I have the coordinates of potentially any points within a 3D arc, representing the path an object will take when launched through the air. X is forward, Z is up, if that is relevant.
Using any of these points, I need to create a bezier curve that follows this arc. The curve requires the tangent values for the very start and end of the curve.
I can't figure out the formula needed to convert this arc into a bezier curve, and I'm hoping someone can help.
I've found plenty of resources discussing circular arcs, but the arc here is not circular.
[Math] Converting points in arc to a bezier curve
3dbezier-curvecurves
Related Solutions
Linear Bézier curve is simply a line given by parametric equation $R(t) = A+t(AB)$ , A being initial point and B being final point.
For Quadratic Bézier curve, take a look at the following picture.
Let the point between $P_1$ and $P_0$ be $Q_1$ and $P_1$ and $P_2$ be $Q_2$. Let our path be traced by $Q_0$. Then from above figure.
$$ \frac{P_0Q_1}{P_0P_1} = \frac{P_1Q_2}{P_1P_2} = \frac{Q_1Q_0}{Q_1Q2} = t \text{ (say)} $$
$$Q_1 = P_0 + t(P_0P_1), Q_2 = P_1 + t(P_1P_2)$$
So we have
$$Q_0 = Q_1 + t(Q_1Q_2) = P_0 + t(P_0P_1) + t(P_1 + t(P_1P_2) - (P_0 + t(P_0P_1)))$$
Have a look at more elaborate article on Wikipedia.
"Justification" for an approximation may be in the eye of the beholder, but:
The true projection is an ellipse inscribed in the indicated quadrilateral.
The circular arcs have their centers chosen so that (by elementary geometry) each arc is tangent to two sides of the parallelogram. In particular, each arc is tangent to its "neighboring" arcs, i.e., the arcs join up smoothly.
An arc of the ellipse "near" the major axis or minor axis is "approximately circular", since it's the image of a circular arc under a linear transformation $T$ and the axes of the ellipse are the eigenspaces of $T$.
The circles' centers are easily found with a straightedge, given the isometric grid.
Best Answer
Assuming that the arc is indeed a parabola, then the procedures outlined at Convert segment of parabola to quadratic bezier curve and http://www.ams.org/samplings/feature-column/fcarc-bezier should provide the appropriate calculations.