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Three.js - rotating an Object3D around an axis

I'm trying to rotate a mesh loaded into an Object3D using OBJMTLLoader.
var obj = new THREE.Object3D();
// loading and stuff
obj.rotation.y += 0.1; //inside the update function
This works as it should, but only for y and z axes. Using the same code, but for the x axis yields the same result as rotating it around the z axis, but clockwise instead of counterclockwise.
Unfortunately, I need to rotate it around the x axis.
var xAxis = new THREE.Vector3(1,0,0);
obj.rotateOnAxis( xAxis, 0.1 );
This does rotate the object around the x axis.
However, I want to tween the rotation of the object, so I need a way to explicitly change the angle an object is rotated by, instead of rotating it for a specific amount.
Any ideas why obj.rotation.y and obj.rotation.z work properly, but obj.rotation.x doesn't?

Upon loading the mesh (before trying to rotate it around the x axis), I rotated it by 90 degrees around the y axis. Because of the default Euler order (which is XYZ), the local axes no longer corresponded to the world axes.
obj.eulerOrder = 'YXZ';
Using the above line of code before rotating the mesh around the y axis solved the problem.
A pretty good explanation of Euler order can be found here.

Three.js camera controls for space game

In a three.js project, I'm using a modified version of PointerLockControls.js for the camera control. I want to modify the rotation functionality so that there is no absolute "up" axis by which the camera rotates, but rather moving the mouse up or down will pitch indefinitely, same for left and right for yawing (and keys for rolling).
I can't seem to get the yawing component working, as it seems to rotate around the same axis regardless of pitch. (ie moving left or right when face facing straight upwards will just rotate the camera)
Any help in the right direction would be great!

I had the same problem recently, so I had a look at some of the THREE.*Controls files, similar to you.
Using those as a basis, I made this:
https://github.com/squarefeet/THREE.ObjectControls
The important bits are the following (see here for context):
var updateTarget = function( dt ) {
var velX = positionVector.x * dt,
velY = positionVector.y * dt,
velZ = positionVector.z * dt;
rotationQuaternion.set(
rotationVector.x * dt,
rotationVector.y * dt,
rotationVector.z * dt,
1
).normalize();
targetObject.quaternion.multiply( rotationQuaternion );
targetObject.translateX( velX );
targetObject.translateY( velY );
targetObject.translateZ( velZ );
};
The rotationVector is probably of most interest to you, so here's what it's doing:
It's using a THREE.Vector3 to describe the rotation, the rotationVector variable in this example.
Each component of the rotationVector (x, y, z) is relative to pitch, yaw, and roll respectively.
Set a quaternion's x, y, and z values to the of the rotation vector, making sure the w component is always 1 (to learn what the w component does, see here, it's a great answer.
Normalizing this quaternion will get us a quaternion of length 1, which is very handy when we come to the next step...
targetObject in this case is an instance of THREE.Object3D (a THREE.Mesh, which inherits from THREE.Object3D), so it has a quaternion we can play with.
So now, it's just a matter of multiplying your targetObject's quaternion by your shiny new rotationQuaternion.
Since our object is now rotated to where we want, we can move it along it's new axis angles by using translateX/Y/Z.
The important thing to note here is that quaternions don't act like Euler vectors. Rather than adding two quaternions together to get a new angle, you multiply them.
Anyway, I hope that helps you somewhat!

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.

How do I calculate the rotation along the y-axis given a 2D direction vector?

