# Geodesic Curvature of a circle from a cone

The circular cone with a vertex angle of $$2\phi$$ that is parametrized by $$x(u,v) = (u\tan\phi \cos v, u\tan \phi \sin v, u)$$ for $$0 \leq u \leq u_0$$ and $$0 \leq v \leq 2\pi$$. I want to find the Geodesic curvature of the circle $$u = u_0$$. Now I know that the equation for this is $$k_g = kN(n \times T)$$. But I am kind of confused, not sure how to make this equations of the circle given this cone. Also in this equation for geodesic curvature isn't $$n$$ the normal to the surface but in our case we are using a curve? How can I proceed on this question.

• But it's a curve lying in a surface. Geodesic/normal curvature only make sense in that situation. You can certainly compute $n$ from the parametrization for the surface. You can also do it by decomposing $kN$ geometrically into a component normal to the surface and a component tangent to the surface. Nov 7, 2018 at 22:58

Set up radial,normal and tangential components $$(B,T,N)$$ at first for bottom circle. The derivation is per standard definition. The following link can be helpful to adopt (as an exercise) a straight meridian in place of a circular meridian:

Geodesic Curvature Sphere Parallels

In general for cone

$$\frac{u}{R_g} =u \kappa_g =\cos \phi$$

The particular case for $$\phi= \pi/4$$

is about the simplest case of $$\kappa_{g}$$ in 3D. $$\phi= \pi/4$$ since $$z=u$$ so $$\kappa_g=\dfrac{u}{\sin \phi}=\dfrac{u}{\cos \phi}.$$ On cone development (center of circle is cone vertex) we note reciprocal of the cone bottom circle radius as $$\kappa_{g}$$. $$R_g= u/\cos \phi = u \sec \phi$$.

• I'm not sure I understand this... Nov 7, 2018 at 20:53