I am facing the challenge of figuring out the visible view width and height of a ThreeJS mesh object in pixel units.
In the screenshot below you can see objects floating in 3D space, on mouse click I need to be able to figure out what view width and height they are occupying in pixels
As I am rather new to ThreeJS it is taking me rather long to find a solution, so I would welcome any kind of assistance.
The below function shows what kind of approaches I have been trying.
getObjectSizeInViewSpace(object){
const size = new THREE.Vector3()
const box = new THREE.Box3().setFromObject(object).getSize(size)
size.project(this.camera)
let halfWidth = window.innerWidth / 2;
let halfHeight = window.innerHeight / 2;
size.x = (size.x*halfWidth)
size.y = (size.y*halfHeight)
return new THREE.Vector2(size.x,size.y)
}
You're looking for Vector3.project(). This basically takes world-space (3D) coordinates, and uses the camera's viewport to convert into normalized device coordinates, which range from [-1, 1]. For example x: -1 is the left side of the screen, and x: 1 is the right side. So you'll have to take the 4 vectors (top left, top right, bottom left, bottom right)` of your plane to calculate their pixel dimensions in your browser:
// Get 3D positions of top left corner (assuming they're not rotated)
vec3 topLeft = new Vector3(
plane.position.x - planeWidth / 2,
plane.position.y - planeHeight / 2,
plane.positon.z
);
// This converts x, y, z to the [-1, 1] range
topLeft.project(camera);
// This converts from [-1, 1] to [0, windowWidth]
const topLeftX = (1 + topLeft.x) / 2 * window.innerWidth;
const topLeftY = (1 - topLeft.y) / 2 * window.innerHeight;
Notice the topLeftY value is inverted, since -y in 3D space goes +y in pixel coordinates. Do this 4 times (once for each corner), and then you can subtract (right - left) to get the width, and the same for the height.
Is there any way to scale drawing elements on a canvas? I'm creating an application where then user can place points on a canvas, but when I try to resize the browser window all the elements disappear.
My initial tentative was to calculate the screen difference before and after the resizing. After I get the percentages, I just sat the scale on the canvas and place the coordinates that was saved from the first time I drew on the canvas, but still doesn't work, if the points appear on the canvas, it is on the same place without scaling. Can someone give me little line of thought?
private calculateCanvasScreenDifference(previousSize, guestScreen) {
return ((controlScreen - currentSize) * 100) / controlScreen;
}
let difWidthPercent = Math.abs(this.calculateCanvasScreenDifference(canvasPreviousWidth, canvasWidth) * 0.01);
let difHeightPercent = Math.abs(this.calculateCanvasScreenDifference(canvasPreviousHeight, canvasHeight) * 0.01);
let scaleX = ((Math.abs(difWidthPercent) <= 1) ? 1.00 - difWidthPercent : difWidthPercent - 1.00);
let scaleY = ((Math.abs(difHeightPercent) <= 1) ? 1.00 - difHeightPercent : difHeightPercent - 1.00);
this.cx.scale(Number(scaleX), Number(scaleY));
...
...
// then start recreating the drawing that was previous saved on an array of object(x, y values)
this.cx.beginPath();
this.cx.arc(coord.x, coord.y, 7, 0, Math.PI * 2, true);
this.cx.stroke();
Keep track of your canvas width and start with a scale factor of 1.
let originalWidth = canvas.width;
let scale = 1;
On resize calculate the new scale factor. And update tracked canvas size.
let scale = newWidth / originalWidth;
originalWidth = newWidth;
Use the scale factor for all drawing at all times. e.g.
context.arc(coord.x * scale, coord.y * scale, radius, 0, Math.PI*2, false);
Note: This approach assumes the original and new canvas sizes are proportional. If not then you will need to track width and height, and calculate separate x and y scale factors.
I'm drawing a line chart with canvas. The chart is responsive, but the line has to have a fixed width.
I made it responsive with css
#myCanvas{
width: 80%;
}
,so the stroke is scaled.
The only solution I have found is to get the value of the lineWidth with the proportion between the width attribute of the canvas and its real width.
To apply it, I clear and draw the canvas on resize.
<canvas id="myCanvas" width="510" height="210"></canvas>
<script type="text/javascript">
var c = document.getElementById("myCanvas");
var ctx = c.getContext("2d");
function draw(){
var canvasattrwidth = $('#myCanvas').attr('width');
var canvasrealwidth = $('#myCanvas').width();
// n sets the line width
var n = 4;
var widthStroke = n * (canvasattrwidth / canvasrealwidth) ;
ctx.lineWidth = widthStroke;
ctx.beginPath();
ctx.moveTo( 0 , 10 );
ctx.lineTo( 200 , 100 );
ctx.stroke();
}
$(window).on('resize', function(){
ctx.clearRect(0, 0, c.width, c.height);
draw();
});
draw();
</script>
This is my first canvas and I think there is an easier way to made the lineWidth fixed (not to clear and draw everytime on resize), but I cannot find it.
