Develop Reference

Pixi.js best way to create a dragable and clickable handle that rotates around a point

I have 8 handle graphics that represent 5 different states (closed, flow rate 1, flow rate 2, flow rate 3, flow rate 4). Handle graphics 6,7, and 8 also represent flow rate 1, 2, and 3. The images depict a buret handle that rotates around a center point. For each handle state, I need to show the matching texture. I need the user to be able to drag the handle and have it move through the different graphics as the mouse moves around the center point. I also need the user to be able to click on the right side to increase the flow rate and click on the left side to decrease the flow rate.
I have looking into using getBounds() from the image and using that as a hit box but that seems like it won't work because i am removing the old texture and adding a new one depending on the mouse position when dragging. not to mention the images all have similar dimensions.
I have also though about creating 16 hit boxes (2 for each of the 8 images, 1 on the left side for decreasing flow rate, one on the right side for increasing flow rate) and adding and removing the hit boxes with the texture but this seems overly tedious and i don't think it will work with dragging.
Let me know if you have any ideas!
Thanks

Drag a rotating switch
Assuming you get a mouse coord that is relative to the valve eg mouse event pageX, pageY properties.
You can create a function that takes the element, number valve steps, and mouse coords and spits out the values you want.
function getValueSetting(x, y, valveSteps, valveElement) {
const bounds = valveElement.getBoundingClientRect();
const centerX = (bounds.left + bounds.right) / 2;
const centerY = (bounds.top + bounds.bottom) / 2;
const left = x < centerX;
const distance = Math.hypot(x - centerX, y - centerY);
const pos = (Math.atan2(y - centerY, x - centerX) + Math.PI) / (Math.PI * 2);
return {
left,
right: !left,
distance,
pos: Math.round(pos * valveSteps - (valveSteps / 4)),
};
}
If the valve positions step by 1 hour on the clock make valveSteps = 12
Call the function const valveState = getValueSetting(mouseEvent.pageX, mouseEvent.pageY, 12, valveElment);
The object returned will have bools for left and right of the center, and pos will be one of 12 positions starting at 12 o'clock pos = 0 to 11 o'clock pos === 11. The distance property is the distance from the valve.
In the function the angle position subtracts (valveSteps / 4) because Math.atan2 return 0 at the 3 o'clock mark. The subtract (valveSteps / 4) rotate back 1 quarter turn to set 0 at 12 o'clock.
Example
The example draws 5 valve positions.
Move the mouse over the valve handle (red) and the cursor will change to a pointer. Click and drag the mouse to turn the valve. Once dragging the mouse will hold the valve until you release the button.
If not over the handle, but near the valve clicks left and right will message appropriate message.
const size = 64; // size of image
const valveSteps = 12; // total number of angle steps
const valveStep = (Math.PI * 2) / valveSteps; // angle steps in radians
const startAngle = -valveStep * 2; // visual start angle of handle
const valveStart = 1; // starting pos of valve
setTimeout(() => {
const valves = [
createValve(64, startAngle),
createValve(64, startAngle + valveStep),
createValve(64, startAngle + valveStep * 2),
createValve(64, startAngle + valveStep * 3),
createValve(64, startAngle + valveStep * 4),
];
setValve(valves[0]);
var dragging = false;
var currentPos = 0;
var level = 0;
mouse.onupdate = () => {
const valveSetting = getValueSetting(mouse.x, mouse.y, valveSteps, valveA);
if (valveSetting.distance < size && valveSetting.pos - valveStart === currentPos) {
document.body.style.cursor = "pointer";
} else {
document.body.style.cursor = "default";
}
if (mouse.button && (valveSetting.distance < size || dragging)) {
if (valveSetting.distance < size / 2 && valveSetting.pos - valveStart === currentPos) {
if (valveSetting.pos >= valveStart && valveSetting.pos < valveStart + valves.length) {
dragging = true;
}
}
console.clear()
if (dragging) {
let pos = valveSetting.pos - valveStart;
pos = pos < 0 ? 0 : pos > valves.length - 1 ? valves.length - 1 : pos
setValve(valves[pos]);
currentPos = pos;
console.log("Valve pos: " + pos);
} else if (valveSetting.left) {
level --;
console.log("Turn down " + level);
mouse.button = false;
} else if (valveSetting.right) {
level ++;
console.log("Turn up " + level);
mouse.button = false;
}
} else {
dragging = false;
}
}
},0);
function setValve(image) {
valveA.innerHTML = "";
$$(valveA, image); // appends image to element valveA
}
function getValueSetting(x, y, valveSteps, valveElement) {
const bounds = valveElement.getBoundingClientRect();
const centerX = (bounds.left + bounds.right) / 2;
const centerY = (bounds.top + bounds.bottom) / 2;
const left = x < centerX;
const distance = Math.hypot(x - centerX, y - centerY);
const pos = (Math.atan2(y - centerY, x - centerX) + Math.PI) / (Math.PI * 2);
return {
left,
right: !left,
distance,
pos: Math.round(pos * valveSteps - (valveSteps / 4)),
};
}
function createValve(size, angle) {
const canvas = $("canvas", {width: size, height: size});
const ctx = canvas.getContext("2d");
const r = size * 0.4;
const c = size / 2;
ctx.strokeStyle = "red";
ctx.lineCap = "round";
ctx.lineWidth = 8;
ctx.beginPath();
ctx.lineTo(Math.cos(angle) * r + c, Math.sin(angle) * r + c);
ctx.lineTo(-Math.cos(angle) * r * 0.2 + c, -Math.sin(angle) * r * 0.2 + c);
ctx.stroke();
ctx.beginPath();
ctx.arc(c, c, 8, 0, Math.PI * 2);
ctx.strokeStyle = "black";
ctx.lineWidth = 2;
ctx.stroke();
return canvas;
}
// Boiler plate
const $ = (tag, props = {}) => Object.assign(document.createElement(tag), props);
const $$ = (p, ...sibs) => sibs.reduce((p,sib) => (p.appendChild(sib), p), p);
const mouse = {x : 0, y : 0, button : false}
function mouseEvents(e){
mouse.x = e.pageX;
mouse.y = e.pageY;
mouse.button = e.type === "mousedown" ? true : e.type === "mouseup" ? false : mouse.button;
mouse.onupdate && mouse.onupdate();
}
["down","up","move"].forEach(name => document.addEventListener("mouse" + name,mouseEvents));
.valveContainer {
position: absolute;
top: 30px;
left 30px;
border: 2px solid white;
}
<div id="valveA" class="valveContainer"></div>

