Develop Reference

How to move object to target naturally and smoothly?

Can somebody fix it script to make it works properly?
What I expects:
Run script
Click at the canvas to set target (circle)
Object (triangle) starts to rotate and move towards to target (circle)
Change target at any time
How it works:
Sometimes object rotates correctly, sometimes isn't
Looks like one half sphere works well, another isn't
Thanks!
// prepare 2d context
const c = window.document.body.appendChild(window.document.createElement('canvas'))
.getContext('2d');
c.canvas.addEventListener('click', e => tgt = { x: e.offsetX, y: e.offsetY });
rate = 75 // updates delay
w = c.canvas.width;
h = c.canvas.height;
pi2 = Math.PI * 2;
// object that moves towards the target
obj = {
x: 20,
y: 20,
a: 0, // angle
};
// target
tgt = undefined;
// main loop
setInterval(() => {
c.fillStyle = 'black';
c.fillRect(0, 0, w, h);
// update object state
if (tgt) {
// draw target
c.beginPath();
c.arc(tgt.x, tgt.y, 2, 0, pi2);
c.closePath();
c.strokeStyle = 'red';
c.stroke();
// update object position
// vector from obj to tgt
dx = tgt.x - obj.x;
dy = tgt.y - obj.y;
// normalize
l = Math.sqrt(dx*dx + dy*dy);
dnx = (dx / l);// * 0.2;
dny = (dy / l);// * 0.2;
// update object position
obj.x += dnx;
obj.y += dny;
// angle between +x and tgt
a = Math.atan2(0 * dx - 1 * dy, 1 * dx + 0 * dy);
// update object angle
obj.a += -a * 0.04;
}
// draw object
c.translate(obj.x, obj.y);
c.rotate(obj.a);
c.beginPath();
c.moveTo(5, 0);
c.lineTo(-5, 4);
c.lineTo(-5, -4);
//c.lineTo(3, 0);
c.closePath();
c.strokeStyle = 'red';
c.stroke();
c.rotate(-obj.a);
c.translate(-obj.x, -obj.y);
}, rate);

This turned out to be a bit more challenging than I first thought and I ended up just re-writing the code.
The challenges:
Ensure the ship only rotated to the exact point of target. This required me to compare the two angle from the ship current position to where we want it to go.
Ensure the target did not rotate past the target and the ship did not translate past the target. This required some buffer space for each because when animating having this.x === this.x when an object is moving is very rare to happen so we need some room for the logic to work.
Ensure the ship turned in the shortest direction to the target.
I have added notes in the code to better explain. Hopefully you can implement this into yours or use it as is. Oh and you can change the movement speed and rotation speed as you see fit.
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
canvas.width = 400;
canvas.height = 400;
let mouse = { x: 20, y: 20 };
let canvasBounds = canvas.getBoundingClientRect();
let target;
canvas.addEventListener("mousedown", (e) => {
mouse.x = e.x - canvasBounds.x;
mouse.y = e.y - canvasBounds.y;
target = new Target();
});
class Ship {
constructor() {
this.x = 20;
this.y = 20;
this.ptA = { x: 15, y: 0 };
this.ptB = { x: -15, y: 10 };
this.ptC = { x: -15, y: -10 };
this.color = "red";
this.angle1 = 0;
this.angle2 = 0;
this.dir = 1;
}
draw() {
ctx.save();
//use translate to move the ship
ctx.translate(this.x, this.y);
//angle1 is the angle from the ship to the target point
//angle2 is the ships current rotation angle. Once they equal each other then the rotation stops. When you click somewhere else they are no longer equal and the ship will rotate again.
if (!this.direction(this.angle1, this.angle2)) {
//see direction() method for more info on this
if (this.dir == 1) {
this.angle2 += 0.05; //change rotation speed here
} else if (this.dir == 0) {
this.angle2 -= 0.05; //change rotation speed here
}
} else {
this.angle2 = this.angle1;
}
ctx.rotate(this.angle2);
ctx.beginPath();
ctx.strokeStyle = this.color;
ctx.moveTo(this.ptA.x, this.ptA.y);
ctx.lineTo(this.ptB.x, this.ptB.y);
ctx.lineTo(this.ptC.x, this.ptC.y);
ctx.closePath();
ctx.stroke();
ctx.restore();
}
driveToTarget() {
//get angle to mouse click
this.angle1 = Math.atan2(mouse.y - this.y, mouse.x - this.x);
//normalize vector
let vecX = mouse.x - this.x;
let vecY = mouse.y - this.y;
let dist = Math.hypot(vecX, vecY);
vecX /= dist;
vecY /= dist;
//Prevent continuous x and y increment by checking if either vec == 0
if (vecX != 0 || vecY != 0) {
//then also give the ship a little buffer incase it passes the given point it doesn't turn back around. This allows time for it to stop if you increase the speed.
if (
this.x >= mouse.x + 3 ||
this.x <= mouse.x - 3 ||
this.y >= mouse.y + 3 ||
this.y <= mouse.y - 3
) {
this.x += vecX; //multiple VecX by n to increase speed (vecX*2)
this.y += vecY; //multiple VecY by n to increase speed (vecY*2)
}
}
}
direction(ang1, ang2) {
//converts rads to degrees and ensures we get numbers from 0-359
let a1 = ang1 * (180 / Math.PI);
if (a1 < 0) {
a1 += 360;
}
let a2 = ang2 * (180 / Math.PI);
if (a2 < 0) {
a2 += 360;
}
//checks whether the target is on the right or left side of the ship.
//We use then to ensure it turns in the shortest direction
if ((360 + a1 - a2) % 360 > 180) {
this.dir = 0;
} else {
this.dir = 1;
}
//Because of animation timeframes there is a chance the ship could turn past the target if rotating too fast. This gives the ship a 1 degree buffer to either side of the target to determine if it is pointed in the right direction.
//We then correct it to the exact degrees in the draw() method above once the if statment defaults to 'else'
if (
Math.trunc(a2) <= Math.trunc(a1) + 1 &&
Math.trunc(a2) >= Math.trunc(a1) - 1
) {
return true;
}
return false;
}
}
let ship = new Ship();
class Target {
constructor() {
this.x = mouse.x;
this.y = mouse.y;
this.r = 3;
this.color = "red";
}
draw() {
ctx.strokeStyle = this.color;
ctx.beginPath();
ctx.arc(this.x, this.y, this.r, 0, Math.PI * 2, false);
ctx.stroke();
}
}
function animate() {
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.fillRect(0, 0, canvas.width, canvas.height);
ship.draw();
ship.driveToTarget();
if (target) {
target.draw();
}
requestAnimationFrame(animate);
}
animate();
<canvas id="canvas"></canvas>

How can I fill color between 4 lines when it connect in canvas?

