For a game project, I'm drawing images using their properties such as fileName, position, scale, rotation.
Here is the part that does the drawing:
this.context.save();
this.context.translate(item.position.x, item.position.y);
if (item.rotation > 0) {
this.context.rotate(item.rotation * (Math.PI / 180));
}
if (item.scale.x !== 1 || item.scale.y !== 1) {
this.context.scale(item.scale.x, item.scale.y);
}
var width = item.imageSize.width * item.scale.x;
var height = item.imageSize.height * item.scale.y;
this.context.drawImage(this.assets.image[item.fileName], -(width / 2), -(height / 2), width, height);
this.context.restore();
(don't mind the strange positioning, it's not important)
This works fine, but there is one thing that I don't understand:
Rotation and scaling can be done in two different ways: first scale and then rotate, or other way around. Logically, one would think that first scale then rotation is correct, but for some reason, it only works correctly if I first rotate then scale.
What is the correct way of doing this?
Where is your point of origin for your objects? Are the x/y the top left? If so that could be causing the issue.
Demo Scaling Then Rotating
ctx.translate(box.x, box.y);
ctx.scale(2,2);
ctx.rotate(box.angle);
Demo Rotating Then Scaling
ctx.translate(box.x, box.y);
ctx.rotate(box.angle);
ctx.scale(2,2);
If you notice both of those demos work fine regardless of when I perform the scaling and rotating. My point of origins however (where I translate to) are in the center of the boxes.
To answer your question (or attempt to) there is no right or wrong way, you can scale first or rotate first, its really just a matter of preference.
The "correct way" really depends on what you're trying to do. For you, it seems like rotating and then scaling results in what you expect.
Here is a fiddle that shows the difference between rotating then scaling and scaling then rotating: http://jsfiddle.net/HYbC7/3/
What's unexpected for me is when you scale then rotate. If you scale(x, y) where x and y are not equal, then rotate, then the rotation along the x-axis will be different than the rotation along the y-axis and the resulting grid will be skewed.
Related
Can someone help me understand how three.js initially determines the size/scale of a sprite?
At the moment I'm working with 4 sprites (PNGs with transparency that are 3000px × 1830px) stacked in 3D space, but I'm having to scale them up between 16x and 22x. In order to keep the sprites from looking squashed, though, I have to scale the y-axis 75% of the x-scale.
Eventually, I want to be able to pull in images systematically, and have them scale appropriately.
It's possible I just haven't set this thing up correctly. It looks right, but it's super hacky-feeling right now. I pretty much changed a bunch of numbers, until it looked right to me. I don't like that. I want to understand.
Here's what I'm working with currently:
http://valorink.com/3d-test/stackoverflow/
Looking into the code of the Sprite class reveals that a simple plane with width and height of 1 is created in the constructor. I wouldn't have expected anything else, because the geometry size is usually not defined by the texture size.
You probably want them to fill the viewport, so you have to scale them. With perspective camera its a bit of math, because the amount of x-scale (or y-scale) to fit the viewport size relates to the distance to the camera. So, it should be something like
var halfHeigt = distanceToCamera / Math.tan( camera.fov/2 * Math.PI / 180 );
y-scale = halfHeight * 2;
And of course you need to consider the aspect ratio in order to not looking squashed. So, x-scale should be y-scale * textureWidth / textureHeight, or the other way round y-scale = x-scale * textureHeight / textureWidth.
I am trying to build a small app where my users can straighten up a tilted face with just 2 clicks
I ask my users to click on the middle of the nose and the middle of the eyebrows of the face within the image.
From there I get 2 points eyebrowMiddle(x1,y1) and noseMiddle (x2,y2).
Is it possible via these 2 points to calculate how much Canvas
rotation I need to have to rotate the image and make the face straight
in relation to the canvas rectangle?
Also, how can I detect and adjust accordingly if the image is tilted
to the left or right?
Here is a more descriptive image to show you what I mean now.
PS:
x1,y1 and x2,y2 are in relation to the canvas perimeter of
course, not the browser window or anything else.
We have tried the line equation such as m = (x2-x1) / (y2-y1) but the
result is always near 1 so I don't think we are following the right
course at the moment.
We don't care if the image looks wrong in the canvas as long as the
face features are parallel in relation to the bottom of the canvas
(they should be looking straight).
To perform such a rotation, you need to decide of the pivot point. Here i choose the eyebrow.
Then you have to choose a point in the target canvas where this pivot point will be hooked. I decided to choose the point at middle x coordinates, and at fourth of the screen in y.
To compute the rotation angle, you have to use atan2, which will nicely give you the angle for a given deltaY / deltaX in between two points ( angle = Math.atan2 ( delta y , delta x ) ) .
Then to draw :
- Translate to the target point.
- rotate by right angle.
- draw the image centering on its pivot.