How to draw herringbone pattern on html canvas

I Have to draw Herringbone pattern on canvas and fill with image
some one please help me I am new to canvas 2d drawing.
I need to draw mixed tiles with cross pattern (Herringbone)
var canvas = this.__canvas = new fabric.Canvas('canvas');
var canvas_objects = canvas._objects;
// create a rectangle with a fill and a different color stroke
var left = 150;
var top = 150;
var x=20;
var y=40;
var rect = new fabric.Rect({
left: left,
top: top,
width: x,
height: y,
angle:45,
fill: 'rgba(255,127,39,1)',
stroke: 'rgba(34,177,76,1)',
strokeWidth:0,
originX:'right',
originY:'top',
centeredRotation: false
});
canvas.add(rect);
for(var i=0;i<15;i++){
var rectangle = fabric.util.object.clone(getLastobject());
if(i%2==0){
rectangle.left = rectangle.oCoords.tr.x;
rectangle.top = rectangle.oCoords.tr.y;
rectangle.originX='right';
rectangle.originY='top';
rectangle.angle =-45;
}else{
fabric.log('rectangle: ', rectangle.toJSON());
rectangle.left = rectangle.oCoords.tl.x;
rectangle.top = rectangle.oCoords.tl.y;
fabric.log('rectangle: ', rectangle.toJSON());
rectangle.originX='left';
rectangle.originY='top';
rectangle.angle =45;
}
//rectangle.angle -90;
canvas.add(rectangle);
}
fabric.log('rectangle: ', canvas.toJSON());
canvas.renderAll();
function getLastobject(){
var last = null;
if(canvas_objects.length !== 0){
last = canvas_objects[canvas_objects.length -1]; //Get last object
}
return last;
}
How to draw this pattern in canvas using svg or 2d,3d method. If any third party library that also Ok for me.
I don't know where to start and how to draw this complex pattern.
some one please help me to draw this pattern with rectangle fill with dynamic color on canvas.
Here is a sample of the output I need: (herringbone pattern)
I tried something similar using fabric.js library here is my JSFiddle

