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Topic: line integrals of cubic splines
Replies: 4   Last Post: Jun 15, 2001 8:12 AM

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 Jive Dadson Posts: 171 Registered: 12/11/04
Re: line integrals of cubic splines
Posted: Jun 13, 2001 10:25 PM

Peter Spellucci wrote:
>
> In article <3B267DA9.AFB88865@ix.netcom.com>,
> |> Say I've got a two or three dimensional cubic spline S (a twice
> |> continuously differentiable map from a time interval [0..T] into 2 or 3
> |> space). Let L be the cumulative line integral of S, e.g., L(t) = length
> |> of the curve from time 0 to t. I need the inverse (D) of L: Given a
> |> distance d, L(d) = t such that L(t) = d.
> |>
> |> Are there any techniques (analytic perhaps) that I should be aware of,
> |> short of brute force? I don't even know what the state-of-the-art
> |> techniques are for arbitrary line integrals. Any pointers would be
> |> appreciated.
> |>
> |> Best regards,
> |> "Jive"
> I don't believe that this can be done analytically. Hence make a sufficiently
> fine table of L(t), t in [0,T]. searching in the table given d, gives you a
> small inclusion interval. use inverse interpolation for the first guess, then use
> zerofinding based on function values only to solve
> L(t)=d for t. e.g. fzero in http://www.netlib.org/fmm/fzero.f
> hope that helps
> peter

Thanks. That's as I had feared and expected.

Now then, any advice on calculating L? Recall that it's made up of
three cubic splines (with the same knot positions, BTW).

I searched the web for a general purpose path integral routine, but I
came up empty. What I did yesterday was to code up an algorithm that
calculates the length of a "connect-the-dots" path for 3 dots, 5 dots,
9, 17, 33, 2^^n+1, et cetera. Let's call the approximation of the
length based on 3 points L3, based on 5 pts L5, and so forth. I a fit a
rational polynomial to the the pts, (1/2,L3) (1/4,L5) (1/8,L9)
(1/16,L17) et cetera, and extrapolate that to zero. Is that a
reasonable method? Are the abscissae 1/2 1/4 1/8 etc. correct?

Is there information on this subject on the web?

Thanks again.

Best regards,
Jive

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