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Is it possible to draw an image on a path in HTML5 canvas?

For instance, say I have the following path.
<canvas id="main" width="500" height="250"></canvas>
var canvas = document.getElementById("main");
var ctx = canvas.getContext("2d");
ctx.beginPath();
ctx.moveTo(20,20);
ctx.lineTo(100,20);
ctx.arcTo(150,20,150,70,50);
ctx.lineTo(150,120);
ctx.lineWidth = 3;
ctx.stroke();
Is it possible to draw images on the arc of the line? If so, How?

Slice an image to draw on curves.
Yes it is possible, though ideally this would be a job for WebGL. The next best solution is a scan line render but that is way to much CPU load for poor Javascript to manage.
The next best I mean "OK sort of." option is a little image slicing.
You simply draw the image in thin slices around the arc. The 2D renderer is not perfect and tries to draw half pixels as best it can. The result is some noise along the edge of each slice where you can see through. To overcome this I draw each slice slightly wider to cover up any holes.
If you need high quality, rendering it all at double the size on an offscreen canvas and then scale down to a onscreen canvas (don't forget smoothing) will make most think it was drawn that way.
As the inner and outer edges of the arc have different circumferences some of the image must be squashed or stretched. In the demo I keep the inner edge of the image to the correct width and stretch the outer edge. It is easy to change but ensure that you use the outer edge to workout how many slices to draw.
WARNING the radius given is for the inner edge. It is vetted to stop the for loop getting too long and blocking the page. You may want to limit the radius so the inner circumference is the same as the image width. radius = radius < img.width / (Math.PI * 2) ? img.width / (Math.PI * 2) : radius;
It is easy to adapt to lines and curves. All you need is the tangent or curve normal (should be unit vector ie length 1) Use this vector to set the transform ctx.setTransform(nx,ny,tx,ty,px,py). THe first two values point out from the bottom of the image to the top, the next two numbers are along the tangent from left to right. The last two are the point on the curve to draw the slice.
// creates a blank image with 2d context
var createImage=function(w,h){var i=document.createElement("canvas");i.width=w;i.height=h;i.ctx=i.getContext("2d");return i;}
// create a canvas and add to dom
var can = createImage(512,512);
document.body.appendChild(can);
var ctx = can.ctx;
// create a image (canvas) to draw on the arc.
const textToDisplay = "<<Image on arc>>"
ctx.font = "64px arial";
var w = ctx.measureText(textToDisplay).width + 8;
var text = createImage(w + 64,84);
text.ctx.fillStyle = "#F90";
text.ctx.strokeStyle = "black";
text.ctx.lineWidth = 16;
text.ctx.fillRect(0,0,text.width,text.height);
text.ctx.strokeRect(0,0,text.width,text.height);
text.ctx.font = "64px arial";
text.ctx.fillStyle = "#0F0";
text.ctx.strokeStyle = "Black";
text.ctx.lineWidth = 4;
text.ctx.strokeText(textToDisplay,38,58);
text.ctx.fillText(textToDisplay,38,58);
// draws image on arc
// img image to render
// x,y center of arc
// radius the inner edge (bottom of image) radius
// fromAng The angle to start drawing the image in radians
// toAng (optional if not given image width will be used to get toAng)
// returns undefined
function drawArcImage(img,x,y,radius,fromAng,toAng){
// WARNING if you let the radius get to small the ratio between the inner and out circumference
// gets very large. This will result in the image being stretched over a quintabazzilon pixels.
// so must vet the radius or you will block the page and upset the browser gods.
radius = Math.abs(radius); // only positive
radius = radius < img.height / 8 ? img.height / 8 : radius;
var outRad = radius + img.height;
var cir = Math.PI * 2 * radius; // get inner circumference
if(toAng === undefined){
var toAng = (img.width / cir) * Math.PI * 2 ; // get the angle the image will cover
}
var cirOut = toAng * outRad; // get the out edge distance in pixels
var imgStep = img.width / cirOut; // the image step per slice
var imgX = 0; // track the image line to draw
var angStep = toAng / cirOut; // the angle steps
// For each pixel on the out edge draw a slice
for(var i = 0; i < toAng; i += angStep){
var dx = Math.cos(fromAng + i);
var dy = Math.sin(fromAng + i);
// set up the transform to draw a slice from the inner to outer edges
ctx.setTransform(dy,-dx,-dx,-dy,dx * radius + x,dy * radius + y);
// get and draw the slice. I stretch it a little (2pix) to cover imperfect rendering
ctx.drawImage(img,imgX,0,imgStep,img.height,-1,-img.height,2,img.height);
// move to next slice
imgX += imgStep;
}
ctx.setTransform(1,0,0,1,0,0); // reset the transform
}
// animate the image to prove it is real.. LOL
var animTick = 0;
var animRate = 0.01;
var pos = 0;
// update function call via RAF
function update(){
animTick += animRate; // update tick
// random anim sin waves.
var rad = Math.sin(animTick) * (256-text.height - 20) + 20;
pos += Math.sin(animTick*10) * 0.02;
pos += Math.sin(animTick/ 3) * 0.02;
pos += Math.sin(animTick/ 7) * 0.05;
// clear
ctx.clearRect(0,0,can.width,can.height)
// draw
drawArcImage(text,256,256,rad,pos)
// do again and again and again
requestAnimationFrame(update);
}
update();

