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Calculating a quaternion from two combined angles

I'm creating a script that rotates a THREE.js camera arround based on a mobile phones gyroscope input. It's currently working pretty well, except that every time I rotate my phone over a quadrant, the camera will turn 180 degrees instead of continuing as intended. This is the code that I currently use:
private onDeviceOrientation = ( event ) => {
if( event.alpha !== null && event.beta !== null && event.gamma !== null ) {
let rotation = [
event.beta,
event.alpha,
event.gamma
],
this.orientation = new THREE.Vector3(rotation[0], rotation[1], rotation[2]);
this.viewer.navigation.setTarget(this.calcPosition());
}
};
private calcPosition = () => {
const camPosition = this.viewer.navigation.getPosition(),
radians = Math.PI / 180,
aAngle = radians * - this.orientation.y,
bAngle = radians * + this.orientation.z,
distance = this.calcDistance();
let medianX = Math.cos(bAngle) * Math.sin(aAngle);
let medianY = Math.cos(bAngle) * Math.cos(aAngle);
let nX = camPosition.x + (medianX * distance),
nY = camPosition.y + (medianY * distance),
nZ = camPosition.z + Math.sin(bAngle) * distance;
return new THREE.Vector3(nX, nY, nZ);
};
window.addEventListener('deviceorientation', this.onDeviceOrientation, false);
Soafter doing some research I found that I need to use a Quaternion prevent the switchen when going into a new quadrant. I have no experience with Quaternions, so I was wondering what the best way would be to combine the two Vector3's in the code above into a singel Quaternion.
[Edit]
I calculate the distance using this method:
private calcDistance = (): number => {
const camPosition = this.viewer.navigation.getPosition();
const curTarget = this.viewer.navigation.getTarget();
let nX = camPosition.x - curTarget.x,
nY = camPosition.y - curTarget.y,
nZ = camPosition.z - curTarget.z;
return Math.sqrt((nX * nX) + (nY * nY) + (nZ * nZ));from squared averages
};
And I follow the MDN conventions when working with the gyroscope.
[Edit #2]
Turns out I had my angle all wrong, I managed to fix it by calculating the final position like this:
let nX = camPosition.x - (Math.cos(zAngle) * Math.sin(yAngle)) * distance,
nY = camPosition.y + (Math.cos(zAngle) * Math.cos(yAngle)) * distance,
nZ = camPosition.z - (Math.cos(xAngle) * Math.sin(zAngle)) * distance;

Here is the closest I can give you to an answer:
First of all, you don't need a quaternion. (If you really find yourself needing to convert between Euler angles and quaternions, it is possible as long as you have all the axis conventions down pat.) The Euler angle orientation information you obtain from the device is sufficient to represent any rotation without ambiguity; if you were calculating angular velocities, I'd agree that you want to avoid Euler angles since there are some orientations in which the rates of change of the Euler angles go to infinity. But you're not, so you don't need it.
I'm going to try to summarize the underlying problem you're trying to solve, and then tell you why it might not be solvable. 🙁
You are given the full orientation of the device with a camera, as yaw, pitch, and roll. Assuming yaw is like panning the camera horizontally, and pitch is like tilting the camera vertically, then roll is a degree of freedom that doesn't change affect direction the camera is pointing, but it does affect the orientation of the images the camera sees. So you are given three coordinates, where two have to do with the direction the camera is pointing, and one does not.
You are trying to output this information to the camera controller but you are only allowed to specify the target location, which is the point in space that the camera is looking. This is to be specified via three Cartesian coordinates, which you can calculate from the direction the camera is pointing (2 degrees of freedom) and the distance to the target object (one degree of freedom).
So you have three inputs and three outputs, but only two of those have anything to do with each other. The target location has no way to represent the roll direction of the camera, and the orientation of the camera has no way to represent the distance to some target object.
Since you don't have a real target object, you can just pick an arbitrary fixed distance (1, for example) and use it. You certainly don't have anything from which to calculate it... if I follow your code, you are defining distance in terms of the target location, which is itself defined in terms of the distance from the previous step. This is extra work for no benefit at best (the distance drifts around some initial value), and numerically unstable at worst (the distance drifts toward zero and you lose precision or get infinities). Just use a fixed value for distance and make it simple.
So now you probably have a system that points a camera in a direction, but you cannot tell it what the roll angle is. That means your camera controller is apparently just going to choose it for you based on the yaw and pitch angles. Let's say it always picks zero degrees (that would be the least crazy thing it could do). This will cause discontinuities when the roll angle and yaw angle line up (when the pitch is at ±90°): Imagine pointing a physical camera at the northern horizon and yawing around westward, past the western horizon, and settling on the southern horizon. The whole time, the roll angle of the camera is 0°, so there's no problem. But now imagine pointing it at the northern horizon, and pitching upward, past the zenith, and continuing to pitch backward until you are facing the southern horizon. Now the camera is upside down; the roll angle is 180°. But if the camera controller doesn't change the roll angle from 0°, then it will do a nonphysical "flip" right when you pass the zenith. The problem is that there really is no way to synthesize a roll angle based purely on position and not have this happen. We've just demonstrated that there are two ways to move your camera from pointing north to pointing south, where the roll angle is completely different at the end.
So you're stuck, I guess. Well, maybe not. Can you rotate the image from the camera based on the roll angle of the device orientation? That is, add the roll back into the displayed image? If so, you may have a solution. Let's say the roll angle of the camera controller is always at zero. Then you just rotate the image by the desired roll angle (something derived from beta I guess?) and you're done. If the camera controller has some other convention for choosing the roll angle, you will need to figure that out, undo it, and add the roll angle back on.
Without the actual system in front of me I probably can't help you debug your way to a solution. So I think this is where my journey with this question must end. Good luck!
Summary:
You don't need a quaternion
Pick a fixed distance to your simulated target
Add the roll angle by rotating the image before displaying it
Good luck!

