Diagonal corner to corner gradient in canvas - javascript

I need to have a diagonal gradient from corner to corner in a canvas, not css.
Check example : http://jsfiddle.net/58y8b/77/
The first box is fine, since it's a square the gradient coordinates are just top left and bottom right.
Is there a way to calculate those coordinates for the next 2 rectangles so they get the gradient from corner to corner like the first box?

Fitting a gradient to 4 corners of Rectangle
This can be done with a little trig.
The solution
The diagram shows what needs to be done.
We need to find the coords of point E and D
We have the rectangle (assume the top left is at (0,0) that has width and height W, H. We find the angle pheta (bottom left) which is the same angle as pheta on the bottom right. We need the length of the line AB which is part of a right triangle that we have the hypot H for and the angle pheta. So AB is H * cos(pheta). We then get the vector at 90 deg of the diagonal line, and set its length to AB then find the center C and subtract the vector to get E and add the vector to the center C for D
To find the angle for the diagonal line use Math.atan2(-H,W).
The code
The code for it is taken from the fiddle you supplied and for rect4
//==========================================================
// NOTE this assumes that the top left of rect is at (0,0)
// if it is not then add the top left coordinate to the
// coordinates of E and D when defining the gradient position
// all else is the same
//-----------------------------------------------------------
// get angle from bottom left to top right
var pheta = Math.atan2(-rect4.height,rect4.width);
// get the length of the line from bottom right to diagonal line
// we got the ang of.
var AB = Math.abs(rect4.height * Math.cos(pheta)); // dont need the abd but cant be bothered
// explaining why its negative
// get vector at 90 deg from found angle
var xdx = Math.cos(pheta + Math.PI/2);
var xdy = Math.sin(pheta + Math.PI/2);
// from the center C of rectangle move AB dist back and forward to find
// points E and D
var x1 = rect4.width/2 - xdx * AB;
var y1 = rect4.height/2 - xdy * AB;
var x2 = rect4.width/2 + xdx * AB;
var y2 = rect4.height/2 + xdy * AB;
// Now create the gradient from E to D
rect4.setGradient('fill', {
type: 'linear',
x1: x1,
y1: y1,
x2: x2,
y2: y2,
colorStops: gradient
});
The result
The image need no words..

Related

Resizing rotated rectangle on HTML canvas

I have an instance of HTML 5 canvas and a rectangle drawn on it.
My drawing function takes a resizing angle into account and uses relative coordinates.
Relative coordinates're based upon three variables: top left rectangle point, rectangle width and rectangle height.
Rectangle width and rectangle height're calculated using two points: top left rectangle point and bottom right rectangle point.
To sum up, drawing function depends on top left rectangle point, bottom right rectangle point and rotation. It's an important point for the following text!
Here's a code snippet:
var canvas = document.getElementById('imageCanvas');
var ctx = canvas.getContext('2d');
var xTopLeft = 550;
var yTopLeft = 200;
var xBottomRight = 750;
var yBottomRight = 450;
var w = Math.max(xTopLeft, xBottomRight) - Math.min(xTopLeft, xBottomRight);
var h = Math.max(yTopLeft, yBottomRight) - Math.min(yTopLeft, yBottomRight);
var r = 1;
function draw() {
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.save()
ctx.translate(xTopLeft + w / 2, yTopLeft + h / 2);
ctx.rotate(r);
ctx.fillStyle = "yellow";
ctx.fillRect(w / 2 * (-1), h / 2 * (-1), w, h);
ctx.restore()
}
Here's my rectangle with a bunch of controls: eight resizing handles (white) and one rotation handle (green):
Rotating works fine.
Resizing works fine.
And I also try to implement resizing after rotation. Here's my approach with a humble illustration:
Grab the coordinates of the red point (it's mouse cursor coordiantes)
Derotate the point using negative angle to get derotated coordinates
function rotatePoint(x, y, center_x, center_y, angle) {
var new_x = (x - cx) * Math.cos(angle) - (y - cy) * Math.sin(angle) + cx;
var new_y = (x - cx) * Math.sin(angle) + (y - cy) * Math.cos(angle) + cy;
return [new_x, new_y]
}
Update xTopLeft, yTopLeft and redraw
Done
The idea behind this approach is simple. Since my drawing function depeneds on top left rectangle point and bottom right rectangle point I just need to get their new coordinates.
