How can I integrate the derivative of my tangent vector in a parametric curve?

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I have the symbolic representation of a parametric curve, of its first and second derivatives and of the Frenet-Serret vectors.

However, what I would like to make is to integrate a normal vector following the curve without any torsion as given by Frenet normal and binormal vectors when curvature is reversed (see picture).enter image description here

To make it clearer : I'm drawing this "pipe" by connecting ellipses aligned on the frenet vectors. However, when curvature reverses, it makes an unnecessary torsade in my "pipe" because the normal and binormal vectors, both turn 180°.

So, I'd like to build 2 other vectors (instead of normal and binormal), which would keep the same orientation.

My tangent vector is the derivative of my position (x(t), y(t), z(t)) => (d(x(t))/dt, d(y(t))/dt, d(z(t))/dt). I have calculated the second derivative (d²(x(t))/dt², d²(y(t))/dt², d²(z(t))/dt²), so I would say that my tangent vector is some integral of the second derivative, and in the same logic, I inferred that I should be able to find another form of the integral my second derivative, which should be able to integrate that function to see what becomes a normal vector at the point t. But at this point, I feel blocked and I can't see how to continue the reasoning.

reply to @Tedshifrin : Those are (approximated) ellipses : Here is what you see with (approximated) circles enter image description here

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Some people suggested in the comments that I might need the Bishop's frame, but since my differential geometry knowledge is very basic, I failed to understand how to build it from the papers I found/received.

However, I managed to build a satisfying frame which solved the torsion problem observed with Frenet-Serret frame. It might not be perfect, but it is much better. The method to build it consists in

  • building a first orthogonal vector as the cross product of the tangent vector with a unitary vertical vector (It's OK, because my function never get purely vertical): m1 = T x (0, 0, 1)
  • norming it: M1 = m1/|m1|
  • building the second vector as the cross product of the tangent vector and M1 : M2 = T x M1

Here is the result : enter image description here enter image description here