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
Related
I'm building a p5js donut chart, but I'm struggling to show the data labels in the middle. I think I have managed to get the boundaries right for it, but how would match the angle that I'm in? Or is there a way of matching just through the colours?
https://i.stack.imgur.com/enTBo.png
I have started by trying to match the boundaries of the chart to the pointer, which I managed to do using mouseX and mouseY. Any suggestions, please?
if(mouseX >= width / 2 - width * 0.2 && mouseY >= height / 2 - width * 0.2
&& mouseX <= width / 2 + width * 0.2 && mouseY <= height / 2 + width * 0.2)
{
//console.log("YAY!!! I'm inside the pie chart!!!");
}
else
{
textSize(14);
text('Hover over to see the labels', width / 2, height / 2);
}
};
[1]: https://i.stack.imgur.com/enTBo.png
While you could theoretically use the get() function to check the color of the pixel under the mouse cursor and correlate that with one of the entries in your dataset, I think you would be much better off doing the math to determine which segment the mouse is currently over. And conveniently p5.js provides helper functions that make it very easy.
In the example you showed you are only checking if the mouse cursor is in a rectangular region. But in reality you want to check if the mouse cursor is within a circle. To do this you can use the dist(x1, y1, x2, y2) function. Once you've established that the mouse cursor is over your pie chart, you'll want to determine which segment it is over. This can be done by finding the angle between a line draw from the center of the chart to the right (or whichever direction is where you started drawing the wedges), and a line drawn from the center of the chart to the mouse cursor. This can be accomplished using the angleBetween() function of p5.Vector.
Here's a working example:
const colors = ['red', 'green', 'blue'];
const thickness = 40;
let segments = {
foo: 34,
bar: 55,
baz: 89
};
let radius = 80, centerX, centerY;
function setup() {
createCanvas(windowWidth, windowHeight);
noFill();
strokeWeight(thickness);
strokeCap(SQUARE);
ellipseMode(RADIUS);
textAlign(CENTER, CENTER);
textSize(20);
centerX = width / 2;
centerY = height / 2;
}
function draw() {
background(200);
let keys = Object.keys(segments);
let total = keys.map(k => segments[k]).reduce((v, s) => v + s, 0);
let start = 0;
// Check the mouse distance and angle
let mouseDist = dist(centerX, centerY, mouseX, mouseY);
// Find the angle between a vector pointing to the right, and the vector
// pointing from the center of the window to the current mouse position.
let mouseAngle =
createVector(1, 0).angleBetween(
createVector(mouseX - centerX, mouseY - centerY)
);
// Counter clockwise angles will be negative 0 to PI, switch them to be from
// PI to TWO_PI
if (mouseAngle < 0) {
mouseAngle += TWO_PI;
}
for (let i = 0; i < keys.length; i++) {
stroke(colors[i]);
let angle = segments[keys[i]] / total * TWO_PI;
arc(centerX, centerY, radius, radius, start, start + angle);
// Check mouse pos
if (mouseDist > radius - thickness / 2 &&
mouseDist < radius + thickness / 2) {
if (mouseAngle > start && mouseAngle < start + angle) {
// If the mouse is the correct distance from the center to be hovering over
// our "donut" and the angle to the mouse cursor is in the range for the
// current slice, display the slice information
push();
noStroke();
fill(colors[i]);
text(`${keys[i]}: ${segments[keys[i]]}`, centerX, centerY);
pop();
}
}
start += angle;
}
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.3.1/p5.js"></script>
I think I know the source of the problem was that #thenewbie experienced: it is the p5 library being used. I was using the p5.min.js and experiencing the same problem. Once I started using the full p5.js library, the issue was resolved and #Paul's script worked.
Here is a link I came across while researching this which put me onto the solution:
https://github.com/processing/p5.js/issues/3973
Thanks Paul for the clear explanations and code above.
I am wondering how you can rotate an image by only using the transform function. From my understanding this is not possible, since the only things you can do with transform are the following:
Horizontal scaling
Horizontal skewing
Vertical skewing
Vertical scaling
Horizontal moving
Vertical moving
Source: https://developer.mozilla.org/en-US/docs/Web/API/Canvas_API/Tutorial/Transformations
And I don't see how any of these would be able to rotate the shape, is this even possible. I assume it must be possible, since rotate is in-fact a type transformation.