For instance, here's a simplified code for B point:
if (point == 'B') {
var newPointB = rotatePoint(mouse.x, mouse.y, center_x, center_y, -r);
xBottomRight = newPointB[0];
yTopLeft = newPointB[1];
}
But it doesn't work as expected: while resizing my rotated rectangle shifts, jumps and totally misbehaves.
In search of insights I've stumbled upon this article. The article covers my problem, but I don't get author's approach and can't implement it.
Why should I always lock the coordinates of the A point? My top left handle is intended to resize the rectangle in a north-west direction, so it would be necessary to change the coordinates of the A point...
Why should we recalculate the center point before derotation? It breaks the idea of uniform matrix transformations...
What's the correct algorithm in my case?
I was also facing same problem. It turned out that the angle I was using was in degree. Try multiplying angle in rotatePoint function with (Math.PI / 180).

Html canvas - rotate individual shapes on center axis

As you can see in the demo the L shape is getting cropped off the top of the screen and should be rotated 180 degrees and flush with the top left corner. I noticed two things that don't work as expected, the first is when I change ctx.translate(x, y) to ctx.moveTo(x, y) and increase the shape position to 100, 100 it moves more than 100px with translate, where as moveTo seems accurate. The second is that using a negative translate after ctx.stroke() has no affect on the shapes position.
var shape = {};
function draw(shape) {
var canvas = document.getElementById('canvas');
if (canvas.getContext) {
var ctx = canvas.getContext('2d');
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.save();
var x = shape.position.x + 0.5;
var y = shape.position.y + 0.5;
ctx.translate(x, y);
ctx.translate(shape.width * shape.scale/2, shape.height * shape.scale/2);
ctx.rotate(shape.orientation * Math.PI/180);
ctx.beginPath();
for (var i = 0; i < shape.points.length; i++) {
x = shape.points[i].x * shape.scale + shape.position.x + 0.5;
y = shape.points[i].y * shape.scale + shape.position.y + 0.5;
ctx.lineTo(x, y);
}
ctx.strokeStyle = '#fff';
ctx.stroke();
ctx.translate(-shape.width * shape.scale/2, -shape.height * shape.scale/2);
ctx.restore();
}
}
// L Shape
shape.points = [];
shape.points.push({ x:0, y:0 });
shape.points.push({ x:0, y:3 });
shape.points.push({ x:2, y:3 });
shape.points.push({ x:2, y:2 });
shape.points.push({ x:1, y:2 });
shape.points.push({ x:1, y:0 });
shape.points.push({ x:0, y:0 });
shape.position = {x: 0, y: 0};
shape.scale = 30;
shape.width = 3;
shape.height = 2;
shape.orientation = 180;
draw(shape);
#canvas {
background: #272B34; }
<canvas id="canvas" width="400" height="600"></canvas>
The easiest way to do 2D tranforms is via the setTransform function which takes 6 numbers, 2 vectors representing the direction and scale of the X and y axis, and one coordinate representing the new origin.
Unlike the other transform functions which are dependent of the current state setTransform is not effected by any transform done before it is called.
To set the transform for a matrix that has a square aspect (x and y scale are the same) and that the y axis is at 90 deg to the x ( no skewing) and a rotation is as follows
// x,y the position of the orign
function setMatrix(x,y,scale,rotate){
var xAx = Math.cos(rotate) * scale; // the x axis x
var xAy = Math.sin(rotate) * scale; // the x axis y
ctx.setTransform(xAx, xAy, -xAy, xAx, x, y);
}
//use
setMatrix(100,100,20,Math.PI / 4);
ctx.strokeRect(-2,-2,4,4); // draw a square centered at 100,100
// scaled 20 times
// and rotate clockwise 45 deg
Update
In response to the questions in the comments.
Why sin and cos?
Can you also explain why you used cos and sin for the axis?
