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We can go further. The next quantity of interest is how much
the plane of curvature "twists". This is measured
by the torsion This derivative can easily be calculated using
the fact that the direction B(t) is normal both to
the tangent vector T(t) and the vector in the plane
of curvature, N(t) that is normal to T(t).
We have just seen that N(t) is in the direction
of the projection of a(t) normal to T(t).
Since a and v are in the plane of curvature,
a Notice further that we can apply the product rule to the cross product to get and the latter term is 0. We also have: We can also use the fact that Putting all this together we get We find then that the torsion is the magnitude of this
and hence of the component of Exercises:15.3 As a test of your manipulative skill, see if you can wade through the steps above yourself and get the correct answer (which may be the one given above). 15.4 Set up a spreadsheet that for a curve you input in three dimensions using a parametric representation, computes the curvature and torsion at representative values of the parameter and the coordinates of the curve point at the same values. Where is the torsion the greatest in your range? |
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