non-parametric quadratic Bézier curve through 3 points in 2D (in R)

Question

This is about explicit/non-parametric quadratic Bézier curves. Normally you can't fit a quadratic Bézier curve to 3 points because the X-variable is also a function (Bézier = parametric function), but when the control points are equidistant, you can: it is called an explicit/non-parametric Bézier function. I want to fit a quadratic bernstein polynomial to 3 random points in a 2D plane, and the x-axis coordinates of the 3 control points have to be equidistant. Also, the (outer) control points don't have to coincide with the 2 outer data points like they normally do.

I guess this requires solving a set of equations, but which ones? And how do I do that in R given the limitations I set (curve through the 3 data points, control points same horizontal distance, and control points not necessarily on the data points?

The quadratic Bézier function is B(t)=(1-t)^2*P0+2*t*(1-t)*P1+t^2*P2,

if you run this in R, you will see what I meant:

# control points are equidistant: here the horizontal distance is 20
cpx<-c(-20,0,20)
# y-values can be random
cpy<-c(0,2,-4)
t<-seq(0,1,len=101)

# the 3 control points
P0<-matrix(data=c(cpx[1],cpy[1]),nrow=1,ncol=2,byrow=FALSE,dimnames=NULL)
P1<-matrix(data=c(cpx[2],cpy[2]),nrow=1,ncol=2,byrow=FALSE,dimnames=NULL)
P2<-matrix(data=c(cpx[3],cpy[3]),nrow=1,ncol=2,byrow=FALSE,dimnames=NULL)

# the quadratic Bernstein polynomial:
B<-(1-t)^2%*%P0+2*t*(1-t)%*%P1+t^2%*%P2

par(mfrow=c(1,1))
plot(cpx,cpy,type="p",pch=20,xlab="",ylab="")
abline(v=c(min(cpx),max(cpx)),lty=3,col='red')
text(cpx[1],cpy[1],"P0",cex=.8,pos=4)
text(cpx[2],cpy[2],"P1",cex=.8,pos=1)
text(cpx[3],cpy[3],"P2",cex=.8,pos=2)
segments(cpx[1],cpy[1],cpx[2],cpy[2],lty=3);segments(cpx[2],cpy[2],cpx[3],cpy[3],lty=3)
lines(B,col="DeepSkyBlue")

# 3 random points on the curve:
pnts<-sort(sample(1:length(t),3,replace=F),decreasing=F)
point1<-pnts[1]
point2<-pnts[2]
point3<-pnts[3]
points(B[point1,1],B[point1,2],col='orange',pch=20)
points(B[point2,1],B[point2,2],col='orange',pch=20)
points(B[point3,1],B[point3,2],col='orange',pch=20)
segments(B[point1,1],B[point1,2],B[point2,1],B[point2,2],lwd=2,col='orange',lty=1)
segments(B[point2,1],B[point2,2],B[point3,1],B[point3,2],lwd=2,col='orange',lty=1)

Here's a similar but not equal topic. Here and here some nice Bézier animations.

Answer 1

Normally you can't fit a quadratic Bézier curve to 3 points

A quadratic Bézier curve has 3 control points, which means it has 3 degrees of freedom and should be able to fit any 3 points without a problem. A linear Bézier curve would have trouble, but a quadratic or higher would be fine. To fit a parametric Bézier you need to specify t values for each of the points that you want to fit, but that is the only tricky bit.

If you want an explicit Bézier, we can probably do that. I assume that the points that you're trying to fit are going to be in your x range [-20..20] ?

If so, then the first thing you need to do is find the x value of each point you're fitting to, and then convert that to a t value. In your example your x range is [-20..20] and your t range is [0..1] if I'm reading your code correctly, so we'll need to normalize. Your t value is (x+20)/40.

Now that you have all 3 t values, you solve 3 equations for 3 unknowns, where the unknowns are the y values of the control points. The 3 equations all have the same form, specifically:

(1-tfit)^2*y0 + 2*tfit*(1-tfit)*y1 + tfit^2*y2 = yfit

where tfit is the t value of the point you're trying to fit, and yfit is the y value of the same point.

Solve the set of equations for y0, y1, and y2.

---

So, let's look at some sample data. We'll take as input your 3 data points:

d0 = (-4.8000, 0.3648)
d1 = (7.2000, -0.9792)
d2 = (8.4000, -1.1928)

We'll also take as input the range of your explicit quadratic bezier curve

rmin = -20
rmax = 20

Now we compute the t values for your data points

t0 = (d0.x - rmin) / (rmax - rmin)
t1 = (d1.x - rmin) / (rmax - rmin)
t2 = (d2.x - rmin) / (rmax - rmin)

More specifically, these are:

t0 = 0.38000000000000000
t1 = 0.68000000000000005
t2 = 0.70999999999999996

We now have 3 equations:

(1-t0)*(1-t0)*y0 + 2*(1-t0)*t0*y1 + t0*t0*y2 = d0.y
(1-t1)*(1-t1)*y0 + 2*(1-t1)*t1*y1 + t1*t1*y2 = d1.y
(1-t2)*(1-t2)*y0 + 2*(1-t2)*t2*y1 + t2*t2*y2 = d2.y

We are solving for y0, y1, and y2, which are the y values of the explicit bezier curve.

I had a LSQ solver handy so I used it, but other methods would work fine.

Here is the matrix I put in (A|b):

0.38440000000000002 0.47120000000000001 0.14440000000000000  | 0.36480000000000001
0.10239999999999996 0.43519999999999998 0.46240000000000009  | -0.97919999999999996
0.084100000000000022 0.41180000000000005 0.50409999999999999 | -1.1928000000000001

and here is the solution it produced:

y0 = -2.6513220228646483e-014
y1 = 2.0000000000000262
y2 = -4.0000000000000169

Hope that helps.. :)