Trippy disco flooring
To get the pattern you need to draw rectangles one horizontal tiled one space left or right for each row down and the same for the vertical rectangle.
The rectangle has an aspect of width 2 time height.
Drawing the pattern is simple.
Rotating is easy as well the harder part is finding where to draw the tiles for the rotation.
To do that I create a inverse matrix of the rotation (it reverses a rotation). I then apply that rotation to the 4 corners of the canvas 0,0, width,0 width,height and 0,height this gives me 4 points in the rotated space that are at the edges of the canvas.
As I draw the tiles from left to right top to bottom I find the min corners for the top left, and the max corners for the bottom right, expand it out a little so I dont miss any pixels and draw the tiles with a transformation set the the rotation.
As I could not workout what angle you wanted it at the function will draw it at any angle. On is animated, the other is at 60deg clockwise.
Warning demo contains flashing content.
Update The flashing was way to out there, so have made a few changes, now colours are a more pleasing blend and have fixed absolute positions, and have tied the tile origin to the mouse position, clicking the mouse button will cycle through some sizes as well.
const ctx = canvas.getContext("2d");
const colours = []
for(let i = 0; i < 1; i += 1/80){
colours.push(`hsl(${Math.floor(i * 360)},${Math.floor((Math.sin(i * Math.PI *4)+1) * 50)}%,${Math.floor(Math.sin(i * Math.PI *8)* 25 + 50)}%)`)
}
const sizes = [0.04,0.08,0.1,0.2];
var currentSize = 0;
const origin = {x : canvas.width / 2, y : canvas.height / 2};
var size = Math.min(canvas.width * 0.2, canvas.height * 0.2);
function drawPattern(size,origin,ang){
const xAx = Math.cos(ang); // define the direction of xAxis
const xAy = Math.sin(ang);
ctx.setTransform(1,0,0,1,0,0);
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.setTransform(xAx,xAy,-xAy,xAx,origin.x,origin.y);
function getExtent(xAx,xAy,origin){
const im = [1,0,0,1]; // inverse matrix
const dot = xAx * xAx + xAy * xAy;
im[0] = xAx / dot;
im[1] = -xAy / dot;
im[2] = xAy / dot;
im[3] = xAx / dot;
const toWorld = (x,y) => {
var point = {};
var xx = x - origin.x;
var yy = y - origin.y;
point.x = xx * im[0] + yy * im[2];
point.y = xx * im[1] + yy * im[3];
return point;
}
return [
toWorld(0,0),
toWorld(canvas.width,0),
toWorld(canvas.width,canvas.height),
toWorld(0,canvas.height),
]
}
const corners = getExtent(xAx,xAy,origin);
var startX = Math.min(corners[0].x,corners[1].x,corners[2].x,corners[3].x);
var endX = Math.max(corners[0].x,corners[1].x,corners[2].x,corners[3].x);
var startY = Math.min(corners[0].y,corners[1].y,corners[2].y,corners[3].y);
var endY = Math.max(corners[0].y,corners[1].y,corners[2].y,corners[3].y);
startX = Math.floor(startX / size) - 2;
endX = Math.floor(endX / size) + 2;
startY = Math.floor(startY / size) - 2;
endY = Math.floor(endY / size) + 2;
// draw the pattern
ctx.lineWidth = size * 0.1;
ctx.lineJoin = "round";
ctx.strokeStyle = "black";
var colourIndex = 0;
for(var y = startY; y <endY; y+=1){
for(var x = startX; x <endX; x+=1){
if((x + y) % 4 === 0){
colourIndex = Math.floor(Math.abs(Math.sin(x)*size + Math.sin(y) * 20));
ctx.fillStyle = colours[(colourIndex++)% colours.length];
ctx.fillRect(x * size,y * size,size * 2,size);
ctx.strokeRect(x * size,y * size,size * 2,size);
x += 2;
ctx.fillStyle = colours[(colourIndex++)% colours.length];
ctx.fillRect(x * size,y * size, size, size * 2);
ctx.strokeRect(x * size,y * size, size, size * 2);
x += 1;
}
}
}
}
// Animate it all
var update = true; // flag to indecate something needs updating
function mainLoop(time){
// if window size has changed update canvas to new size
if(canvas.width !== innerWidth || canvas.height !== innerHeight || update){
canvas.width = innerWidth;
canvas.height = innerHeight
origin.x = canvas.width / 2;
origin.y = canvas.height / 2;
size = Math.min(canvas.width, canvas.height) * sizes[currentSize % sizes.length];
update = false;
}
if(mouse.buttonRaw !== 0){
mouse.buttonRaw = 0;
currentSize += 1;
update = true;
}
// draw the patter
drawPattern(size,mouse,time/2000);
requestAnimationFrame(mainLoop);
}
requestAnimationFrame(mainLoop);
mouse = (function () {
function preventDefault(e) { e.preventDefault() }
var m; // alias for mouse
var mouse = {
x : 0, y : 0, // mouse position
buttonRaw : 0,
over : false, // true if mouse over the element
buttonOnMasks : [0b1, 0b10, 0b100], // mouse button on masks
buttonOffMasks : [0b110, 0b101, 0b011], // mouse button off masks
bounds : null,
eventNames : "mousemove,mousedown,mouseup,mouseout,mouseover".split(","),
event(e) {
var t = e.type;
m.bounds = m.element.getBoundingClientRect();
m.x = e.pageX - m.bounds.left - scrollX;
m.y = e.pageY - m.bounds.top - scrollY;
if (t === "mousedown") { m.buttonRaw |= m.buttonOnMasks[e.which - 1] }
else if (t === "mouseup") { m.buttonRaw &= m.buttonOffMasks[e.which - 1] }
else if (t === "mouseout") { m.over = false }
else if (t === "mouseover") { m.over = true }
e.preventDefault();
},
start(element) {
if (m.element !== undefined) { m.remove() }
m.element = element === undefined ? document : element;
m.eventNames.forEach(name => document.addEventListener(name, mouse.event) );
document.addEventListener("contextmenu", preventDefault, false);
},
}
m = mouse;
return mouse;
})();
mouse.start(canvas);
canvas {
position : absolute;
top : 0px;
left : 0px;
}
<canvas id=canvas></canvas>
Un-animated version at 60Deg
const ctx = canvas.getContext("2d");
const colours = ["red","green","yellow","orange","blue","cyan","magenta"]
const origin = {x : canvas.width / 2, y : canvas.height / 2};
var size = Math.min(canvas.width * 0.2, canvas.height * 0.2);
function drawPattern(size,origin,ang){
const xAx = Math.cos(ang); // define the direction of xAxis
const xAy = Math.sin(ang);
ctx.setTransform(1,0,0,1,0,0);
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.setTransform(xAx,xAy,-xAy,xAx,origin.x,origin.y);
function getExtent(xAx,xAy,origin){
const im = [1,0,0,1]; // inverse matrix
const dot = xAx * xAx + xAy * xAy;
im[0] = xAx / dot;
im[1] = -xAy / dot;
im[2] = xAy / dot;
im[3] = xAx / dot;
const toWorld = (x,y) => {
var point = {};
var xx = x - origin.x;
var yy = y - origin.y;
point.x = xx * im[0] + yy * im[2];
point.y = xx * im[1] + yy * im[3];
return point;
}
return [
toWorld(0,0),
toWorld(canvas.width,0),
toWorld(canvas.width,canvas.height),
toWorld(0,canvas.height),
]
}
const corners = getExtent(xAx,xAy,origin);
var startX = Math.min(corners[0].x,corners[1].x,corners[2].x,corners[3].x);
var endX = Math.max(corners[0].x,corners[1].x,corners[2].x,corners[3].x);
var startY = Math.min(corners[0].y,corners[1].y,corners[2].y,corners[3].y);
var endY = Math.max(corners[0].y,corners[1].y,corners[2].y,corners[3].y);
startX = Math.floor(startX / size) - 4;
endX = Math.floor(endX / size) + 4;
startY = Math.floor(startY / size) - 4;
endY = Math.floor(endY / size) + 4;
// draw the pattern
ctx.lineWidth = 5;
ctx.lineJoin = "round";
ctx.strokeStyle = "black";
for(var y = startY; y <endY; y+=1){
for(var x = startX; x <endX; x+=1){
ctx.fillStyle = colours[Math.floor(Math.random() * colours.length)];
if((x + y) % 4 === 0){
ctx.fillRect(x * size,y * size,size * 2,size);
ctx.strokeRect(x * size,y * size,size * 2,size);
x += 2;
ctx.fillStyle = colours[Math.floor(Math.random() * colours.length)];
ctx.fillRect(x * size,y * size, size, size * 2);
ctx.strokeRect(x * size,y * size, size, size * 2);
x += 1;
}
}
}
}
canvas.width = innerWidth;
canvas.height = innerHeight
origin.x = canvas.width / 2;
origin.y = canvas.height / 2;
size = Math.min(canvas.width * 0.2, canvas.height * 0.2);
drawPattern(size,origin,Math.PI / 3);
canvas {
position : absolute;
top : 0px;
left : 0px;
}
<canvas id=canvas></canvas>

