I'm stuck on a problem :
Let's consider a square on a surface that is moving (in a video). So remember it's not always a plane surface, it can be skewed, rotated, etc.
Right now, I'm detecting it with Aruco JS and getting the coordinates (x,y) of its 4 corners. I'm pretty sure that, starting with this coordinates, I can render the transformation using transform: matrix3d();.
The thing is : I have about 0 knowledge in maths stuff and especially trigonometry. And I really would learn / understand that.
To sum up : With the coordinates of the corners of a square in real time, how can I apply the transformation to another element using CSS transform property ?
Here's some things I've done so far (this code is executed inside a window.requestAnimationFrame) :
console.log('rotation : ', rotation[0], rotation[1], rotation[2]);
console.log('translation : ', translation);
var dimensions = {
width: lineDistance(corners[0], corners[1]),
height: lineDistance(corners[0], corners[3])
}, center = {
x: corners[0].x + (corners[1].x - corners[0].x)/2,
y: corners[0].y + (corners[3].y - corners[0].y)/2
}, rotateAngle = angle(corners[0].x, corners[0].y, corners[1].x, corners[1].y),
rotateXangle = parseInt(rotation[1][2]);
img.style.top = corners[0].y;
img.style.left = corners[0].x;
img.style.width = dimensions.width;
img.style.height = dimensions.height;
img.style.transform = 'rotate('+rotateAngle+'deg) rotateX('+Math.asin(-rotateXangle)+'deg) rotateY('+-Math.atan2(rotation[0][2], rotation[2][2])+'deg) rotateZ('+Math.atan2(rotation[1][0], rotation[1][1])+'deg)';
The core of this question has been asked on Math Stack Exchange, titled Finding the Transform matrix from 4 projected points (with Javascript). My answer there should serve your needs as well. Plus the better math type setting there will make things easier to read.
Related
TLDR; Given a set of triangle vertices and a normal vector (all in unit space), how do I calculate X, Y, Z Euler rotation angles of the triangle in world space?
I am attemping to display a 3D model in HTML - with actual HTML tags and CSS transforms. I've already loaded an OBJ file into a Javascript class instance.
The model is triangulated. My first aim is just to display the triangles as planes (HTML elements are rectangular) - I'll be 'cutting out' the triangle shapes with CSS clip-path later on.
I am really struggling to understand and get the triangles of the model rotated correctly.
I thought a rotation matrix could help me out, but my only experience with those is where I already have the rotation vector and I need to convert and send that to WebGL. This time there is no WebGL (or tutorials) to make things easier.
The following excerpt shows the face creation/'rendering' of faces. I'm using the face normal as the rotation but I know this is wrong.
for (const face of _obj.faces) {
const vertices = face.vertices.map(_index => _obj.vertices[_index]);
const center = [
(vertices[0][0] + vertices[1][0] + vertices[2][0]) / 3,
(vertices[0][1] + vertices[1][1] + vertices[2][1]) / 3,
(vertices[0][2] + vertices[1][2] + vertices[2][2]) / 3
];
// Each vertex has a normal but I am just picking the first vertex' normal
// to use as the 'face normal'.
const normals = face.normals.map(_index => _obj.normals[_index]);
const normal = normals[0];
// HTML element creation code goes here; reference is 'element'.
// Set face position (unit space)
element.style.setProperty('--posX', center[0]);
element.style.setProperty('--posY', center[1]);
element.style.setProperty('--posZ', center[2]);
// Set face rotation, converting to degrees also.
const rotation = [
normal[0] * toDeg,
normal[1] * toDeg,
normal[2] * toDeg,
];
element.style.setProperty('--rotX', rotation[0]);
element.style.setProperty('--rotY', rotation[1]);
element.style.setProperty('--rotZ', rotation[2]);
}
The CSS first translates the face on X,Y,Z, then rotates it on X,Y,Z in that order.
I think I need to 'decompose' my triangles' rotation into separate axis rotations - i.e rotate on X, then on Y, then on Z to get the correct rotation as per the model face.
I realise that the normal vector gives me an orientation but not a rotation around itself - I need to calculate that. I think I have to determine a vector along one triangle side and cross it with the normal, but this is something I am not clear on.
I have spent hours looking at similar questions on SO but I'm not smart enough to understand or make them work for me.
Is it possible to describe what steps to take without Latex equations? I'm good with pseudo code but my Math skills are severely lacking.
The full code is here: https://whoshotdk.co.uk/cssfps/ (view HTML source)
The mesh building function is at line 422.
The OBJ file is here: https://whoshotdk.co.uk/cssfps/data/model/test.obj
The Blender file is here: https://whoshotdk.co.uk/cssfps/data/model/test.blend
The mesh is just a single plane at an angle, displayed in my example (wrongly) in pink.
The world is setup so that -X is left, -Y is up, -Z is into the screen.
Thank You!