2D transform basics
6 values as 3 vectors
The transform is a set of 6 numbers. The 6 numbers as 3 pairs represent the direction and scale of the x axis, the direction and scale of the y axis, and the position of the origin.
Default transform AKA Identity matrix
The default transform (called the identity matrix) has the values ctx.setTransform(1, 0, 0, 1, 0, 0) meaning that
the x axis is 1 transformed pixel per CSS pixel in the direction {x: 1, y: 0} left to right
the y axis is 1 transformed pixel per CSS pixel in the direction {x: 0, y: 1} top to bottom
the origin is at pixel location {x: 0, y: 0} top left corner
Scaling
If we scale the transform we increase the length of the first two vectors. To scale by 2 the transform is ctx.setTransform(2, 0, 0, 2, 0, 0);
the x axis is 1 transformed pixel for every 2 CSS pixel in the x direction {x: 2, y: 0} left to right
the y axis is 1 transformed pixel for every 2 CSS pixel in the y direction {x: 0, y: 2} top to bottom
the origin is still top left {x: 0, y: 0}
Rotate and translation
If we want to rotate by 90deg a square 256 by 256 image then the transform is ctx.setTransform(0, 1, -1, 0, 256, 0)
the x axis is 1 transformed pixel per CSS pixel down in the y direction {x: 0, y: 1}
the y axis is 1 transformed pixel per CSS pixel across in the negative x direction {x: -1, y: 0} right to left
the origin (where the image 0, 0 will be on the canvas) is {x: 256, y: 0}
Thus if we run
ctx.setTransform(0, 1, -1, 0, 256, 0);
ctx.drawImage(myImage, 0, 0, 256, 256); // will draw image rotated 90deg CW
We get a rotated image.
A vector
A vector is two values that have a x and y value. The vector defines a direction and length.
Create a rotated unit vector
To convert a direction to a vector we use sin and cos
const myDirection = angle;
const myDirectionAsRadians = angle * (Math.PI / 180); // convert angle to radians
const x = Math.cos(myDirectionAsRadians)
const y = Math.sin(myDirectionAsRadians)
If we set myDirection to 90 (deg) then x = 0 and y = 1 pointing down the canvas
Using sin and cos creates a vector in any direction. It has a special property in that its length is always 1. We call such a vector a Unit vector. You may sometimes see a vector being normalized. This converts a vector of any length to a unit vector. It is done by dividing the vector x and y by its length.
function normalize(vector) {
const length = Math.hypot(vector.x, vector.y);
vector.x /= length;
vector.y /= length;
}
NOTE a vector with zero length eg x: 0, y:0 can not be normalized. Not because it has no length (the length is 0) but because it has no direction.
Scale a rotated vector
We can define an angle and a scale
const myDirection = -90;
const myDirectionAsRadians = -90 * (Math.PI / 180); // -90 as radians
const myScale = 2;
const x = Math.cos(myDirectionAsRadians) * myScale
const y = Math.sin(myDirectionAsRadians) * myScale
Now for -90 deg the vector is x = 0 and y = -2 pointing up and two CSS pixels long.
Quick rotate vector 90deg CW
For a uniform scale and rotation (the image is always square) all we need is a single vector. For example from the above. x = 0 and y = -2 (pointing up) can be rotated 90 CW by swapping the two components and negating the new x. eg xx = -y and y = x to get xx = 2 and y = 0 2 CSS pixels from left two right. Thus we have the direction and scale of both the x and y axis. With the y axis always 90 CW from the x.
Using the transform to draw
Create rotated and scaled transform
To create a transform that rotates any angle and scales by any amount
function scaleAndRotate(scale, rotate) { // rotate is in radians
// get direction and length of x axis
const xAX = Math.cos(rotate) * scale;
const xAY = Math.sin(rotate) * scale;
// get direction and length of y axis that is 90 deg CW of x axis and same length
const [yAX, yAY] = [-xAY, xAX]; // swap and negate new x
// set the transform
ctx.setTransform(xAX, xAY, yAX, yAY, 0, 0);
}
Drawing an image
Lets create a function that will draw an image anywhere on the canvas that is rotated and scaled uniformly. We will use the center of the image as the reference point
function drawImageScaleRotate(img, x, y, scale, rotate) {
// define the direction and scale of x axis
const xAX = Math.cos(rotate) * scale;
const xAY = Math.sin(rotate) * scale;
// create the transform with yaxis at 90 CW of x axis and origin at x, y
ctx.setTransform(xAX, xAY, -xAY, xAX, x, y);
// Draw the image so that its center is at the new origin x, y
ctx.drawImage(img, -img.width / 2, -img.height / 2);
}
There is much more
When we set the transform with ctx.setTranform we replace the existing transform. This transform remains current. If we use ctx.transform, ctx.rotate, ctx.scale, ctx.translate the transforms are applied to the current transform, you build a transform in stages.