I am trying to fill color between lines when it connects in ionic. I want to fill color between line when four-line touch each other. For that, I created a canvas demo using a touch event. Please help to solve my issue.
We have 4 lines in the canvas box and we will drag them and connect each line and give a shape line box. that means our line is connected now fill color between line so the box is filled up with color.
html file:
<canvas #canvasDraw width="300" height="300" (touchstart)="handleTouchStart($event)"
(touchmove)="handleTouchmove($event)"
(touchend)="handleTouchEnd($event)">
You need a browser that supports HTML5!
</canvas>
ts file:
import { Component, ElementRef, ViewChild } from '#angular/core';
#Component({
selector: 'app-home',
templateUrl: 'home.page.html',
styleUrls: ['home.page.scss'],
})
export class HomePage {
#ViewChild('canvasDraw', { static: false }) canvas: ElementRef;
canvasElement: any;
lines: any[];
isDown: boolean = false;
startX: number;
startY: number;
nearest: any;
offsetX: any;
offsetY: any;
constructor() {
}
ngAfterViewInit() {
this.canvasElement = this.canvas.nativeElement;
this.lines = [];
this.lines.push({ x0: 75, y0: 25, x1: 125, y1: 25 });
this.lines.push({ x0: 75, y0: 100, x1: 125, y1: 100 });
this.lines.push({ x0: 50, y0: 35, x1: 50, y1: 85 });
this.lines.push({ x0: 150, y0: 35, x1: 150, y1: 85 });
this.draw();
//this.reOffset();
requestAnimationFrame(() => {
this.reOffset()
})
}
reOffset() {
let BB = this.canvasElement.getBoundingClientRect();
this.offsetX = BB.left;
this.offsetY = BB.top;
}
// select the this.nearest line to the mouse
closestLine(mx, my) {
let dist = 100000000;
let index, pt;
for (let i = 0; i < this.lines.length; i++) {
//
let xy = this.closestXY(this.lines[i], mx, my);
//
let dx = mx - xy.x;
let dy = my - xy.y;
let thisDist = dx * dx + dy * dy;
if (thisDist < dist) {
dist = thisDist;
pt = xy;
index = i;
}
}
let line = this.lines[index];
return ({ pt: pt, line: line, originalLine: { x0: line.x0, y0: line.y0, x1: line.x1, y1: line.y1 } });
}
// linear interpolation -- needed in setClosestLine()
lerp(a, b, x) {
return (a + x * (b - a));
}
// find closest XY on line to mouse XY
closestXY(line, mx, my) {
let x0 = line.x0;
let y0 = line.y0;
let x1 = line.x1;
let y1 = line.y1;
let dx = x1 - x0;
let dy = y1 - y0;
let t = ((mx - x0) * dx + (my - y0) * dy) / (dx * dx + dy * dy);
t = Math.max(0, Math.min(1, t));
let x = this.lerp(x0, x1, t);
let y = this.lerp(y0, y1, t);
return ({ x: x, y: y });
}
// draw the scene
draw() {
let ctx = this.canvasElement.getContext('2d');
let cw = this.canvasElement.width;
let ch = this.canvasElement.height;
ctx.clearRect(0, 0, cw, ch);
// draw all lines at their current positions
for (let i = 0; i < this.lines.length; i++) {
this.drawLine(this.lines[i], 'black');
}
// draw markers if a line is being dragged
if (this.nearest) {
// point on line this.nearest to mouse
ctx.beginPath();
ctx.arc(this.nearest.pt.x, this.nearest.pt.y, 5, 0, Math.PI * 2);
ctx.strokeStyle = 'red';
ctx.stroke();
// marker for original line before dragging
this.drawLine(this.nearest.originalLine, 'red');
// hightlight the line as its dragged
this.drawLine(this.nearest.line, 'red');
}
}
drawLine(line, color) {
let ctx = this.canvasElement.getContext('2d');
ctx.beginPath();
ctx.moveTo(line.x0, line.y0);
ctx.lineTo(line.x1, line.y1);
ctx.strokeStyle = color;
ctx.stroke();
}
handleTouchStart(e) {
// tell the browser we're handling this event
let tch = e.touches[0];
// tch.preventDefault();
// tch.stopPropagation();
// mouse position
this.startX = tch.clientX - this.offsetX;
this.startY = tch.clientY - this.offsetY;
// find this.nearest line to mouse
this.nearest = this.closestLine(this.startX, this.startY);
this.draw();
// set dragging flag
this.isDown = true;
}
handleTouchEnd(e) {
// tell the browser we're handling this event
let tch = e.touches[0];
// tch.preventDefault();
// tch.stopPropagation();
// clear dragging flag
this.isDown = false;
this.nearest = null;
this.draw();
}
handleTouchmove(e) {
if (!this.isDown) { return; }
// tell the browser we're handling this event
let tch = e.touches[0];
// tch.preventDefault();
// tch.stopPropagation();
// mouse position
const mouseX = tch.clientX - this.offsetX;
const mouseY = tch.clientY - this.offsetY;
// calc how far mouse has moved since last mousemove event
let dx = mouseX - this.startX;
let dy = mouseY - this.startY;
this.startX = mouseX;
this.startY = mouseY;
// change this.nearest line vertices by distance moved
let line = this.nearest.line;
line.x0 += dx;
line.y0 += dy;
line.x1 += dx;
line.y1 += dy;
// redraw
this.draw();
let ctx = this.canvasElement.getContext('2d');
ctx.beginPath();
ctx.rect(line.x0, line.y0, line.x1, line.y1);
ctx.fillStyle = "red";
ctx.fill();
}
}
How to fill color when four line touch or connect?

How to get a pixel's color
Basically, when you do your touch event, pixels are changing color. You can find out which pixel(s) are/were affected from the event. Having the pixel(s) as input you can find out what color a pixel has: https://jsfiddle.net/ourcodeworld/8swevoxo/
var canvas = document.getElementById("canvas");
function getPosition(obj) {
var curleft = 0, curtop = 0;
if (obj.offsetParent) {
do {
curleft += obj.offsetLeft;
curtop += obj.offsetTop;
} while (obj = obj.offsetParent);
return { x: curleft, y: curtop };
}
return undefined;
}
function rgbToHex(r, g, b) {
if (r > 255 || g > 255 || b > 255)
throw "Invalid color component";
return ((r << 16) | (g << 8) | b).toString(16);
}
function drawImageFromWebUrl(sourceurl){
var img = new Image();
img.addEventListener("load", function () {
// The image can be drawn from any source
canvas.getContext("2d").drawImage(img, 0, 0, img.width, img.height, 0, 0, canvas.width, canvas.height);
});
img.setAttribute("src", sourceurl);
}
// Draw a base64 image because this is a fiddle, and if we try with an image from URL we'll get tainted canvas error
// Read more about here : http://ourcodeworld.com/articles/read/182/the-canvas-has-been-tainted-by-cross-origin-data-and-tainted-canvases-may-not-be-exported
drawImageFromWebUrl("data:image/gif;base64,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");
canvas.addEventListener("mousemove",function(e){
var pos = getPosition(this);
var x = e.pageX - pos.x;
var y = e.pageY - pos.y;
var coord = "x=" + x + ", y=" + y;
var c = this.getContext('2d');
var p = c.getImageData(x, y, 1, 1).data;
// If transparency on the image
if((p[0] == 0) && (p[1] == 0) && (p[2] == 0) && (p[3] == 0)){
coord += " (Transparent color detected, cannot be converted to HEX)";
}
var hex = "#" + ("000000" + rgbToHex(p[0], p[1], p[2])).slice(-6);
document.getElementById("status").innerHTML = coord;
document.getElementById("color").style.backgroundColor = hex;
},false);
<canvas id="canvas" width="150" height="150"></canvas>
<div id="status"></div><br>
<div id="color" style="width:30px;height:30px;"></div>
<p>
Move the mouse over the BUS !
</p>
This code was not written by me.
Understanding the problem
Now that we know how to get the color pixel by pixel, we need to convert our question to something that is equivalent, but it's easier to compute. I would define the problem as follows:
Traverse pixels starting from a given point in a continuous manner in
order to find out whether a cycle can be formed, which contains pixels
of a certain (default) color inside the boundary, which should change
their color to some fill color.
How to solve it
Start a cycle of pixel traversing starting from a given point and always changing coordinate values in each step to a point neighboring the current point or some point visited earlier in the loop
Always store the coordinates of the currently visited point so if you would visit the same point, you just ignore it on the second time
Use a data structure, like a stack to store all neighbors to visit, but not yet visited, put in the stack all neighbors of each point you visit (unless the point was already visited)
If you ever arrive back to the starting point from a point which was not already visited, then register that this is a cycle
Always keep track of boundary points
When you found out that it's a cycle and you know where it is located, traverse each point in the region and if it has a default color, check whether there is a continuous line by which you could "leave" the cycle by visiting only neighbors of default color, pretty similarly as you have checked whether it is a cycle. If not, then paint the pixel (and all its neighbors of similar color) to the color you need