This is an answer to a similar question:
You could, in the draw loop implement a "line drawing algorithm" that does not exactly draw a line but draws an item at a place where that point would be. Except, replace the line algorithm here to draw an arc instead.
function line(x0, y0, x1, y1){
var dx = Math.abs(x1-x0);
var dy = Math.abs(y1-y0);
var sx = (x0 < x1) ? 1 : -1;
var sy = (y0 < y1) ? 1 : -1;
var err = dx-dy;
while(true){ // put draw loop here.
drawImage(image,x0,y0);//setPixel(x0,y0); // Do what you need to for this
if ((x0==x1) && (y0==y1)) break;
var e2 = 2*err;
if (e2 >-dy){ err -= dy; x0 += sx; }
if (e2 < dx){ err += dx; y0 += sy; }
}
}
code taken from: Bresenham algorithm in Javascript
I would suggest using a library like p5.js to do something like this. http://p5js.org

Image Manipulation - add image with corners in exact positions

I have an image which is a background containing a boxed area like this:
I know the exact positions of the corners of that shape, and I'd like to place another image within it. (So it appears to be inside the box).
I'm aware of the drawImage method for HTML5 canvas, but it seems to only support x, y, width, height parameters rather than exact coordinates. How might I draw an image onto a canvas at a specific set of coordinates, and ideally have the browser itself handle stretching the image.