There is a question with the similar problem
html5 canvas prevent linewidth scaling
but with the method scale(), so I cannot use that solution.
There is no way to get a real world dimension of details for the canvas such as millimeters or inches so you will have to do it in pixels.
As the canvas resolution decreases the pixel width of a line needs to decrease as well. The limiting property of line width is a pixel. Rendering a line narrower than a pixel will only approximate the appearance of a narrower line by reducing the opacity (this is done automatically)
You need to define the line width in terms of the lowest resolution you will expect, within reason of course and adjust that width as the canvas resolution changes in relation to this selected ideal resolution.
If you are scaling the chart by different amounts in the x and y directions you will have to use the ctx.scale or ctx.setTransform methods. As you say you do not want to do this I will assume that your scaling is always with a square aspect.
So we can pick the lowest acceptable resolution. Say 512 pixels for either width or height of the canvas and select the lineWidth in pixels for that resolution.
Thus we can create two constants
const NATIVE_RES = 512; // the minimum resolution we reasonably expect
const LINE_WIDTH = 1; // pixel width of the line at that resolution
// Note I Capitalize constants, This is non standard in Javascript
Then to calculate the actual line width is simply the actual canvas.width divided by the NATIVE_RES then multiply that result by the LINE_WIDTH.
var actualLineWidth = LINE_WIDTH * (canvas.width / NATIVE_RES);
ctx.lineWidth = actualLineWidth;
You may want to limit that size to the smallest canvas dimension. You can do that with Math.min or you can limit it in the largest dimension with Math.max
For min dimention.
var actualLineWidth = LINE_WIDTH * (Math.min(canvas.width, canvas.height) / NATIVE_RES);
ctx.lineWidth = actualLineWidth;
For max dimension
var actualLineWidth = LINE_WIDTH * (Math.max(canvas.width, canvas.height) / NATIVE_RES);
ctx.lineWidth = actualLineWidth;
You could also consider the diagonal as the adjusting factor that would incorporate the best of both x and y dimensions.
// get the diagonal resolution
var diagonalRes = Math.sqrt(canvas.width * canvas.width + canvas.height * canvas.height)
var actualLineWidth = LINE_WIDTH * (diagonalRes / NATIVE_RES);
ctx.lineWidth = actualLineWidth;
And finally you may wish to limit the lower range of the line to stop strange artifacts when the line gets smaller than 1 pixel.
Set lower limit using the diagonal
var diagonalRes = Math.sqrt(canvas.width * canvas.width + canvas.height * canvas.height)
var actualLineWidth = Math.max(1, LINE_WIDTH * (diagonalRes / NATIVE_RES));
ctx.lineWidth = actualLineWidth;
This will create a responsive line width that will not go under 1 pixel if the canvas diagonal resolution goes under 512.
The method you use is up to you. Try them out a see what you like best. The NATIVE_RES I picked "512" is also arbitrary and can be what ever you wish. You will just have to experiment with the values and method to see which you like best.
If your scaling aspect is changing then there is a completely different technique to solve that problem which I will leave for another question.
Hope this has helped.
I'm building an app in which I present some planes with textures. However, I would like to calculate the radius of the helix (which I use in my calculations to create a helix), dynamically based on the frustum width and the camera position.
The helix is positioned at the center of the screen x=0, y=0, z=0.
I would like this to take under consideration the screen orientation (landscape/ portrait).So far this is the code I have but it seems that I'm missing something because the planes at the left and the right are not inside the viewport.
App.prototype.calculateHelixRadius = function(){
// plane width = height = 512;
var friend = this.getFriend();
var vFOV = friend.camera.fov * Math.PI / 180;
var dist = utils.getAbsPointsDistance3D(friend.camera.position, friend.scene.position);
var aspect = friend.settings.container.clientWidth / friend.settings.container.clientHeight;
var frustumHeight = 2.0 * dist * Math.tan(0.5 * vFOV);
var frustumWidth = frustumHeight * aspect;
return utils.isLandscape() ? frustumHeight / 2 : frustumWidth / 2 ;
};
What am I doing wrong and why are the planes at the edges of the screen not inside?
Also for reference here is the code of getAbsPointsDistance3D
var utils = {
// other helpers...
getAbsPointsDistance3D: function(p1, p2) {
var xd = p2.x - p1.x;
var yd = p2.y - p1.y;
var zd = p2.z - p1.z;
return Math.sqrt(xd * xd + yd * yd + zd * zd);
}
};
update
I tried decreasing the dist parameter but the results are not consistent...