The best way to approach this is to examine the pattern and analyse its symmetry and how it repeats.
You can look at this several ways. For example, you could rotate the patter 45 degrees so that the tiles are plain orthogonal rectangles. But let's just look at it how it is. I am going to assume you are happy with it with 45deg tiles.
Like the tiles themselves, it turns out the pattern has a 2:1 ratio. If we repeat this pattern horizontally and vertically, we can fill the canvas with the completed pattern.
We can see there are five tiles that overlap with our pattern block. However we don't need to draw them all when we draw each pattern block. We can take advantage of the fact that blocks are repeated, and we can leave the drawing of some tiles to later rows and columns.
Let's assume we are drawing the pattern blocks from left to right and top to bottom. Which tiles do we need to draw, at a minimum, to ensure this pattern block gets completely drawn (taking into account adjacent pattern blocks)?
Since we will be starting at the top left (and moving right and downwards), we'll need to draw tile 2. That's because that tile won't get drawn by either the block below us, or the block to the right of us. The same applies to tile 3.
It turns out those two are all we'll need to draw for each pattern block. Tile 1 and 4 will be drawn when the pattern block below us draws their tile 2 and 3 respectively. Tile 5 will be drawn when the pattern block to the south-east of us draws their tile 1.
We just need to remember that we may need to draw an extra column on the right-hand side, and at the bottom, to ensure those end-of-row and end-of-column pattern blocks get completely drawn.
The last thing to work out is how big our pattern blocks are.
Let's call the short side of the tile a and the long side b. We know that b = 2 * a. And we can work out, using Pythagoras Theorem, that the height of the pattern block will be:
h = sqrt(a^2 + a^2)
= sqrt(2 * a^2)
= sqrt(2) * a
The width of the pattern block we can see will be w = 2 * h.
Now that we've worked out how to draw the pattern, let's implement our algorithm.
const a = 60;
const b = 120;
const h = 50 * Math.sqrt(2);
const w = h * 2;
const h2 = h / 2; // How far tile 1 sticks out to the left of the pattern block
// Set of colours for the tiles
const colours = ["red","cornsilk","black","limegreen","deepskyblue",
"mediumorchid", "lightgrey", "grey"]
const canvas = document.getElementById("herringbone");
const ctx = canvas.getContext("2d");
// Set a universal stroke colour and width
ctx.strokeStyle = "black";
ctx.lineWidth = 4;
// Loop through the pattern block rows
for (var y=0; y < (canvas.height + h); y+=h)
{
// Loop through the pattern block columns
for (var x=0; x < (canvas.width + w); x+=w)
{
// Draw tile "2"
// I'm just going to draw a path for simplicity, rather than
// worrying about drawing a rectangle with rotation and translates
ctx.beginPath();
ctx.moveTo(x - h2, y - h2);
ctx.lineTo(x, y - h);
ctx.lineTo(x + h, y);
ctx.lineTo(x + h2, y + h2);
ctx.closePath();
ctx.fillStyle = colours[Math.floor(Math.random() * colours.length)];
ctx.fill();
ctx.stroke();
// Draw tile "3"
ctx.beginPath();
ctx.moveTo(x + h2, y + h2);
ctx.lineTo(x + w - h2, y - h2);
ctx.lineTo(x + w, y);
ctx.lineTo(x + h, y + h);
ctx.closePath();
ctx.fillStyle = colours[Math.floor(Math.random() * colours.length)];
ctx.fill();
ctx.stroke();
}
}
<canvas id="herringbone" width="500" height="400"></canvas>

JavaScript Point Collision with Regular Hexagon

I'm making an HTML5 canvas hexagon grid based system and I need to be able to detect what hexagonal tile in a grid has been clicked when the canvas is clicked.
Several hours of searching and trying my own methods led to nothing, and porting implementations from other languages has simply confused me to a point where my brain is sluggish.
The grid consists of flat topped regular hexagons like in this diagram:
Essentially, given a point and the variables specified in this image as the sizing for every hexagon in the grid (R, W, S, H):
I need to be able to determine whether a point is inside a hexagon given.
An example function call would be pointInHexagon(hexX, hexY, R, W, S, H, pointX, pointY) where hexX and hexY are the coordinates for the top left corner of the bounding box of a hexagonal tile (like the top left corner in the image above).
Is there anyone who has any idea how to do this? Speed isn't much of a concern for the moment.

Simple & fast diagonal rectangle slice.
Looking at the other answers I see that they have all a little over complicated the problem. The following is an order of magnitude quicker than the accepted answer and does not require any complicated data structures, iterators, or generate dead memory and unneeded GC hits. It returns the hex cell row and column for any related set of R, H, S or W. The example uses R = 50.
Part of the problem is finding which side of a rectangle a point is if the rectangle is split diagonally. This is a very simple calculation and is done by normalising the position of the point to test.
Slice any rectangle diagonally
Example a rectangle of width w, and height h split from top left to bottom right. To find if a point is left or right. Assume top left of rectangle is at rx,ry
var x = ?;
var y = ?;
x = ((x - rx) % w) / w;
y = ((y - ry) % h) / h;
if (x > y) {
// point is in the upper right triangle
} else if (x < y) {
// point is in lower left triangle
} else {
// point is on the diagonal
}
If you want to change the direction of the diagonal then just invert one of the normals
x = 1 - x; // invert x or y to change the direction the rectangle is split
if (x > y) {
// point is in the upper left triangle
} else if (x < y) {
// point is in lower right triangle
} else {
// point is on the diagonal
}
Split into sub cells and use %
The rest of the problem is just a matter of splitting the grid into (R / 2) by (H / 2) cells width each hex covering 4 columns and 2 rows. Every 1st column out of 3 will have diagonals. with every second of these column having the diagonal flipped. For every 4th, 5th, and 6th column out of 6 have the row shifted down one cell. By using % you can very quickly determine which hex cell you are on. Using the diagonal split method above make the math easy and quick.
And one extra bit. The return argument retPos is optional. if you call the function as follows
var retPos;
mainLoop(){
retPos = getHex(mouse.x, mouse.y, retPos);
}
the code will not incur a GC hit, further improving the speed.
Pixel to Hex coordinates
From Question diagram returns hex cell x,y pos. Please note that this function only works in the range 0 <= x, 0 <= y if you need negative coordinates subtract the min negative pixel x,y coordinate from the input
// the values as set out in the question image
var r = 50;
var w = r * 2;
var h = Math.sqrt(3) * r;
// returns the hex grid x,y position in the object retPos.
// retPos is created if not supplied;
// argument x,y is pixel coordinate (for mouse or what ever you are looking to find)
function getHex (x, y, retPos){
if(retPos === undefined){
retPos = {};
}
var xa, ya, xpos, xx, yy, r2, h2;
r2 = r / 2;
h2 = h / 2;
xx = Math.floor(x / r2);
yy = Math.floor(y / h2);
xpos = Math.floor(xx / 3);
xx %= 6;
if (xx % 3 === 0) { // column with diagonals
xa = (x % r2) / r2; // to find the diagonals
ya = (y % h2) / h2;
if (yy % 2===0) {
ya = 1 - ya;
}
if (xx === 3) {
xa = 1 - xa;
}
if (xa > ya) {
retPos.x = xpos + (xx === 3 ? -1 : 0);
retPos.y = Math.floor(yy / 2);
return retPos;
}
retPos.x = xpos + (xx === 0 ? -1 : 0);
retPos.y = Math.floor((yy + 1) / 2);
return retPos;
}
if (xx < 3) {
retPos.x = xpos + (xx === 3 ? -1 : 0);
retPos.y = Math.floor(yy / 2);
return retPos;
}
retPos.x = xpos + (xx === 0 ? -1 : 0);
retPos.y = Math.floor((yy + 1) / 2);
return retPos;
}
Hex to pixel
And a helper function that draws a cell given the cell coordinates.
// Helper function draws a cell at hex coordinates cellx,celly
// fStyle is fill style
// sStyle is strock style;
// fStyle and sStyle are optional. Fill or stroke will only be made if style given
function drawCell1(cellPos, fStyle, sStyle){
var cell = [1,0, 3,0, 4,1, 3,2, 1,2, 0,1];
var r2 = r / 2;
var h2 = h / 2;
function drawCell(x, y){
var i = 0;
ctx.beginPath();
ctx.moveTo((x + cell[i++]) * r2, (y + cell[i++]) * h2)
while (i < cell.length) {
ctx.lineTo((x + cell[i++]) * r2, (y + cell[i++]) * h2)
}
ctx.closePath();
}
ctx.lineWidth = 2;
var cx = Math.floor(cellPos.x * 3);
var cy = Math.floor(cellPos.y * 2);
if(cellPos.x % 2 === 1){
cy -= 1;
}
drawCell(cx, cy);
if (fStyle !== undefined && fStyle !== null){ // fill hex is fStyle given
ctx.fillStyle = fStyle
ctx.fill();
}
if (sStyle !== undefined ){ // stroke hex is fStyle given
ctx.strokeStyle = sStyle
ctx.stroke();
}
}

