What unit is the first derivative of a quadratic Bézier curve expressed in?

Question

I'm using quadratic Bézier curves to determine the velocity vector at the endpoints of a path - I know only discrete points, not the velocity in those points. The velocity vector is supposed to be the first derivative with respect to the parameter t for the Bézier curve. Now t is determined from the ratio of two time differences, so that it reduces the real time interval of the path to the interval [0,1] between those endpoints. Strictly the resultant t then becomes a dimensionless ratio, and the calculated velocity vector is thus no longer expressed in units of length per units of time, but merely in units of length ?! Can someone shed some light on this ?

Answer 1

Let's see how things behave under refinement, i.e., when you replace the curve $(A,B,C)$ from $A$ to $C$ with helper point $B$, which corresponts to duration $\Delta t$, with the $(A',B',C')=(A,\frac{A+B}2,\frac{A+2B+C}4)$ and $(\frac{A+2B+C}4,\frac{B+C}2,C)$, each corresponding to $\frac 12 \Delta t$. Notably, $B'-A'$ is exactly half of $B-A$ and we halved our time interval, so we suspect the initial velocity to be proportional to $\frac{B-A}{\Delta t}$. If we repeat the subdivision process, the points become more and more collinear and equidistant and in the collinear equidistant limit case we clearly have $$v_0=\frac{B-A}{2\Delta t}$$ which must therefore hold in general.