I use Math.sin and Math.cos to calculate the X axis and thus the Y axis as well (because y is at 90 deg to x) because it is slightly quicker than adding the rotation as a separate transform.
When you use any of the transform functions apart from setTransform you are doing a matrix multiplication. The next snippet is the JS equivalent minimum calculations done when using ctx.rotate, ctx.scale, ctx.translate, or ctx.transform
// mA represent the 2D context current transform
mA = [1,0,0,1,0,0]; // default transform
// mB represents the transform to apply
mB = [0,1,-1,0,0,0]; // Rotate 90 degree clockwise
// m is the resulting matrix
m[0] = mA[0] * mB[0] + mA[2] * mB[1];
m[1] = mA[1] * mB[0] + mA[3] * mB[1];
m[2] = mA[0] * mB[2] + mA[2] * mB[3];
m[3] = mA[1] * mB[2] + mA[3] * mB[3];
m[4] = mA[0] * mB[0] + mA[2] * mB[1] + mA[4];
m[5] = mA[1] * mB[0] + mA[3] * mB[1] + mA[5];
As you can see there are 12 multiplications and 6 additions plus the need for memory to hold the intermediate values and if the call was to ctx.rotation the sin and cos of the angle would also be done. This is all done in native code in the JavaScript engine so is quicker than doing in JS, but side stepping the matrix multiplication by calculating the axis in JavaScript results in less work. Using setTransform simply replaces the current matrix and does not require a matrix multiplication to be performed.
The alternative to the answer's setMatrix function can be
function setMatrix(x,y,scale,rotate){
ctx.setTransform(scale,0,0,scale, x, y); // set current matrix
ctx.rotate(rotate); // multiply current matrix with rotation matrix
}
which does the same and does look cleaner, though is slower and when you want to do things like games where performance is very important often called functions should be as quick as possible.
To use the setMatrix function
So how would I use this for custom shapes like the L in my demo?
Replacing your draw function. BTW you should be getting the context outside any draw function.
// assumes ctx is the 2D context in scope for this function.
function draw(shape) {
var i = 0;
setMatrix(shape.position.x, shape.position.y, shape.scale, shape.orientation); // orientation is in radians
ctx.strokeStyle = '#fff';
ctx.beginPath();
ctx.moveTo(shape.points[i].x, shape.points[i++].y)
while (i < shape.points.length) {
ctx.lineTo(shape.points[i].x, shape.points[i++].y);
}
ctx.closePath(); // draw line from end to start
ctx.stroke();
}
In your code you have the line stored such that its origin (0,0) is at the top left. When defining shapes you should define it in terms of its local coordinates. This will define the point of rotation and scaling and represents the coordinate that will be at the transforms origin (position x,y).
Thus you should define your shape at its origin
function createShape(originX, originY, points){
var i;
const shape = [];
for(i = 0; i < points.length; i++){
shape.push({
x : points[i][0] - originX,
y : points[i][1] - originY,
});
}
return shape;
}
const shape = {};
shape.points = createShape(
1,1.5, // the local origin relative to the coords on next line
[[0,0],[0,3],[2,3],[2,2],[1,2],[1,0]] // shape coords
);

HTML5 canvas translate back after scale and rotation

I'm trying to do a few thing with canvas. First I have a user upload an image, if the image is larger than I want I need to scale it down. That part is working just fine. Recently we ran into an issue with iPhone users uploading images. These have orientation issues. I've figured out how to get the orientation extracted, my issue is what happens when I manipulate the image in the canvas.
This is what I need to do: Get the image, translate(), scale(), rotate(), translate() <- get it back to its original position, drawImage().