Quadrilateral transform
One way to go about this is to use Quadrilateral transforms. They are different than 3D transforms and would allow you to draw to a canvas in case you want to export the result.
The example shown here is simplified and uses basic sub-divison and "cheats" on the rendering itself - that is, it draws in a small square instead of the shape of the sub-divided cell but because of the small size and the overlap we can get away with it in many non-extreme cases.
The proper way would be to split the shape into two triangles, then scan pixel wise in the destination bitmap, map the point from destination triangle to source triangle. If the position value was fractional you would use that to determine pixel interpolation (f.ex. bi-linear 2x2 or bi-cubic 4x4).
I do not intend to cover all this in this answer as it would quickly become out of scope for the SO format, but the method would probably be suitable in this case unless you need to animate it (it is not performant enough for that if you want high resolution).
Method
Lets start with an initial quadrilateral shape:
The first step is to interpolate the Y-positions on each bar C1-C4 and C2-C3. We're gonna need current position as well as next position. We'll use linear interpolation ("lerp") for this using a normalized value for t:
y1current = lerp( C1, C4, y / height)
y2current = lerp( C2, C3, y / height)
y1next = lerp(C1, C4, (y + step) / height)
y2next = lerp(C2, C3, (y + step) / height)
This gives us a new line between and along the outer vertical bars.
Next we need the X positions on that line, both current and next. This will give us four positions we will fill with current pixel, either as-is or interpolate it (not shown here):
p1 = lerp(y1current, y2current, x / width)
p2 = lerp(y1current, y2current, (x + step) / width)
p3 = lerp(y1next, y2next, (x + step) / width)
p4 = lerp(y1next, y2next, x / width)
x and y will be the position in the source image using integer values.
We can use this setup inside a loop that will iterate over each pixel in the source bitmap.
Demo
The demo can be found at the bottom of the answer. Move the circular handles around to transform and play with the step value to see its impact on performance and result.
The demo will have moire and other artifacts, but as mentioned earlier that would be a topic for another day.
Snapshot from demo:
Alternative methods
You can also use WebGL or Three.js to setup a 3D environment and render to canvas. Here is a link to the latter solution:
Three.js
and an example of how to use texture mapped surface:
Three.js texturing (instead of defining a cube, just define one place/face).
Using this approach will enable you to export the result to a canvas or an image as well, but for performance a GPU is required on the client.
If you don't need to export or manipulate the result I would suggest to use simple CSS 3D transform as shown in the other answers.
/* Quadrilateral Transform - (c) Ken Nilsen, CC3.0-Attr */
var img = new Image(); img.onload = go;
img.src = "https://i.imgur.com/EWoZkZm.jpg";
function go() {
var me = this,
stepEl = document.querySelector("input"),
stepTxt = document.querySelector("span"),
c = document.querySelector("canvas"),
ctx = c.getContext("2d"),
corners = [
{x: 100, y: 20}, // ul
{x: 520, y: 20}, // ur
{x: 520, y: 380}, // br
{x: 100, y: 380} // bl
],
radius = 10, cPoint, timer, // for mouse handling
step = 4; // resolution
update();
// render image to quad using current settings
function render() {
var p1, p2, p3, p4, y1c, y2c, y1n, y2n,
w = img.width - 1, // -1 to give room for the "next" points
h = img.height - 1;
ctx.clearRect(0, 0, c.width, c.height);
for(y = 0; y < h; y += step) {
for(x = 0; x < w; x += step) {
y1c = lerp(corners[0], corners[3], y / h);
y2c = lerp(corners[1], corners[2], y / h);
y1n = lerp(corners[0], corners[3], (y + step) / h);
y2n = lerp(corners[1], corners[2], (y + step) / h);
// corners of the new sub-divided cell p1 (ul) -> p2 (ur) -> p3 (br) -> p4 (bl)
p1 = lerp(y1c, y2c, x / w);
p2 = lerp(y1c, y2c, (x + step) / w);
p3 = lerp(y1n, y2n, (x + step) / w);
p4 = lerp(y1n, y2n, x / w);
ctx.drawImage(img, x, y, step, step, p1.x, p1.y, // get most coverage for w/h:
Math.ceil(Math.max(step, Math.abs(p2.x - p1.x), Math.abs(p4.x - p3.x))) + 1,
Math.ceil(Math.max(step, Math.abs(p1.y - p4.y), Math.abs(p2.y - p3.y))) + 1)
}
}
}
function lerp(p1, p2, t) {
return {
x: p1.x + (p2.x - p1.x) * t,
y: p1.y + (p2.y - p1.y) * t}
}
/* Stuff for demo: -----------------*/
function drawCorners() {
ctx.strokeStyle = "#09f";
ctx.lineWidth = 2;
ctx.beginPath();
// border
for(var i = 0, p; p = corners[i++];) ctx[i ? "lineTo" : "moveTo"](p.x, p.y);
ctx.closePath();
// circular handles
for(i = 0; p = corners[i++];) {
ctx.moveTo(p.x + radius, p.y);
ctx.arc(p.x, p.y, radius, 0, 6.28);
}
ctx.stroke()
}
function getXY(e) {
var r = c.getBoundingClientRect();
return {x: e.clientX - r.left, y: e.clientY - r.top}
}
function inCircle(p, pos) {
var dx = pos.x - p.x,
dy = pos.y - p.y;
return dx*dx + dy*dy <= radius * radius
}
// handle mouse
c.onmousedown = function(e) {
var pos = getXY(e);
for(var i = 0, p; p = corners[i++];) {if (inCircle(p, pos)) {cPoint = p; break}}
}
window.onmousemove = function(e) {
if (cPoint) {
var pos = getXY(e);
cPoint.x = pos.x; cPoint.y = pos.y;
cancelAnimationFrame(timer);
timer = requestAnimationFrame(update.bind(me))
}
}
window.onmouseup = function() {cPoint = null}
stepEl.oninput = function() {
stepTxt.innerHTML = (step = Math.pow(2, +this.value));
update();
}
function update() {render(); drawCorners()}
}
body {margin:20px;font:16px sans-serif}
canvas {border:1px solid #000;margin-top:10px}
<label>Step: <input type=range min=0 max=5 value=2></label><span>4</span><br>
<canvas width=620 height=400></canvas>

