We Keep Coding

How can I do a 2D morphing circle in three.js that moves in a liquid way?

I have a mesh which is a circle geometry. I would like to animate it like in this example from two.js, a 2D library:
https://two.js.org/examples/physics.html
For now I look at this example and put the camera on the top of the shape but I'm sure there's a more simple way for my needs: https://threejs.org/examples/#webgl_gpgpu_water
Does anyone know how I can do that?

You simply need to shift the vertices positions according some sin() or cos() value according X and Y coordinates and a incremental phase (time) to animate.
your vertex shader could include something like this, where phase is incrementing with time (typically, a clock).
glPosition.x = vertex.x + sin((phase*frequency) + vertex.y) * amplitude;
glPosition.y = vertex.y + sin((phase*frequency) + vertex.x) * amplitude;
The basic concept is here, but you have to adapt the components yourself by testing the result. You probably should adjust frequency, amplitude, adding some more factors to add asymetry and randomness.

Cubic Bezier Curve between two points on a sphere in three.js

I'm letting the user click on two points on a sphere and I would then like to draw a line between the two points along the surface of the sphere (basically on the great circle). I've been able to get the coordinates of the two selected points and draw a QuadraticBezierCurve3 between the points, but I need to be using CubicBezierCurve3. The problem is is that I have no clue how to find the two control points.
Part of the issue is everything I find is for circular arcs and only deals with [x,y] coordinates (whereas I'm working with [x,y,z]). I found this other question which I used to get a somewhat-working solution using QuadraticBezierCurve3. I've found numerous other pages with math/code like this, this, and this, but I really just don't know what to apply. Something else I came across mentioned the tangents (to the selected points), their intersection, and their midpoints. But again, I'm unsure of how to do that in 3D space (since the tangent can go in more than one direction, i.e. a plane).
An example of my code: http://jsfiddle.net/GhB82/
To draw the line, I'm using:
function drawLine(point) {
var middle = [(pointA['x'] + pointB['x']) / 2, (pointA['y'] + pointB['y']) / 2, (pointA['z'] + pointB['z']) / 2];
var curve = new THREE.QuadraticBezierCurve3(new THREE.Vector3(pointA['x'], pointA['y'], pointA['z']), new THREE.Vector3(middle[0], middle[1], middle[2]), new THREE.Vector3(pointB['x'], pointB['y'], pointB['z']));
var path = new THREE.CurvePath();
path.add(curve);
var curveMaterial = new THREE.LineBasicMaterial({
color: 0xFF0000
});
curvedLine = new THREE.Line(path.createPointsGeometry(20), curveMaterial);
scene.add(curvedLine);
}
Where pointA and pointB are arrays containing the [x,y,z] coordinates of the selected points on the sphere. I need to change the QuadraticBezierCurve3 to CubicBezierCurve3, but again, I'm really at a loss on finding those control points.

I have a description on how to fit cubic curves to circular arcs over at http://pomax.github.io/bezierinfo/#circles_cubic, the 3D case is essentially the same in that you need to find out the (great) circular cross-section your two points form on the sphere, and then build the cubic Bezier section along that circle.
Downside: Unless your arc is less than or equal to roughly a quarter circle, one curve is not going to be enough, you'll need two or more. You can't actually model true circular curves with Bezier curves, so using cubic instead of quadratic just means you can approximate a longer arc segment before it starts to look horribly off.
So on a completely different solution note: if you have an arc command available, much better to use that than to roll your own (and if three.js doesn't support them, definitely worth filing a feature request for, I'd think)

How do I 'wrap' a plane over a sphere with three.js?