I wonder if the following explains your clipping.
You calculate your frustum characteristics, then calculate the helix radius using, say, the frustum width (width or height depending on the screen aspect...I may be getting some of the particulars wrong here because your question does not completely explain the details, but the general concepts still hold). The image below is a top view of the scenario which shows a circle representing the cylinder that encloses the helix. I believe you have calculated radius1. If so, note that there will be clipping of the cylinder (the shaded area), and thus the helix, in "front" of the cylinder centre.
Instead you need to calculate the cylinder/helix radius as shown in the second image, i.e. you need radius2. If the large angle at the image left is fov (again, vFOV? or hFOV?, etc., depending on whether your helix is going up-down or side-to-side, etc.), then its half angle is fov/2. This is the same angle shown in the centre of the cylinder. Thus, you need to decrease your helix radius as follows: radius2 = radius1 * cos(fov/2).
Basically I'm asking this question for JavaScript: Calculate Bounding box coordinates from a rotated rectangle
In this case:
iX = Width of rotated (blue) HTML element
iY = Height of rotated (blue) HTML element
bx = Width of Bounding Box (red)
by = Height of Bounding Box (red)
x = X coord of Bounding Box (red)
y = Y coord of Bounding Box (red)
iAngle/t = Angle of rotation of HTML element (blue; not shown but
used in code below), FYI: It's 37 degrees in this example (not that it matters for the example)
How does one calculate the X, Y, Height and Width of a bounding box (all the red numbers) surrounding a rotated HTML element (given its width, height, and Angle of rotation) via JavaScript? A sticky bit to this will be getting the rotated HTML element (blue box)'s original X/Y coords to use as an offset somehow (this is not represented in the code below). This may well have to look at CSS3's transform-origin to determine the center point.
I've got a partial solution, but the calculation of the X/Y coords is not functioning properly...
var boundingBox = function (iX, iY, iAngle) {
var x, y, bx, by, t;
//# Allow for negetive iAngle's that rotate counter clockwise while always ensuring iAngle's < 360
t = ((iAngle < 0 ? 360 - iAngle : iAngle) % 360);
//# Calculate the width (bx) and height (by) of the .boundingBox
//# NOTE: See https://stackoverflow.com/questions/3231176/how-to-get-size-of-a-rotated-rectangle
bx = (iX * Math.sin(iAngle) + iY * Math.cos(iAngle));
by = (iX * Math.cos(iAngle) + iY * Math.sin(iAngle));
//# This part is wrong, as it's re-calculating the iX/iY of the rotated element (blue)
//# we want the x/y of the bounding box (red)
//# NOTE: See https://stackoverflow.com/questions/9971230/calculate-rotated-rectangle-size-from-known-bounding-box-coordinates
x = (1 / (Math.pow(Math.cos(t), 2) - Math.pow(Math.sin(t), 2))) * (bx * Math.cos(t) - by * Math.sin(t));
y = (1 / (Math.pow(Math.cos(t), 2) - Math.pow(Math.sin(t), 2))) * (-bx * Math.sin(t) + by * Math.cos(t));
//# Return an object to the caller representing the x/y and width/height of the calculated .boundingBox
return {
x: parseInt(x), width: parseInt(bx),
y: parseInt(y), height: parseInt(by)
}
};
I feel like I am so close, and yet so far...
Many thanks for any help you can provide!
TO HELP THE NON-JAVASCRIPTERS...
Once the HTML element is rotated, the browser returns a "matrix transform" or "rotation matrix" which seems to be this: rotate(Xdeg) = matrix(cos(X), sin(X), -sin(X), cos(X), 0, 0); See this page for more info.
I have a feeling this will enlighten us on how to get the X,Y of the bounding box (red) based solely on the Width, Height and Angle of the rotated element (blue).
New Info
Humm... interesting...
Each browser seems to handle the rotation differently from an X/Y perspective! FF ignores it completely, IE & Opera draw the bounding box (but its properties are not exposed, ie: bx & by) and Chrome & Safari rotate the rectangle! All are properly reporting the X/Y except FF. So... the X/Y issue seems to exist for FF only! How very odd!
Also of note, it seems that $(document).ready(function () {...}); fires too early for the rotated X/Y to be recognized (which was part of my original problem!). I am rotating the elements directly before the X/Y interrogation calls in $(document).ready(function () {...}); but they don't seem to update until some time after(!?).
When I get a little more time, I will toss up a jFiddle with the example, but I'm using a modified form of "jquery-css-transform.js" so I have a tiny bit of tinkering before the jFiddle...
So... what's up, FireFox? That ain't cool, man!
The Plot Thickens...
Well, FF12 seems to fix the issue with FF11, and now acts like IE and Opera. But now I am back to square one with the X/Y, but at least I think I know why now...