You can use CSS Transforms to make your image look like that box. For example:
img {
margin: 50px;
transform: perspective(500px) rotateY(20deg) rotateX(20deg);
}
<img src="https://via.placeholder.com/400x200">
Read more about CSS Transforms on MDN.

This solution relies on the browser performing the compositing. You put the image that you want warped in a separate element, overlaying the background using position: absolute.
Then use CSS transform property to apply any perspective transform to the overlay element.
To find the transform matrix you can use the answer from: How to match 3D perspective of real photo and object in CSS3 3D transforms

plotting random points in circle sector

I have a pie chart in canvas and I wanted to plot random points in each sector of that pie.
I have got the area of each sector. using the arc sector
var arcsector = Math.PI * (2 * sector / total);
var startAngle = (lastend - offset) * (radius/Math.PI);
var endAngle = (lastend + arcsector - offset) * (radius/Math.PI);
var sectorAngle = arcsector * (radius/Math.PI);
var sectorArea = .5 * (sectorAngle*Math.PI/180) * (radius*radius);
How can I randomly plot points within that area?

A pie is a part of a circle, which, with your notations, starts at startAngle and ends at endAngle.
Most simple way to get a random point is to build a random angle (between
startAngle and endAngle) and a random radius, then you have your point with those lines :
var randAngle = startAngle + Math.random()*( endAngle - startAngle );
var randRadius = Math.random()*radius;
var randX = centerX + randRadius * Math.cos(randAngle);
var randY = centerY + randRadius * Math.sin(randAngle);
ctx.fillRect ( randX, randY, 1, 1 ) ;
repeat the number of times required !

The simple approach is to:
Create a temporary arc shape on path
Create a random point
Hit-test the point against the shape and plot if inside
You can create a temporary arc path by doing something like this (adjust to match your situation) (and no need to stroke/fill):
ctx.beginPath();
ctx.moveTo(cx, cy);
ctx.arc(cx, cy, radius, startAngle, endAngle);
ctx.closePath();
Then create random points within the bounds of that arc, or just use a very basic approach (which is probably fast enough in most case unless you would need a lot of points) - and the spread is even compared to using a radius based approach:
var randomX = cx + radius * 2 * Math.random() - radius;
var randomY = cy + radius * 2 * Math.random() - radius;
and finally hit-test:
if (ctx.isPointInPath(randomX, randomY)) {
// plot point, count etc.
}
FIDDLE
Update
An even more efficient way to generate random points in the arc shape (and spread them more even) is to draw directly to an off-screen canvas without using any bound checking and no cos/sin operations, which are expensive, and finally composite that on top of your arc shape (or use arc as clip).
// create off-screen canvas
var ocanvas = document.createElement('canvas');
var octx = ocanvas.getContext('2d');
var d;
d = ocanvas.width = ocanvas.height = 300;
octx.fillStyle = '#fff';
while(count) {
var randomX = d * Math.random();
var randomY = d * Math.random();
octx.fillRect(randomX - 1, randomY - 1, 2, 2);
count--;
}
// composite random points with main arc
ctx.globalCompositeOperation = 'source-atop';
ctx.drawImage(ocanvas, 0, 0);
ctx.globalCompositeOperation = 'source-over';
It can be optimized further by having the off-screen canvas represent only the bounds of the arc shape.
FIDDLE