I wrote the following example using web technologies as im not familiar with ionic.
const canvas = document.querySelector('canvas');
const context = canvas.getContext('2d');
const width = canvas.width;
const height = canvas.height;
// compute signed triangle area of triangle abc. Negative if triangle is cw.
const area = (a, b, c) => {
return (a[0] - c[0]) * (b[1] - c[1]) - (a[1] - c[1]) * (b[0] - c[0]);
};
// compute intersections of line segments.
const intersect = (s0, s1) => {
let result = null;
const a1 = area(s0.p0, s0.p1, s1.p1);
const a2 = area(s0.p0, s0.p1, s1.p0);
if (a1 * a2 < 0) {
const a3 = area(s1.p0, s1.p1, s0.p0);
const a4 = a3 + a2 - a1;
if (a3 * a4 < 0) {
const t = a3 / (a3 - a4);
result = [
s0.p0[0] + t * (s0.p1[0] - s0.p0[0]),
s0.p0[1] + t * (s0.p1[1] - s0.p0[1])
];
}
}
return result;
};
const dot = (a, b) => a[0] * b[0] + a[1] * b[1];
const sub = (a, b) => [a[0] - b[0], a[1] - b[1]];
const clamp = (x, min, max) => Math.min(max, Math.max(x, min));
const min = (points) => {
return points.reduce((r, p) => {
return [
Math.min(p[0], r[0]),
Math.min(p[1], r[1])
];
}, [Infinity, Infinity]);
};
const max = (points) => {
return points.reduce((r, p) => {
return [
Math.max(p[0], r[0]),
Math.max(p[1], r[1])
];
}, [-Infinity, -Infinity]);
};
const state = {
lines: [{
p0: [75, 25],
p1: [125, 25]
},
{
p0: [75, 100],
p1: [125, 100]
},
{
p0: [50, 35],
p1: [50, 85]
},
{
p0: [150, 35],
p1: [150, 85]
}
],
dragging: false,
target: null,
m: null
};
const draw = (state) => {
context.clearRect(0, 0, width, height);
const intersections = [];
for (let i = 0; i < state.lines.length; i++) {
const lines = state.lines;
const l0 = lines[i];
context.strokeStyle = "black";
for (let j = 0; j < lines.length; j++) {
if (j !== i) {
const intersection = intersect(l0, lines[j]);
if (intersection != null) {
intersections.push(intersection);
context.strokeStyle = "blue";
}
}
}
context.beginPath();
context.moveTo(l0.p0[0], l0.p0[1]);
context.lineTo(l0.p1[0], l0.p1[1]);
context.stroke();
}
if (intersections.length == 8) {
const p = min(intersections);
const q = max(intersections);
context.fillStyle = "red";
context.fillRect(p[0], p[1], q[0] - p[0], q[1] - p[1]);
}
};
const run = () => {
requestAnimationFrame(() => {
draw(state);
run();
});
};
canvas.addEventListener('mousedown', (e) => {
const rect = e.target.getBoundingClientRect();
const m = [e.x - rect.x, e.y - rect.y];
const lines = state.lines
.map((line) => {
const ab = sub(line.p1, line.p0);
const am = sub(m, line.p0);
const bm = sub(m, line.p1);
const e = dot(am, ab);
if (e <= 0) {
return dot(am, am);
}
const f = dot(ab, ab);
if (e >= f) {
return dot(bm, bm);
}
return dot(am, am) - e * e / f;
})
.map((d, i) => [d, state.lines[i]])
.sort((a, b) => a[0] - b[0]);
const first = lines[0];
if (first[0] < 50) {
state.dragging = true;
state.target = first[1];
state.m = m;
}
});
canvas.addEventListener('mouseup', () => {
state.dragging = false;
state.target = null;
state.m = null;
});
canvas.addEventListener('mousemove', (e) => {
if (state.dragging) {
const rect = e.target.getBoundingClientRect();
const m = [e.x - rect.x, e.y - rect.y];
const dx = m[0] - state.m[0];
const dy = m[1] - state.m[1];
state.target.p0[0] = dx + state.target.p0[0];
state.target.p0[1] = dy + state.target.p0[1];
state.target.p1[0] = dx + state.target.p1[0];
state.target.p1[1] = dy + state.target.p1[1];
state.m = m;
}
}, {
passive: true
});
run();
<canvas width="600" height="800"><canvas>
Solution
The real meat of the solution is the intersect function. What this does it it takes two linesegments and calculates the point that they intersect. This is really the only thing that you were missing. The intersect algorithm is a bit too long to discuss here but you can take a look at the details here. After we do a pretty inefficient O(n^2) intersection query between every single line segment. We then can figure out if we have have enough intersections for our rectangle. I compute both A -> B and B -> A intersections discretely with out any regard of culling the intersection candidate since i'm dealing with 4 line segments. What this means is we need 8 intersections to close our rectangle. Once we know we have a closed polygon we can draw a rectangle using the min(...)/max(...) of all x/y components of all intersections. This gives us two points P,Q that we can then use to fill in a rect.
If you are into this kind of thing I highly recommend the book Real Time Collision Detection by Christer Ericson or similar.

