how to draw smooth curve through N points using javascript HTML5 canvas?

The problem with joining subsequent sample points together with disjoint "curveTo" type functions, is that where the curves meet is not smooth. This is because the two curves share an end point but are influenced by completely disjoint control points. One solution is to "curve to" the midpoints between the next 2 subsequent sample points. Joining the curves using these new interpolated points gives a smooth transition at the end points (what is an end point for one iteration becomes a control point for the next iteration.) In other words the two disjointed curves have much more in common now.

This solution was extracted out of the book "Foundation ActionScript 3.0 Animation: Making things move". p.95 - rendering techniques: creating multiple curves.

Note: this solution does not actually draw through each of the points, which was the title of my question (rather it approximates the curve through the sample points but never goes through the sample points), but for my purposes (a drawing application), it's good enough for me and visually you can't tell the difference. There is a solution to go through all the sample points, but it is much more complicated (see http://www.cartogrammar.com/blog/actionscript-curves-update/)

Here is the the drawing code for the approximation method:

// move to the first point
   ctx.moveTo(points[0].x, points[0].y);


   for (i = 1; i < points.length - 2; i ++)
   {
      var xc = (points[i].x + points[i + 1].x) / 2;
      var yc = (points[i].y + points[i + 1].y) / 2;
      ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
   }
 // curve through the last two points
 ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x,points[i+1].y);

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A bit late, but for the record.

You can achieve smooth lines by using cardinal splines (aka canonical spline) to draw smooth curves that goes through the points.

I made this function for canvas - it's split into three function to increase versatility. The main wrapper function looks like this:

function drawCurve(ctx, ptsa, tension, isClosed, numOfSegments, showPoints) {

    showPoints  = showPoints ? showPoints : false;

    ctx.beginPath();

    drawLines(ctx, getCurvePoints(ptsa, tension, isClosed, numOfSegments));

    if (showPoints) {
        ctx.stroke();
        ctx.beginPath();
        for(var i=0;i<ptsa.length-1;i+=2) 
                ctx.rect(ptsa[i] - 2, ptsa[i+1] - 2, 4, 4);
    }
}

To draw a curve have an array with x, y points in the order: x1,y1, x2,y2, ...xn,yn.

Use it like this:

var myPoints = [10,10, 40,30, 100,10]; //minimum two points
var tension = 1;

drawCurve(ctx, myPoints); //default tension=0.5
drawCurve(ctx, myPoints, tension);

The function above calls two sub-functions, one to calculate the smoothed points. This returns an array with new points - this is the core function which calculates the smoothed points:

function getCurvePoints(pts, tension, isClosed, numOfSegments) {

    // use input value if provided, or use a default value   
    tension = (typeof tension != 'undefined') ? tension : 0.5;
    isClosed = isClosed ? isClosed : false;
    numOfSegments = numOfSegments ? numOfSegments : 16;

    var _pts = [], res = [],    // clone array
        x, y,           // our x,y coords
        t1x, t2x, t1y, t2y, // tension vectors
        c1, c2, c3, c4,     // cardinal points
        st, t, i;       // steps based on num. of segments

    // clone array so we don't change the original
    //
    _pts = pts.slice(0);

    // The algorithm require a previous and next point to the actual point array.
    // Check if we will draw closed or open curve.
    // If closed, copy end points to beginning and first points to end
    // If open, duplicate first points to befinning, end points to end
    if (isClosed) {
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.push(pts[0]);
        _pts.push(pts[1]);
    }
    else {
        _pts.unshift(pts[1]);   //copy 1. point and insert at beginning
        _pts.unshift(pts[0]);
        _pts.push(pts[pts.length - 2]); //copy last point and append
        _pts.push(pts[pts.length - 1]);
    }

    // ok, lets start..

    // 1. loop goes through point array
    // 2. loop goes through each segment between the 2 pts + 1e point before and after
    for (i=2; i < (_pts.length - 4); i+=2) {
        for (t=0; t <= numOfSegments; t++) {

            // calc tension vectors
            t1x = (_pts[i+2] - _pts[i-2]) * tension;
            t2x = (_pts[i+4] - _pts[i]) * tension;

            t1y = (_pts[i+3] - _pts[i-1]) * tension;
            t2y = (_pts[i+5] - _pts[i+1]) * tension;

            // calc step
            st = t / numOfSegments;

