Countdown animation circular - CreateJS / EaselJS / TweenJS - javascript

I am quite new with createJS - I want to achieve like a countdown timer animation:
I stumbled upon this issue which have this fiddle - I want to achive something like this but I want to create an arc shape:
I tried adjusting the code and changed the point values but it only gave me a diamond one instead of a perfect curve.
Do I need to point every values to achieve a perfect circle like:
points = [{x: 50, y: 0}, {x: 51, y: 1}, {x:52, y: 2}, ...etc]
Or is there other way (maybe a function or plugin) that does this stuff? I wasn't able to find anything on their documentation

You might be interested in using an arc() graphic, along with the EaselJS "Command" approach to graphics:
1) Create a full arc
var s = new createjs.Shape().set({x:100,y:100});
s.strokeCmd = s.graphics.s("red")
.ss(10,"round")
.arc(0,0,80,0,Math.PI*2)
2) Store off the last "command"
var cmd = s.command; // This will be the arc command
3) Set the command endAngle to 0. You can do this in the arc() method too
cmd.endAngle = 0;
4) In your animation, increment the endAngle over time. In this example I normalize 100 to mean 100% of the radius (Math.PI*2)
var index = 0;
function tick(event) {
index += 1; // Fake Percent
cmd.endAngle = index/100 * Math.PI*2;
stage.update(event);
}
Here is a quick fiddle: https://jsfiddle.net/lannymcnie/pgeut9cr/
If instead you just want to animate the circle over a fixed period, you can tween the endAngle value. Here is an example using TweenJS: https://jsfiddle.net/lannymcnie/pgeut9cr/2/
createjs.Tween.get(cmd)
.to({endAngle:Math.PI*2}, 2000, createjs.Ease.quadInOut);
Cheers,

Related

Weird rotation behavior with hammerjs and threejs

I want to rotate an object3D with hammerjs gestures.
Basically the rotation work, but with two issues I can't figure out.
The direction of the rotation changes randomly. It turns left. I stop and than move my fingers in the same diction again and suddenly instead of continuing to rotate left its going right. Happens sometimes but not always.
Once the rotation is started it only rotates to the same direction, despite me moving my fingers to different directions.
Here is how I handle the rotation:
public rotateObject3D (e: HammerInput): void {
if (rotationEnabled) {
const translation = new THREE.Vector3();
const rotation = new THREE.Quaternion();
const scale = new THREE.Vector3();
const rotateMatrix = new THREE.Matrix4();
const deltaRotationQuaternion = new THREE.Quaternion();
this._myObject.matrix.decompose(translation, rotation, scale);
this._deltaRot = (e.rotation * 0.01);
deltaRotationQuaternion.setFromEuler(new THREE.Euler(
0,
this._deltaRot * (Math.PI / 180),
0,
'XYZ'
));
deltaRotationQuaternion.multiplyQuaternions(deltaRotationQuaternion, rotation);
this._myObject.matrix = rotateMatrix.compose(translation, deltaRotationQuaternion, scale);
}
}
And this is the call of it:
this._hammerManager.on('rotate', (e) => {
this._arTools.rotateObject3D(e);
});
Is there anything I am missing?
It looks like you're using the absolute rotation value, instead of the change in rotation. For instance, consider the following scenario:
Rotation1: 12°
Rotation2: 10°
Rotation3: 5°
By multiplying your quaternions, your object is being rotated to 12°, then 22°, then finally 27°, even though you were turning back towards 0. This is because you're adding the new rotation to the last rotation on each event.
What you should do is save the previous rotation value, and subtract it from the new one, to get the rotation delta:
var previousRot = 0;
var newRot = 0;
rotateObject3D(e) {
newRot = e.rotation - previousRot;
// You could use newRot for your quaternion calculations
// but modifying .rotation.y is simpler
myObject.rotation.y += newRot
// Save value to be used on next event
previousRot = newRot;
}
With this method, the scenario above will give you the change in rotation. Your object will first rotate by +12°, then -2°, then -5°, for a more natural behavior.
Just make sure to reset previousRot = 0 on HammerJS' rotateend event so you don't use the value from previous gestures when a new one begins.