When I do that part of the image is off in the abyss.
if (dimensions[0] > 480 || dimensions[1] > 853) {
// Scale the image.
var horizontal = width > height;
if (horizontal) {
scaledHeight = 480;
scaleRatio = scaledHeight / height;
scaledWidth = width * scaleRatio;
} else {
scaledWidth = 640;
scaleRatio = scaledWidth / width;
scaledHeight = height * scaleRatio;
}
canvas['width'] = scaledWidth;
canvas['height'] = scaledHeight;
ctx['drawImage'](image, 0, 0, width, height, 0, 0, scaledWidth, scaledHeight);
/* Rotate Image */
orientation = 8; //manual orientation -> on the site we use loadImage to get the orientation
if(orientation != 1){
switch(orientation){
case 8:
case 6:
canvas.width = scaledHeight;
canvas.height = scaledWidth;
break;
}
var halfScaledWidth = scaledWidth/2;
var halfScaledheight = scaledHeight/2;
ctx.save(); //<- SAVE
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.translate(halfScaledWidth,halfScaledheight);
switch(orientation){
case 8: //rotate left
ctx.scale(scaleRatio,scaleRatio);
ctx.rotate(-90*Math.PI/180);
ctx.translate(-1380,-1055); // <-Manuial numbers
break;
case 3: //Flip upside down
ctx.scale(scaleRatio,scaleRatio);
ctx.rotate(180*Math.PI/180);
ctx.translate(-925,-595); //<-Manuial numbers
break;
case 6: //rotate right
ctx.scale(scaleRatio,scaleRatio);
ctx.rotate(90*Math.PI/180);
ctx.translate(-462,-130); //<-Manuial numbers
break;
}
//re-translate and draw image
//ctx.translate(-halfScaledWidth,-halfScaledheight);
ctx.drawImage(image,-halfScaledWidth, -halfScaledheight);
ctx.restore(); //<- RESTORE
}
/* Rotate Image */
}
I have orientation set manually so I can see how it looks in each position Im worried about. If its a portrait orientation I flip the canvas.
I've tried save() and restore(). I've tried translate(x,y) then translate(-x,-y).. My guess is that because of the scale the grid is off and x and y need to be multiplied. I tried doing that against the scaleRatio and still didn't work.
As you can see I manually set the translate back but that only works with the image size I am working with, so not a good solution at all!
Here is the code:
JSFiddle If I do a normal rotate right it all works.
Thanks!
Transformations
For the easy answer if you are not interested in the how skip to the bottom where you will find an alternative approch to your problem. It is all commented. I have made a bit of a guess as to what you wanted.
If you are interested in what I consider a simpler way to use the 2D transformation functions read the rest.
Matrix Math
When you use translate, scale, and rotate via the canvas 2D API what you are doing is multiplying the existing matrix with one created with each function.
Basically when you do
ctx.rotate(Math.PI); // rotate 180 deg
the API creates a new rotation matrix and multiplies the existing matrix with it.
Unlike normal math multiplication matrix multiplication will change the result depending on what order you multiply. In normal math multiplication A * B = B * A but this does not hold true for matrices mA * mB != mB * mA (Note the not equals)
This becomes more problematic when you need to apply several different transformations.
ctx.scale(2,2);
ctx.rotate(Math.PI/2);
ctx.translate(100,100);
Does not give the same result as
ctx.scale(2,2);
ctx.translate(100,100);
ctx.rotate(Math.PI/2);
The order that you need to apply the tranforms depends on what you are trying to achieve. Using the API this way is very handy for complex linked animations. Unfortunately it is also a source of endless frustration if you are not aware of matrix math. It also forces many to use the save and restore functions to restore the default transformation, that in some situations can be very costly in GPU performance.
setTransform()
We are in luck though as the 2D API also has the function ctx.setTransform(a, b, c, d, e, f) which is all you really should ever need. This function replaces the existing transform with the one supplied. Most of the documentation is rather vague as to the meaning of the a,b,c,d,e,f but contained in them is the rotation, scale, and translation.
One handy use of the function is to set the default transform rather than use save and restore.
