I am working on this "simple" problem for hours with no success, although I tried many ways to solve it using all kind of solutions suggested in SO.
My problem is as follows:
I have a point on a canvas, which when I click on it my app does something, after identifying the point by comparing the mouse click coordinates to the stored position of the point.
After zooming into the point, using the mouse wheel, I click on the point again but the mouse coordinates no longer fits the stored position of the point.
I need to either transform the mouse coordinates to it's coordinates before the zoom, so I will be able to compare to the stored position, or to transform the stored position to the new canvas so it can be compare to the coordinates of the mouse. Any of the solution is fine by me.
I know the following data:
The "scale" value,
The size of the canvas (top, left, width, height),
The new origin of the canvas (top, left)
I would like a solution using java script.
Finally figured it out and it is quite simple, once I understood the concept.
Save the new canvas origin after doing the zoom (in JS it is calling ctx.translate() and ctx.scale(), where ctx is the canvas context.
When need to calculate the mouse position in the old coordinate system, one has to add back the moved origin of the canvas, and multiply by the scale factor.
seat.x = (-new_org.x + pos.x) / scale;
seat.y = (-new_org.y + pos.y) / scale;
where pos is the calculated mouse pointer
pos.x = event.clientX - .getBoundingClientRect().left;
pos.y = event.clientY - .getBoundingClientRect().top;
Related
I've already asked this same question months ago, but no one was able to answer me, even after making a fully functional example on Plunker, then, I am going to ask it again, and yes, I still have the same problem.
My problem: find the centre of an element who have some rotation in it, after resizing it, to use it as the new pivot of rotation.
In my practical example, it is possible to see the problem in action; I have created two circles to show the problem better. After rotating and resizing the element, it's possible to see how the red and blue circles are apart from each other.
Blue Circle: the "correct" position of the centre, achieved by setting the cx/cy coordinates as the calculated element centre, plus, applying the transform rotate in it. The transform translates the circle to the correct position.
Red Circle: same as the blue circle, minus the transform rotate, these values are the ones used as the rotation pivot for the transform rotate().
My assumptions until here: By applying the transform rotate() in the blue circle, I'm considering the rotation angle in the calculated centre, so all I have to do is replicate the matrix calculations made by the rotate() function. I'm already doing this with the four handles that the user can click to make a rotation, what could go wrong?
My goal: Resize an element with rotation keeping the pivot of rotation in the centre.
I think this answer gave me some info, the math here helped me with the rotation handles starting position, but still, I can't find the right way to calculate the new centre after the resize.
The example was made using D3js + AngularJS v1. I work actively with both, but I am new to the geometry math world.
Again, this is the project on Plunker.
To get the centre of the transformed and rotated element, the most accurate way would probably be to get the browser to calculate it for you.
First create an SVGPoint object to hold our original centre point.
var centre = svg.createSVGPoint();
Initialize this point with the centre of the original object. You can get that by calling getBBox() on the element, and performing a smiple calculation.
var bbox = obj.getBBox();
centre.x = bbox.x + bbox.width / 2;
centre.y = bbox.y + bbox.height / 2;
Next, get the transform matrix from the transform attribute of the transformed object
var matrix = transformedObj.transform.baseVal.consolidate().matrix
Now we can transform our SVGPoint object with this matrix.
var transformedCentre = centre.matrixTransform(matrix);
After this, the x and y properties of transformedCentre should be your transformed centre point.
This should work, but I haven't tested it.
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'm currently making a game, and i need to zoom into canvas.
I've read a lot about how to zoom in canvas, with the ctx.scale() propriety, the thing is, I want to zoom with both fingers.
I already have the zoom, but it's zooming from the top/left canvas, and not on the middle of my fingers.
I have the middle point between finger 1 and finger 2, but i don't know how to zoom into that specific middle point !
This exemple pretty sums up what i need (I just need the zoom) :
Zoom Canvas to Mouse Cursor
It's working really fine, but with the wheel !
If any of you as any ideas, I'd be really glad to talk ! :)
Thanks everyone !
I added the touch events ... now you need to find the middle of all points in the evt.touches array (each one has a clientX and clientY, among other properties)
You will also need to keep track the distance beetwen those points in order to change the zoom level.
This bin might help you (check line 46) http://jsbin.com/dived/1/edit?js,output
If you have the center point from which you want to scale, you could use the translation tool provided by the CanvasRenderingContext2D:
void translate(
in float x,
in float y
);
So when you have your canvas context in a variable ctx do
ctx.translate(x, y);
And then
ctx.scale(x, y);
The ctx.translate(x, y) places the origin to x, y coordinates. By default, the origin of the CanvasRenderingContext2D is at 0, 0 (http://www.w3.org/TR/2dcontext/) so that's why the scaling originates from the top left corner.
Source https://developer.mozilla.org/en/docs/Web/API/CanvasRenderingContext2D#translate()
I am working on a page where I can view images. I want to create a rotation tool. I've done that, but, it's not working consistently. When I set up the centre point to rotate by, the image jumps slightly, and it gets worse each time. I was experimenting, and, I have code to add a wedge to the top left corner of my top level group ( so, at 0,0 ). If I rotate the image by 45 degrees and drag it so that half of it is off the left edge of my canvas, then I call getAbsolutePosition on the wedge and on the group, I get these values:
layer.getAbsolutePosition()
Object {x: 104.66479545850302, y: 279.2748571151325}
wedge.getAbsolutePosition()
Object {x: 180.2684127179338, y: -73.48773356791764}
I think this means my y position is actually the bottom of the image, which is off screen.
What I want to do, is calculate the absolute position of the middle of my image, when the mouse moves over it, regardless of it's rotation. I have some code that works out points with rotation, which seems like it works at first, almost, but it just gets more and more broken the more I use the tool. I feel like there's something about how Kinetic is tracking these things and what it's reporting, that I am missing. Any hints would be most appreciated. Tutorials I can read are even better ( yes, I've read everything linked from the KineticJS site and searched the web ).
In a nutshell, the question is, if I have an image inside a group, and it's rotated, how do I work out the centre point of the image, taking the rotation in to account, and how do I set the offset so it will rotate from that point, and stay in the same place ?
Thanks
As you've discovered about KinetiJS:
rotation is easy
dragging is easy
dragging+rotation is difficult
After you drag your image you must reset its rotation point (offsetX/offsetY).
KineticJS makes dragging+rotation more difficult than it has to be.
Resetting the offset points of your image will cause KineticJS to automatically move your image (Noooo!!).
That's what's causing your jumping.
The solution to the "jumping" problem:
When you reset the image's rotation point (offsetX/OffsetY) you must also reset the image's X/Y position.
This code resets both XY and Offsets for an image after dragging:
A Demo: http://jsfiddle.net/m1erickson/m9Nw7/
// calc new position and offset
var pos=rect.getPosition();
var size=rect.getSize();
var offset=rect.getOffset();
var newX=pos.x-offset.x+size.width/2;
var newY=pos.y-offset.y+size.height/2;
// reset both position and offset
rect.setPosition([newX,newY]);
rect.setOffset(size.width/2,size.height/2);
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)