2
$\begingroup$

Where does the equation $\ddot\gamma=k_n N+k_g N \times\dot \gamma$ come from?

What is the intuition behind it?

I know $k_n$ is the normal curvature and $k_g$ is the geodesic curvature.

Futhermore I know $N$, $\dot\gamma$ and $N \times \dot\gamma$ are all mutually perpendicular to eachother.

$\endgroup$

1 Answer 1

2
$\begingroup$

Since the three vectors are mutually perpendicular, $\ddot \gamma$ is a linear combination of them.

But $\ddot\gamma$ is also perpendicular to $\dot \gamma$ (when $\gamma$ unit speed, differentiate $\dot \gamma \cdot \dot\gamma = 1$ to see that $\ddot\gamma \cdot \dot\gamma$ vanish).

So the coefficient of $\ddot \gamma$ along $\dot \gamma$ must be zero, and $\ddot \gamma = \kappa_n N + \kappa_g N \times \dot\gamma$ for some $\kappa_n, \kappa_g$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .