I'm aiming to solve t for a 3D cubic Bézier curve define through 4 points (start pt, 1st handle, 2nd handle, end pt) at which the curve has a given tangent vector (or better, a parallel one:)
Background: I have a profile of an object (archaeological stuff), drawn through Bézier curves, and do have to operate some transformations on the segments, but wish to be able to preserve the extremities unchanged -> I do have to split/slice the curves, but wish to be able to specify the 'slope' at which the curves has to be sliced. Yes, the derivative of the Bézier function...
So far, I have no success for this case with the FindRoot
, Solve
etc., but I'm a newbie and... please be tolerant, first question ever asked :)
Here a exemple:
bezierCurve = {{0., 0., 0.}, {1.62, 0., 0.}, {3.96, 0., -0.18}, {4.42, 0., -0.64}}
(Upper quarter of the front profile drawing of a French Neolithic copper axe blade, if you ask...)
f = BezierFunction[bezierCurve]
f'[1]
(* {1.38, 0., -1.38} *)
which, for me, is equivalent to -Pi/4
or (-45°
) on the x-z axes. So the tangent of this curve is running from {1, 0, 0}
to {1/Sqrt[2], 0, -1/Sqrt[2]}
Hence the question: how do I solve the t
at which the derivative is parallel to a given vector?
THANKS A LOT