            // calc cardinals
            c1 =   2 * Math.pow(st, 3)  - 3 * Math.pow(st, 2) + 1; 
            c2 = -(2 * Math.pow(st, 3)) + 3 * Math.pow(st, 2); 
            c3 =       Math.pow(st, 3)  - 2 * Math.pow(st, 2) + st; 
            c4 =       Math.pow(st, 3)  -     Math.pow(st, 2);

            // calc x and y cords with common control vectors
            x = c1 * _pts[i]    + c2 * _pts[i+2] + c3 * t1x + c4 * t2x;
            y = c1 * _pts[i+1]  + c2 * _pts[i+3] + c3 * t1y + c4 * t2y;

            //store points in array
            res.push(x);
            res.push(y);

        }
    }

    return res;
}

And to actually draw the points as a smoothed curve (or any other segmented lines as long as you have an x,y array):

function drawLines(ctx, pts) {
    ctx.moveTo(pts[0], pts[1]);
    for(i=2;i<pts.length-1;i+=2) ctx.lineTo(pts[i], pts[i+1]);
}

var ctx = document.getElementById("c").getContext("2d"); function drawCurve(ctx, ptsa, tension, isClosed, numOfSegments, showPoints) { ctx.beginPath(); drawLines(ctx, getCurvePoints(ptsa, tension, isClosed, numOfSegments)); if (showPoints) { ctx.beginPath(); for(var i=0;i

This results in this:

https://i.stack.imgur.com/xWCkp.jpg

You can easily extend the canvas so you can call it like this instead:

ctx.drawCurve(myPoints);

Add the following to the javascript:

if (CanvasRenderingContext2D != 'undefined') {
    CanvasRenderingContext2D.prototype.drawCurve = 
        function(pts, tension, isClosed, numOfSegments, showPoints) {
       drawCurve(this, pts, tension, isClosed, numOfSegments, showPoints)}
}

You can find a more optimized version of this on NPM (npm i cardinal-spline-js) or on GitLab.

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The first answer will not pass through all the points. This graph will exactly pass through all the points and will be a perfect curve with the points as [{x:,y:}] n such points.

var points = [{x:1,y:1},{x:2,y:3},{x:3,y:4},{x:4,y:2},{x:5,y:6}] //took 5 example points
ctx.moveTo((points[0].x), points[0].y);

for(var i = 0; i < points.length-1; i ++)
{

  var x_mid = (points[i].x + points[i+1].x) / 2;
  var y_mid = (points[i].y + points[i+1].y) / 2;
  var cp_x1 = (x_mid + points[i].x) / 2;
  var cp_x2 = (x_mid + points[i+1].x) / 2;
  ctx.quadraticCurveTo(cp_x1,points[i].y ,x_mid, y_mid);
  ctx.quadraticCurveTo(cp_x2,points[i+1].y ,points[i+1].x,points[i+1].y);
}

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I decide to add on, rather than posting my solution to another post. Below are the solution that I build, may not be perfect, but so far the output are good.

Important: it will pass through all the points!

If you have any idea, to make it better, please share to me. Thanks.

Here are the comparison of before after:

https://i.stack.imgur.com/M9adH.png

Save this code to HTML to test it out.

Your browser does not support the HTML5 canvas tag.

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As Daniel Howard points out, Rob Spencer describes what you want at http://scaledinnovation.com/analytics/splines/aboutSplines.html.

Here's an interactive demo: http://jsbin.com/ApitIxo/2/

Here it is as a snippet in case jsbin is down.

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I found this to work nicely

function drawCurve(points, tension) {
    ctx.beginPath();
    ctx.moveTo(points[0].x, points[0].y);

    var t = (tension != null) ? tension : 1;
    for (var i = 0; i < points.length - 1; i++) {
        var p0 = (i > 0) ? points[i - 1] : points[0];
        var p1 = points[i];
        var p2 = points[i + 1];
        var p3 = (i != points.length - 2) ? points[i + 2] : p2;

        var cp1x = p1.x + (p2.x - p0.x) / 6 * t;
        var cp1y = p1.y + (p2.y - p0.y) / 6 * t;

        var cp2x = p2.x - (p3.x - p1.x) / 6 * t;
        var cp2y = p2.y - (p3.y - p1.y) / 6 * t;

        ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, p2.x, p2.y);
    }
    ctx.stroke();
}

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Give KineticJS a try - you can define a Spline with an array of points. Here's an example:

Old url: http://www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/

See archive url: https://web.archive.org/web/20141204030628/http://www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/

