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.

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 7, 2018 at 22:58

1 Answer 1


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

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

enter image description here

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 $\kappa_{g}$. $$ R_g= u/\cos \phi = u \sec \phi $$. enter image description here

  • $\begingroup$ I'm not sure I understand this... $\endgroup$
    – Sasha
    Nov 7, 2018 at 20:53

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