Javascript canvas draw line strange behavior using algorithm

There are plenty of examples on how to draw lines on canvas, in js.
But for only educational purposes i want to draw line using algorithm. basically method gets two Vector2 points, from them it finds middle point, then it continues like that recursively until minimum distance of 2 pixels is reached.
I have DrawPoint method to basically draw 1 point on canvas, and DrawLine method that does all the job.
For now I have 2 problems:
1: points are not colored red, as they should be.
2:
It doesnt look like a line.
For Vector2 i used "Victor.js" plugin, and it seems to be working well.
this is code i have:
JS:
var point2 = new Victor(100, 100);
var point3 = new Victor(150, 150);
DrawLine(point2, point3);
function DrawLine(vec0, vec1)
{
var point0 = new Victor(vec0.x, vec0.y);
var point1 = new Victor(vec1.x, vec1.y);
var dist = point1.distance(point0);
if (dist < 2)
return;
//this is how it should look like in c# var middlePoint = point0 + (point1 - point0)/2; But looks like i cant just divide by 2 using victor js because i can only divide vector by vector.
var middlePoint = point0.add(point1.subtract(point0).divide(new Victor(2,2)));
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
function DrawPoint(point){
var c = document.getElementById("screen");
var ctx = c.getContext("2d");
ctx.fillStyle = "FF0000";
ctx.fillRect(point.x, point.y, 3,1);
}
I really appreciate any help you can provide.
The victor.js documentation shows that most functions of Victors do not return new Victors, but operate on the current instance. In a way, v1.add(v2) is semantically more like v1 += v2 and not v1 + v2.
The problem is with calculating the midpoint. You could use the mix() method, which blends two vectors with a weight. You must clone() the Victor first, otherwise point0will be midofied:
var middlePoint = point0.clone().mix(point1, 0.5);
If you don't change the original Vectors, you don't need to create new instances of Victors from the arguments, you can use the arguments directly:
function DrawLine(point0, point1)
{
var dist = point1.distance(point0);
if (dist < 2) return;
var middlePoint = point0.clone().mix(point1, 0.5);
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
Finally, as Sven the Surfer has already said in a comment, "FF0000" isn't a valid colour. Use "#FF0000", note the hash mark, or one of the named web colours such as "crimson".

How do you animate a line using EaselJS and TweenJS

My goal is to have a line animate from point A to point B using the Tween function.
The drawing library I am using is EaselJS and for Tweening I am using TweenJS.
Is this possible using the moveTo function to animate a straight line from point A to point B?
My current setup is as follows:
var line = new createjs.Shape();
line.beginStroke('cyan');
line.setStrokeStyle(3);
line.moveTo(100, 100);
My goal is to animate this path from x100 y100 to x0 y0.
Animation Example:
I have tried this using the following and nothing happens:
var line = new createjs.Shape();
line.beginStroke('cyan');
line.setStrokeStyle(3);
line.moveTo(100, 100);
createjs.Tween.get(line).to({x: 0, y: 0}, 1000, createjs.Ease.sineInOut);
Nothing happens.
Draw Example:
However if I use this I get the line as expected but it is not animated:
var line = new createjs.Shape();
line.beginStroke('cyan');
line.setStrokeStyle(3);
line.moveTo(100, 100);
line.lineTo(0, 0);
This draws a line as expected.
Is there some way to use the lineTo method with the tweening method I am missing? I have checked the documentation of both Easel and TweenJS and can't find an example or any clear instructions on how to do this or if this is not possible.
Any help is appreciated.
The easiest approach is to use a Graphics command. The .command property returns the last created graphics command, which you can then manipulate with a tween:
var cmd = line.graphics.lineTo(100,100).command;
createjs.Tween.get(cmd).to({x:0}, 1000);
Here's a working example: https://jsfiddle.net/gskinner/17Lk8a9s/1/
Demo: http://jsfiddle.net/thpbr1vt/3/
Attention! Performance.
var stage = new createjs.Stage("canvas");
var vpoint = new createjs.Point( 100, 100);
var line = new createjs.Graphics();
line.beginStroke( 'cyan' );
line.moveTo( vpoint.x, vpoint.y );
var s = new createjs.Shape(line);
stage.addChild(s);
createjs.Tween.get(vpoint).to({x: 0, y: 0}, 1000, createjs.Ease.sineInOut);
createjs.Ticker.addEventListener("tick", tick);
function tick() {
line.lineTo( vpoint.x, vpoint.y );
stage.update();
}