I think you need something like this~
EDITED
I did some maths and here you have it. This is not a perfect version but probably will help you...
Ah, you only need a R parameter because based on it you can calculate H, W and S. That is what I understand from your description.
// setup canvas for demo
var canvas = document.getElementById('canvas');
canvas.width = 300;
canvas.height = 275;
var context = canvas.getContext('2d');
var hexPath;
var hex = {
x: 50,
y: 50,
R: 100
}
// Place holders for mouse x,y position
var mouseX = 0;
var mouseY = 0;
// Test for collision between an object and a point
function pointInHexagon(target, pointX, pointY) {
var side = Math.sqrt(target.R*target.R*3/4);
var startX = target.x
var baseX = startX + target.R / 2;
var endX = target.x + 2 * target.R;
var startY = target.y;
var baseY = startY + side;
var endY = startY + 2 * side;
var square = {
x: startX,
y: startY,
side: 2*side
}
hexPath = new Path2D();
hexPath.lineTo(baseX, startY);
hexPath.lineTo(baseX + target.R, startY);
hexPath.lineTo(endX, baseY);
hexPath.lineTo(baseX + target.R, endY);
hexPath.lineTo(baseX, endY);
hexPath.lineTo(startX, baseY);
if (pointX >= square.x && pointX <= (square.x + square.side) && pointY >= square.y && pointY <= (square.y + square.side)) {
var auxX = (pointX < target.R / 2) ? pointX : (pointX > target.R * 3 / 2) ? pointX - target.R * 3 / 2 : target.R / 2;
var auxY = (pointY <= square.side / 2) ? pointY : pointY - square.side / 2;
var dPointX = auxX * auxX;
var dPointY = auxY * auxY;
var hypo = Math.sqrt(dPointX + dPointY);
var cos = pointX / hypo;
if (pointX < (target.x + target.R / 2)) {
if (pointY <= (target.y + square.side / 2)) {
if (pointX < (target.x + (target.R / 2 * cos))) return false;
}
if (pointY > (target.y + square.side / 2)) {
if (pointX < (target.x + (target.R / 2 * cos))) return false;
}
}
if (pointX > (target.x + target.R * 3 / 2)) {
if (pointY <= (target.y + square.side / 2)) {
if (pointX < (target.x + square.side - (target.R / 2 * cos))) return false;
}
if (pointY > (target.y + square.side / 2)) {
if (pointX < (target.x + square.side - (target.R / 2 * cos))) return false;
}
}
return true;
}
return false;
}
// Loop
setInterval(onTimerTick, 33);
// Render Loop
function onTimerTick() {
// Clear the canvas
canvas.width = canvas.width;
// see if a collision happened
var collision = pointInHexagon(hex, mouseX, mouseY);
// render out text
context.fillStyle = "Blue";
context.font = "18px sans-serif";
context.fillText("Collision: " + collision + " | Mouse (" + mouseX + ", " + mouseY + ")", 10, 20);
// render out square
context.fillStyle = collision ? "red" : "green";
context.fill(hexPath);
}
// Update mouse position
canvas.onmousemove = function(e) {
mouseX = e.offsetX;
mouseY = e.offsetY;
}
#canvas {
border: 1px solid black;
}
<canvas id="canvas"></canvas>
Just replace your pointInHexagon(hexX, hexY, R, W, S, H, pointX, pointY) by the var hover = ctx.isPointInPath(hexPath, x, y).
This is for Creating and copying paths
This is about the Collision Detection
var canvas = document.getElementById("canvas");
var ctx = canvas.getContext("2d");
var hexPath = new Path2D();
hexPath.lineTo(25, 0);
hexPath.lineTo(75, 0);
hexPath.lineTo(100, 43);
hexPath.lineTo(75, 86);
hexPath.lineTo(25, 86);
hexPath.lineTo(0, 43);
function draw(hover) {
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.fillStyle = hover ? 'blue' : 'red';
ctx.fill(hexPath);
}
canvas.onmousemove = function(e) {
var x = e.clientX - canvas.offsetLeft, y = e.clientY - canvas.offsetTop;
var hover = ctx.isPointInPath(hexPath, x, y)
draw(hover)
};
draw();
<canvas id="canvas"></canvas>