I am relatively new to three.js and am trying to position and manipulate a plane object to have the effect of laying over the surface of a sphere object (or any for that matter), so that the plane takes the form of the object surface. The intention is to be able to move the plane on the surface later on.
I position the plane in front of the sphere and index through the plane's vertices casting a ray towards the sphere to detect the intersection with the sphere. I then try to change the z position of said vertices, but it does not achieve the desired result. Can anyone give me some guidance on how to get this working, or indeed suggest another method?
This is how I attempt to change the vertices (with an offset of 1 to be visible 'on' the sphere surface);
planeMesh.geometry.vertices[vertexIndex].z = collisionResults[0].distance - 1;
Making sure to set the following before rendering;
planeMesh.geometry.verticesNeedUpdate = true;
planeMesh.geometry.normalsNeedUpdate = true;
I have a fiddle that shows where I am, here I cast my rays in z and I do not get intersections (collisions) with the sphere, and cannot change the plane in the manner I wish.
http://jsfiddle.net/stokewoggle/vuezL/
You can rotate the camera around the scene with the left and right arrows (in chrome anyway) to see the shape of the plane. I have made the sphere see through as I find it useful to see the plane better.
EDIT: Updated fiddle and corrected description mistake.

Sorry for the delay, but it took me a couple of days to figure this one out. The reason why the collisions were not working was because (like we had suspected) the planeMesh vertices are in local space, which is essentially the same as starting in the center of the sphere and not what you're expecting. At first, I thought a quick-fix would be to apply the worldMatrix like stemkoski did on his github three.js collision example I linked to, but that didn't end up working either because the plane itself is defined in x and y coordinates, up and down, left and right - but no z information (depth) is made locally when you create a flat 2D planeMesh.
What ended up working is manually setting the z component of each vertex of the plane. You had originaly wanted the plane to be at z = 201, so I just moved that code inside the loop that goes through each vertex and I manually set each vertex to z = 201; Now, all the ray start-positions were correct (globally) and having a ray direction of (0,0,-1) resulted in correct collisions.
var localVertex = planeMesh.geometry.vertices[vertexIndex].clone();
localVertex.z = 201;
One more thing was in order to make the plane-wrap absolutely perfect in shape, instead of using (0,0,-1) as each ray direction, I manually calculated each ray direction by subtracting each vertex from the sphere's center position location and normalizing the resulting vector. Now, the collisionResult intersection point will be even better.
var directionVector = new THREE.Vector3();
directionVector.subVectors(sphereMesh.position, localVertex);
directionVector.normalize();
var ray = new THREE.Raycaster(localVertex, directionVector);
Here is a working example:
http://jsfiddle.net/FLyaY/1/
As you can see, the planeMesh fits snugly on the sphere, kind of like a patch or a band-aid. :)
Hope this helps. Thanks for posting the question on three.js's github page - I wouldn't have seen it here. At first I thought it was a bug in THREE.Raycaster but in the end it was just user (mine) error. I learned a lot about collision code from working on this problem and I will be using it later down the line in my own 3D game projects. You can check out one of my games at: https://github.com/erichlof/SpacePong3D
Best of luck to you!
-Erich

Your ray start position is not good. Probably due to vertex coordinates being local to the plane. You start the raycast from inside the sphere so it never hits anything.
I changed the ray start position like this as a test and get 726 collisions:
var rayStart = new THREE.Vector3(0, 0, 500);
var ray = new THREE.Raycaster(rayStart, new THREE.Vector3(0, 0, -1));
Forked jsfiddle: http://jsfiddle.net/H5YSL/
I think you need to transform the vertex coordinates to world coordinates to get the position correctly. That should be easy to figure out from docs and examples.

Three.js - Rotating a sphere around a certain axis

I have a problem. In Three.js, I want to rotate a sphere (Earth) around axis tilted by 23.5 degs. I found sphere.rotation.x, sphere.rotation.y and sphere.rotation.z, but when I combine them in the correct ratio, the sphere's rotation is quite weird - it has no permanent rotation axis. I think I need a function like sphere.rotation.vector(1,0,-1). Does anyone know how this function is called and how the correct syntax is?
Many thanks for answers!