ET VOILA, it works :-)
function rotate() {
ctx.save();
// go to default center position
ctx.translate(eyeBrowTargetPosition.x, eyeBrowTargetPosition.y);
// compute angle
var yDelta = noseMiddle.y - eyebrowMiddle.y;
var xDelta = noseMiddle.x - eyebrowMiddle.x ;
var angle = Math.atan2 (yDelta ,xDelta);
// compensate for angle
ctx.rotate(angle);
//draw image centering input on eyebrow
ctx.drawImage(face, -eyebrowMiddle.x, -eyebrowMiddle.y);
ctx.restore();
};
jsbin is here :
http://jsbin.com/wavokaku/2/edit?js,output
result with an approximation of the existing green dots :
I have a rounded rectangle at a specific x, y, w, h on a canvas. I first do a context.translate to get the object where I want it, then when it comes to rotating it, this is where I'm having issues working out the math needed.
I can do a simple context.rotate(Math.PI/180 * 25) to rotate it 25degs but it rotates from the x,y. I really want to shift the rotating point to like x + (w/2) and y + (w/2).
I'm not sure how to tell the rotate method to rotate it around a different point. I think I have to rotate it like normal but recalculate x,y perhaps based on the rotation maybe?
The canvas always rotates about the origin (0,0). The ctx.translate command can be thought of as shifting the origin, so you must translate by (x+w/2, y+h/2) before you rotate if you wish to rotate about the center of the rectangle.
(and of course, translate back after, or use save and restore)
I'm having serious trouble understanding how to compute the coordinates of rotation/scale pivots (e.g. rotation point) for SVG transformations, using Raphael.js. In short, if you apply a transformation such as image.transform("S1.5R45"), the transformations are applied in relation to the default rotation & scale pivot, which I'm not sure how to calculate.
To exemplify, I've put together a fiddle (jsfiddle.net/GVEqf/), where the aim is to have exactly the same output in both the viewports, for a couple of transformations on an image object. In the first viewport, I don't specify the rotation point, but in the second one I do. However, I can't get the same results. What I need is to input the rotation coordinates for the second viewport, so that the output is identical with the first case.
Please help.
When not specified the pivot is the center of the element.
Here you have to take care of the position you have applied to the images and the scaling that will be done. Since in this case your scaling is relative to the top left corner of the image, we can just multiply the center coordinate by it.
// Compute rotation pivot coordinates
var scaling = 1.5;
rx = (x + (img_width / 2)) * scaling;
ry = (y + (img_height / 2)) * scaling;
// Apply transformations
image1.transform("S1.5,1.5,0,0R45");
image2.transform("S1.5,1.5,0,0R45,"+rx+","+ry);
http://jsfiddle.net/TYCJ7/
I'm using canvas for a project and I have a number of elements that I'm skewing. I'm only skewing on the y value and just want to know what the new width of the image is after skewing (so I can align it with another canvas element). Check out the code below to see what I mean
ctx.save();
//skew the context
ctx.transform(1,0,1.3,0,0,0);
//draw two images with different heights/widths
ctx.drawImage(image,0,0,42,60);
ctx.drawImage(image,0,0,32,25);
The goal would be to know that the 42 by 60 image was now a X by 60 image so I could do some translating before drawing it at 0,0. It's easy enough to measure each image individually, but I have different skew values and heights/widths throughout the project that need to be align. Currently I use this code (works decently for images between 25 and 42 widths):
var skewModifier = imageWidth*(8/6)+(19/3);
var skewAmount = 1.3; //this is dynamic in my app
var width = (skewModifier*skewAmount)+imageWidth;
As images get wider though this formula quickly falls apart (I think it's a sloping formula not a straight value like this one). Any ideas on what canvas does for skews?
You should be able to derive it mathematically. I believe:
Math.atan(skewAmount) is the angle, in radians, that something is skewed with respect to the origin.
So 1.3 would skew the object by 0.915 radians or 52 degrees.
So here's a red unskewed object next to the same object skewed (painted green). So you have a right triangle:
We know the origin angle (0.915 rads) and we know the adjacent side length, which is 60 and 25 for your two images. (red's height).
The hypotenuse is the long side thats being skewed.
And the opposite side is the triangle bottom - how much its been skewed!
Tangent gets us opposite / adjacent if I recall, so for the first one:
tan(0.915) = opposite / 60, solving for the opposite in JavaScript code we have:
opposite = Math.tan(0.915)*60
So the bottom side of the skewed object starts about 77 pixels away from the origin. Lets check our work in the canvas:
http://jsfiddle.net/LBzUt/
Looks good to me!
The triangle in question of course is the canvas origin, that black dot I painted, and the bottom-left of the red rectangle, which is the original position that we're searching for before skewing.
That was a bit of a haphazard explanation. Any questions?
Taking Simon's fiddle example one step further, so you can simply enter the degrees:
Here's the fiddle
http://jsfiddle.net/LBzUt/33/