I have to find points (say 10 points) in Bezier curve with 2 control points such that they are at equidistant positions in the curve. Currently I am using the following formula which gives me points but they are not equidistant.

t = ………..from (1/10 to 10/10);
q1 = t * t * t * -1 + t * t *  3 + t * -3 + 1;
q2 = t * t * t *  3 + t * t * -6 + t *  3;
q3 = t * t * t * -3 + t * t *  3;
q4 = t * t * t;

coordinates of a point:

x = q1 * startPointxValue + q2 * controlPoint1xValue + q3 * controlPoint2xValue + q4 * endPointxValue;
y = q1 * startPointyValue + q2 * controlPoint1yValue + q3 * controlPoint2yValue + q4 * endPointyValue;

I tried to use this but due my bad understanding, still unable to correct the values :(. Please help me in finding the values.

attaching sample image of required points (here 11): enter image description here

  • $\begingroup$ How is distance supposed to be measured? The usual way (chordlengths, in a straight line), or by arclength along the curve. $\endgroup$
    – bubba
    Mar 5, 2013 at 10:30
  • 1
    $\begingroup$ Just out of curiosity -- why do you want equidistant points? I've been working with Bezier curves for 35 years, and I can't remember any time I ever wanted to sprinkle one with equidistant points. Intriguing. $\endgroup$
    – bubba
    Mar 5, 2013 at 10:40
  • $\begingroup$ The answer you linked to seems good, to me. It looks easy enough to understand and code. You said you don't see how the control points are used. The "Babylonian method" measures distances between points like $B(t_7)$ and $B(t_8)$. The "B" is the equation of the Bezier curve, as a function of t. So, to get $B(t_7)$, for example, you have to plug $t=t_7$ into your formula for $q1,q2,q3,q4$, and then plug those $q$ values into your formulae for $x$ and $y$. After that's all done, $(x,y) = B(t_7)$. The control points are used to calculate the $B(t)$ values (the points on the curve). $\endgroup$
    – bubba
    Mar 5, 2013 at 10:53
  • $\begingroup$ @bubba should be arc length distance, so that they look like the distance between every point looks same from next point, okay,, mathematics is REALLY used in daily life, never heard about Bezier during my study of mathematics, i am an iOS developer where i am required to plot milestones at equal distances on a Road (Bezier curve) as shown above :D so i need those coordinates $\endgroup$
    – Kenpachi
    Mar 5, 2013 at 11:07
  • $\begingroup$ so if my understanding is correct, the method i am using to calculate the coordinates is right, and using Babylonian method I just re-calculate the "t" so that I get correct equidistant coordinates... right? $\endgroup$
    – Kenpachi
    Mar 5, 2013 at 11:14

2 Answers 2


The brute force approach is to just use a polyline approximation. If this is just for graphics, it will certainly be accurate enough. Even for more rigorous applications, it will probably be OK.

Compute a large number of points $n$ along the curve. I'd recommend somewhere in the range $n=50$ to $n=1000$, depending on your accuracy requirements and processor speed. Specifically, let the step-length be $s = 1/n$, and let $t_i = i*s$, and let $P_i = C(t_i)$ be a point on the curve, for $i = 0,1, \ldots , n$. Also, let $c_i = \Vert P_i - P_{i-1} \Vert$ be the $i$-th chord length, and let

$$d_i = \frac{\sum_{j=0}^i c_j}{\sum_{j=0}^n c_j}$$

So, in words, $d_i$ is the fractional distance along the polyline $P_0P_1 \ldots P_n$ at the $i$-th point $P_i$.

Now suppose you want to distribute $m+1$ points $Q_0, \ldots , Q_m$ at equal arclengths along the curve. Here's how you locate the $r$-th point $Q_r$.

Let $d = r/m$. Find $i$ such that $d_{i} \le d \le d_{i+1}$. Then let

$$ u = \frac{d - d_i}{d_{i+1} - d_i} $$

$$ t = t_i + u*s = (i+u)*s$$

$$ Q_r = C(t)$$

Getting all the indexing correct is a chore -- there are lots of opportunities for off-by-one errors. But, other than that, it's straightforward. The basic idea is that distances along the curve can be approximated by distances measured along the polyline $P_0P_1 \ldots P_n$.

You can use your existing code to calculate any point $(x,y) = C(t)$ on the curve.

Here's an implementation in C#:

class Program
   static void Main()
      Position endpt0, endpt1, control0, control1;
      endpt0 = new Position(100,100);    control0 = new Position(200,200);
      endpt1 = new Position(100,200);    control1 = new Position(400,200);
      BezierCurve curve = new BezierCurve(endpt0, control0, control1, endpt1);

      // Construct polyline with large number of points
      int n = 1000;
      double s = 1.0/(n-1);
      double[] tt = new double[n];
      Position[] PP = new Position[n];
      double[] cc = new double[n];
      for (int i = 0 ; i < n ; i++) 
         tt[i] = i*s;
         PP[i] = curve.Position(tt[i]);
         if (i > 0) cc[i] = Distance(PP[i], PP[i-1]);

      // Get fractional arclengths along polyline
      double[] dd = new double[n];
      dd[0] = 0;
      for (int i = 1 ; i < n ; i++) dd[i] = dd[i-1] + cc[i];
      for (int i = 1 ; i < n ; i++) dd[i] = dd[i]/dd[n-1];

      // Number of points to place on curve
      int m = 10;
      double step = 1.0/(m-1);
      Position[] QQ = new Position[m];

      for (int r = 0 ; r < m ; r++)
         double d = r*step;
         int i = FindIndex(dd, d);
         double u = (d - dd[i]) / (dd[i+1] - dd[i]);
         double t = (i + u)*s;
         QQ[r] = curve.Position(t);

   // Find index i such that dd[i] < d < dd[i+1]
   public static int FindIndex(double[] dd, double d)
      int i = 0;
      for (int j = 0 ; j < dd.Length ; j++)
         if (d > dd[j]) i = j;
      return i;

As said by Bubba, an option is to flatten the curve (sample it as a polyline) then approximate the curvilinear abscissas by accumulation of chord lengths.

Another option if to solve the ODE

ds / dt = √(x'² + y'²)

for constant increments of s. So we formulate as

dt = √(x'(t)² + y'(t)²) ds

and integrate for example by Runge-Kutta.

Note that the first approach does the same thing, but integrating by a first order method.


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