I see this type of thing a lot. (The Example 1, referenced further down)
// transform for image 1
ctx.save(); // save state
ctx.scale(2,2);
ctx.rotate(Math.PI/2);
ctx.translate(100,100);
// draw the image
ctx.drawImage(img1, -img1.width / 2, -img1.height / 2);
ctx.restore(); // restore saved state
// transform for image 2
ctx.save(); // save state agian
ctx.scale(1,1);
ctx.rotate(Math.PI);
ctx.translate(100,100);
// draw the image
ctx.drawImage(img2, -img2.width / 2, -img2.height / 2);
ctx.restore(); // restore saved state
An easier way is to just drop the save and restores and reset the transform manually by setting it to the Identity matrix
ctx.scale(2,2);
ctx.rotate(Math.PI/2);
ctx.translate(100,100);
// draw the image
ctx.drawImage(img1, -img1.width / 2, -img1.height / 2);
ctx.setTransform(1,0,0,1,0,0); // restore default transform
// transform for image 2
ctx.scale(1,1);
ctx.rotate(Math.PI);
ctx.translate(100,100);
// draw the image
ctx.drawImage(img2, -img2.width / 2, -img2.height / 2);
ctx.setTransform(1,0,0,1,0,0); // restore default transform
Now then I am sure you are still wondering what are these numbers being passed to setTransform and what do they mean?
The easiest way to remember them is as 2 vectors and 1 coordinate. The two vectors describe the direction and scale of a single pixel, the coordinate is simply the x,y pixel location of the origin (the location that drawing at 0,0 will be on the canvas).
A Pixel and its axis
Imagine a single pixel, this is the abstract transformed pixel that can be scaled and rotated by the current transformation. It has two axis, X and Y. To describe each axis we need two numbers ( a vector) these describes the screen (untransformed) direction and scale of the top and left side of the pixel. So for a normal pixel that matches the screen pixels the X axis is across the top from left to right and is one pixel long. The vector is (1,0) one pixel across, no pixels down. For the Y axis that goes down the screen the vector is (0,1) no pixels across, one pixel down. The origin is the top right screen pixel which is at coordinate (0,0).
Thus we get the Identity Matrix, the default matrix for the 2D API (and many other APIs) The X axis (1,0), Y axis (0,1) and the origin (0,0) which match the six arguments for setTransform(1,0,0,1,0,0).
Now say we want to scale the pixel up. All we do is increase the size of the X and Y Axis setTransform(2,0,0,2,0,0) is the same as scale(2,2) (from the default transform) Our pixel's top is now two pixels long across the top and two pixels long down the left side. To scale down setTransform(0.5,0,0,0.5,0,0) our pixel is now half a pixel across and down.
These two axis vectors (a,b) & (c,d) can point in any direction, are completely independent of each other , they don't have to be at 90 deg to each other so can skew the pixel, nor do they require that they be the same length so you can change the pixel aspect. The origin is also independent and is just the canvas absolute coordinates in pixels of the origin and can be set to anywhere on or off the canvas.
Now say we want to rotate the transform 90Deg clockwise, scale up both axes by 2 and position the origin at the center of the canvas. We want the X axis (top) of our pixel to be 2 pixels long and pointing down the screen. The vector is (0,2) 0 across and two down. We want the left side of our pixel to 2 long and point to the left of the screen (-2,0) Negative two across and none down. And the origin at the center is (canvas.width / 2, canvas.height / 2) to get the final matrix that is setTransform(0,2,-2,0,canvas.width / 2, canvas.height / 2)
Rotate the other way is setTransform(0,-2,2,0,canvas.width / 2, canvas.height / 2)
Easy Rotate 90deg
You may notice that rotating 90 degrees is just swapping the vectors and changing a sign.
The vector (x,y) rotated 90 degrees clockwise is (-y,x).
The vector (x,y) rotated 90 degrees anti-clockwise is (y,-x).
Swap the x, and y and negate the y for clockwise or negate the x for the anticlockwise rotation.
For 180 it is starting at 0 deg vector (1,0)
// input vector
var x = 1;
var y = 0;
// rotated vector
var rx90 = -y; // swap y to x and make it negative
var ry90 = x; // x to y as is
// rotate again same thing
var rx180 = -ry90;
var rx180 = rx90;
// Now for 270
var rx270 = -ry180; // swap y to x and make it negative
var rx270 = rx180;
Or all in terms of just x and y
0 deg (x,y)
90deg (-y,x)
180deg (-x,-y)
270deg (y,-x)
and back to 360 (x,y).