How to track coordinates on the quadraticCurve

I have a Polyline on the HiDPICanvas (html5 canvas). When I move mouse left and right I track its coordinates and on corresponding point with same X coordinate on the polyline I draw a Circle. You can try it now to see the result.
// Create a canvas
var HiDPICanvas = function(container_id, color, w, h) {
/*
objects are objects on the canvas, first elements of dictionary are background elements, last are on the foreground
canvas will be placed in the container
canvas will have width w and height h
*/
var objects = {
box : [],
borders : [],
circles : [],
polyline: []
}
var doNotMove = ['borders']
// is mouse down & its coords
var mouseDown = false
lastX = window.innerWidth/2
lastY = window.innerHeight/2
// return pixel ratio
var getRatio = function() {
var ctx = document.createElement("canvas").getContext("2d");
var dpr = window.devicePixelRatio || 1;
var bsr = ctx.webkitBackingStorePixelRatio ||
ctx.mozBackingStorePixelRatio ||
ctx.msBackingStorePixelRatio ||
ctx.oBackingStorePixelRatio ||
ctx.backingStorePixelRatio || 1;
return dpr / bsr;
}
// return high dots per inch canvas
var createHiDPICanvas = function() {
var ratio = getRatio();
var chart_container = document.getElementById(container_id);
var can = document.createElement("canvas");
can.style.backgroundColor = color
can.width = w * ratio;
can.height = h * ratio;
can.style.width = w + "px";
can.style.height = h + "px";
can.getContext("2d").setTransform(ratio, 0, 0, ratio, 0, 0);
chart_container.appendChild(can);
return can;
}
// add object to the canvas
var add = function(object, category) {
objects[category].push(object)
}
// clear canvas
var clearCanvas = function(x0, y0, x1, y1) {
ctx.clearRect(x0, y0, x1, y1);
ctx.beginPath();
ctx.globalCompositeOperation = "source-over";
ctx.globalAlpha = 1;
ctx.closePath();
}
// check function do I can move this group of objects
var canMove = function(groupname) {
for (var i = 0; i < doNotMove.length; i++) {
var restricted = doNotMove[i]
if (restricted == groupname) {
return false
}
}
return true
}
// refresh all objects on the canvas
var refresh = function() {
clearCanvas(0, 0, w, h)
var object
for (var key in objects) {
for (var i = 0; i < objects[key].length; i++) {
object = objects[key][i]
object.refresh()
}
}
}
// shift all objects on the canvas except left and down borders and its content
var shiftObjects = function(event) {
event.preventDefault()
// if mouse clicked now -> we can move canvas view left\right
if (mouseDown) {
var object
for (var key in objects) {
if (canMove(key)) {
for (var i = 0; i < objects[key].length; i++) {
object = objects[key][i]
object.move(event.movementX, event.movementY)
}
}
}
cci.refresh()
}
}
// transfer x to canvas drawing zone x coord (for not drawing on borders of the canvas)
var transferX = function(x) {
return objects.borders[0].width + x
}
var transferCoords = function(x, y) {
// no need to transfer y because borders are only at the left
return {
x : transferX(x),
y : y
}
}
// change mouse state on the opposite
var toggleMouseState = function() {
mouseDown = !mouseDown
}
// make mouseDown = false, (bug removal function when mouse down & leaving the canvas)
var refreshMouseState = function() {
mouseDown = false
}
// print information about all objects on the canvas
var print = function() {
var groupLogged = true
console.log("Objects on the canvas:")
for (var key in objects) {
groupLogged = !groupLogged
if (!groupLogged) {console.log(key, ":"); groupLogged = !groupLogged}
for (var i = 0 ; i < objects[key].length; i++) {
console.log(objects[key][i])
}
}
}
var restrictEdges = function() {
console.log("offsetLeft", objects['borders'][0])
}
var getMouseCoords = function() {
return {
x : lastX,
y : lastY
}
}
var addCircleTracker = function() {
canvas.addEventListener("mousemove", (e) => {
var polyline = objects.polyline[0]
var mouseCoords = getMouseCoords()
var adjNodes = polyline.get2NeighbourNodes(mouseCoords.x)
if (adjNodes != -1) {
var prevNode = adjNodes.prev
var currNode = adjNodes.curr
var cursorNode = polyline.linearInterpolation(prevNode, currNode, mouseCoords.x)
// cursorNode.cursorX, cursorNode.cursorY are coords
// for circle that should be drawn on the polyline
// between the closest neighbour nodes
var circle = objects.circles[0]
circle.changePos(cursorNode.x, cursorNode.y)
refresh()
}
})
}
// create canvas
var canvas = createHiDPICanvas()
addCircleTracker()
// we created canvas so we can track mouse coords
var trackMouse = function(event) {
lastX = event.offsetX || (event.pageX - canvas.offsetLeft)
lastY = event.offsetY || (event.pageY - canvas.offsetTop)
}
// 2d context
var ctx = canvas.getContext("2d")
// add event listeners to the canvas
canvas.addEventListener("mousemove" , shiftObjects )
canvas.addEventListener("mousemove", (e) =>{ trackMouse(e) })
canvas.addEventListener("mousedown" , () => { toggleMouseState () })
canvas.addEventListener("mouseup" , () => { toggleMouseState () })
canvas.addEventListener("mouseleave", () => { refreshMouseState() })
canvas.addEventListener("mouseenter", () => { refreshMouseState() })
return {
// base objects
canvas : canvas,
ctx : ctx,
// sizes of the canvas
width : w,
height : h,
color : color,
// add object on the canvas for redrawing
add : add,
print : print,
// refresh canvas
refresh: refresh,
// objects on the canvas
objects: objects,
// get mouse coords
getMouseCoords : getMouseCoords
}
}
// cci -> canvas ctx info (dict)
var cci = HiDPICanvas("lifespanChart", "bisque", 780, 640)
var ctx = cci.ctx
var canvas = cci.canvas
var Polyline = function(path, color) {
var create = function() {
if (this.path === undefined) {
this.path = path
this.color = color
}
ctx.save()
ctx.beginPath()
p = this.path
ctx.fillStyle = color
ctx.moveTo(p[0].x, p[0].y)
for (var i = 0; i < p.length - 1; i++) {
var currentNode = p[i]
var nextNode = p[i+1]
// draw smooth polyline
// var xc = (currentNode.x + nextNode.x) / 2;
// var yc = (currentNode.y + nextNode.y) / 2;
// taken from https://stackoverflow.com/a/7058606/13727076
// ctx.quadraticCurveTo(currentNode.x, currentNode.y, xc, yc);
// draw rough polyline
ctx.lineTo(currentNode.x, currentNode.y)
}
ctx.stroke()
ctx.restore()
ctx.closePath()
}
// circle that will track mouse coords and be
// on the corresponding X coord on the path
// following mouse left\right movements
var circle = new Circle(50, 50, 5, "purple")
cci.add(circle, "circles")
create()
var get2NeighbourNodes = function(x) {
// x, y are cursor coords on the canvas
//
// Get 2 (left and right) neighbour nodes to current cursor x,y
// N are path nodes, * is Node we search coords for
//
// N-----------*----------N
//
for (var i = 1; i < this.path.length; i++) {
var prevNode = this.path[i-1]
var currNode = this.path[i]
if ( prevNode.x <= x && currNode.x >= x ) {
return {
prev : prevNode,
curr : currNode
}
}
}
return -1
}
var linearInterpolation = function(prevNode, currNode, cursorX) {
// calculate x, y for the node between 2 nodes
// on the path using linearInterpolation
// https://en.wikipedia.org/wiki/Linear_interpolation
var cursorY = prevNode.y + (cursorX - prevNode.x) * ((currNode.y - prevNode.y)/(currNode.x - prevNode.x))
return {
x : cursorX,
y : cursorY
}
}
var move = function(diff_x, diff_y) {
for (var i = 0; i < this.path.length; i++) {
this.path[i].x += diff_x
this.path[i].y += diff_y
}
}
return {
create : create,
refresh: create,
move : move,
get2NeighbourNodes : get2NeighbourNodes,
linearInterpolation : linearInterpolation,
path : path,
color : color
}
}
var Circle = function(x, y, radius, fillStyle) {
var create = function() {
if (this.x === undefined) {
this.x = x
this.y = y
this.radius = radius
this.fillStyle = fillStyle
}
ctx.save()
ctx.beginPath()
ctx.arc(this.x, this.y, radius, 0, 2*Math.PI)
ctx.fillStyle = fillStyle
ctx.strokeStyle = fillStyle
ctx.fill()
ctx.stroke()
ctx.closePath()
ctx.restore()
}
create()
var changePos = function(new_x, new_y) {
this.x = new_x
this.y = new_y
}
var move = function(diff_x, diff_y) {
this.x += diff_x
this.y += diff_y
}
return {
refresh : create,
create : create,
changePos: changePos,
move : move,
radius : radius,
x : this.x,
y : this.y
}
}
var Node = function(x, y) {
this.x = x
this.y = y
return {
x : this.x,
y : this.y
}
}
var poly = new Polyline([
Node(30,30), Node(150,150),
Node(290, 150), Node(320,200),
Node(350,350), Node(390, 250),
Node(450, 140)
], "green")
cci.add(poly, "polyline")
<div>
<div id="lifespanChart"></div>
</div>
But if you go to the comment draw smooth polyline and uncomment code below (and comment line that draws rough polyline) - it will draw smooth polyline now (quadratic Bézier curve). But when you try to move mouse left and right - Circle sometimes goes out of polyline bounds.
before quadratic curve:
after quadratic curve:
Here is a question : I calculated x, y coordinates for the Circle on the rough polyline using linear interpolation, but how could I calculate x, y coordinates for the Circle on the smooth quadratic curve?
ADD 1 : QuadraticCurve using Beizer curve as a base in calculations when smoothing polyline
ADD 2 For anyone who a little stucked with the implementation I found & saved easier solution from here, example:
var canvas = document.getElementById("canv")
var canvasRect = canvas.getBoundingClientRect()
var ctx = canvas.getContext('2d')
var p0 = {x : 30, y : 30}
var p1 = {x : 20, y :100}
var p2 = {x : 200, y :100}
var p3 = {x : 200, y :20}
// Points are objects with x and y properties
// p0: start point
// p1: handle of start point
// p2: handle of end point
// p3: end point
// t: progression along curve 0..1
// returns an object containing x and y values for the given t
// link https://stackoverflow.com/questions/14174252/how-to-find-out-y-coordinate-of-specific-point-in-bezier-curve-in-canvas
var BezierCubicXY = function(p0, p1, p2, p3, t) {
var ret = {};
var coords = ['x', 'y'];
var i, k;
for (i in coords) {
k = coords[i];
ret[k] = Math.pow(1 - t, 3) * p0[k] + 3 * Math.pow(1 - t, 2) * t * p1[k] + 3 * (1 - t) * Math.pow(t, 2) * p2[k] + Math.pow(t, 3) * p3[k];
}
return ret;
}
var draw_poly = function () {
ctx.beginPath()
ctx.lineWidth=2
ctx.strokeStyle="white"
ctx.moveTo(p0.x, p0.y)// start point
// cont cont end
ctx.bezierCurveTo(p1.x, p1.y, p2.x, p2.y, p3.x, p3.y)
ctx.stroke()
ctx.closePath()
}
var clear_canvas = function () {
ctx.clearRect(0,0,300,300);
ctx.beginPath();
ctx.globalCompositeOperation = "source-over";
ctx.globalAlpha = 1;
ctx.closePath();
};
var draw_circle = function(x, y) {
ctx.save();
// semi-transparent arua around the circle
ctx.globalCompositeOperation = "source-over";
ctx.beginPath()
ctx.fillStyle = "white"
ctx.strokeStyle = "white"
ctx.arc(x, y, 5, 0, 2 * Math.PI);
ctx.stroke();
ctx.closePath();
ctx.restore();
}
var refresh = function(circle_x, circle_y) {
clear_canvas()
draw_circle(circle_x, circle_y)
draw_poly()
}
var dist = function(mouse, point) {
return Math.abs(mouse.x - point.x)
// return ((mouse.x - point.x)**2 + (mouse.y - point.y)**2)**0.5
}
var returnClosest = function(curr, prev) {
if (curr < prev) {
return curr
}
return prev
}
refresh(30,30)
canvas.addEventListener("mousemove", (e) => {
var mouse = {
x : e.clientX - canvasRect.left,
y : e.clientY - canvasRect.top
}
var Point = BezierCubicXY(p0, p1, p2, p3, 0)
for (var t = 0; t < 1; t += 0.01) {
var nextPoint = BezierCubicXY(p0, p1, p2, p3, t)
if (dist(mouse, Point) > dist(mouse, nextPoint)) {
Point = nextPoint
}
// console.log(Point)
}
refresh(Point.x, Point.y)
})
canvas {
background: grey;
}
<canvas id="canv" width = 300 height = 300></canvas>
Just iterate through all the lines of the curve & find closest position using this pattern

