Quadratic Bezier Curve: Calculate t given x

4
votes

Good day. I am using a Quadratic Bezier Curve with the following configurations:

Start Point P1 = (1, 2) Anchor Point P2 = (1, 8) End Point P3 = (10, 8)

I know that given a t, I know I can solve for x and y using the following equation: 

t = 0.5; // given example value
x = (1 - t) * (1 - t) * P1.x + 2 * (1 - t) * t * P2.x + t * t * P3.x;
y = (1 - t) * (1 - t) * P1.y + 2 * (1 - t) * t * P2.y + t * t * P3.y;

where P1.x is the x coordinate of P1, and so on.

What I've tried now is that given an x value, I calculate for t using wolframalpha and then I plug that t in to the y equation and I get a my x and y point.

However, I want to automate finding t and then y. I have a formula to get x and y given a t. However, I don't have a formula to get t based on x. I'm a bit rusty with my algebra and expanding the first equation to isolate t doesn't look too easy.

Does anyone have a formula to get t based on x? My google search skills are failing me as of now.

I think it's also worth noting that my Bezier curve faces right.

Any help will be very much appreciated. Thanks.

3
mathpointbezier
As a Beziér curve is not a function, you can have several x for one y and several y for one x coordinate. - karatedog
A quadratic Bezier curve is a (parametric) function, and, unless it's a straight line, it intersects another line at at-most 2 points. I gave the straightforward solution below for intersections with a vertical line (x = constant); did it not work? - dwn

3 Answers

1
votes

problem is that what you want to solve is not function in general

  • for any t is just one (x,y) pair
  • but for any x there can be 0,1,2,+inf solutions of t

I would do this iteratively

you already can get any point p(t)=Bezier(t) so use iteration of t to minimize distance |p(t).x-x|

  1. for(t=0.0,dt=0.1;t<=1.0;t+=dt)

  2. find all local mins of d=|p(t).x-x|

    so when d start rising again set dt*=-0.1 and stop if |dt|<1e-6 or any other threshold. Stop if t is out of interval <0,1> and remember the solution to some list. Restore original t,dt and reset the local min search variables

  3. process all local mins

    eliminate all that has bigger distance then some threshold/accuracy compute y and do what you need with the point ...

It is much slower then algebraic approach but you can use this for any curvature not just quadratic

Usually cubic curves are used and do this algebraically with them is a nightmare.

0
votes

Look at your Bernstein polynomials B[i]; you have...

x = SUM_i ( B[i](t) * P[i].x )

...where...

B[0](t) = t^2 - 2*t + 1
B[1](t) = -2*t^2 + 2*t
B[2](t) = t^2

...so you can rearrange (assuming I did this right)...

0 = (P[0].x - 2*P[1].x + P[2].x) * t^2 + (-2*P[0].x + 2*P[1].x) * t + P[0].x - x

Now you should just be able to use the quadratic formula to find if the solutions for t exist (i.e., are real, not complex), and what they are.

0
votes
import numpy as np
import matplotlib.pyplot as plt
#Control points
p0=(1000,2500); p1=(2000,-1500); p2=(5000,3000)
#x-coordinates to fit
xcoord = [1750., 2750., 3950.,4760., 4900.]

# t variable with as few points as needed, considering accuracy. I found 30 is good enough 
t = np.linspace(0,1,30)

# calculate coordinates of quadratic Bezier curve
x = (1 - t) * (1 - t) * p0[0] + 2 * (1 - t) * t * p1[0] + t * t * p2[0];
y = (1 - t) * (1 - t) * p0[1] + 2 * (1 - t) * t * p1[1] + t * t * p2[1];

# find the closest points to each x-coordinate. Interpolate y-coordinate
ycoord=[]
for ind in xcoord:
    for jnd in range(len(x[:-1])):
        if ind >= x[jnd] and ind <= x[jnd+1]:
            ytemp = (ind-x[jnd])*(y[jnd+1]-y[jnd])/(x[jnd+1]-x[jnd]) + y[jnd]
            ycoord.append(ytemp)


plt.figure()
plt.xlim(0, 6000)
plt.ylim(-2000, 4000)
plt.plot(p0[0],p0[1],'kx', p1[0],p1[1],'kx', p2[0],p2[1],'kx')
plt.plot((p0[0],p1[0]),(p0[1],p1[1]),'k:', (p1[0],p2[0]),(p1[1],p2[1]),'k:')
plt.plot(x,y,'r', x, y, 'k:')
plt.plot(xcoord, ycoord, 'rs')
plt.show()