This is a very handy attribute of a vector that we can exploit to simplify the creation of our transformation matrix. In most situations we do not want to skew our image thus we know that the Y axis is always 90Deg clockwise from the x axis. Now we only need to describe the x axis and by applying the 90deg rotation to that vector we have the y axis.
So the vars x and y are the scale and direction of the top of our pixel (x axis), ox, oy are the location of the origin on the canvas (translation) .
var x = 1; // one pixel across
var y = 0; // none down
var ox = canvas.width / 2; // center of canvas
var oy = canvas.height / 2;
Now to create the transform is
ctx.setTransform(x, y, -y, x, ox, oy);
Note that the y axis is at 90 degs to the x axis.
Trig and the Unit vector
All well and easy when the axis are aligned to the top and sides, how do you get the vector for a axis at an arbitrary angle such as is supplied by the argument for ctx.rotate(angle) For that we need a tiny bit of trig. The Math function Math.cos(angle) returns the x component of the angle, angle and Math.sin(angle) gives us the Y component. For zero deg cos(0) = 1 and sin(0) = 0 for 90 deg (Math.PI/2 radians) cos(PI/2) = 0 and sin(PI/2) = 1.
The beauty of using sin and cos is that the two numbers that we get for our vector always give us a vector that is 1 unit (pixel) long (this is called a normalised vector or a unit vector) thus cos(a)2 + sin(a)2 = 1
Why does this matter? because it makes scaling very easy. Assuming that we always keep the aspect square we only need one number for the scale. To scale a vector you simply multiply it by the scale
var scale = 2; // scale of 2
var ang = Math.random() * 100; // any random angle
var x = Math.cos(ang); // get the x axis as a unit vector.
var y = Math.sin(ang);
// scale the axis
x *= scale;
y *= scale;
the vector x,y is now two units long.
Better than using save, restore, rotate, scale, translate... :(
Now put it all together to create a matrix with an arbitrary rotation, scale and translation (origin)
// ctx is the 2D context,
// originX, and originY is the origin, same as ctx.translate(originX,originY)
// rotation is the angle in radians same as ctx.rotate(rotation)
// scale is the scale of x and y axis same as ctx.scale(scale,scale)
function createTransform(ctx,originX,originY,rotation,scale){
var x, y;
x = Math.cos(rotation) * scale;
y = Math.sin(rotation) * scale;
ctx.setTransform(x, y, -y, x, originX, originY);
}
Now to apply that to the example (1) given above
// dont need ctx.save(); // save state
// dont need ctx.scale(2,2);
// dont need ctx.rotate(Math.PI/2);
// dont need ctx.translate(100,100);
createMatrix(ctx, 100, 100, Math.PI/2, 2)
// draw the image normally
ctx.drawImage(img1, -img1.width / 2, -img1.height / 2);
// dont need ctx.restore(); // restore saved state
// transform for image 2
// dont need ctx.save(); // save state agian
// dont need ctx.scale(1,1);
// dont need ctx.rotate(Math.PI);
// dont need ctx.translate(100,100);
// we don't have to reset the default transform because
// ctx.setTransform completely replaces the current transform
createMatrix(ctx, 100, 100, Math.PI/2, 2)
// draw the image
ctx.drawImage(img2, -img2.width / 2, -img2.height / 2);
// dont need ctx.restore(); // restore saved state
And that is how you use setTransform to simplify transforming the canvas, rather than guessing, trial and error, scale, rotates, and translates back and forth within a sea of save and restores.
Using that to simplify your code
The answer
And now to your question
I am not entirely sure what you are after, I presume you a dont mind scaling the canvas to accommodate the image, that the image is always in the center and that the aspect remains the same. As the rotations are aligned to the screen I will set the transforms manualy
// this is in set up code
const MAX_SIZE_WIDTH = 640;
const MAX_SIZE_HEIGHT = 480;
orientationData = [];
orientationData[6] = Math.PI/2; // xAxis pointing down
orientationData[8] = -Math.PI/2; // xAxis pointing up
orientationData[3] = Math.PI; //xAxis pointing to left
// in your code
var orient,w,h,iw,ih,scale,ax,ay; // w and h are canvas size
// assume image is the loaded image
iw = image.width; // get the image width and height so I dont have to type as much.
