I have a question that I'm working on, and I have solved the question on my own terms, however I don't understand the solution that was provided, and would like to understand it. The question reads:
Let $T$ be a Torus parametrized by:
$$\phi(\theta, \varphi) = ((R+r\cos(\theta))\cos(\varphi), (R+r\cos(\theta))\sin(\varphi), r\sin(\theta))$$
Find the Geodesic Curvature when $\theta=\text{constant}$ and $\varphi=\text{constant}$.
Now, the solution reads:
$$\phi_{\theta} = (-r\sin(\theta)\sin(\varphi), -r\sin(\theta)\sin(\varphi), r\cos(\theta))$$
$$\phi_{\varphi} = (-(R+r\cos(\theta))\sin(\varphi), (R+r\cos(\theta))\cos(\varphi), 0)$$
$$\vec N =(-\cos(\theta)\cos(\varphi), \cos(\theta)\sin(\varphi), \sin(\theta))$$
We have circles that have Radius $(R+r\cos(c))$ when $\theta=c\in\mathbb{R}$. Then, the Curves when $\theta=c$ have curvature $\kappa_C = (R+r\cos(c))^{-1}$.
Projecting the Normal Vector of the Curve onto the Tangent Plane gives that the Geodesic Curvature is given by:
$$\kappa_G=\sin(c)\cdot(R+r\cos(c))^{-1}$$
I am wondering, what Tangent Plane is he talking about? Furthermore, how did he just know that projecting the Normal Vector of the Curve onto the Tangent Plane would be $\sin(c)$ times the Curve Curvature? If anyone had any tips to add to this I would really appreciate it. Thank you.
Best Answer
At a point $p$ of your torus, the normal field to a curve in the sense of the Frenet frame (green, scaled for visibility) need not be perpendicular to the tangent plane of the torus, and so can have a non-zero tangential component (blue), whose magnitude at $\phi(c, \varphi)$ is $\sin c$ times the magnitude of the normal vector.