The transform functions are relatively expensive in terms of CPU cycles. That is way using sin and cos to build the matrix is much faster than using ctx.scale, ctx.rotate, ctx.translate to do the same thing starting from default.
Building transforms can become tricky as we need to keep track of what stage we are at.
We generally only use these function not to transform a single image (text, path, or what ever) but to create linked transforms.
For example a game object like a tank. The body of the tank is transformed (rotated and positioned) then the turret which is rotated with the body but has an additional independent rotation by using ctx.rotate. Full explanation is beyond the scope of this question.
The final function
From all this we can create a simplified function that will draw an image with its center at any location, that is uniformly scaled and rotated
function drawImageScaleRotate(img, x, y, scale, rotate) {
const xAX = Math.cos(rotate) * scale;
const xAY = Math.sin(rotate) * scale;
ctx.setTransform(xAX, xAY, -xAY, xAX, x, y);
ctx.drawImage(img, -img.width / 2, -img.height / 2);
}
To reset the transform to the default use ctx.resetTransform NOTE not fully supported yet or use ctx.setTransform(1,0,0,1,0,0);
Animated transforms faster than CSS + HTML or SVG
Using the above function is the 2nd fastest way to draw animated rotated scaled images, faster than CSS + HTML or SVG. You can literally fill the screen with animated images.
Demo
var w,h;
var image = new Image;
image.src = "https://i.stack.imgur.com/C7qq2.png?s=328&g=1";
var canvas = document.createElement("canvas");
var ctx = canvas.getContext("2d");
canvas.style.position = "absolute";
canvas.style.top = "0px";
canvas.style.left = "0px";
document.body.appendChild(canvas);
const resize = () => { w = canvas.width = innerWidth; h = canvas.height = innerHeight;}
const rand = (min,max) => Math.random() * (max ?(max-min) : min) + (max ? min : 0);
const DO = (count,callback) => { while (count--) { callback(count) } }
resize();
addEventListener("resize",resize);
const sprites = [];
DO(500,()=>{
sprites.push({
xr : rand(w), yr : rand(h),
x : 0, y : 0, // actual position of sprite
r : rand(Math.PI * 2),
scale : rand(0.1,0.25),
dx : rand(-2,2), dy : rand(-2,2),
dr : rand(-0.2,0.2),
});
});
function drawImage(image, spr){
const xAX = Math.cos(spr.r) * spr.scale;
const xAY = Math.sin(spr.r) * spr.scale;
ctx.setTransform(xAX, xAY, -xAY, xAX, spr.x, spr.y);
ctx.drawImage(image, -image.width / 2, -image.height / 2);
}
function update(){
var ihM,iwM;
ctx.setTransform(1,0,0,1,0,0);
ctx.clearRect(0,0,w,h);
if(image.complete){
var iw = image.width;
var ih = image.height;
for(var i = 0; i < sprites.length; i ++){
var spr = sprites[i];
spr.xr += spr.dx;
spr.yr += spr.dy;
spr.r += spr.dr;
// keeps images in canvas adds space to all sides so that image
// can move completely of the canvas befor warping to other side
// I do this to prevent images visualy popping in and out at edges
iwM = iw * spr.scale * 2 + w;
ihM = ih * spr.scale * 2 + h;
spr.x = ((spr.xr % iwM) + iwM) % iwM - iw * spr.scale;
spr.y = ((spr.yr % ihM) + ihM) % ihM - ih * spr.scale;
drawImage(image,spr);
}
}
requestAnimationFrame(update);
}
requestAnimationFrame(update);
If you are wondering which is the fastest way to draw animated content. That is via webGL. The above can draw 1000 scaled rotated images on most devices at a good frame rate. WebGL can easily draw 10000 (with extra features eg colored) in the same time.
You can rotate your image easily. You can specify angles by which you wants to apply rotation.
