Hello Garbonzo,
Thanks for the kind words. Let me recap and expand (and correct!) a
little bit on the answer.
First - the values you provided are the X / Y coordinates for the
control points of a Quadratic Bezier Curve as described at
http://en.wikipedia.org/wiki/Bezier_curve
[apparently GA doesn't reformat the URL with the é properly]
As illustrated at the same page (scroll down), a Quadratic Bezier
Curve has slope limiting as follows:
at t=0, slope of P0 -> P1
at t=1, slope of P1 -> P2
Hmm. It appears my spreadsheet is in error - let me fix it. [...]
OK. With the corrected worksheet, there is still an infinite number of
solutions where the first condition appears to be true.
The corrected solution is where Y0 = 0 and X0 is in the range (-1, 1).
This is a condition where
p1 (P0, P1, P2) has P1 -> P2 has slope zero.
p2 (P0, P1, P2) has P0 -> P1 has slope zero.
Correcting my previous message, making the second derivative equal,
the simplest solution is where
X0 = 0, Y0 = 0
and
X2 = 2, Y2 = 1
In this case, the resulting curves p1 and p2 are symmetric around the
origin, and thus the second derivative has to be equal on both sides
of Y=0. Note in this case that both (X0, Y0) and (X2, Y2) must be
specified for the condition to be true.
Reviewing the formal analysis I described before, I found at least one
more mistake. The corrected versions are:
B(t) = t^2*(P0-2*P1+P2) + t*(-2*P0+2*P1) + (P0)
The first derivative of B(t) in terms of t is:
B'(t) = 2*t*(P0-2*P1+P2) + (-2*P0+2*P1)
and second derivative is:
B''(t) = 2*(P0 - 2*P1 + P2)
I suggest you confirm this with your own analysis. It looks odd to me
that the order of the three points
P0, P1, P2
or
P2, P1, P0
doesn't change the acceleration - so I may have made an additional
mistake [I'll think about it some more & post a clarification if I
find it]
However, based on the above and doing a little testing with the
worksheet, it appears that the condition of:
Y0 = 0 and X0 is in the range (-1, 1)
is still OK when X2=2, Y2=-1. Taking a look at equating B''(t) for the
two splines, the point (X0, Y0) falls out during simplification and
the other points are symmetric - so the acceleration is still
"constant" and equal for both splines.
If any part of the answer is unclear or if you need additional
information on some part of the answer, please make a clarification
request. I will be glad to help further.
Good luck with your upcoming test.
--Maniac |