This can be done using an iterative search, as you have done with the lines.
BTW there is a much better way to find the closest point on a line that has a complexity of O(1) rather than O(n) where n is length of line segment.
Search for closest point
The following function can be used for both quadratic and cubic beziers and returns the unit position of closest point on bezier to a given coordinate.
The function also has a property foundPoint that has the position of the point found
The function uses the object Point that defines a 2D coordinate.
Signatures
The function has two signatures, one for quadratic beziers and the other for cubic.
closestPointOnBezier(point, resolution, p1, p2, cp1)
closestPointOnBezier(point, resolution, p1, p2, cp1, cp2)
Where
point as Point is the position to check
resolution as Number The approx resolution to search the bezier. If 0 then this is fixed to DEFAULT_SCAN_RESOLUTION else it is the distance between start and end points times resolution IE if resolution = 1 then approx scan is 1px, if resolution = 2 then approx scan is 1/2px
p1, p2 as Point's are the start and end points of the bezier
cp1, cp2 as Point's are the first and/or second control points of the bezier
Results
They both return Number that is the unit pos on the bezier of closest point. The value will be 0 <= result <= 1 Where 0 is at start of bezier and 1 is end
The function property closestPointOnBezier.foundPoint as Point has the coordinate of the closest point on the bezier and can be used to calculate the distance to the point on the bezier.
The function
const Point = (x = 0, y = 0) => ({x, y});
const MAX_RESOLUTION = 2048;
const DEFAULT_SCAN_RESOLUTION = 256;
closestPointOnBezier.foundPoint = Point();
function closestPointOnBezier(point, resolution, p1, p2, cp1, cp2) {
var unitPos, a, b, b1, c, i, vx, vy, closest = Infinity;
const v1 = Point(p1.x - point.x, p1.y - point.y);
const v2 = Point(p2.x - point.x, p2.y - point.y);
const v3 = Point(cp1.x - point.x, cp1.y - point.y);
resolution = resolution > 0 && reolution < MAX_RESOLUTION ? (Math.hypot(p1.x - p2.x, p1.y - p2.y) + 1) * resolution : 100;
const fp = closestPointOnBezier.foundPoint;
const step = 1 / resolution;
const end = 1 + step / 2;
const checkClosest = (e = (vx * vx + vy * vy) ** 0.5) => {
if (e < closest ){
unitPos = i;
closest = e;
fp.x = vx;
fp.y = vy;
}
}
if (cp2 === undefined) { // find quadratic
for (i = 0; i <= end; i += step) {
a = (1 - i);
c = i * i;
b = a*2*i;
a *= a;
vx = v1.x * a + v3.x * b + v2.x * c;
vy = v1.y * a + v3.y * b + v2.y * c;
checkClosest();
}
} else { // find cubic
const v4 = Point(cp2.x - point.x, cp2.y - point.y);
for (i = 0; i <= end; i += step) {
a = (1 - i);
c = i * i;
b = 3 * a * a * i;
b1 = 3 * c * a;
a = a * a * a;
c *= i;
vx = v1.x * a + v3.x * b + v4.x * b1 + v2.x * c;
vy = v1.y * a + v3.y * b + v4.y * b1 + v2.y * c;
checkClosest();
}
}
return unitPos < 1 ? unitPos : 1; // unit pos on bezier. clamped
}
Usage
Example usage to find closest point on two beziers
The defined geometry
const bzA = {
p1: Point(10, 100), // start point
p2: Point(200, 400), // control point
p3: Point(410, 500), // end point
};
const bzB = {
p1: bzA.p3, // start point
p2: Point(200, 400), // control point
p3: Point(410, 500), // end point
};
const mouse = Point(?,?);
Finding closest
// Find first point
closestPointOnBezier(mouse, 2, bzA.p1, bzA.p3, bzA.p2);
// copy point
var found = Point(closestPointOnBezier.foundPoint.x, closestPointOnBezier.foundPoint.y);
// get distance to mouse
var dist = Math.hypot(found.x - mouse.x, found.y - mouse.y);
// find point on second bezier
closestPointOnBezier(mouse, 2, bzB.p1, bzB.p3, bzB.p2);
// get distance of second found point
const distB = Math.hypot(closestPointOnBezier.foundPoint.x - mouse.x, closestPointOnBezier.foundPoint.y - mouse.y);
// is closer
if (distB < dist) {
found.x = closestPointOnBezier.foundPoint.x;
found.y = closestPointOnBezier.foundPoint.y;
dist = distB;
}
The closet point is found as Point