If you have a plane and want to rotate it to be in the same direction as some normal, you need to figure out the angles between that plane's normal vector and the normal vector you want. The Euler angles between two 3D vectors can be complicated, but in this case the initial plane normal should always be the same, so I'll assume the plane normal starts pointing towards positive X to make the maths simpler.
You also probably want to rotate before you translate, so that everything is easier since you'll be rotating around the origin of the coordinate system.
By taking the general 3D rotation matrix (all three 3D rotation matrices multiplied together, you can find it on the Wikipedia page) and applying it to the vector (1,0,0) you can then get the equations for the three angles a, b, and c needed to rotate that initial vector to the vector (x,y,z). This results in:
x = cos(a)*cos(b)
y = sin(a)*cos(b)
z = -sin(b)
Then rearranging these equations to find a, b and c, which will be the three angles you need (the three values of the rotation array, respectively):
a = atan(y/x)
b = asin(-z)
c = 0
So in your code this would look like:
const rotation = [
Math.atan2(normal[1], normal[0]) * toDeg,
Math.asin(-normal[2]) * toDeg,
0
];
It may be that you need to use a different rotation matrix (if the order of the rotations is not what you expected) or a different starting vector (although you can just use this method and then do an extra 90 degree rotation if each plane actually starts in the positive Y direction, for example).
I'm new to javascript and three.js, i'm trying to figure out how to get the 3d positions on a webpage.
for ex. i want to set a point light at (50,20,10) x,y,z values. how can I know that where this x,y,z values will come on a webpage?.I have seen code like below.
var light2 = new THREE.PointLight(0xffffff, 10)
light2.position.set(500, 100, 0)
scene.add(light2)
I have googled but I didn't get enough information to sort out the things properly, can somebody help me with a good explanation or some article/tutorial link?
Just a little background first...
3D space is infinite.
The canvas is a viewport into that space.
The size of the canvas on the page has no direct relation to anything in 3D space. You can double the size of the canvas, and all it does is make the rendered image bigger.
To your question...
That's not to say you can't figure out where on the canvas a 3D thing might appear. You can project any point in 3D space into Normalized Device Coordinates with Vector3.project.
var light2 = new THREE.PointLight(0xffffff, 10)
light2.position.set(500, 100, 0)
scene.add(light2)
// Where is it in NDC?
var ndc = new THREE.Vector3().copy( light2.position ).project( camera );
Normalized Device Coordinates range from -1 to 1, and represent the normalized width and height of the viewport with (0,0) at the center of the rendered image. So, you will need to convert them from NDC into pixel values on your canvas. (Also, you can ignore the z component, because the screen has no depth.)
var canvasX = ( ndc.x + 1 ) * canvasWidth / 2;
var canvasY = ( ndc.y + 1 ) * canvasHeight / -2;
Working on a little "zombies" or "tag you're it" or "ew! you got cooties"-styled game where each AI object (a person, basically) runs around randomly. There is an initial object that is "it" or "infected" and as it moves about the screen and touches/overlaps/collides with another object it should change the touched object to the same color as the object that touched it. Newly infected objects can continue to infect other objects they randomly collide with, until - in principle - the whole population is the same color as the first infected object. (I'll worry about fancier AI where infected actively hunt nearby objects or healthy objects can avoid infected objects, later).
But after looking at various similar questions in StackOverflow that generally deal with 2 DIVs colliding, or use some sort of jQuery draggable detection trick, I'm still at a bit of a loss as to how to build upon those ideas to scale up a simple "if I am touching/overlapping/colliding with another object it should get infected too" that can be applied to a large number of elements on the page, say... less than 100 so as not to drag the browser down.
I basically get as far as determining position and widths/heights of the objects so that I know how much space they take, but then the brain goes 'bzzzzt' when trying to develop a function that checks over all the population for collisions.
Got the population moving around randomly without trouble - see JSFiddle https://jsfiddle.net/digitalmouse/5tvyjhjL/1/ for the related code. Affected function should be in the 'animateDiv()', seen below to make the stackoverflow question asking editor happy that I included some code in my question. :)
function animateDiv($target) {
var newq = makeNewPosition($target.parent());
var oldq = $target.offset();
var speed = calcSpeed([oldq.top, oldq.left], newq);
// I believe collision should be dealt with here,
// just before moving an object
$target.animate({
top: newq[0],
left: newq[1]
}, speed, function () {
animateDiv($target);
});
}
Any hints, tricks, adaptations, or code snippets that push me in the right direction are appreciated.
a quick, down and dirty solution (there are more complex algorithms) would be to use:
document.elementFromPoint(x, y);
It gets the element at the position specified. The full spec can be found here.
Assuming your 'zombies' are rectangular, you could call this for each corner, and if you get a hit, that isn't the background or the element you're checking, you've got a collision...
EDIT:
An alternate method, even 'downer and dirtier' than above, but stupidly quick, would be to get the centre points of the two objects to check, then find their absolute displacements in X and Y. If the differences are less than the sum of half their widths and heights then they are overlapping. It's by no means pix perfect, but it should be able to handle a large number objects really quickly.
