Quadratic Bézier Curve: Calculate Points

Question

I\'d like to calculate a point on a quadratic curve. To use it with the canvas element of HTML5.

When I use the quadraticCurveTo() function in JavaScript, I have a s

Answer 1

I edited talkhabis answer (cubic curve) so the curve is displayed with the right coordinates. (Couldn't comment) The Y-coordinates needed to be changed (-p[].y+150). (A new variable for that might be a nicer and more efficient solution, but you get the idea)

// Apply points to SVG and create the curve and controllers :

var path  =  document.getElementById('path'),
    ctrl1 =  document.getElementById('ctrl1'),
    ctrl2 =  document.getElementById('ctrl2'),
    D = 'M ' + p0.x + ' ' + (-p0.y+150) +
    'C ' + c0.x + ' ' + (-c0.y+150) +', ' + c1.x + ' ' + (-c1.y+150) + ', ' + p1.x + ' ' + (-p1.y+150);

path.setAttribute('d',D);
ctrl1.setAttribute('d','M'+p0.x+','+(-p0.y+150)+'L'+c0.x+','+(-c0.y+150));
ctrl2.setAttribute('d','M'+p1.x+','+(-p1.y+150)+'L'+c1.x+','+(-c1.y+150));

// Lets test the "Bezier Function" 

var t = 0, point = document.getElementById('point');

setInterval(function(){

  var p = Bezier(p0,c0,c1,p1,t);
  point.setAttribute('cx',p.x);
  point.setAttribute('cy',-p.y+150);

  t += 0.01;
  if(t>=1) t=0;

},50);


// OK ... Now tring to get "y" on cruve based on mouse "x" : 

var svg = document.getElementById('svg'),
    point2 = document.getElementById('point2');

svg.onmousemove = function(e){

    var x = (e.pageX - 50)/2,  
        y = (e.pageY - 50)/2;
   // "-50" because of "50px margin" on the left side 
   // and "/2" because the svg width is 300 units and 600 px => 300 = 600/2    

  // Get the x,y by mouse x
  var p = YBX(p0,c0,c1,p1,x); 

  point2.setAttribute('cx',p.x);
  point2.setAttribute('cy',-p.y+150);  
} 

http://jsfiddle.net/u214gco8/1/

I also created some C-Code to test the results for the cubic curve. Just enter the X and Y coordinates in the main function.

#include <stdio.h>
#include <stdlib.h> 
#include <math.h> 

void bezierCurve(int x[] , int y[]) 
{ 
    double xu = 0.0 , yu = 0.0 , u = 0.0 ; 
    int i = 0 ; 
    for(u = 0.0 ; u <= 1.0 ; u += 0.05) 
    { 
        xu = pow(1-u,3)*x[0]+3*u*pow(1-u,2)*x[1]+3*pow(u,2)*(1-u)*x[2] 
             +pow(u,3)*x[3]; 
        yu = pow(1-u,3)*y[0]+3*u*pow(1-u,2)*y[1]+3*pow(u,2)*(1-u)*y[2] 
            +pow(u,3)*y[3]; 
        printf("X: %i   Y: %i \n" , (int)xu , (int)yu) ; 
    } 
} 

int main(void) {
    int x[] = {0,75,50,300};
    int y[] = {0,2,140,100};
    bezierCurve(x,y);
    return 0;
}

https://ideone.com/glLXcB

Answer 2

Use the quadratic Bézier formula, found, for instance, on the Wikipedia page for Bézier Curves:

In pseudo-code, that's

t = 0.5; // given example value
x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;

p[0] is the start point, p[1] is the control point, and p[2] is the end point. t is the parameter, which goes from 0 to 1.

Answer 3

I created this demo :

// x = a * (1-t)³ + b * 3 * (1-t)²t + c * 3 * (1-t)t² + d * t³
//------------------------------------------------------------
// x = a - 3at + 3at² - at³ 
//       + 3bt - 6bt² + 3bt³
//             + 3ct² - 3ct³
//                    + dt³
//--------------------------------
// x = - at³  + 3bt³ - 3ct³ + dt³
//     + 3at² - 6bt² + 3ct²
//     - 3at + 3bt
//     + a
//--------------------------------
// 0 = t³ (-a+3b-3c+d) +  => A
//     t² (3a-6b+3c)   +  => B
//     t  (-3a+3b)     +  => c
//     a - x              => D
//--------------------------------

var A = d - 3*c + 3*b - a,
    B = 3*c - 6*b + 3*a,
    C = 3*b - 3*a,
    D = a-x;

// So we need to solve At³ + Bt² + Ct + D = 0 

Full example here

may help someone.

Answer 4

Just a note: If you are using the usual formulas presented here then don't expect t = 0.5 to return the point at half of the curve's length.. In most cases it won't.

More on this here under "§23 — Tracing a curve at fixed distance intervals" and here.

Answer 5

In case somebody needs the cubic form:

        //B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3

        x = (1-t)*(1-t)*(1-t)*p0x + 3*(1-t)*(1-t)*t*p1x + 3*(1-t)*t*t*p2x + t*t*t*p3x;
        y = (1-t)*(1-t)*(1-t)*p0y + 3*(1-t)*(1-t)*t*p1y + 3*(1-t)*t*t*p2y + t*t*t*p3y;