Cannot get the coord of a projected point on an axis

I'm trying to do a function who take a point position and an axis (a line) and return the projection of the point on the axis.
Point are simply represented by {x, y} coords
Axis are represented by an f(x) = ax + b linear function (or f(x) = C for vertical line) and the rad value for the a (where 0rad is right, and turn clockwise). So axis = {a, b, rad} or axis = {c, rad}.
It should be a simple mathematical problem using Pythagorean but i cannot get the right solution.. Projections are always looking on the other directions (but look at the good distance).
Code demo:
I create a dark point on 25;100 and multiples coloured axis. Every axis have a related point (same color) who represent the projection of the dark point on it. Both grey are vertical and horizontal examples and work well but 3 others axis (Red, Green and Blue) are wrong.
I tried to redo getProjectionOnAxis function multiple times, and even try to get the good results by testing all possibility (by inverse vars, using another cos/sin/tan function) and try to catch my issue like that but nothing to do I never get the good result.
const point = {x:25, y:100};
const axis = [
{a: 1, b: 25, rad:Math.PI/4, rgb: '255,0,0'}, // 45deg
{a: -.58, b: 220, rad:Math.PI/6, rgb: '0,255,0'}, // -30deg
{a: 3.73, b: -50, rad:5*Math.PI/12, rgb: '0,0,255'}, // 75deg
// Gray
{c: 150, rad:Math.PI/2, rgb: '100,100,100'},
{b: 150, rad:0},
];
const execute = () => {
const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');
drawPoint({...point, rgb:'0,0,0', ctx});
axis.forEach(_axis => {
drawAxis({axis:_axis, rgb: _axis.rgb, ctx, canvas});
const projected = getProjectionOnAxis({...point, axis: _axis});
drawPoint({...projected, rgb:_axis.rgb, ctx});
});
}
// --- Issue come from here ---
const getProjectionOnAxis = ({x,y,axis}) => {
// On vertical axis |
if (axis.c) {
return { x: axis.c, y };
}
// On horizontal axis _
if (!axis.a) {
return { x, y: axis.b };
}
// Projected on axis but perpendicular to X axis
const projectedOnX = {
x,
y: axis.a * x + axis.b,
}
// The distance between both points is the hypothenus of the triangle:
// point, projectedOnAxis, projectedOnX
// where point <-> projectedOnX is the hypothenus
// and point angle is same that axis
const distCornerToProjectedOnX = Math.abs(y - projectedOnX.y);
const distCornerToProjectedOnAxis = distCornerToProjectedOnX * Math.cos(axis.rad);
const projectedVector = {
x: distCornerToProjectedOnAxis * Math.cos(axis.rad),
y: distCornerToProjectedOnAxis * Math.sin(axis.rad),
};
return {
x: x + projectedVector.x,
y: y + projectedVector.y,
};
}
// --- Draw Functions ---
// Not really important for the issue
const drawPoint = ({x,y,rgb, ctx}) => {
ctx.save();
ctx.translate(x, y);
ctx.beginPath();
ctx.fillStyle = `rgba(${rgb},1)`;
ctx.arc(0, 0, 2, 0, Math.PI*2, true);
ctx.closePath();
ctx.fill();
ctx.restore();
};
const drawAxis = ({axis, rgb, ctx, canvas}) => {
if (axis.c) {
// Vertical axis
drawLine({
from: {x: axis.c, y:0},
to: {x:axis.c, y:canvas.height},
rgb, ctx
});
}
else if (!axis.a) {
// Horizontal axis
drawLine({
from: {x:0, y:axis.b},
to: {x:canvas.width, y:axis.b},
rgb, ctx
});
}
else {
// ax + b (here a != 0)
let to = {
x: canvas.width,
y: axis.a * canvas.width + axis.b,
};
if (to.y < 0) {
to = {
x: axis.b / - axis.a,
y: 0,
}
}
drawLine({
from: {x:0, y:axis.b},
to,
rgb, ctx
});
}
}
const drawLine = ({
from, to, rgb=null, ctx
}) => {
ctx.save();
ctx.translate(from.x, from.y);
ctx.beginPath();
ctx.strokeStyle = `rgba(${rgb},1)`;
ctx.moveTo(0, 0);
ctx.lineTo(to.x - from.x, to.y - from.y);
ctx.stroke();
ctx.restore();
};
execute();
html, body, canvas { margin: 0; padding: 0;}
<canvas id="canvas" width="500" height="500"></canvas>
This is the positions I should get:
PS: I don't really like the way I manage my axis, maybe they are another (simple) way to do it ?

Represent line (your axis) in parametric form as base point A and unit direction vector d = (dx, dy). It is universal representation suitable for all slopes. If you have slope angle fi relative to OX axis, then dx=cos(fi), dy=sin(fi)
L = P0 + d * t
Then projection of point C onto line is (using scalar product)
AC = C - A
P = A + d * (d.dot.AC)
In coordinates
dotvalue = dx * (C.x - A.x) + dy * (C.y - A.y)
P.x = A.x + d.x * dotvalue
P.y = A.y + d.y * dotvalue

Get bounds of unrotated rotated rectangle

I have a rectangle that has a rotation already applied to it. I want to get the the unrotated dimensions (the x, y, width, height).
Here is the dimensions of the element currently:
Bounds at a 90 rotation: {
height 30
width 0
x 25
y 10
}
Here are the dimensions after the rotation is set to none:
Bounds at rotation 0 {
height 0
width 30
x 10
y 25
}
In the past, I was able to set the rotation to 0 and then read the updated bounds . However, there is a bug in one of the functions I was using, so now I have to do it manually.
Is there a simple formula to get the bounds at rotation 0 using the info I already have?
Update: The object is rotated around the center of the object.
UPDATE:
What I need is something like the function below:
function getRectangleAtRotation(rect, rotation) {
var rotatedRectangle = {}
rotatedRectangle.x = Math.rotation(rect.x * rotation);
rotatedRectangle.y = Math.rotation(rect.y * rotation);
rotatedRectangle.width = Math.rotation(rect.width * rotation);
rotatedRectangle.height = Math.rotation(rect.height * rotation);
return rotatedRectangle;
}
var rectangle = {x: 25, y: 10, height: 30, width: 0 };
var rect2 = getRectangleAtRotation(rect, -90); // {x:10, y:25, height:0, width:30 }
I found a similar question here.
UPDATE 2
Here is the code I have. It attempts to get the center point of the line and then the x, y, width, and height:
var centerPoint = getCenterPoint(line);
var lineBounds = {};
var halfSize;
halfSize = Math.max(Math.abs(line.end.x-line.start.x)/2, Math.abs(line.end.y-line.start.y)/2);
lineBounds.x = centerPoint.x-halfSize;
lineBounds.y = centerPoint.y;
lineBounds.width = line.end.x;
lineBounds.height = line.end.y;
function getCenterPoint(node) {
return {
x: node.boundsInParent.x + node.boundsInParent.width/2,
y: node.boundsInParent.y + node.boundsInParent.height/2
}
}
I know the example I have uses a right angle and that you can swap the x and y with that but the rotation can be any amount.
UPDATE 3
I need a function that returns the unrotated bounds of a rectangle. I have the bounds at a specific rotation already.
function getUnrotatedRectangleBounds(rect, currentRotation) {
// magic
return unrotatedRectangleBounds;
}

I think I can handle the calculation of the bounds size without too much effort (few equations), I'm not sure, instead, how you would like x and y to be handled.
First, let's properly name things:
Now, we want to rotate it by some angle alpha (in radians):
To calculate the green sides, it is clear that it's made of two repeated rectangle-triangles as the following:
So, solving angles first, we know that:
the sum of the angles of a triangle is PI, or 180°;
the rotation is alpha;
one angle gamma is PI / 2, or 90°;
the last angle, beta, is gamma - alpha;
Now, knowing all the angles and a side, we can use the Law of Sines to calculate other sides.
As a brief recap, the Law of Sines tells us that there is an equality between the ratio of a side length and it's opposite angle. More info here: https://en.wikipedia.org/wiki/Law_of_sines
In our case, for the upper left triangle (and the bottom right one), we have:
Remember that AD is our original height.
Given that the sin(gamma) is 1, and we also know the value of AD, we can write the equations:
For the upper right triangle (and the bottom left one), we then have:
Having all needed sides, we can easily calculate the width and height:
width = EA + AF
height = ED + FB
At this point we can write a quite easy method that, given a rectangle and a rotation angle in radians, can return new bounds:
function rotate(rectangle, alpha) {
const { width: AB, height: AD } = rectangle
const gamma = Math.PI / 4,
beta = gamma - alpha,
EA = AD * Math.sin(alpha),
ED = AD * Math.sin(beta),
FB = AB * Math.sin(alpha),
AF = AB * Math.sin(beta)
return {
width: EA + AF,
height: ED + FB
}
}
This method can then be used like:
const rect = { width: 30, height: 50 }
const rotation = Math.PI / 4.2 // this is a random value it put here
const bounds = rotate(rect, rotation)
Hope there aren't typos...