ih = image.height;
if(orientation != 1){
var orient = orientationData[orientation];
if(orient === undefined){
return; // bad data so return
}
// get scale and resize canvas to suit
// is the image on the side
if(orientation === 6 || orientation === 8){
// on side so swap width and height
// get the height and width scales for the image, dont scale
// if the image is smaller than the dimension
scale = Math.min(1,
MAX_SIZE_WIDTH / ih,
MAX_SIZE_HEIGHT / iw
);
w = canvas.width = scale * ih;
h = canvas.height = scale * iw;
}else{
// for normal orientation
scale = Math.min(1,
MAX_SIZE_WIDTH / iw,
MAX_SIZE_HEIGHT / ih
);
h = canvas.height = scale * ih;
w = canvas.width = scale * iw;
}
// Do you really need to clear the canvas if the image is filling it??
// ensure that the default transform is set
ctx.setTransform(1, 0, 0, 1, 0, 0);
// clear the canvas
ctx.clearRect(0, 0, w, h);
// now create the transformation matrix to
// position the image in the center of the screen
// first get the xAxis and scale it
ax = Math.cos(orient) * scale;
ay = Math.sin(orient) * scale;
// now set the transform, the origin is always the canvas center
// and the Y axis is 90 deg clockwise from the xAxis and same scale
ctx.setTransform(ax, ay, -ay, ax, w / 2, h / 2);
// now draw the image offset by half its width and height
// so that it is centered on the canvas
ctx.drawImage(image,-iw / 2, -ih / 2);
// restore the default transform
ctx.setTransform(1, 0, 0, 1, 0, 0);
} // done.
Try this one https://jsfiddle.net/uLdf4paL/2/. Your code is correct, but you have to change orientation variable when you try to rotate and scale image (if it is what you wanna get).
My solution for this was using two canvases
one for the rotation and scaling (around the center point of the image)
then using a new canvas for translating and drawing the prev canvas on it. it worked perfectly for me:
const root = document.querySelector("#root");
const img = document.querySelector('#img');
const canvas = document.querySelector('canvas');
const ctx = canvas.getContext("2d");
const tempCanv = document.createElement('canvas')
const tempCtx = tempCanv.getContext('2d');
canvas.height = img.height
canvas.width = img.width
tempCanv.width = img.width
tempCanv.height = img.height
const position = [0.25, 0.25];
const rotation = -40;
const scale = [37.222, 37.222];
const x = img.width / 2;
const y = img.height / 2;
// Set the origin to the center
tempCtx.translate(x, y);
// now rotate and scale around the center
tempCtx.rotate(Math.PI / 180 * rotation);
tempCtx.scale(scale[0] / 100, scale[1] / 100);
// Translate back to origin
tempCtx.translate(-x, -y);
// Draw temp
tempCtx.drawImage(img,0,0);
// Translate the final canvas
ctx.translate(position[0] * canvas.width, position[1] * canvas.height);
ctx.drawImage(tempCanv, 0, 0);
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
<style>
#root{
display: flex;
gap: 2rem;
padding: 1rem;
overflow:visible
position relative;
}
.image-container{
}
#img{
transform: rotate(-40deg) scale(0.37222, 0.37222);
top: 25%;
left: 25%;
position: relative;
}
</style>
</head>
<body>
<div id="root">
<div class="image-container">
<img id="img" src="https://media.istockphoto.com/photos/cat-world-picture-id1311993425?b=1&k=20&m=1311993425&s=170667a&w=0&h=vFvrS09vrSeKH_u2XZVmjuKeFiIEjTkwr9KQdyOfqvg=" alt="red">
</div>
<canvas id="canvas"></canvas>
</div>
</body>
</html>

javascript canvas draw multiple rectangles around a circle facing the centre

I am looking for a way to draw multiple rectangles around a circle at an angle facing the centre. What i have so far, currently just drawing the rectangles around a circle facing one direction lacking the angle inclination towards the centre -
https://thysultan.com/projects/thyplayer/
What i want is for the rectangles to incline at an angle such that each rectangle is facing the centre of the circle at it's designated position.