Demo: http://jsfiddle.net/jv6nP/3/
it's not perfect that points are at border and thus their radius being bigger than zero makes them overlap onto other parts of pie. And this also results in them going over black border.
var can = $('#can')[0].getContext('2d'),
border=2,
x=100,
y=75,
r=60,
sRadius= 0,
leadAngle=null,
points= [],
dotRadius=2,
data = {
water:[30,'#5CC5FA'],
earth:[60,'#F0A71F'],
air:[10,'#26EDE3']
};
function reDraw(){
//making border...
can.beginPath();
can.arc(x,y,r+border,0,2*Math.PI);
can.fillStyle='black';
can.fill();
var newAngle=null;
for (var k in data) { //making piechart..
leadAngle = (2*Math.PI)*(data[k][0]/100);
newAngle = sRadius+leadAngle;
calPoints(sRadius,leadAngle,k);
can.beginPath();
can.arc(x,y,r,sRadius,newAngle);
can.lineTo(x,y);
can.fillStyle=data[k][1];
can.fill();
sRadius= newAngle;
}
//calculating points..
function calPoints(s,e,name) {
if (name!='water') return;
var py,px,rAngle,rRad;
for (var i=0; i<15; i++) {
rAngle=s+Math.random()*(e);
rRad = Math.random()*r;
px = (Math.cos(rAngle) * rRad)+x;
py = (Math.sin(rAngle) * rRad)+y;
points.push([px,py]);
}
}
//plotting dots from data...
points.forEach(function(v){
can.beginPath();
can.arc(v[0],v[1],dotRadius,0,2*Math.PI);
can.fillStyle='fff';
can.fill();
});
points=[];
requestAnimationFrame(reDraw);
}
reDraw();

Finding coordinates after canvas Rotation

0 to 0,-70 by this :
ctx.strokeStyle = "red";
ctx.lineWidth = 2;
ctx.rotate(Math.PI/-10;);
ctx.beginPath();
ctx.moveTo(0,0);
ctx.lineTo(0,-70);
ctx.stroke();
And I can rotate that by 'PI/-10', and that works.
How i can get the x,y points of this after using rotate?

Your x and y points will still be 0 and -70 as they are relative to the translation (rotation). It basically means you would need to "reverse engineer" the matrix to get the resulting position you see on the canvas.
If you want to calculate a line which goes 70 pixels at -10 degrees you can use simple trigonometry to calculate it yourself instead (which is easier than going sort of backwards in the matrix).
You can use a function like this that takes your context, the start position of the line (x, y) the length (in pixels) and angle (in degrees) of the line you want to draw. It draw the line and returns an object with x and y for the end point of that line:
function lineToAngle(ctx, x1, y1, length, angle) {
angle *= Math.PI / 180;
var x2 = x1 + length * Math.cos(angle),
y2 = y1 + length * Math.sin(angle);
ctx.moveTo(x1, y1);
ctx.lineTo(x2, y2);
ctx.stroke();
return {x: x2, y: y2};
}
Then just call it as:
var pos = lineToAngle(ctx, 0, 0, 70, -10);
//show result of end point
console.log('x:', pos.x.toFixed(2), 'y:', pos.y.toFixed(2));
Result:
x: 68.94 y: -12.16
Or you can instead extend the canvas' context by doing this:
if (typeof CanvasRenderingContext2D !== 'undefined') {
CanvasRenderingContext2D.prototype.lineToAngle =
function(x1, y1, length, angle) {
angle *= Math.PI / 180;
var x2 = x1 + length * Math.cos(angle),
y2 = y1 + length * Math.sin(angle);
this.moveTo(x1, y1);
this.lineTo(x2, y2);
return {x: x2, y: y2};
}
}
And then use it directly on your context like this:
var pos = ctx.lineToAngle(100, 100, 70, -10);
ctx.stroke(); //we stroke separately to allow this being used in a path
console.log('x:', pos.x.toFixed(2), 'y:', pos.y.toFixed(2));
(0 degrees will point to the right).