I've made a solution for you that demonstrates the point in triangle approach to this problem.
http://codepen.io/spinvector/pen/gLROEp
maths below:
isPointInside(point)
{
// Point in triangle algorithm from http://totologic.blogspot.com.au/2014/01/accurate-point-in-triangle-test.html
function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y)
{
var denominator = ((y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3));
var a = ((y2 - y3)*(x - x3) + (x3 - x2)*(y - y3)) / denominator;
var b = ((y3 - y1)*(x - x3) + (x1 - x3)*(y - y3)) / denominator;
var c = 1 - a - b;
return 0 <= a && a <= 1 && 0 <= b && b <= 1 && 0 <= c && c <= 1;
}
// A Hex is composite of 6 trianges, lets do a point in triangle test for each one.
// Step through our triangles
for (var i = 0; i < 6; i++) {
// check for point inside, if so, return true for this function;
if(pointInTriangle( this.origin.x, this.origin.y,
this.points[i].x, this.points[i].y,
this.points[(i+1)%6].x, this.points[(i+1)%6].y,
point.x, point.y))
return true;
}
// Point must be outside.
return false;
}

Here is a fully mathematical and functional representation of your problem. You will notice that there are no ifs and thens in this code other than the ternary to change the color of the text depending on the mouse position. This whole job is in fact nothing more than pure simple math of just one line;
(r+m)/2 + Math.cos(a*s)*(r-m)/2;
and this code is reusable for all polygons from triangle to circle. So if interested please read on. It's very simple.
In order to display the functionality I had to develop a mimicking model of the problem. I draw a polygon on a canvas by utilizing a simple utility function. So that the overall solution should work for any polygon. The following snippet will take the canvas context c, radius r, number of sides s, and the local center coordinates in the canvas cx and cy as arguments and draw a polygon on the given canvas context at the right position.
function drawPolgon(c, r, s, cx, cy){ //context, radius, sides, center x, center y
c.beginPath();
c.moveTo(cx + r,cy);
for(var p = 1; p < s; p++) c.lineTo(cx + r*Math.cos(p*2*Math.PI/s), cy + r*Math.sin(p*2*Math.PI/s));
c.closePath();
c.stroke();
}
We have some other utility functions which one can easily understand what exactly they are doing. However the most important part is to check whether the mouse is floating over our polygon or not. It's done by the utility function isMouseIn. It's basically calculating the distance and the angle of the mouse position to the center of the polygon. Then, comparing it with the boundaries of the polygon. The boundaries of the polygon can be expressed by simple trigonometry, just like we have calculated the vertices in the drawPolygon function.
We can think of our polygon as a circle with an oscillating radius at the frequency of number of sides. The oscillation's peak is at the given radius value r (which happens to be at the vertices at angle 2π/s where s is the number of sides) and the minimum m is r*Math.cos(Math.PI/s) (each shows at at angle 2π/s + 2π/2s = 3π/s). I am pretty sure the ideal way to express a polygon could be done by the Fourier transformation but we don't need that here. All we need is a constant radius component which is the average of minimum and maximum, (r+m)/2 and the oscillating component with the frequency of number of sides, s and the amplitude value maximum - minimum)/2 on top of it, Math.cos(a*s)*(r-m)/2. Well of course as per Fourier states we might carry on with smaller oscillating components but with a hexagon you don't really need further iteration while with a triangle you possibly would. So here is our polygon representation in math.
(r+m)/2 + Math.cos(a*s)*(r-m)/2;
Now all we need is to calculate the angle and distance of our mouse position relative to the center of the polygon and compare it with the above mathematical expression which represents our polygon. So all together our magic function is orchestrated as follows;
function isMouseIn(r,s,cx,cy,mx,my){
var m = r*Math.cos(Math.PI/s), // the min dist from an edge to the center
d = Math.hypot(mx-cx,my-cy), // the mouse's distance to the center of the polygon
a = Math.atan2(cy-my,mx-cx); // angle of the mouse pointer
return d <= (r+m)/2 + Math.cos(a*s)*(r-m)/2;
}
So the following code demonstrates how you might approach to solve your problem.
// Generic function to draw a polygon on the canvas
function drawPolgon(c, r, s, cx, cy){ //context, radius, sides, center x, center y
c.beginPath();
c.moveTo(cx + r,cy);
for(var p = 1; p < s; p++) c.lineTo(cx + r*Math.cos(p*2*Math.PI/s), cy + r*Math.sin(p*2*Math.PI/s));
c.closePath();
c.stroke();
}
// To write the mouse position in canvas local coordinates
function writeText(c,x,y,msg,col){
c.clearRect(0, 0, 300, 30);
c.font = "10pt Monospace";
c.fillStyle = col;
c.fillText(msg, x, y);
}
// Getting the mouse position and coverting into canvas local coordinates
function getMousePos(c, e) {
var rect = c.getBoundingClientRect();
return { x: e.clientX - rect.left,
y: e.clientY - rect.top
};
}
// To check if mouse is inside the polygone
function isMouseIn(r,s,cx,cy,mx,my){
var m = r*Math.cos(Math.PI/s),
d = Math.hypot(mx-cx,my-cy),
a = Math.atan2(cy-my,mx-cx);
return d <= (r+m)/2 + Math.cos(a*s)*(r-m)/2;
}
// the event listener callback
function mouseMoveCB(e){
var mp = getMousePos(cnv, e),
msg = 'Mouse at: ' + mp.x + ',' + mp.y,
col = "black",
inside = isMouseIn(radius,sides,center[0],center[1],mp.x,mp.y);
writeText(ctx, 10, 25, msg, inside ? "turquoise" : "red");
}
// body of the JS code
var cnv = document.getElementById("myCanvas"),
ctx = cnv.getContext("2d"),
sides = 6,
radius = 100,
center = [150,150];
cnv.addEventListener('mousemove', mouseMoveCB, false);
drawPolgon(ctx, radius, sides, center[0], center[1]);
#myCanvas { background: #eee;
width: 300px;
height: 300px;
border: 1px #ccc solid
}
<canvas id="myCanvas" width="300" height="300"></canvas>

At the redblog there is a full explanation with math and working examples.
The main idea is that hexagons are horizontally spaced by $3/4$ of hexagons size, vertically it is simply $H$ but the column needs to be taken to take vertical offset into account. The case colored red is determined by comparing x to y at 1/4 W slice.

RGB Sliders wiggle when changing value

I'm trying to implement my own color wheel picker, and I'm using this color wheel, as the base. I (finally) successfully added RGB sliders. So now when you change the color wheels color, the RGB sliders change dynamically, and when you change the RGB sliders, the color wheels color updates too.
The problem is, when I slide a slider from the RGB sliders, the other sliders tend to move a little bit too. For example, if I slide the green value, the red and blue values change a little bit.
I'm not exactly sure what's wrong. How can I get the other sliders to not move when I move one slider? (Obviously if I don't set the sliders value in redraw(), the sliders won't change when I move 1 slider, but I'm trying to find the core issue.)
JSFiddle
var b = document.body;
var colorWheelDiv = document.getElementById('colorWheelDiv');
var colorWheel = document.createElement('canvas');
var colorWheelOverlay = document.createElement('div');
var a = colorWheel.getContext('2d');
var label = document.getElementById('label');
var input = document.getElementById('input');
var redInput = document.getElementById('red');
var greenInput = document.getElementById('green');
var blueInput = document.getElementById('blue');
var alphaInput = document.getElementById('alpha');
var rgbInput = document.getElementsByClassName('rgbInput');
document.body.clientWidth; // fix bug in webkit: http://qfox.nl/weblog/218
// Jquery Elements
var $redSlider = $('#red');
var $greenSlider = $('#green');
var $blueSlider = $('#blue');
var $alphaSlider = $('#alpha');
(function() {
// Declare constants and variables to help with minification
// Some of these are inlined (with comments to the side with the actual equation)
var doc = document;
doc.colorWheel = doc.createElement;
b.a = b.appendChild;
// Add the colorWheel and the colorWheelOverlay
colorWheelDiv.appendChild(colorWheelOverlay);
colorWheelDiv.appendChild(colorWheel);
colorWheelOverlay.id = 'colorWheelOverlay';
colorWheel.id = 'colorWheel';
var width = colorWheel.width = colorWheel.height = colorWheelDiv.clientHeight,
imageData = a.createImageData(width, width),
pixels = imageData.data,
oneHundred = input.value = input.max = 100,
circleOffset = 10,
diameter = width - circleOffset * 2,
radius = diameter / 2,
radiusPlusOffset = radius + circleOffset,
radiusSquared = radius * radius,
two55 = 255,
currentY = oneHundred,
currentX = -currentY,
center = radius / 2,
wheelPixel = circleOffset * 4 * width + circleOffset * 4;
// Math helpers
var math = Math,
PI = math.PI,
PI2 = PI * 2,
sqrt = math.sqrt,
atan2 = math.atan2;
// Load color wheel data into memory.
for (y = input.min = 0; y < width; y++) {
for (x = 0; x < width; x++) {
var rx = x - radius,
ry = y - radius,
d = rx * rx + ry * ry,
rgb = colorWheel_hsvToRgb(
(atan2(ry, rx) + PI) / PI2, // Hue
sqrt(d) / radius, // Saturation
1 // Value
);
// Print current color, but hide if outside the area of the circle
pixels[wheelPixel++] = rgb[0];
pixels[wheelPixel++] = rgb[1];
pixels[wheelPixel++] = rgb[2];
pixels[wheelPixel++] = d > radiusSquared ? 0 : two55;
}
}
// Bind Event Handlers
input.oninput = redraw;
colorWheel.onmousedown = doc.onmouseup = function(e) {
// Unbind mousemove if this is a mouseup event, or bind mousemove if this a mousedown event
doc.onmousemove = /p/.test(e.type) ? 0 : (redraw(e), redraw);
}
$(".rgbInput").not($alphaSlider).slider({
range: "max",
min: 0,
max: 255,
value: 0,
slide: function(event, ui) {
redrawRGB();
}
});
function redrawRGB() {
var red = $('#red').slider('value');
var green = $('#green').slider('value');
var blue = $('#blue').slider('value');
var hsv = colorWheel_rgbToHsv(red, green, blue);
var newD = math.round(math.pow(radius * hsv.s, 2));
var newTheta = (hsv.h * PI2) - PI;
currentX = math.round(math.sqrt(newD) * math.cos(newTheta));
currentY = math.round(math.sqrt(newD) * math.sin(newTheta));
input.value = math.round(hsv.v * 100);
redraw(0);
}
// Handle manual calls + mousemove event handler + input change event handler all in one place.
function redraw(e) {
// Only process an actual change if it is triggered by the mousemove or mousedown event.
// Otherwise e.pageX will be undefined, which will cause the result to be NaN, so it will fallback to the current value
currentX = e.pageX - colorWheelDiv.offsetLeft - colorWheel.offsetLeft - radiusPlusOffset || currentX;
currentY = e.pageY - colorWheelDiv.offsetTop - colorWheel.offsetTop - radiusPlusOffset || currentY;
// Scope these locally so the compiler will minify the names. Will manually remove the 'var' keyword in the minified version.
var theta = atan2(currentY, currentX),
d = currentX * currentX + currentY * currentY;
// If the x/y is not in the circle, find angle between center and mouse point:
// Draw a line at that angle from center with the distance of radius
// Use that point on the circumference as the draggable location
if (d > radiusSquared) {
currentX = radius * math.cos(theta);
currentY = radius * math.sin(theta);
theta = atan2(currentY, currentX);
d = currentX * currentX + currentY * currentY;
}
var vValue = parseInt(input.value, 10);
var rgb = colorWheel_hsvToRgb(
(theta + PI) / PI2, // Current hue (how many degrees along the circle)
sqrt(d) / radius, // Current saturation (how close to the middle)
vValue / oneHundred // Current value (input type="range" slider value)
)
label.textContent = b.style.background = rgb[3];
colorWheelOverlay.style.opacity = ((vValue + 100 - 15) - (vValue * 2)) / oneHundred;
// Set slider Position \\
$redSlider.slider("value", math.round(rgb[0]));
$greenSlider.slider("value", math.round(rgb[1]));
$blueSlider.slider("value", math.round(rgb[2]));
// Reset to color wheel and draw a spot on the current location.
a.putImageData(imageData, 0, 0);
// Draw the current spot.
// I have tried a rectangle, circle, and heart shape.
/*
// Rectangle:
a.fillStyle = '#000';
a.fillRect(currentX+radiusPlusOffset,currentY+radiusPlusOffset, 6, 6);
*/
// Circle:
a.beginPath();
a.strokeStyle = '#000';
a.arc(~~currentX + radiusPlusOffset, ~~currentY + radiusPlusOffset, 4, 0, PI2);
a.stroke();
// Heart:
/* a.font = "1em arial";
a.fillText("♥", currentX+radiusPlusOffset-4,currentY+radiusPlusOffset+4);*/
}
// Created a shorter version of the HSV to RGB conversion function in TinyColor
// https://github.com/bgrins/TinyColor/blob/master/tinycolor.js
function colorWheel_hsvToRgb(h, s, v) {
h *= 6;
var i = ~~h,
f = h - i,
p = v * (1 - s),
q = v * (1 - f * s),
t = v * (1 - (1 - f) * s),
mod = i % 6,
r = [v, q, p, p, t, v][mod] * two55,
g = [t, v, v, q, p, p][mod] * two55,
b = [p, p, t, v, v, q][mod] * two55;
return [r, g, b, "rgb(" + math.round(r) + "," + math.round(g) + "," + math.round(b) + ")"];
}
function colorWheel_rgbToHsv(r, g, b) {
r = r / two55;
g = g / two55;
b = b / two55;
var max = math.max(r, g, b),
min = math.min(r, g, b);
var h, s, v = max;
var d = max - min;
s = max === 0 ? 0 : d / max;
if (max == min) {
h = 0; // achromatic
} else {
switch (max) {
case r:
h = (g - b) / d + (g < b ? 6 : 0);
break;
case g:
h = (b - r) / d + 2;
break;
case b:
h = (r - g) / d + 4;
break;
}
h /= 6;
}
return {
h: h,
s: s,
v: v
};
}
// Kick everything off
redraw(0);
})();
#colorWheelDiv {
width: 400px;
height: 400px;
position: relative;
}
#colorWheelOverlay {
background-color: black;
position: absolute;
pointer-events: none;
}
#colorWheelDiv,
#colorWheelOverlay,
#colorWheel {
border-radius: 50%;
}
#colorWheelOverlay,
#colorWheel {
width: 100%;
height: 100%;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/themes/smoothness/jquery-ui.css">
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>
R:
<div id="red" class="rgbInput"></div>
<br />G:
<div id="green" class="rgbInput"></div>
<br />B:
<div id="blue" class="rgbInput"></div>
<br />A:
<div id="alpha" class="rgbInput"></div>
<br />
<div id=colorWheelDiv></div>
<p id='label' style="font-style: normal; font-variant: normal; font-weight: normal; font-stretch: normal; font-size: 2em; line-height: normal; font-family: courier;">rgb(239,183,131)</p>
<input id="input" max="100" type="range" min="0">
Update
I tried that answer, and it didn't help. I still get the same results. The title may be the same, but I don't think the actual question is the same

