Wednesday, August 15, 2012
Bending Energy Continued - Finding K
In the previous post I showed how to find F'(s) of a a curve F(t) = <X(t),Y(t),Z(t)>
Now according to the definition of bending energy , it is the integrated sum of squares of curvature over length of the curve. That is we need to find the curvature K(s). It is always like this , Just getting better and better! .
Curvature is the rate of change magnitude of Unit tangent vector with respect to curve length
That is , it is
K(s) = || dF'(t) / ds ||
Lets find dF'(t) / ds
This is (dF'(t)/ dt) * (dt/ ds) using chain rule of diff.
= || dF'(t) / dt ||
|| ds/ dt || we know ds/dt = || F'(t) || from previous post. and dF'(t)/dt is F''(t) .
K(t) = || F''(t) || / || F'(t) ||
we calculated the value of K(t). For curves like circle , K(t) is same everywhere ,so just we can remove parameter t for circle. Also for circle there is exists an easier fomula : K = 1 / radius.