Source : https://www.w3schools.com/cssref/css3_pr_transform.asp
<style>
img.a {
transform: rotate(180deg);
}
</style>
<img class="a" src="https://picsum.photos/id/237/200/300"/>
use transform "rotate".and use it as below
p{
color:red;
font-size:12px;
text-align:center;
}
.rotate1{
transform:rotate(45deg);
margin-top:40px;
}
.rotate2{
transform:rotate(90deg);
margin-top:40px;
}
.rotate3{
transform:rotate(180deg);
margin-top:40px;
}
<p class="rotate1">ROTATE1</p>
<p class="rotate2">ROTATE2</p>
<p class="rotate3">ROTATE3</p>
You have ctx.rotate(radians) function.
Read below:
https://www.w3schools.com/tags/canvas_rotate.asp
I was trying to do a perspective grid on my canvas and I've changed the function from another website with this result:
function keystoneAndDisplayImage(ctx, img, x, y, pixelHeight, scalingFactor) {
var h = img.height,
w = img.width,
numSlices = Math.abs(pixelHeight),
sliceHeight = h / numSlices,
polarity = (pixelHeight > 0) ? 1 : -1,
heightScale = Math.abs(pixelHeight) / h,
widthScale = (1 - scalingFactor) / numSlices;
for(var n = 0; n < numSlices; n++) {
var sy = sliceHeight * n,
sx = 0,
sHeight = sliceHeight,
sWidth = w;
var dy = y + (sliceHeight * n * heightScale * polarity),
dx = x + ((w * widthScale * n) / 2),
dHeight = sliceHeight * heightScale,
dWidth = w * (1 - (widthScale * n));
ctx.drawImage(img, sx, sy, sWidth, sHeight,
dx, dy, dWidth, dHeight);
}
}
It creates almost-good perspective grid, but it isn't scaling the Height, so every square has got the same height. Here's a working jsFiddle and how it should look like, just below the canvas. I can't think of any math formula to distort the height in proportion to the "perspective distance" (top).
I hope you understand. Sorry for language errors. Any help would be greatly appreciatedRegards
There is sadly no proper way besides using a 3D approach. But luckily it is not so complicated.
The following will produce a grid that is rotatable by the X axis (as in your picture) so we only need to focus on that axis.
To understand what goes on: We define the grid in Cartesian coordinate space. Fancy word for saying we are defining our points as vectors and not absolute coordinates. That is to say one grid cell can go from 0,0 to 1,1 instead of for example 10,20 to 45, 45 just to take some numbers.
At the projection stage we project these Cartesian coordinates into our screen coordinates.
The result will be like this:
ONLINE DEMO
Ok, lets dive into it - first we set up some variables that we need for projection etc:
fov = 512, /// Field of view kind of the lense, smaller values = spheric
viewDist = 22, /// view distance, higher values = further away
w = ez.width / 2, /// center of screen
h = ez.height / 2,
angle = -27, /// grid angle
i, p1, p2, /// counter and two points (corners)
grid = 10; /// grid size in Cartesian
To adjust the grid we don't adjust the loops (see below) but alter the fov and viewDist as well as modifying the grid to increase or decrease the number of cells.
Lets say you want a more extreme view - by setting fov to 128 and viewDist to 5 you will get this result using the same grid and angle:
The "magic" function doing all the math is as follows:
function rotateX(x, y) {
var rd, ca, sa, ry, rz, f;
rd = angle * Math.PI / 180; /// convert angle into radians
ca = Math.cos(rd);
sa = Math.sin(rd);
ry = y * ca; /// convert y value as we are rotating
rz = y * sa; /// only around x. Z will also change
/// Project the new coords into screen coords
f = fov / (viewDist + rz);
x = x * f + w;
y = ry * f + h;
return [x, y];
}
And that's it. Worth to mention is that it is the combination of the new Y and Z that makes the lines smaller at the top (at this angle).
Now we can create a grid in Cartesian space like this and rotate those points directly into screen coordinate space:
/// create vertical lines
for(i = -grid; i <= grid; i++) {
p1 = rotateX(i, -grid);
p2 = rotateX(i, grid);
ez.strokeLine(p1[0], p1[1], p2[0], p2[1]); //from easyCanvasJS, see demo
}
/// create horizontal lines
for(i = -grid; i <= grid; i++) {
p1 = rotateX(-grid, i);
p2 = rotateX(grid, i);
ez.strokeLine(p1[0], p1[1], p2[0], p2[1]);
}
Also notice that position 0,0 is center of screen. This is why we use negative values to get out on the left side or upwards. You can see that the two center lines are straight lines.