I'm making a 3D game, and I need the player mesh always facing the back of the camera. I already figured out how to get a 2D speed vector (direction along the x-z plane), but now I need to rotate the mesh in the speed vector's direction...
Basically, every mesh has a .rotation property, and that property is a 3D vector. I am only interested in rotation over the y-axis, that's the one that is perpendicular to the surface (x-z) plane.
The rotation doesn't use degrees, but radians, so I thought it would be something like this:
mesh.rotation.y = (mesh.direction.x - mesh.direction.z)*Math.PI*2;
But this doesn't seem to cut it...
The direction/speed is a, as a said, 2D vector, and it consist of real numbers between -1 and 1. At all times sqrt(x*x + y*y) == 1, so it forms a "circle", this is because speed needs to be equal in all directions, obviously.
The speed vector changes only when I drag the mouse over the screen, and so should the rotation, and it is calculated like this:
var c = Math.sqrt(cameraPos.x*cameraPos.x + cameraPos.z*cameraPos.z); //This is the distance from the camera to the mesh, which is at (0, 0) for simplicity of this presentation.
var rat = 1/c;
mesh.direction.x = cameraPos.x*rat; //Direction vector = the speed vector
mesh.direction.z = cameraPos.z*rat;

If I understand correctly, atan2 will do the trick:
mesh.rotation.y = Math.atan2(mesh.direction.z, mesh.direction.x)
Result is in radians. It basically calculates the angle between the vector and X axis. You might need to switch parameters or use minus operator here or there.

How can a large canvas have an animated 'viewable area'

The question title may be vague. Basically, imagine a racing game built in canvas. The track takes up 10,000 x 10,000 pixels of screen space. However the browser window is 500 x 500 pixels. The car should stay centered in the browser and the 'viewable' area of the 10,000 x 10,000 canvas will change. Otherwise the car would just drive off the edge at disappear.
Does this technique have a name?
What are the basic principles to make this happen?

If the car should stay at the same position (relative to the canvas' position), then you should not move the car. Instead, move the background picture/track/map to the other side.
Causing your eyes to think the car moves right can be done by either moving the car to the right, or by moving the map to the left. The second option seems to be what you want, since the car won't move whereas the viewable area (i.e. the map) will.
This is a quick demo from scratch: http://jsfiddle.net/vXsqM/.
It comes down to altering the map's position the other way round:
$("body").on("keydown", function(e) {
if(e.which === 37) pos.x += speed; // left key, so move map to the right
if(e.which === 38) pos.y += speed;
if(e.which === 39) pos.x -= speed;
if(e.which === 40) pos.y -= speed;
// make sure you can't move the map too far.
// clamp does: if x < -250 return -250
// if x > 0 return 0
// else it's allowed, so just return x
pos.x = clamp(pos.x, -250, 0);
pos.y = clamp(pos.y, -250, 0);
draw();
});
You can then draw the map with the position saved:
ctx.drawImage(img, pos.x, pos.y);
If you're looking for a way to actually move the car when the map cannot be moved any further (because you're driving the car close to a side of the map), then you'd have to extend the clamping and also keep track of when the car should be moved and how far: http://jsfiddle.net/vXsqM/1/.
// for x coordinate:
function clamp2(x, y, a, b) { // x = car x, y = map x, a = min map x, b = max map x
return y > b ? -y : y < a ? a - y : x;
}
The position clamping then becomes a little more complex:
// calculate how much car should be moved
posCar.x = clamp2(posCar.x, posMap.x, -250, 0);
posCar.y = clamp2(posCar.y, posMap.y, -250, 0);
// also don't allow the car to be moved off the map
posCar.x = clamp(posCar.x, -100, 100);
posCar.y = clamp(posCar.y, -100, 100);
// calculate where the map should be drawn
posMapReal.x = clamp(posMap.x, -250, 0);
posMapReal.y = clamp(posMap.y, -250, 0);
// keep track of where the map virtually is, to calculate car position
posMap.x = clamp(posMap.x, -250 - 100, 0 + 100);
posMap.y = clamp(posMap.y, -250 - 100, 0 + 100);
// the 100 is because the car (circle in demo) has a radius of 25 and can
// be moved max 100 pixels to the left and right (it then hits the side)