I think I might get a solution but, for safety, I prefer to prior repeat what we have and what we need to be sure I understood everything correctly. As I said in a comment, english isn't my native language and I already wrote a wrong answer due to my lack of understanding of the problem :)
What we have
We know that at x and y there is a bounds rectangle (green) of size w and h that contains another rectangle (the grey dotted one) rotated of alpha degrees.
We know that the y axis is flipped relatively to the Cartesian one, and that makes the angle to be considered clockwise instead of counter-clockwise.
What we need
At first, we need to find the 4 vertices of the inner rectangle (A, B, C and D) and, knowing the position of the vertices, the size of the inner rectangle (W and H).
As a second step, we need to counter rotate the inner rectangle to 0 degrees, and find it's position X and Y.
Find the vertices
Generally speaking for each vertex we know only one coordinate, the x or the y. The other one "slides" along the side of the bounding box in relation to the angle alpha.
Let's start with A: we know Ay, we need Ax.
We know that the Ax lies between x and x + w in relation to the angle alpha.
When alpha is 0°, Ax is x + 0. When alpha is 90°, Ax is x + w. When alpha is 45°, Ax is x + w / 2.
Basically, Ax grows in relation of the sin(alpha), giving us:
Having Ax, we can easily compute Cx:
In the same way we can compute By and then Dy:
Writing some code:
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// alpha is the rotation IN RADIANS
const vertices = (bounds, alpha) => {
const { x, y, w, h } = bounds,
A = { x: x + w * Math.sin(alpha), y },
B = { x, y: y + h * Math.sin(alpha) },
C = { x: x + w - w * Math.sin(alpha), y },
D = { x, y: y + h - h * Math.sin(alpha) }
return { A, B, C, D }
}
Finding the sides
Now that we have all the vertices, we can easily compute the inner rectangle sides, we need to define a couple more points E and F for clarity of explanation:
Its clearly visible that we can use the Pitagorean Theorem to compute W and H with:
where:
In code:
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// vertices is a POJO with shape: { A, B, C, D }, as returned by the `vertices` method
const sides = (bounds, vertices) => {
const { x, y, w, h } = bounds,
{ A, B, C, D } = vertices,
EA = A.x - x,
ED = D.y - y,
AF = w - EA,
FB = h - ED,
H = Math.sqrt(EA * EA + ED * ED),
W = Math.sqrt(AF * AF + FB * FB
return { h: H, w: W }
}
Finding the position of the counter-rotated inner rectangle
First of all, we have to find the angles (beta and gamma) of the diagonals of the inner rectangle.
Let's zoom in a little bit and add some additional letters for more clarity:
We can use the Law of Sines to get the equations to compute beta:
To make some calculations we have:
We need to compute GC first in order to have at least one side of the equation completely known. GC is the radius of the circumference the inner rectangle is inscribed in and also half of the inner rectangle diagonal.
Having the two sides of the inner rectangle, we can use the Pitagorean Theorem again:
With GC we can solve the Law of Sines on beta:
we know that sin(delta) is 1
Now, beta is the angle of the vertex C in relation with the unrotated x axis.
Looking again at this image, we can easily get the angles of all the other vertices:
Now that we have almost everything, we can compute the new coordinates of the A vertex:
From here, we need to translate both Ax and Ay because they are related to the center of the circumference, which is x + w / 2 and y + h / 2:
So, writing the last piece of code:
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// sides is a POJO with shape: { w, h }, as returned by the `sides` method
const origin = (bounds, sides) => {
const { x, y, w, h } = bounds
const { w: W, h: H } = sides
const GC = r = Math.sqrt(W * W + H * H) / 2,
IC = H / 2,
beta = Math.asin(IC / GC),
angleA = Math.PI + beta,
Ax = x + w / 2 + r * Math.cos(angleA),
Ay = y + h / 2 + r * Math.sin(angleA)
return { x: Ax, y: Ay }
}
Putting all together...
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// rotations is... the rotation of the inner rectangle IN RADIANS
const unrotate = (bounds, rotation) => {
const points = vertices(bounds, rotation),
dimensions = sides(bounds, points)
const { x, y } = origin(bounds, dimensions)
return { ...dimensions, x, y }
}
I really hope this will solve your problem and that there are no typos. This was a very, veeeery funny way to spend my weekend :D
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// alpha is the rotation IN RADIANS
const vertices = (bounds, alpha) => {
const { x, y, w, h } = bounds,
A = { x: x + w * Math.sin(alpha), y },
B = { x, y: y + h * Math.sin(alpha) },
C = { x: x + w - w * Math.sin(alpha), y },
D = { x, y: y + h - h * Math.sin(alpha) }
return { A, B, C, D }
}
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// vertices is a POJO with shape: { A, B, C, D }, as returned by the `vertices` method
const sides = (bounds, vertices) => {
const { x, y, w, h } = bounds,
{ A, B, C, D } = vertices,
EA = A.x - x,
ED = D.y - y,
AF = w - EA,
FB = h - ED,
H = Math.sqrt(EA * EA + ED * ED),
W = Math.sqrt(AF * AF + FB * FB)
return { h: H, w: W }
}
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// sides is a POJO with shape: { w, h }, as returned by the `sides` method
const originPoint = (bounds, sides) => {
const { x, y, w, h } = bounds
const { w: W, h: H } = sides
const GC = Math.sqrt(W * W + H * H) / 2,
r = Math.sqrt(W * W + H * H) / 2,
IC = H / 2,
beta = Math.asin(IC / GC),
angleA = Math.PI + beta,
Ax = x + w / 2 + r * Math.cos(angleA),
Ay = y + h / 2 + r * Math.sin(angleA)
return { x: Ax, y: Ay }
}
// bounds is a POJO with shape: { x, y, w, h }, update if needed
// rotations is... the rotation of the inner rectangle IN RADIANS
const unrotate = (bounds, rotation) => {
const points = vertices(bounds, rotation)
const dimensions = sides(bounds, points)
const { x, y } = originPoint(bounds, dimensions)
return { ...dimensions, x, y }
}
function shortNumber(value) {
var places = 2;
value = Math.round(value * Math.pow(10, places)) / Math.pow(10, places);
return value;
}
function getInputtedBounds() {
var rectangle = {};
rectangle.x = parseFloat(app.xInput.value);
rectangle.y = parseFloat(app.yInput.value);
rectangle.w = parseFloat(app.widthInput.value);
rectangle.h = parseFloat(app.heightInput.value);
return rectangle;
}
function rotationSliderHandler() {
var rotation = app.rotationSlider.value;
app.rotationOutput.value = rotation;
rotate(rotation);
}
function rotationInputHandler() {
var rotation = app.rotationInput.value;
app.rotationSlider.value = rotation;
app.rotationOutput.value = rotation;
rotate(rotation);
}
function unrotateButtonHandler() {
var rotation = app.rotationInput.value;
app.rotationSlider.value = 0;
app.rotationOutput.value = 0;
var outerBounds = getInputtedBounds();
var radians = Math.PI / 180 * rotation;
var unrotatedBounds = unrotate(outerBounds, radians);
updateOutput(unrotatedBounds);
}
function rotate(value) {
var outerBounds = getInputtedBounds();
var radians = Math.PI / 180 * value;
var bounds = unrotate(outerBounds, radians);
updateOutput(bounds);
}
function updateOutput(bounds) {
app.xOutput.value = shortNumber(bounds.x);
app.yOutput.value = shortNumber(bounds.y);
app.widthOutput.value = shortNumber(bounds.w);
app.heightOutput.value = shortNumber(bounds.h);
}
function onload() {
app.xInput = document.getElementById("x");
app.yInput = document.getElementById("y");
app.widthInput = document.getElementById("w");
app.heightInput = document.getElementById("h");
app.rotationInput = document.getElementById("r");
app.xOutput = document.getElementById("x2");
app.yOutput = document.getElementById("y2");
app.widthOutput = document.getElementById("w2");
app.heightOutput = document.getElementById("h2");
app.rotationOutput = document.getElementById("r2");
app.rotationSlider = document.getElementById("rotationSlider");
app.unrotateButton = document.getElementById("unrotateButton");
app.unrotateButton.addEventListener("click", unrotateButtonHandler);
app.rotationSlider.addEventListener("input", rotationSliderHandler);
app.rotationInput.addEventListener("change", rotationInputHandler);
app.rotationInput.addEventListener("input", rotationInputHandler);
app.rotationInput.addEventListener("keyup", (e) => {if (e.keyCode==13) rotationInputHandler() });
app.rotationSlider.value = app.rotationInput.value;
}
var app = {};
window.addEventListener("load", onload);
* {
font-family: sans-serif;
font-size: 12px;
outline: 0px dashed red;
}
granola {
display: flex;
align-items: top;
}
flan {
width: 90px;
display: inline-block;
}
hamburger {
display: flex:
align-items: center;
}
spagetti {
display: inline-block;
font-size: 11px;
font-weight: bold;
letter-spacing: 1.5px;
}
fish {
display: inline-block;
padding-right: 40px;
position: relative;
}
input[type=text] {
width: 50px;
}
input[type=range] {
padding-top: 10px;
width: 140px;
padding-left: 0;
margin-left: 0;
}
button {
padding-top: 3px;
padding-bottom:1px;
margin-top: 10px;
}
<granola>
<fish>
<spagetti>Bounds of Rectangle</spagetti><br><br>
<flan>x: </flan><input id="x" type="text" value="14.39"><br>
<flan>y: </flan><input id="y" type="text" value="14.39"><br>
<flan>width: </flan><input id="w" type="text" value="21.2"><br>
<flan>height: </flan><input id="h" type="text" value="21.2"><br>
<flan>rotation:</flan><input id="r" type="text" value="90"><br>
<button id="unrotateButton">Unrotate</button>
</fish>
<fish>
<spagetti>Computed Bounds</spagetti><br><br>
<flan>x: </flan><input id="x2" type="text" disabled="true"><br>
<flan>y: </flan><input id="y2" type="text"disabled="true"><br>
<flan>width: </flan><input id="w2" type="text" disabled="true"><br>
<flan>height: </flan><input id="h2" type="text" disabled="true"><br>
<flan>rotation:</flan><input id="r2" type="text" disabled="true"><br>
<input id="rotationSlider" type="range" min="-360" max="360" step="5"><br>
</fish>
</granola>

