# Finding the Geodesic Curvatures of a Torus

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.

• You mean to find the geodesic curvature of the curves $\theta=\text{constant}$ and $\phi=\text{constant}$? The curves $\phi=\text{constant}$ are geodesics, so have geodesic curvature. With regard to your other question, look at definitions. The geodesic curvature comes from the part of the vector $\kappa n$ ($n$ principal normal) lying in the tangent plane of the surface. You need to draw some pictures and use some basic right-angle trigonometry. – Ted Shifrin Apr 24 '17 at 21:36
• yes, I have edited it accordingly – Felicio Grande Apr 24 '17 at 21:42
• I'm confused because along $\theta=\textbf{constant}$, the Normal vector of this curve is always perpendicular to the Tangent Plane on this curve no? – Felicio Grande Apr 24 '17 at 21:43
• No, that's for $\phi=\text{constant}$ (the little circles), isn't it? – Ted Shifrin Apr 24 '17 at 21:49
• Tangent plane of the surface at the point $P$ where you're computing geodesic curvature of the relevant curve!!! – Ted Shifrin Apr 24 '17 at 21:57

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.
• How do you know that it's $\sin(c)$ times the magnitude of the normal vector? I'm confused by how to go about this question – Felicio Grande Apr 26 '17 at 0:35
• The parameter $\theta$ is "latitude": $\theta = 0$ on the "outer equator", $\theta = \pi$ on the "inner equator", $\theta = \pi/2$ on the top circle, and so forth. In the plane, draw the unit circle with $\theta$ as polar coordinate (thinking of this circle as the translated meridian $\varphi = 0$), and consider the (green) vector $(-1, 0)$ and its (blue) projection to the tangent line to the circle. The length of the blue projection is $|\sin\theta|$ times the magnitude of the green vector by elementary trigonometry. :) – Andrew D. Hwang Apr 26 '17 at 10:44