Animating the drawing of a line in the canvas with fabric.js

I can draw a line in the canvas using:
var myLine = new fabric.Polyline([{x:200,y:200},{x:200,y:200}])
var canvas = new fabric.Canvas('c');
canvas.add(myLine);
However, I want to animate the drawing. I tried:
myLine.animate("points","[{x:200,y:200},{x:10,y:10}]",{onChange: canvas.renderAll.bind(canvas)})
But it's not working, and I couldn't see any way to animate the drawing of the line using fabric.js - I know I can use canvas methods directly but I am curious is fabric.js offers something more concise.
I made a jsFiddle based on http://fabricjs.com/polygon-animation/ and I change it into a fabricjs Polyline. You can set the start and end values from here:
var startPoints = [
{x: 1, y: 1},
{x: 2, y: 2}
];
var endPoints = [
{x: 1, y: 1},
{x: 200, y: 200}
];
No solution suited me so far so here a js-fiddle with what i came up with. Its based on the prev solution by Nistor Christian:
I Made a simple function which accepts the Canvas (in case you want to use this on more than one canvas), color, the original Line-Koordinates (StartXY, EndXY) and the new Line-Koordinates (NewStartX,NewStartY).
function animateLine(canvasInstance,color,
startX,startY,endX,endY,
newStartX, newStartY,newEndX,newEndY)