You are converting the RGB values to HSV and then converting them back to RGB to set the other slider values. The conversion process has finite precision, so that's why you see the wiggling of the other sliders. The solution would be to set the RGB slider values directly from the original RGB values, not from the computed HSV values.
UPDATE
You can accomplish this as follows:
// Set slider Position \\
if(e){
$redSlider.slider( "value", math.round(rgb[0]));
$greenSlider.slider( "value", math.round(rgb[1]));
$blueSlider.slider( "value", math.round(rgb[2]));
}

Canvas - Draw only when hovering over new tile instead of whole canvas

Let's say I have a canvas that is split into a 15x10 32-pixel checkboard. Thus, I have this:
var canvas = document.getElementById('canvas');
var context = canvas.getContext('2d');
var tileSize = 32;
var xCoord
var yCoord
var tilesX = 15; // tiles across
var tilesY = 10; // tiles up and down
var counted = 1; // for drawing purpose for checkerboard for visual guidance
var mouseSel = new Image()
mouseSel.src = 'http://i.imgur.com/vAA03NB.png' // mouse selection
mouseSel.width = 32
mouseSel.height = 32
function isOdd(num) {
return num % 2;
}
function getMousePos(canvas, evt) {
// super simple stuff here
var rect = canvas.getBoundingClientRect();
return {
x: evt.clientX - rect.left,
y: evt.clientY - rect.top
};
}
drawCanvas(); // upon intilization... draw
function drawCanvas() {
for (var y = 0; y <= 10; y++) {
for (var x = 0; x <= 15; x++) {
if (isOdd(counted)) {
context.fillStyle = '#dedede'
context.fillRect(x * 32, y * 32, 32, 32);
// checkboard drawn complete.
}
counted++;
} // end first foor loop
counted++;
} // end last for loop
if (counted >= 176) counted = 1 // once all tiles (16x11) are drawn... reset counter for next instance
}
canvas.addEventListener('mousemove', function (evt) {
context.clearRect(0, 0, canvas.width, canvas.height); // clear canvas so mouse isn't stuck
drawCanvas(); // draw checkboard
// get the actual x,y position of 15x10 32-pixel checkboard
var mousePos = getMousePos(canvas, evt);
mousePos.xCoord = Math.floor(mousePos.x / tileSize)
mousePos.yCoord = Math.floor(mousePos.y / tileSize)
// draw the mouse selection
context.drawImage(mouseSel, (mousePos.xCoord * 32), (mousePos.yCoord * 32), 32, 32) // draw mouse selection
// debug
var message = ' (' + mousePos.xCoord + ',' + mousePos.yCoord + ') | (' + mousePos.x + ',' + mousePos.y + ')';
var textarea = document.getElementById('debug');
textarea.scrollTop = textarea.scrollHeight;
$('#debug').append(message + '\n');
}, false);
canvas#canvas {
background: #ABABAB;
position: relative;
z-index: 1;
float: left;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<canvas id="canvas" height="352" width="512" tabindex="0"></canvas>
<textarea name="" id="debug" cols="30" rows="35"></textarea>
**NOTE: ** Make sure to scroll down in that preview pane so you can see the debug textarea.
As you can see, the event of "drawing" fires EVERY single time it moves. That means every pixel.
I am trying to figure out how to make the drawing fire ONLY when a new x,y coord has changed. Because it'd be useless to redraw the mouse selection when it's only moved 5 pixels across and it's still going to be drawn at the same place.
My suggestion
Upon entering, have a temporary value and when that is passed, to redraw again?

Make a temporary value and update that if it was different from before. Then put the code in an if statement where either have changed.
var tempX, tempY;
var newX = 100;
var newY = 100;
tempX = mousePos.xCoord;
tempY = mousePos.yCoord;
if (newX !== tempX || newY !== tempY) {
// code here
}
if (tempX !== newX) newX = mousePos.xCoord;
if (tempY !== newY) newY = mousePos.yCoord;
JSFiddle: http://jsfiddle.net/weka/bvnma354/8/