And that's all there is to it. To color a cell you simply select the Cartesian coordinate and then convert it by calling rotateX() and you will have the coordinates you need for the corners.
For example - a random cell number is picked (between -10 and 10 on both X and Y axis):
c1 = rotateX(cx, cy); /// upper left corner
c2 = rotateX(cx + 1, cy); /// upper right corner
c3 = rotateX(cx + 1, cy + 1); /// bottom right corner
c4 = rotateX(cx, cy + 1); /// bottom left corner
/// draw a polygon between the points
ctx.beginPath();
ctx.moveTo(c1[0], c1[1]);
ctx.lineTo(c2[0], c2[1]);
ctx.lineTo(c3[0], c3[1]);
ctx.lineTo(c4[0], c4[1]);
ctx.closePath();
/// fill the polygon
ctx.fillStyle = 'rgb(200,0,0)';
ctx.fill();
An animated version that can help see what goes on.
I've been recently adding shadows to a project. I've ended up with something that I like, but the shadows are a solid transparent color throughout. I would prefer them to be a fading gradient as they go further.
What I currently have:
What I'd like to achieve:
Right now I'm using paths to draw my shadows on a 2D Canvas. The code that is currently in place is the following:
// Check if edge is invisible from the perspective of origin
var a = points[points.length - 1];
for (var i = 0; i < points.length; ++i, a = b)
{
var b = points[i];
var originToA = _vec2(origin, a);
var normalAtoB = _normal(a, b);
var normalDotOriginToA = _dot(normalAtoB, originToA);
// If the edge is invisible from the perspective of origin it casts
// a shadow.
if (normalDotOriginToA < 0)
{
// dot(a, b) == cos(phi) * |a| * |b|
// thus, dot(a, b) < 0 => cos(phi) < 0 => 90° < phi < 270°
var originToB = _vec2(origin, b);
ctx.beginPath();
ctx.moveTo(a.x, a.y);
ctx.lineTo(a.x + scale * originToA.x,
a.y + scale * originToA.y);
ctx.lineTo(b.x + scale * originToB.x,
b.y + scale * originToB.y);
ctx.lineTo(b.x, b.y);
ctx.closePath();
ctx.globalAlpha = _shadowIntensity / 2;
ctx.fillStyle = 'black';
ctx.fillRect(_innerX, _innerY, _innerWidth, _innerHeight);
ctx.globalAlpha = _shadowIntensity;
ctx.fill();
ctx.globalAlpha = 1;
}
}
Suggestions on how I could go about achieving this? Any and all help is highly appreciated.
You can use composition + the new filter property on the context which takes CSS filters, in this case blur.
You will have to do it in several steps - normally this falls under the 3D domain, but we can "fake" it in 2D as well by rendering a shadow-map.
Here we render a circle shape along a line represented by length and angle, number of iterations, where each iteration increasing the blur radius. The strength of the shadow is defined by its color and opacity.
If the filter property is not available in the browser it can be replaced by a manual blur (there are many out there such as StackBoxBlur and my own rtblur), or simply use a radial gradient.
For multiple use and speed increase, "cache" or render to an off-screen canvas and when done composite back to the main canvas. This will require you to calculate the size based on max blur radius as well as initial radius, then render it centered at angle 0°. To draw use drawImage() with a local transform transformed based on start of shadow, then rotate and scale (not shown below as being a bit too broad).
In the example below it is assumed that the main object is drawn on top after the shadow has been rendered.
The main function takes the following arguments:
renderShadow(ctx, x, y, radius, angle, length, blur, iterations)
// ctx - context to use
// x/y - start of shadow
// radius - shadow radius (assuming circle shaped)
// angle - angle in radians. 0° = right
// length - core-length in pixels (radius/blur adds to real length)
// blur - blur radius in pixels. End blur is radius * iterations
// iterations - line "resolution"/quality, also affects total end blur
Play around with shape, shadow color, blur radius etc. to find the optimal result for your scene.