EDIT 2:
First off, we need to get the centres of each object (to check)
// Values for main object
// pop these in vars as we'll need them again in a sec...
hw = object.style.width >> 1; // half width of object
hh = object.style.height >> 1; // (bit shift is faster than / 2)
cx = object.style.left + hw; // centre point in x
cy = object.style.top + hh; // and in y
// repeat for secondary object
If you don't know / store the width and height you can use:
object.getBoundingClientRect();
which returns a 'rect' object with the fields left, top, right and bottom.
Now we check proximity...
xDif = Math.abs(cx - cx1); // where cx1 is centre of object to check against
if(xDif > hw + hw1) return false; // there is no possibility of a collision!
// if we get here, there's a possible collision, so...
yDif = Math.abs(cy - cy1);
if(yDif > hh + hh1) return false; // no collision - bug out.
else {
// handle collision here...
}
Danny
I have an application with many draggable objects that can also be rotated in 90 degree increments. I'm trying to figure out how to stop the user from dragging the objects outside the Raphael paper (canvas).
This is fairly simple for unrotated objects. I can simply see if the current x and y coordinates are less than 0 and set them to 0 instead. I can adjust similarly by checking if they are outside the canvas width and height.
However, a problem arises when the object is rotated because for some odd reason the coordinate plane rotates as well. Is there an easy way to keep objects inside the canvas? Or is there an example of some this somewhere?
I have spent many hours fiddling with this and I can't seem to make sense of the rotated coordinate plane in order to adjust my calculations. Even when debugging the current coordinates, they seem to shift oddly if I drag an object, release it, and then drag the object again.
Any help is greatly appreciated.
Thanks,
Ryan
I had a similar problem, I needed to move a shape within the boundaries of another shape, so what I did was:
element.drag(onstart, onmove, onend);
...
onStart: function(x,y,e){
// Initialize values so it doesn't recalculate per iteration
// this allows to resume dragging from the point it were left
App.oldX = 0;
App.oldY = 0;
App.currentCircleX = App.fingerPath.attr('cx');
App.currentCircleY = App.fingerPath.attr('cy');
},
onMove: function(dx,dy,x,y,e){
App.setDirection(dx,dy);
},
onEnd: function(e){
// nothing to do here for now
},
// this function tells the element to move only if it's within the bound area
setDirection: function(dx, dy){
var isXYinside;
this.newX = this.currentCircleX - (this.oldX - dx);
this.newY = this.currentCircleY - (this.oldY - dy);
// HERE is the key, this method receives your bounding path and evaluates the positions given and then returns true or false
isXYinside = Raphael.isPointInsidePath(this.viewportPath, this.newX, this.newY);
this.oldX = dx;
this.oldY = dy;
// so if it is within the bound area, will move, otherwise will just stay there
if (isXYinside) {
this.fingerPath.attr({
"cx": this.newX,
"cy": this.newY
});
this.currentCircleX = this.newX;
this.currentCircleY = this.newY;
}
}
I know this is an old one, but I stumbled upon this question when trying to figure out a way to do it. So here's my 2 cents in case someone has this problem.
Reference:
Raphael.isPointInsidePath
Have you tried Element.getBBox()
There Are 2 flavones which give the result before rotation and after rotation
You should toggle the Boolean argument and test it
I am wanting to create a simple abstract pattern using the html5 canvas tag and javascript. I have worked out the guts of what I want it to do using some variables, functions and objects, but with the boundary detection that I have employed I am wanting each particular shape to go back to its starting position when it goes out of the screen (and thus loop the animation).
So with that being my question, here is my code. Also any other structure tips are appreciated as I am new to OO in Javascript.
See my progress here: http://helloauan.com/apps/test/
Cheers!
I'm not really sure if what you mean exactly is, once the big white diagnal lines are all the way off the top right corner of page, that's when you want them to start back at the bottom left ? right?
What you need to do is check if the line is beyond the width and height of the canvas, and in your case, the window itself since the canvas fills the browser window. So you need to do a series of conditionals. You check if the line x + line width is > canvas width and line.y + line height is > canvas height. If both are true then set the x and y of the line to - what it is at that time. So something like:
if( line.x + line.width > canvas.width && line.y + line.height < 0) {
line.x = -0;
line.y = canvasHeight + line.height;
}
This is how I recycle circles that come in from the right side of the screen and once they exit the left side they start over on the right.
if( d.x + d.radius < 0 ) {
d.radius = 5+(Math.random()*10);
d.x = cwidth + d.radius;
d.y = Math.floor(Math.random()*cheight);
d.vX = -5-(Math.random()*5);
}
The first thing is just psuedo, you should take a look at a thing I made to use as a starting point for things like this. The structure of your code could use some more organization, canvas gets real complex real quick.
Using the arrow keys, move the square off any one of the 4 sides and see it come in on opposite side.
http://anti-code.com/games/envy/envy.html
Fork if you want: https://github.com/jaredwilli/envy