So you're asking "after I set a transform, how can I run points through that transform"?
In that case, see HTML5 Canvas get transform matrix? . The question and answers are somewhat old, but seem up-to-date. I can't find anything in the current HTML5 spec that lets you access and use the transform matrix. (I see that it's theoretically accessable through context.currentTransform, but I don't see any functionality to let you use the matrix - you'd have to multiply your point through the matrix yourself, or fake it by creating a full SVGMatrix for your point vector.)
The top answer shows a transform class you can use. Track your changes through that, and use their transformPoint function to get the point you want transformed to its endpoint.

Spliting a quadratic slice into 3 smaller slices HTML5 CANVAS JS

I have a quadratic curve that I use to create a slice of a piechart. The slice is situated in an axis of x and y, with the center point at (0,0). The radius is variable at radiusX and radiusY. This slice travels 90 degrees.
I need to split this slice into 3 seperate slices (each having 30 degree angle) and have them match whatever curve their parent slice had.
The following images show possible examples of the slice. Black circles adjust the size/shape of the slice:
Here is the function I've made but it's just not working correctly:
//globalPosX and globalPosY equal whatever position each of the two large black circles have repectively.
var canvas = document.getElementById('CV_slices');
var context = canvas.getContext('2d');
var cenX = canvas.width/2;
var cenY = canvas.height/2;
var blackDotX = globalPosX - cenX;
var blackDotY = cenY - globalPosY;
var endX;
var endY;
var controlX;
var controlY;
//set first slice
var startCoOrds = {
x: cenX ,
y: globalPosY
};
for (i=1; i < 4; i++) {
//make end(x,y) of previous slice the start(x,y) for the next slice.
endX = startCoOrds.x - (blackDotX*Math.sin(30));
endY = startCoOrds.y + (blackDotY*Math.cos(30));
//set position of control point using position of start/end positions (at the moment only adjustibng using +10 -10 at end)
controlX = ((endX - startCoOrds.x) /2) + (startCoOrds.x) + 10;
controlY = ((endY - startCoOrds.y) / 2) + (startCoOrds.y) - 10;
// draw slice
context.save();
context.beginPath();
context.moveTo(cenX, cenY);
context.lineTo(startCoOrds.x, startCoOrds.y);
context.quadraticCurveTo(controlX, controlY, endX, endY);
context.lineTo(cenX, cenY);
//make end(x,y) of previous slice the start(x,y) for the next slice
startCoOrds.x = endX;
startCoOrds.y = endY;
context.closePath();
context.globalAlpha = 0.1;
context.fillStyle = "#333333";
context.fill();
context.lineWidth = 2;
context.strokeStyle = "#ffffff";
context.stroke();
context.restore();
}

Use the closest "blackDot" as the radius of a circle,
using the circle, divide your quadrant into 3 (see wiki)
then scale the points by a ratio of the distance between 0,0 and the "blackDot"'s
In effect your arc is a quadrant of a circle that has been scaled in the x or y axis.