Two things:
Canvas transformation methods
First, the canvas transformation methods (along with context.save() and context.restore() are your friends and will greatly simplify the math needed to view a portion of a large 'world`. If you use this approach, you can get the desired behavior just by specifying the portion of the world that is visible and the world-coordinates of everything you want to draw.
This is not the only way of doing things* but the transformation methods are meant for exactly this kind of problem. You can use them or you can reinvent them by manually keeping track of where your background should be drawn, etc., etc.
Here's an example of how to use them, adapted from a project of mine:
function(outer, inner, ctx, drawFunction) {
//Save state so we can return to a clean transform matrix.
ctx.save();
//Clip so that we cannot draw outside of rectangle defined by `outer`
ctx.beginPath();
ctx.moveTo(outer.left, outer.top);
ctx.lineTo(outer.right, outer.top);
ctx.lineTo(outer.right, outer.bottom);
ctx.lineTo(outer.left, outer.bottom);
ctx.closePath();
//draw a border before clipping so we can see our viewport
ctx.stroke();
ctx.clip();
//transform the canvas so that the rectangle defined by `inner` fills the
//rectangle defined by `outer`.
var ratioWidth = (outer.right - outer.left) / (inner.right - inner.left);
var ratioHeight = (outer.bottom - outer.top) / (inner.bottom - inner.top);
ctx.translate(outer.left, outer.top);
ctx.scale(ratioWidth, ratioHeight);
ctx.translate(-inner.left, -inner.top);
//here I assume that your drawing code is a function that takes the context
//and draws your world into it. For performance reasons, you should
//probably pass `inner` as an argument too; if your draw function knows what
//portion of the world it is drawing, it can ignore things outside of that
//region.
drawFunction(ctx);
//go back to the previous canvas state.
ctx.restore();
};
If you are clever, you can use this to create multiple viewports, picture-in-pictures, etc. of different sizes and zoom in and out on stuff.
Performance
Second, as I commented in the code, you should make sure your drawing code knows what portion of your larger 'world' will be visible so that you don't do a lot of work trying to draw things that will not be visible.
The canvas transformation methods are meant for solving exactly this kind of problem. Use 'em!
*You will likely have problems if your world is so large that its coordinates cannot fit in an appropriate integer. You'll hit that problem roughly when your world exceeds billion (10^9) or a long trillion (10^18) pixels in any dimension, depending on whether the integers are 32- or 64-bit. If your world isn't measured in pixels but in 'world units', you'll run into problems when your world's total size and smallest feature scale lead to floating point inaccuracies. In that case, you will need to do extra work to keep track of things... but you'll probably still want to use the canvas transformation methods!

My very first game was a racing game where I moved the background instead of the car and although I want to think now that I had my reasons to make it so... I just didn't know better.
There are a few techniques that you need to know to achieve this well.
Tiled background. You need to make your track out of smaller pieces that tiled. To To draw 10,000 x 10,000 pixels is 100MPix image usually such image will have 32bit depth (4 bytes) this will end up being 400MB in memory. Compressions like PNG, JPEG won't help you since these are made to store and transfer images. They cant be rendered to a canvas without decompressing.
Move the car along your track. There is nothing worst then moving the BG under the car. If you need to add more features to your game like AI cars... now they will have to move along the map and to implement car collisions you need to make some not hard but strange spacial transformations.
Add camera entity. The camera needs to have position and viewport size (this is the size of your canvas). The camera will make or break your game. This is the entity that will give you the sense of speed in the game... You can have a camera shake for collisions, if you have drifts if your game the camera can slide pass the desired position and center back to the car, etc. Of course the most important thing will be tracking the car. Some simple suggestions I can give you are to not put the car in dead center of the camera. put the car a little behind so you can see a bit more what's in front of your. The faster the car moves the more you should offset the camera. Also you can't just compute the position of the camera instead compute desired position and slowly per frame move the current camera position to the desired position.
Now when you have camera and a large tiled map, when you draw the tiles you have to subtrack the camera position. You can also compute which tiles are not visible and skip them. This technique will allow you do extend your game with even larger maps or you can stream your map where you don't have all the tiles loaded and load in advance on background (AJAX) what will be visible soon.