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Lengths of curves
By Rolando Martinez on Monday, November 11, 2002 - 02:51 pm:

Hi. I need to find a point on a curve. Let's say it's a sine. I have a beginning point on this curve, point A, as well as a length, I called it L. I need to find the point (P) I would get after traveling the the length L along the curve, starting from A.

That is to say, if we take the length integral from A to point P then we get L.

Please help.

By Julian Pulman on Monday, November 11, 2002 - 03:40 pm:

The answer to this question depends really on the curve you want to measure along, for reasons I shall explain in due course it may be impossible by analytical means to derive an exact answer, bu t you can always find one using iterative methods.

Consider a small length of the curve, ds, then by pythagoras ds2 = dx2 + dy2
Rearrange this to give,
int
s will be the length along the curve, since we are summing each small length element w.r.t x.
We require that the we can find a closed form of the integral since we need a closed form function in terms of p so that we can solve this to find p.
If we call the integral f(x), then:
f(p) = s + f(a)
We already know s and f(a) by your starting conditions hence we only require that we can analytically find f-1(x), that is, the inverse function, such that p = f-1(s + f(a))
If f is something simple like x2, then f-1(y) = y1/2 and we can find p relatively easily.

In your example, we have f(x) = òa p (1 - cos2x)1/2dx = òa p sinxdx = -cosp + cosa
So, p = cos-1[cosa - s]

if f(x) is something like cosx + x then you'll find it very hard to solve this analytically - you can use the newton raphson method or another iteration method to find the intersection of the line y=s - cosa - a - x and y=cosx, the answer will be p... If you'd like me to explain how to do such iterative methods to find p then just say and I'll write up the general idea.

I apologise if I covered anything too fast, I'm in a bit of a rush at the moment.
If there's anything you want me to reexplain then don't hesitate to reply.


Julian

By Rolando Martinez on Monday, November 11, 2002 - 07:15 pm:

Thank you. I believe you have solved my problem.