Demo
Result if browser supports filter:
var ctx = c.getContext("2d");
// render shadow
renderShadow(ctx, 30, 30, 30, Math.PI*0.25, 300, 2.5, 20);
// show main shape
ctx.beginPath();
ctx.moveTo(60, 30);
ctx.arc(30, 30, 30, 0, 6.28);
ctx.fillStyle = "rgb(0,140,200)";
ctx.fill();
function renderShadow(ctx, x, y, radius, angle, length, blur, iterations) {
var step = length / iterations, // calc number of steps
stepX = step * Math.cos(angle), // calc angle step for x based on steps
stepY = step * Math.sin(angle); // calc angle step for y based on steps
for(var i = iterations; i > 0; i--) { // run number of iterations
ctx.beginPath(); // create some shape, here circle
ctx.moveTo(x + radius + i * stepX, y + i * stepY); // move to x/y based on step*ite.
ctx.arc(x + i * stepX, y + i * stepY, radius, 0, 6.28);
ctx.filter = "blur(" + (blur * i) + "px)"; // set filter property
ctx.fillStyle = "rgba(0,0,0,0.5)"; // shadow color
ctx.fill();
}
ctx.filter = "none"; // reset filter
}
<canvas id=c width=450 height=350></canvas>
How can I detect when the user clicks inside the red bubble?
It should not be like a square field. The mouse must be really inside the circle:
Here's the code:
<canvas id="canvas" width="1000" height="500"></canvas>
<script>
var canvas = document.getElementById("canvas")
var ctx = canvas.getContext("2d")
var w = canvas.width
var h = canvas.height
var bubble = {
x: w / 2,
y: h / 2,
r: 30,
}
window.onmousedown = function(e) {
x = e.pageX - canvas.getBoundingClientRect().left
y = e.pageY - canvas.getBoundingClientRect().top
if (MOUSE IS INSIDE BUBBLE) {
alert("HELLO!")
}
}
ctx.beginPath()
ctx.fillStyle = "red"
ctx.arc(bubble.x, bubble.y, bubble.r, 0, Math.PI*2, false)
ctx.fill()
ctx.closePath()
</script>
A circle, is the geometric position of all the points whose distance from a central point is equal to some number "R".
You want to find the points whose distance is less than or equal to that "R", our radius.
The distance equation in 2d euclidean space is d(p1,p2) = root((p1.x-p2.x)^2 + (p1.y-p2.y)^2).
Check if the distance between your p and the center of the circle is less than the radius.
Let's say I have a circle with radius r and center at position (x0,y0) and a point (x1,y1) and I want to check if that point is in the circle or not.
I'd need to check if d((x0,y0),(x1,y1)) < r which translates to:
Math.sqrt((x1-x0)*(x1-x0) + (y1-y0)*(y1-y0)) < r
In JavaScript.
Now you know all these values (x0,y0) being bubble.x and bubble.y and (x1,y1) being x and y.
To test if a point is within a circle, you want to determine if the distance between the given point and the center of the circle is smaller than the radius of the circle.
Instead of using the point-distance formula, which involves the use of a (slow) square root, you can compare the non-square-rooted (or still-squared) distance between the points. If that distance is less than the radius squared, then you're in!
// x,y is the point to test
// cx, cy is circle center, and radius is circle radius
function pointInCircle(x, y, cx, cy, radius) {
var distancesquared = (x - cx) * (x - cx) + (y - cy) * (y - cy);
return distancesquared <= radius * radius;
}
(Not using your code because I want to keep the function general for onlookers who come to this question later)
This is slightly more complicated to comprehend, but its also faster, and if you intend on ever checking point-in-circle in a drawing/animation/object moving loop, then you'll want to do it the fastest way possible.
Related JS perf test:
http://jsperf.com/no-square-root
Just calculate the distance between the mouse pointer and the center of your circle, then decide whether it's inside:
var dx = x - bubble.x,
dy = y - bubble.y,
dist = Math.sqrt(dx * dx + dy * dy);
if (dist < bubble.r) {
alert('hello');
}
Demo
As mentioned in the comments, to eliminate Math.sqrt() you can use:
var distsq = dx * dx + dy * dy,
rsq = bubble.r * bubble.r;
if (distsq < rsq) {
alert('HELLO');
}
An alternative (not always useful meaning it will only work for the last path (re)defined, but I bring it up as an option):
x = e.pageX - canvas.getBoundingClientRect().left
y = e.pageY - canvas.getBoundingClientRect().top
if (ctx.isPointInPath(x, y)) {
alert("HELLO!")
}
Path can btw. be any shape.
For more details:
http://www.w3.org/TR/2dcontext/#dom-context-2d-ispointinpath