mouse position to isometric tile including height

Struggeling translating the position of the mouse to the location of the tiles in my grid. When it's all flat, the math looks like this:
this.position.x = Math.floor(((pos.y - 240) / 24) + ((pos.x - 320) / 48));
this.position.y = Math.floor(((pos.y - 240) / 24) - ((pos.x - 320) / 48));
where pos.x and pos.y are the position of the mouse, 240 and 320 are the offset, 24 and 48 the size of the tile. Position then contains the grid coordinate of the tile I'm hovering over. This works reasonably well on a flat surface.
Now I'm adding height, which the math does not take into account.
This grid is a 2D grid containing noise, that's being translated to height and tile type. Height is really just an adjustment to the 'Y' position of the tile, so it's possible for two tiles to be drawn in the same spot.
I don't know how to determine which tile I'm hovering over.
edit:
Made some headway... Before, I was depending on the mouseover event to calculate grid position. I just changed this to do the calculation in the draw loop itself, and check if the coordinates are within the limits of the tile currently being drawn. creates some overhead tho, not sure if I'm super happy with it but I'll confirm if it works.
edit 2018:
I have no answer, but since this ha[sd] an open bounty, help yourself to some code and a demo
The grid itself is, simplified;
let grid = [[10,15],[12,23]];
which leads to a drawing like:
for (var i = 0; i < grid.length; i++) {
for (var j = 0; j < grid[0].length; j++) {
let x = (j - i) * resourceWidth;
let y = ((i + j) * resourceHeight) + (grid[i][j] * -resourceHeight);
// the "+" bit is the adjustment for height according to perlin noise values
}
}
edit post-bounty:
See GIF. The accepted answer works. The delay is my fault, the screen doesn't update on mousemove (yet) and the frame rate is low-ish. It's clearly bringing back the right tile.
Source
Intresting task.
Lets try to simplify it - lets resolve this concrete case
Solution
Working version is here: https://github.com/amuzalevskiy/perlin-landscape (changes https://github.com/jorgt/perlin-landscape/pull/1 )
Explanation
First what came into mind is:
Just two steps:
find an vertical column, which matches some set of tiles
iterate tiles in set from bottom to top, checking if cursor is placed lower than top line
Step 1
We need two functions here:
Detects column:
function getColumn(mouseX, firstTileXShiftAtScreen, columnWidth) {
return (mouseX - firstTileXShiftAtScreen) / columnWidth;
}
Function which extracts an array of tiles which correspond to this column.
Rotate image 45 deg in mind. The red numbers are columnNo. 3 column is highlighted. X axis is horizontal
function tileExists(x, y, width, height) {
return x >= 0 & y >= 0 & x < width & y < height;
}
function getTilesInColumn(columnNo, width, height) {
let startTileX = 0, startTileY = 0;
let xShift = true;
for (let i = 0; i < columnNo; i++) {
if (tileExists(startTileX + 1, startTileY, width, height)) {
startTileX++;
} else {
if (xShift) {
xShift = false;
} else {
startTileY++;
}
}
}
let tilesInColumn = [];
while(tileExists(startTileX, startTileY, width, height)) {
tilesInColumn.push({x: startTileX, y: startTileY, isLeft: xShift});
if (xShift) {
startTileX--;
} else {
startTileY++;
}
xShift = !xShift;
}
return tilesInColumn;
}
Step 2
A list of tiles to check is ready. Now for each tile we need to find a top line. Also we have two types of tiles: left and right. We already stored this info during building matching tiles set.
function getTileYIncrementByTileZ(tileZ) {
// implement here
return 0;
}
function findExactTile(mouseX, mouseY, tilesInColumn, tiles2d,
firstTileXShiftAtScreen, firstTileYShiftAtScreenAt0Height,
tileWidth, tileHeight) {
// we built a set of tiles where bottom ones come first
// iterate tiles from bottom to top
for(var i = 0; i < tilesInColumn; i++) {
let tileInfo = tilesInColumn[i];
let lineAB = findABForTopLineOfTile(tileInfo.x, tileInfo.y, tiles2d[tileInfo.x][tileInfo.y],
tileInfo.isLeft, tileWidth, tileHeight);
if ((mouseY - firstTileYShiftAtScreenAt0Height) >
(mouseX - firstTileXShiftAtScreen)*lineAB.a + lineAB.b) {
// WOHOO !!!
return tileInfo;
}
}
}
function findABForTopLineOfTile(tileX, tileY, tileZ, isLeftTopLine, tileWidth, tileHeight) {
// find a top line ~~~ a,b
// y = a * x + b;
let a = tileWidth / tileHeight;
if (isLeftTopLine) {
a = -a;
}
let b = isLeftTopLine ?
tileY * 2 * tileHeight :
- (tileX + 1) * 2 * tileHeight;
b -= getTileYIncrementByTileZ(tileZ);
return {a: a, b: b};
}
Please don't judge me as I am not posting any code. I am just suggesting an algorithm that can solve it without high memory usage.
The Algorithm:
