A quadratic Bezier curve is given in parametric form by:
$$C(t) = (1-t)^2P_0 + 2(1-t)tP_1 + t^2P_2.$$
My points are: $(1,1)$, $(2,2)$ and $(3,3)$. How do I show that this curve has cusps?
Best regards,
Sergey
Cusps on quadratic Bezier curve
bezier-curve
Related Solutions
The simplest heuristic approximation method is the one proposed by Tiller and Hanson. They just offset the legs of the control polygon in perpendicular directions:
The blue curve is the original one, and the three blue lines are the legs of its control polygon. We offset these three lines, and intersect/trim them to get the points A and B. The red curve is the offset. It's only an approximation of the true offset, of course, but it's often adequate.
If the approximation is not good enough for your purposes, you split the curve into two, and approximate the two halves individually. Keep splitting until you're happy. You will certainly have to split the original curves at inflexion points, if any, for example.
Here is an example where the approximation is not very good, and splitting would probably be needed:
There is a long discussion of the Tiller-Hanson algorithm plus possible improvements on this web page.
The Tiller-Hanson approach is compared with several others (most of which are more complex) in this paper.
Another good reference, with more up-to-date materials, is this section from the Patrikalakis-Maekawa-Cho book.
For even more references, you can search for "offset" in this bibliography.
I'll keep using your notation $t$ for time, $d$ for distance, $u$ for initial velocity, $v$ for final velocity, and $a$ for acceleration. Note that $a$ will be a negative constant.
The vehicle stops when $v=u+at=0$, i.e. at $t_f=-\frac ua$. The distance covered is $d=ut+\frac 12at^2$, which at the stopping point will be $d_f=-\frac{u^2}{2a}$. If we then scale these onto a $x$-$y$ graph from $(0,0)$ to $(1,1)$ by letting $x=\frac t{t_f}$ and $y=\frac d{d_f}$ the equation $d=ut+\frac 12at^2$ becomes
$$y=2x-x^2$$
Note that you get the same equation in $x$ and $y$ after scaling, no matter what the values of $t$, $d$, and $u$ are. That makes your job simpler but more boring: you need to approximate the parabola $y=2x-x^2$ for $0\le x\le 1$ by a Bezier curve.
(I made a mistake in my first analysis, corrected brilliantly by @fang in a comment. The following is my corrected analysis.)
You get that curve by using the cubic Bezier curve with starting point $P_0(0,0)$, first control point $P_1(\frac 13,\frac 23)$, second control point $P_2(\frac 23,1)$, and endpoint $P_3(1,1)$. This reproduces the desired parabola exactly. In fact, if you slow down the drawing by increasing the Bezier curve's parameter at a constant rate, the moving point will be at each correct place on the graph at the correct time.
Best Answer
Compute curvature with standard parametric form. Zero radius of curvature gives a cusp.
Primed with respect to $t$,
$$\kappa= \dfrac{(x'y''-y'x'')}{ (x'^2+y'^2)^{\frac32} }\to \infty $$
At a cusp slope $\dfrac{y'}{ x'} $ remains stationary or maximum/minimum.