It seems that even though the X/Y is being reported correctly by the browsers for the rotated object, a "ghost" X/Y still exists on the un-rotated version. It seems as though this is the order of operations:
Starting with an un-rotated element at an X,Y of 20,20
Rotate said element, resulting in the reporting of X,Y as 15,35
Move said element via JavaScript/CSS to X,Y 10,10
Browser logically un-rotates element back to 20,20, moves to 10,10 then re-rotates, resulting in an X,Y of 5,25
So... I want the element to end up at 10,10 post rotation, but thanks to the fact that the element is (seemingly) re-rotated post move, the resulting X,Y differs from the set X,Y.
This is my problem! So what I really need is a function to take the desired destination coords (10,10), and work backwards from there to get the starting X,Y coords that will result in the element being rotated into 10,10. At least I know what my problem is now, as thanks to the inner workings of the browsers, it seems with a rotated element 10=5!
I know this is a bit late, but I've written a fiddle for exactly this problem, on an HTML5 canvas:
http://jsfiddle.net/oscarpalacious/ZdQKg/
I hope somebody finds it useful!
I'm actually not calculating your x,y for the upper left corner of the container. It's calculated as a result of the offset (code from the fiddle example):
this.w = Math.sin(this.angulo) * rotador.h + Math.cos(this.angulo) * rotador.w;
this.h = Math.sin(this.angulo) * rotador.w + Math.cos(this.angulo) * rotador.h;
// The offset on a canvas for the upper left corner (x, y) is
// given by the first two parameters for the rect() method:
contexto.rect(-(this.w/2), -(this.h/2), this.w, this.h);
Cheers
Have you tried using getBoundingClientRect() ?
This method returns an object with current values of "bottom, height, left, right, top, width" considering rotations
Turn the four corners into vectors from the center, rotate them, and get the new min/max width/height from them.
EDIT:
I see where you're having problems now. You're doing the calculations using the entire side when you need to be doing them with the offsets from the center of rotation. Yes, this results in four rotated points (which, strangely enough, is exactly as many points as you started with). Between them there will be one minimum X, one maximum X, one minimum Y, and one maximum Y. Those are your bounds.
My gist can help you
Bounding box of a polygon (rectangle, triangle, etc.):
Live demo https://jsfiddle.net/Kolosovsky/tdqv6pk2/
let points = [
{ x: 125, y: 50 },
{ x: 250, y: 65 },
{ x: 300, y: 125 },
{ x: 175, y: 175 },
{ x: 100, y: 125 },
];
let minX = Math.min(...points.map(point => point.x));
let minY = Math.min(...points.map(point => point.y));
let maxX = Math.max(...points.map(point => point.x));
let maxY = Math.max(...points.map(point => point.y));
let pivot = {
x: maxX - ((maxX - minX) / 2),
y: maxY - ((maxY - minY) / 2)
};
let degrees = 90;
let radians = degrees * (Math.PI / 180);
let cos = Math.cos(radians);
let sin = Math.sin(radians);
function rotatePoint(pivot, point, cos, sin) {
return {
x: (cos * (point.x - pivot.x)) - (sin * (point.y - pivot.y)) + pivot.x,
y: (sin * (point.x - pivot.x)) + (cos * (point.y - pivot.y)) + pivot.y
};
}
let boundingBox = {
x1: Number.POSITIVE_INFINITY,
y1: Number.POSITIVE_INFINITY,
x2: Number.NEGATIVE_INFINITY,
y2: Number.NEGATIVE_INFINITY,
};
points.forEach((point) => {
let rotatedPoint = rotatePoint(pivot, point, cos, sin);
boundingBox.x1 = Math.min(boundingBox.x1, rotatedPoint.x);
boundingBox.y1 = Math.min(boundingBox.y1, rotatedPoint.y);
boundingBox.x2 = Math.max(boundingBox.x2, rotatedPoint.x);
boundingBox.y2 = Math.max(boundingBox.y2, rotatedPoint.y);
});
Bounding box of an ellipse:
Live demo https://jsfiddle.net/Kolosovsky/sLc7ynd1/
let centerX = 350;
let centerY = 100;
let radiusX = 100;
let radiusY = 50;
let degrees = 200;
let radians = degrees * (Math.PI / 180);
let radians90 = radians + Math.PI / 2;
let ux = radiusX * Math.cos(radians);
let uy = radiusX * Math.sin(radians);
let vx = radiusY * Math.cos(radians90);
let vy = radiusY * Math.sin(radians90);
let width = Math.sqrt(ux * ux + vx * vx) * 2;
let height = Math.sqrt(uy * uy + vy * vy) * 2;
let x = centerX - (width / 2);
let y = centerY - (height / 2);