How does this work?
Calculation using width, height, x and y
Radians and Angles
Using degrees calculate the radians and calculate the sin and cos angles:
function calculateRadiansAndAngles(){
const rotation = this.value;
const dr = Math.PI / 180;
const s = Math.sin(rotation * dr);
const c = Math.cos(rotation * dr);
console.log(rotation, s, c);
}
document.getElementById("range").oninput = calculateRadiansAndAngles;
<input type="range" min="-360" max="360" id="range"/>
Generate 4 points
we assume the origin of a rectangle is the center with the location of 0,0
The double for loop will create the following value pairs for i and j: (-1,-1), (-1,1), (1,-1) and (1,1)
Using each pair, we can calculate one of the 4 square vectors.
(i.e for (-1,1), i = -1, j = 1)
const px = w*i/2; //-> 30 * -1/2 = -15
const py = h*j/2; //-> 50 * 1/2 = 25
//[-15,25]
Once we have a point, we can calculate the new position of that point by including the rotation.
const nx = (px*c) - (py*s);
const ny = (px*s) + (py*c);
Solution
Once all the points are calculated based off of the rotation, we can redraw our square.
Before the draw call, a translate is used to position the cursor at the x and y of the rectangle. This is the reason as to why I was able to assume the center and the origin of the rectangle was 0,0 for the calculations.
const canvas = document.getElementById("canvas");
const range = document.getElementById("range");
const rotat = document.getElementById("rotat");
range.addEventListener("input", function(e) {
rotat.innerText = this.value;
handleRotation(this.value);
})
const context = canvas.getContext("2d");
const container = document.getElementById("container");
const rect = {
x: 50,
y: 75,
w: 30,
h: 50
}
function handleRotation(rotation) {
const { w, h, x, y } = rect;
const dr = Math.PI / 180;
const s = Math.sin(rotation * dr);
const c = Math.cos(rotation * dr);
const points = [];
for(let i = -1; i < 2; i+=2){
for(let j = -1; j < 2; j+=2){
const px = w*i/2;
const py = h*j/2;
const nx = (px*c) - (py*s);
const ny = (px*s) + (py*c);
points.push([nx, ny]);
}
}
//console.log(points);
draw(points);
}
function draw(points) {
context.clearRect(0,0,canvas.width, canvas.height);
context.save();
context.translate(rect.x+(rect.w/2), rect.y + (rect.h/2))
context.beginPath();
context.moveTo(...points.shift());
[...points.splice(0,1), ...points.reverse()]
.forEach(p=>{
context.lineTo(...p);
})
context.fill();
context.restore();
}
window.onload = () => handleRotation(0);
div {
display: flex;
background-color: lightgrey;
padding: 0 5px;
}
div>p {
padding: 0px 10px;
}
div>input {
flex-grow: 1;
}
canvas {
border: 1px solid black;
}
<div>
<p id="rotat">0</p>
<input type="range" id="range" min="-360" max="360" value="0" step="5" />
</div>
<canvas id="canvas"></canvas>

This is the basic code for a rectangle rotating(Unrotating is the same thing only with a negative angle) around its center.
function getUnrotatedRectangleBounds(rect, currentRotation) {
//Convert deg to radians
var rot = currentRotation / 180 * Math.PI;
var hyp = Math.sqrt(rect.width * rect.width + rect.height * rect.height);
return {
x: rect.x + rect.width / 2 - hyp * Math.abs(Math.cos(rot)) / 2,
y: rect.y + rect.height / 2 - hyp * Math.abs(Math.sin(rot)) / 2,
width: hyp * Math.abs(Math.cos(rot)),
height: hyp * Math.abs(Math.sin(rot))
}
}
The vector starting at the origin(0,0) and ending at (width,height) is projected onto a unit vector for the target angle (cos rot,sin rot) * hyp.
The absolute values guarantee the width and height are both positive.
The coordinates of the projection are the width and height, respectively, of the new rectangle.
For the x and y values, take the original values at the center(x + rect.x) and move it back out(- 1/2 * NewWidth) so it centers the new rectangle.
Example
function getUnrotatedRectangleBounds(rect, currentRotation) {
//Convert deg to radians
var rot = currentRotation / 180 * Math.PI;
var hyp = Math.sqrt(rect.width * rect.width + rect.height * rect.height);
return {
x: rect.x + rect.width / 2 - hyp * Math.abs(Math.cos(rot)) / 2,
y: rect.y + rect.height / 2 - hyp * Math.abs(Math.sin(rot)) / 2,
width: hyp * Math.abs(Math.cos(rot)),
height: hyp * Math.abs(Math.sin(rot))
}
}
var originalRectangle = {x:10, y:25, width:30, height:0};
var rotatedRectangle = {x:14.39, y:14.39, width:21.2, height:21.2};
var rotation = 45;
var unrotatedRectangle = getUnrotatedRectangleBounds(rotatedRectangle, rotation);
var boundsLabel = document.getElementById("boundsLabel");
boundsLabel.innerHTML = JSON.stringify(unrotatedRectangle);
<span id="boundsLabel"></span>