HTML5 canvas translate back after scale and rotati

Question

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!

Answer 1

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.

Answer 2

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).