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Incredibly late but inspired by Homan's brilliantly simple answer, allow me to post a more general solution (general in the sense that Homan's solution crashes on arrays of points with less than 3 vertices):

function smooth(ctx, points)
{
    if(points == undefined || points.length == 0)
    {
        return true;
    }
    if(points.length == 1)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[0].x, points[0].y);
        return true;
    }
    if(points.length == 2)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[1].x, points[1].y);
        return true;
    }
    ctx.moveTo(points[0].x, points[0].y);
    for (var i = 1; i < points.length - 2; i ++)
    {
        var xc = (points[i].x + points[i + 1].x) / 2;
        var yc = (points[i].y + points[i + 1].y) / 2;
        ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
    }
    ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x, points[i+1].y);
}

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This code is perfect for me:

this.context.beginPath();
this.context.moveTo(data[0].x, data[0].y);
for (let i = 1; i < data.length; i++) {
  this.context.bezierCurveTo(
    data[i - 1].x + (data[i].x - data[i - 1].x) / 2,
    data[i - 1].y,
    data[i - 1].x + (data[i].x - data[i - 1].x) / 2,
    data[i].y,
    data[i].x,
    data[i].y);
}

you have correct smooth line and correct endPoints NOTICE! (y = "canvas height" - y);

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A slightly different answer to the original question;

If anyone is desiring to draw a shape:

that is described by a series of points

where the line has a small curve at the points

the line doesn't necessarily have to pass through the points (i.e. passes slightly "inside", of them)

Then hopefully the below function of mine could help

Your browser does not support the HTML5 canvas tag.

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Bonjour

I appreciate the solution of user1693593 : Hermite polynomials seems the best way to control what will be drawn, and the most satisfying from a mathematical point of view. The subject seems to be closed for a long time but may be some latecomers like me are still interested in it. I've looked for a free interactive plot builder which could allow me to store the curve and reuse it anywhere else, but didn't find this kind of thing on the web : so I made it on my own way, from the wikipedia source mentionned by user1693593. It's difficult to explain how it works here, and the best way to know if it is worth while is to look at https://sites.google.com/view/divertissements/accueil/splines.

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To add to K3N's cardinal splines method and perhaps address T. J. Crowder's concerns about curves 'dipping' in misleading places, I inserted the following code in the getCurvePoints() function, just before res.push(x);

if ((y < _pts[i+1] && y < _pts[i+3]) || (y > _pts[i+1] && y > _pts[i+3])) {
    y = (_pts[i+1] + _pts[i+3]) / 2;
}
if ((x < _pts[i] && x < _pts[i+2]) || (x > _pts[i] && x > _pts[i+2])) {
    x = (_pts[i] + _pts[i+2]) / 2;
}

This effectively creates a (invisible) bounding box between each pair of successive points and ensures the curve stays within this bounding box - ie. if a point on the curve is above/below/left/right of both points, it alters its position to be within the box. Here the midpoint is used, but this could be improved upon, perhaps using linear interpolation.

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If you want to determine the equation of the curve through n points then the following code will give you the coefficients of the polynomial of degree n-1 and save these coefficients to the coefficients[] array (starting from the constant term). The x coordinates do not have to be in order. This is an example of a Lagrange polynomial.

var xPoints=[2,4,3,6,7,10]; //example coordinates
var yPoints=[2,5,-2,0,2,8];
var coefficients=[];
for (var m=0; m<xPoints.length; m++) coefficients[m]=0;
    for (var m=0; m<xPoints.length; m++) {
        var newCoefficients=[];
        for (var nc=0; nc<xPoints.length; nc++) newCoefficients[nc]=0;
        if (m>0) {
            newCoefficients[0]=-xPoints[0]/(xPoints[m]-xPoints[0]);
            newCoefficients[1]=1/(xPoints[m]-xPoints[0]);
    } else {
        newCoefficients[0]=-xPoints[1]/(xPoints[m]-xPoints[1]);
        newCoefficients[1]=1/(xPoints[m]-xPoints[1]);
    }
    var startIndex=1; 
    if (m==0) startIndex=2; 
    for (var n=startIndex; n<xPoints.length; n++) {
        if (m==n) continue;
        for (var nc=xPoints.length-1; nc>=1; nc--) {
        newCoefficients[nc]=newCoefficients[nc]*(-xPoints[n]/(xPoints[m]-xPoints[n]))+newCoefficients[nc-1]/(xPoints[m]-xPoints[n]);
        }
        newCoefficients[0]=newCoefficients[0]*(-xPoints[n]/(xPoints[m]-xPoints[n]));
    }    
    for (var nc=0; nc<xPoints.length; nc++) coefficients[nc]+=yPoints[m]*newCoefficients[nc];
}

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