You do not have to understand how Euler angles or quaternions work to do what you want. You can use
Object3D.rotateOnAxis( axis, angle );
Object3D.rotateOnWorldAxis( axis, angle );
Make sure axis is a unit vector (has length 1), and angle is in radians.
Object3D.rotateOnAxis( axis, angle ) rotates on an axis in object space.
Object3D.rotateOnWorldAxis( axis, angle ) rotates on an axis in world space.
three.js r.104

You need to use quaternions for this. This video explains what quaternions are and how they are used in 3D graphics.
You can construct a quaternion like this:
quaternion = new THREE.Quaternion().setFromAxisAngle( axisOfRotation, angleOfRotation );
Then you apply it to your object by:
object.rotation.set( new THREE.Euler().setFromQuaternion( quaternion ) );
You can also achieve this by using object hierarchies. For example, you can make an Object3D() instance and tilt it by 23.5 degs, then create a sphere (Earth) and add it to the tilted object. The sphere will then rotate around the tilted Y axis. Quaternions however, are the best tool for solving this.

var quaternion = new THREE.Quaternion();
var object = scene.getObjectByName('xxx');
function render(){
quaternion.setFromAxisAngle(new THREE.Vector3(0, 1, 0).normalize(), 0.005);
object.position.applyQuaternion(quaternion);
}
three.js version is 86, see full example on codepen.

You can rotate your sphere using th 'ObjectControls' module for ThreeJS that allows you to rotate a single OBJECT (or a Group), and not the SCENE.
Include the libary:
then
var controls = new THREE.ObjectControls(camera, renderer.domElement, yourMesh);
You can find here a live demo here: https://albertopiras.github.io/threeJS-object-controls/
Here is the repo: https://github.com/albertopiras/threeJS-object-controls.
Hope this helps

How to Draw line in 3D rectangle based on x,y and z?

I would like draw 3D points represented in image to 3D rectangle. Any idea how could I represent these in x,y and z axis
Here projection type is orthographic.
Thanks

