"Karl " <kNsOtSaPhAlM@stanford.edu> wrote in message <gd89hh$rhs$1@fred.mathworks.com>...
> Thanks! This is very helpful. Alas, in trying to follow your derivation, I found the dreaded words: "It can be shown that...". Can you point me to a reference where the derivation is spelled out step by step?

First of all, let me apologize for the confusion of notation in that 7/20/08 article. In the formulas for dx/ds and dy/ds where I used s1 and s2, I meant to use s12 and s23, respectively, as defined by the two previous square root formulas. I must have been in too much of a hurry at the time.
With s denoting the (approximate) arc length along the curve, we can call s1, s2, and s3 the (ascending) values of s at the points P1, P2, and P3. We would then have s12 = s2s1 and s23 = s3s2.
To express x as a quadratic function of s which assumes the values x1, x2, and x3 at the arc length values s1, s2, and s3, we can use the Lagrange interpolation formula
x = x1*(ss2)*(ss3)/(s1s2)/(s1s3) + ...
x2*(ss1)*(ss3)/(s2s1)/(s2s3) + ...
x3*(ss1)*(ss2)/(s3s1)/(s3s2)
If we take the derivative of this x with respect to s and set s equal to s2 to get the derivative at the middle point P2, we get (after a little manipulation):
dx/ds (at s equal to s2) =
x1*(s3s2)/(s2s1)/(s3s1) ...
x2*(2*s2s1s3)/(s2s1/(s3s2) ...
+x3*(s2s1)/(s3s1)/(s3s2)
= ((x2x1)*(s3s2)^2+(x3x2)*(s2s1)^2) / ...
((s2s1)*(s3s2)*(s3s1))
= ((x2x1)*s23^2+(x3x2)*s12^2) / ...
(s12*s23*(s12+s23))
This last is the expression I derived earlier (with s12 and s23 replacing the former confusing s1 and s2.)
The derivation of dy/ds with y1, y2, and y3 replacing x1, x2, and x3 is done the same way.
Roger Stafford