How would one do that?
Trigonometry approach
To animate the lines towards the center you can use either transforms or trigonometry math. Personally I find it easier with math in cases like this so here is such an example.
See #markE's answer for an example on how to use transforms (transforms can be easier on the eye and in the code in general).
Some prerequisites:
We know canvas is oriented so that a 0° angle points towards right. This is essential if you want to mark the frequency range somehow.
We need to calculate the length of the line from outer radius towards center (inner radius)
We need to calculate the end points of the line based on the angle.
An easy solution can be to make a function that calculates a single line at a center (cx, cy) at a certain angle (in radians), with t as length, t being in the range [0, 1] (as the FFT floating point buffer bins). We also provide an outer and inner radius to enable limiting the line:
function getLine(cx, cy, angle, t, oRadius, iRadius) {
var radiusDiff = oRadius - iRadius, // calc radius diff to get max length
length = radiusDiff * t; // now we have the line length
return {
x1: oRadius * Math.cos(angle), // x1 point (outer)
y1: oRadius * Math.sin(angle), // y1 point (outer)
x2: (oRadius - length) * Math.cos(angle), // x2 point (inner)
y2: (oRadius - length) * Math.sin(angle) // y2 point (inner)
}
}
All we need to do now is to feed it data from the FFT analyzer.
NB: Since all lines points towards center, you will have a crowded center area. Something to have in mind when determine the line widths and inner radius as well as number of bins to use.
Example demo
For the example I will just use some random data for the "FFT" and plot 64 bins.
// angle - in radians
function getLine(cx, cy, angle, t, oRadius, iRadius) {
var radiusDiff = oRadius - iRadius, // calc radius diff to get max length
length = radiusDiff * t; // now we have the line length
return {
x1: cx + oRadius * Math.cos(angle), // x1 point (outer)
y1: cy + oRadius * Math.sin(angle), // y1 point (outer)
x2: cx + (oRadius - length) * Math.cos(angle), // x2 point (inner)
y2: cy + (oRadius - length) * Math.sin(angle) // y2 point (inner)
}
}
// calculate number of steps based on bins
var ctx = document.querySelector("canvas").getContext("2d"),
fftLength = 64,
angleStep = Math.PI * 2 / fftLength,
angle = 0,
line;
ctx.beginPath(); // not needed in demo, but when animated
while(angle < Math.PI*2) {
// our line function in action:
line = getLine(250, 250, angle, getFFT(), 240, 50);
ctx.moveTo(line.x1, line.y1); // add line to path
ctx.lineTo(line.x2, line.y2);
angle += angleStep // get next angle
}
ctx.lineWidth = 5; // beware of center area
ctx.stroke(); // stroke all lines at once
// to smooth the "FFT" random data
function getFFT() {return Math.random() * 0.16 + 0.4}
<canvas width=500 height=500></canvas>

Calc position to draw a circle

I have to draw 2 circles where I give the draw function the center point of each circle and the x, y.
I first circle is static, will be on the same x and y.
The second circle is smaller, and will follow the mouse position, but it's center point can't go outside the First circle.
Considering that I have this variables:
Circle1.x, Circle1.y, Circle.radius
Mouse.x, Mouse.y
How can I get the x and y for the circle2?
Here is a pick to show What I need:
Black : Circle1
Red : Circle2
Green : Mouse
Subtract the center of the circle from the mouse position. This results into a vector (dx,dy) pointing from the center of the circle to the mouse position. This vectors length should be equal or smaller than the circles radius. If the length is smaller than the radius of the circle: the mouse position is within the circle and stays the same, If the length is larger than the radius of the circle: normalize the vector and multiply it with the radius, adding the resulting vector to the center of the circle will result in the new position.
Psuedo code:
dx = Mouse.x-Circle1.x
dy = Mouse.y-Circle1.y
length = sqrt(dx*dx+dy*dy)
if (length > Circle.radius)
ratio = Circle1.radius/length
Circle2.x = Circle1.x + dx*ratio
Circle2.y = Circle1.y + dy*ratio
else
Circle2.x = Mouse.x
Circle2.y = Mouse.y
end

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