Actually to determine which tile is on mouse hover we don't need to check all the tiles. At first we think the surface is 2D and find which tile the mouse pointer goes over with the formula OP posted. This is the farthest probable tile mouse cursor can point at this cursor position.
This tile can receive mouse pointer if it's at 0 height, by checking it's current height we can verify if this is really at the height to receive pointer, we mark it and move forward.
Then we find the next probable tile which is closer to the screen by incrementing or decrementing x,y grid values depending on the cursor position.
Then we keep on moving forward in a zigzag fashion until we reach a tile which cannot receive pointer even if it is at it's maximum height.
When we reach this point the last tile found that were at a height to receive pointer is the tile that we are looking for.
In this case we only checked 8 tiles to determine which tile is currently receiving pointer. This is very memory efficient in comparison to checking all the tiles present in the grid and yields faster result.
One way to solve this would be to follow the ray that goes from the clicked pixel on the screen into the map. For that, just determine the camera position in relation to the map and the direction it is looking at:
const camPos = {x: -5, y: -5, z: -5}
const camDirection = { x: 1, y:1, z:1}
The next step is to get the touch Position in the 3D world. In this certain perspective that is quite simple:
const touchPos = {
x: camPos.x + touch.x / Math.sqrt(2),
y: camPos.y - touch.x / Math.sqrt(2),
z: camPos.z - touch.y / Math.sqrt(2)
};
Now you just need to follow the ray into the layer (scale the directions so that they are smaller than one of your tiles dimensions):
for(let delta = 0; delta < 100; delta++){
const x = touchPos.x + camDirection.x * delta;
const y = touchPos.y + camDirection.y * delta;
const z = touchPos.z + camDirection.z * delta;
Now just take the tile at xz and check if y is smaller than its height;
const absX = ~~( x / 24 );
const absZ = ~~( z / 24 );
if(tiles[absX][absZ].height >= y){
// hanfle the over event
}
I had same situation on a game. first I tried with mathematics, but when I found that the clients wants to change the map type every day, I changed the solution with some graphical solution and pass it to the designer of the team. I captured the mouse position by listening the SVG elements click.
the main graphic directly used to capture and translate the mouse position to my required pixel.
https://blog.lavrton.com/hit-region-detection-for-html5-canvas-and-how-to-listen-to-click-events-on-canvas-shapes-815034d7e9f8
https://code.sololearn.com/Wq2bwzSxSnjl/#html
Here is the grid input I would define for the sake of this discussion. The output should be some tile (coordinate_1, coordinate_2) based on visibility on the users screen of the mouse:
I can offer two solutions from different perspectives, but you will need to convert this back into your problem domain. The first methodology is based on coloring tiles and can be more useful if the map is changing dynamically. The second solution is based on drawing coordinate bounding boxes based on the fact that tiles closer to the viewer like (0, 0) can never be occluded by tiles behind it (1,1).
Approach 1: Transparently Colored Tiles
The first approach is based on drawing and elaborated on here. I must give the credit to #haldagan for a particularly beautiful solution. In summary it relies on drawing a perfectly opaque layer on top of the original canvas and coloring every tile with a different color. This top layer should be subject to the same height transformations as the underlying layer. When the mouse hovers over a particular layer you can detect the color through canvas and thus the tile itself. This is the solution I would probably go with and this seems to be a not so rare issue in computer visualization and graphics (finding positions in a 3d isometric world).
Approach 2: Finding the Bounding Tile
This is based on the conjecture that the "front" row can never be occluded by "back" rows behind it. Furthermore, "closer to the screen" tiles cannot be occluded by tiles "farther from the screen". To make precise the meaning of "front", "back", "closer to the screen" and "farther from the screen", take a look at the following:
.
Based on this principle the approach is to build a set of polygons for each tile. So firstly we determine the coordinates on the canvas of just box (0, 0) after height scaling. Note that the height scale operation is simply a trapezoid stretched vertically based on height.
