if I'm wrong, i don't know much about curvature but i see that constant torsion means the wrapping angle (or the angle measured around and/or against the torus circular cross-section) 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: Arc len=(2*pi*R2)/cos(wrapping_angle), from here i can find my wrapping angle, where R2 is torus minor radius. This is valid only for 1 turn spiral helix and ONLY if i have constant torsion (meaning the wrapping angle remains constant). The thing is i don't know the arc length of one turn, but i do know R1, R2 and the azimuth angle (or step-angle or the angle measured around torus central axis). I don't know how to integrate. Can you help?
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Re: Find wrapping angle of helix on a torus with constnt tor

(can't edit my previous post) 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][1]) 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][2]). 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?

[1]: https://i.stack.imgur.com/WVkOD.jpg
[2]: https://i.stack.imgur.com/pRzID.jpg

Niculae George

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