Okay. Let's look at a simple example of what you are trying to accomplish it, and why this is such a complicated problem.
First, lets look a some projection functions. You need a way to mathematically describe how to transform a 3D (or higher dimensional) point into a 2D space (your monitor), or a projection.
The simpiest to understand is a very simple dimetric projection. Something like:
x' = x + z/2;
y' = y + z/4;
What does this mean? Well, x' is you x coordinate 2D projection: for every unit you move backwards in space, the projection will move that point half that many units to the right. And y' represents that same projection for your y coordinate: for every unit you move backwards in space, the projection will move that point a quarter unit up.
So a point at [0,0,0] will get projected to a 2d point of [0,0]. A point at [0,0,4] will get projected to a 2d point of [2,1].
Implemented in JavaScript, it would look something like this:
// Dimetric projection functions
var dimetricTx = function(x,y,z) { return x + z/2; };
var dimetricTy = function(x,y,z) { return y + z/4; };
Once you have these projection functions -- or ways to translate from 3D space into 2D space -- you can use them to start draw your image. A simple example of that using js canvas. First, some context stuff:
var c = document.getElementById("cnvs");
var ctx = c.getContext("2d");
Now, lets make a little helper to draw a 3D point:
var drawPoint = (function(ctx,tx,ty, size) {
return function(p) {
size = size || 3;
// Draw "point"
ctx.save();
ctx.fillStyle="#f00";
ctx.translate(tx.apply(undefined, p), ty.apply(undefined,p));
ctx.beginPath();
ctx.arc(0,0,size,0,Math.PI*2);
ctx.fill();
ctx.restore();
};
})(ctx,dimetricTx,dimetricTy);
This is pretty simple function, we are injecting the canvas context as ctx, as well as our tx and ty functions, which in this case our the dimetric functions we saw earlier.
And now a polygon drawer:
var drawPoly = (function(ctx,tx,ty) {
return function() {
var args = Array.prototype.slice.call(arguments, 0);
// Begin the path
ctx.beginPath();
// Move to the first point
var p = args.pop();
if(p) {
ctx.moveTo(tx.apply(undefined, p), ty.apply(undefined, p));
}
// Draw to the next point
while((p = args.pop()) !== undefined) {
ctx.lineTo(tx.apply(undefined, p), ty.apply(undefined, p));
}
ctx.closePath();
ctx.stroke();
};
})(ctx, dimetricTx, dimetricTy);
With those two functions, you could effectively draw the kind of graph you are looking for. For example:
// The array of points
var points = [
// [x,y,z]
[20,30,40],
[100,70,110],
[30,30,75]
];
(function(width, height, depth, points) {
var c = document.getElementById("cnvs");
var ctx = c.getContext("2d");
// Set some context
ctx.save();
ctx.scale(1,-1);
ctx.translate(0,-c.height);
ctx.save();
// Move our graph
ctx.translate(100,20);
// Draw the "container"
ctx.strokeStyle="#999";
drawPoly([0,0,depth],[0,height,depth],[width,height,depth],[width,0,depth]);
drawPoly([0,0,0],[0,0,depth],[0,height,depth],[0,height,0]);
drawPoly([width,0,0],[width,0,depth],[width,height,depth],[width,height,0]);
drawPoly([0,0,0],[0,height,0],[width,height,0],[width,0,0]);
ctx.stroke();
// Draw the points
for(var i=0;i<points.length;i++) {
drawPoint(points[i]);
}
})(150,100,150,points);
However, you should now be able to start to see some of the complexity of your actual question emerge. Namely, you asked about rotation, in this example we are using an extremely simple projection (our dimetric projection) which doesn't take much other than an oversimplified relationship between depth and its influences on x,y position. As the projections become more complex, you need to know more about your relationship/orientation in 3D space in order to create a reasonable 2D projection.
A working example of the above code can be found here. The example also includes isometric projection functions that can be swapped out for the dimetric ones to see how that changes the way the graph looks. It also does some different visualization stuff that I didn't include here, like drawing "shadows" to help "visualize" the actual orientation -- the limitations of 3D to 2D projections.
It's complicated, and even a superficial discussion is kind of beyond the scope of this stackoverflow. I recommend you read more into the mathematics behind 3D, there are plenty of resources, both online and in print form. Once you have a more solid understanding of the basics of how the math works then return here if you have a specific implementation question about it.

What you want to do is impossible to do using the method you've stated - this is because a box - when rotated in 3 dimensions won't look anything like that diagram of yours. It will also vary based on the type of projection you need. You can, however get started using three.js which is a 3D drawing library for Javascript.
Hope this helps.

How to Draw 3D Rectangle?
posted in: Parallelogram | updated on: 14 Sep, 2012
To sketch 3 - Dimensional Rectangle means we are dealing with the figures which are different from 2 – D figures, which would need 3 axes to represent them. So, how to draw 3D rectangle?
To start with, first make two lines, one vertical and another horizontal in the middle of the paper such that they represent a “t” letter of English. This is what we need to draw for temporary use and will be removed later after the construction of the 3 – D rectangle is complete. Next we draw a Square whose measure of each side is 1 inch. Square must be perfect in Geometry so that 90 degree angles that are formed at respective corners are exact in measure. Now starting from upper right corner of the square we draw a line segment that will be stretched to a measure of 2 inches in the direction at an angle of 45 degrees. Similarly, we repeat the procedure by drawing another Line Segment from the upper left corner of the square and stretching it to 2 inches length in the direction at an angle of 45 degrees. These 2 line segments are considered to be the diagonals with respect to the horizontal line that we drew temporarily in starting. Also these lines will be parallel to each other. Next we draw a line that joins the end Point of these two diagonals.
Next starting from the very right of the 2 inch diagonal end point, draw a line of measure 1 inch that is supposed to be perpendicular to the temporary horizontal line. Next we need to join the lower left corner of the square with end point of the last 1’’ line we drew in 4th step and finally we get our 3 - D rectangular. Now we can erase our initial “t”. This 3- D rectangle resembles a Cuboid.