Wednesday, August 15, 2012
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.
Sunday, August 12, 2012
What is bending Energy ? The precise definition is "it is the sum of squares of curvature of the curve function parameterized over curve length". Bending energy gives the energy stored in the curve. We know any bented objects will store some energy. Bending energy formulation helps to find a value proportional to the energy stored in the curve. For a straight line bending energy is Zero.
Before look into it we need to understand how we can parametrize curve over length
The tricky part is how we can parameterize over curve length.
Consider a vector valued function F(t) = < X(t),Y(t),Z(t) > , t is the parameter , ranges between some values.
F'(t) can be found easily by applying partial differentiation on F with respect to 't' .
Lets now find the length (length function) of this curve
it has been shown by Kennedy, John (2011) in his paper, how to derive expression for F'(s)
Differentiating this with respect to s , we will get
1 = || F'(s) || (of-course s is based on t)
That means after changing the parameter from t to s(length param) the length of F'(s) is getting 1. It is very interesting concept, that means on moving through function F(S) we are moving exactly by unit length. If you imagine this it seems true. Because no matter where the curve is going its length get incremented in equal length.
So how we can find the equation for F'(s) ? Intuitively we can think like this. Anyway F(s) and F(t) represents same curve , only different is magnitude of F'(s) is 1 , But F'(t) may not be 1. But both these vectors points to the same direction. right ? so we(I)can conclude like this
F'(s) = F'(t) / || F'(t) ||
Other-way is like this.
Now Differentiating both sides with respect to t.
ds/dt = || F'(t) ||
If we differentiate function F(t) with respect to s (length) we get
= F'(t) * dt/ds (chain rule of differentiation)
= F'(t) / (ds/dt)
= F'(t) / || F'(t) || ( we know ds/dt = || F'(t) || )
that is dF(t)/ds = F'(t) / || F'(t) which is eqult to F'(s). This is fantastic. :)
Now Next step is finding F''(s) , I will explain that in next post. It is time to sleep.
Nowadays days I am getting more into mathematics than software engineering. To really understand things you need to have great patience and curiosity. After all these years , I am still a novice.