Then we determine the coordinates on the canvas of boxes (1, 0), (0, 1), (1, 1) after height scaling (we would need to subtract anything from those polygons which overlap with the polygon (0, 0)).
Proceed to build each boxes bounding coordinates by subtracting any occlusions from polygons closer to the screen, to eventually get coordinates of polygons for all boxes.
With these coordinates and some care you can ultimately determine which tile is pointed to by a binary search style through overlapping polygons by searching through bottom rows up.
It also matters what else is on the screen. Maths attempts work if your tiles are pretty much uniform. However if you are displaying various objects and want the user to pick them, it is far easier to have a canvas-sized map of identifiers.
function poly(ctx){var a=arguments;ctx.beginPath();ctx.moveTo(a[1],a[2]);
for(var i=3;i<a.length;i+=2)ctx.lineTo(a[i],a[i+1]);ctx.closePath();ctx.fill();ctx.stroke();}
function circle(ctx,x,y,r){ctx.beginPath();ctx.arc(x,y,r,0,2*Math.PI);ctx.fill();ctx.stroke();}
function Tile(h,c,f){
var cnv=document.createElement("canvas");cnv.width=100;cnv.height=h;
var ctx=cnv.getContext("2d");ctx.lineWidth=3;ctx.lineStyle="black";
ctx.fillStyle=c;poly(ctx,2,h-50,50,h-75,98,h-50,50,h-25);
poly(ctx,50,h-25,2,h-50,2,h-25,50,h-2);
poly(ctx,50,h-25,98,h-50,98,h-25,50,h-2);
f(ctx);return ctx.getImageData(0,0,100,h);
}
function put(x,y,tile,image,id,map){
var iw=image.width,tw=tile.width,th=tile.height,bdat=image.data,fdat=tile.data;
for(var i=0;i<tw;i++)
for(var j=0;j<th;j++){
var ijtw4=(i+j*tw)*4,a=fdat[ijtw4+3];
if(a!==0){
var xiyjiw=x+i+(y+j)*iw;
for(var k=0;k<3;k++)bdat[xiyjiw*4+k]=(bdat[xiyjiw*4+k]*(255-a)+fdat[ijtw4+k]*a)/255;
bdat[xiyjiw*4+3]=255;
map[xiyjiw]=id;
}
}
}
var cleanimage;
var pickmap;
function startup(){
var water=Tile(77,"blue",function(){});
var field=Tile(77,"lime",function(){});
var tree=Tile(200,"lime",function(ctx){
ctx.fillStyle="brown";poly(ctx,50,50,70,150,30,150);
ctx.fillStyle="forestgreen";circle(ctx,60,40,30);circle(ctx,68,70,30);circle(ctx,32,60,30);
});
var sheep=Tile(200,"lime",function(ctx){
ctx.fillStyle="white";poly(ctx,25,155,25,100);poly(ctx,75,155,75,100);
circle(ctx,50,100,45);circle(ctx,50,80,30);
poly(ctx,40,70,35,80);poly(ctx,60,70,65,80);
});
var cnv=document.getElementById("scape");
cnv.width=500;cnv.height=400;
var ctx=cnv.getContext("2d");
cleanimage=ctx.getImageData(0,0,500,400);
pickmap=new Uint8Array(500*400);
var tiles=[water,field,tree,sheep];
var map=[[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[1,1],[1,2],[3,2],[1,1]],
[[0,0],[1,1],[2,2],[3,2],[1,1]],
[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[0,0],[0,0],[0,0],[0,0]]];
for(var x=0;x<5;x++)
for(var y=0;y<5;y++){
var desc=map[y][x],tile=tiles[desc[0]];
put(200+x*50-y*50,200+x*25+y*25-tile.height-desc[1]*20,
tile,cleanimage,x+1+(y+1)*10,pickmap);
}
ctx.putImageData(cleanimage,0,0);
}
var mx,my,pick;
function mmove(event){
mx=Math.round(event.offsetX);
my=Math.round(event.offsetY);
if(mx>=0 && my>=0 && mx<cleanimage.width && my<cleanimage.height && pick!==pickmap[mx+my*cleanimage.width])
requestAnimationFrame(redraw);
}
function redraw(){
pick=pickmap[mx+my*cleanimage.width];
document.getElementById("pick").innerHTML=pick;
var ctx=document.getElementById("scape").getContext("2d");
ctx.putImageData(cleanimage,0,0);
if(pick!==0){
var temp=ctx.getImageData(0,0,cleanimage.width,cleanimage.height);
for(var i=0;i<pickmap.length;i++)
if(pickmap[i]===pick)
temp.data[i*4]=255;
ctx.putImageData(temp,0,0);
}
}
startup(); // in place of body.onload
<div id="pick">Move around</div>
<canvas id="scape" onmousemove="mmove(event)"></canvas>
Here the "id" is a simple x+1+(y+1)*10 (so it is nice when displayed) and fits into a byte (Uint8Array), which could go up to 15x15 display grid already, and there are wider types available too.
(Tried to draw it small, and it looked ok on the snippet editor screen but apparently it is still too large here)
Computer graphics is fun, right?
This is a special case of the more standard computational geometry "point location problem". You could also express it as a nearest neighbour search.
To make this look like a point location problem you just need to express your tiles as non-overlapping polygons in a 2D plane. If you want to keep your shapes in a 3D space (e.g. with a z buffer) this becomes the related "ray casting problem".
One source of good geometry algorithms is W. Randolf Franklin's website and turf.js contains an implementation of his PNPOLY algorithm.
For this special case we can be even faster than the general algorithms by treating our prior knowledge about the shape of the tiles as a coarse R-tree (a type of spatial index).

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