How to get Rotation around world axis Three js

I would like to get the rotation values form an Object3D according to world axises such as
roation.x, rotation.y, rotation.z
currently when I call
object.rotation.x
I get the rotation value according to the objects local axises.
Thanks.

Sorry for the thread necro.
I had the same problematic, and couldn't find a solution until I read some ThreeJS doc about Euler angles. It became clear all of a sudden.
Three.js uses intrinsic (Tait-Bryan) ordering, also known as yaw, pitch and roll. This means that rotations are performed with respect to the local coordinate system. That is, for order 'XYZ', the rotation is first around world-X, then around local-Y (which may now be different from the world Y-axis), then local-Z (which may be different from the world Z-axis).
Some implementations may use extrinsic (proper) ordering, in which case rotations are performed with respect to the world coordinate system, so that for order 'XYZ', the rotations are around world-X, world-Y, and world-Z.
Converting between the two types is relatively straightforward, you just need to reverse the order and the rotation, so that an intrinsic (three.js) Euler rotation of angles a, b, c about XYZ will be equivalent to to an extrinsic Euler rotation of angles c, b, a about ZYX.
If you need to get the world rotations in XYZ order I suppose you can do the following :
var worldRot = new THREE.Euler();
obj.getWorldRotation().copy(worldRot);
worldRot.reorder("ZYX");
// use worldRot.x, worldRot.y, worldRot.z

To rotate an object in world axis, you can use below function
var rtWorldMatrix;
function rotateAroundWorldAxis(object, axis, radians)
{
rtWorldMatrix= new THREE.Matrix4();
rtWorldMatrix.makeRotationAxis(axis.normalize(), radians);
rtWorldMatrix.multiplySelf(object.matrix);
object.matrix = rtWorldMatrix;
object.rotation.getRotationFromMatrix(object.matrix, object.scale);
}
And call
rotateAroundWorldAxis(objectToRotate, new THREE.Vector3(1,0,0), 90 * Math.PI/180);

Determining the point and angle at which two circles intersect.

I have recently started playing with canvas after seeing how easy it can be. My first project was just to keep a circle in its boundaries as it moves around. I made a few more things involving the movement of circles and now...
I'm currently working on bouncing two circles off of each other when they hit. You can see the example of that here: http://jsfiddle.net/shawn31313/QQMgm/7/
However, I would like to use a little more real world physics. At the moment, when the circles hit each other they just reverse their path.
As shown here:
// Dont be confused, this is just the Distance Formula
// We compare the distance of the two circles centers to the sum of the radii of the two
// circles. This is because we want to check when they hit each other on the surface
// and not the center.
var distance = Math.sqrt(Math.pow(c1.x - c2.x, 2) + Math.pow(c1.y - c2.y, 2));
var r1 = c1.rad;
var r2 = c2.rad;
if (distance < r1 + r2) {
// Change the slope of both circle
// I would like to figure out a more effecience way of bouncing the circles back
// However, I have no idea how to determine the angle the ball was struck,
// and with that information bounce it off at that angle
c1.xi = -c1.xi; // path is reversed
c1.yi = -c1.yi;
c2.xi = -c1.xi;
c2.yi = -c1.yi;
}
However, I would like the circles to go in opposite direction determined by the point and angle of intersection.
I am only in the 9th grade and not sure how the formula for something like this would look. But I know that it is possible because this kind of physics is present in many games. An example would be an 8-ball game. When the balls hit each other, they move across the table according to how the balls hit each other.
I would appreciate any tips on how to do this or if I should wait until I have a stronger understanding of Physics and Math in general.

too bad we can't draw a very simple scheme.
As far as physics is concerned, you know that the total momentum is conserved, see
http://en.wikipedia.org/wiki/Momentum
There is a good illustration and formulas here http://en.wikipedia.org/wiki/Elastic_collision#Two-_and_three-dimensional
You can simplify formulas if the two object have the same weight.
so now, let's consider the reference frame in which c2 is fixed and center in (0,0).
c1 velocity in this reference would be :
c1.xfi=c1.xi-c2.xi
c1.yfi=c1.yi-c2.yi
Now you have a collision when the distance between the two is the sum of radius. Consider the tangent plane of the two circles.
You now have to decompose the velocity of c1 into a tangent component, which is conserved, and a perpendicular (following the line between c1 and c2), which is transfered to c2.
Then you need to go back to your original reference frame.
(sorry i didn't give you the exact formulas but they are on the links I provided)

If I were doing this myself, I would implement the motion using Newtons law of restitution. Essentially this is a coefficient that relates approach and separation speed of 2 particles before/after impact and it has a value that depends on the material properties of your particles.
Your analysis will essentially amount to identifying the point of impact, then breaking down the approach velocities into components that are parallel and perpendicular to the line of centres of the circle at the point of impact.
The momentum of the particles is conserved perpendicular to the line of centres (so the velocities in that direction remain unchanged by the collision) and the law of restitution applies to the velocities parallel to the line of centres. Thus if you fix the coefficient of restitution (it has to be between 0 and 1) to some value of your choice you can use this law to calculate the separation speeds along the line of centres of your particles after collision using the value of the approach speeds.
If your particles are all of the same mass and radius then the calculations become simpler. You can model elastic collisions by setting the coefficient to 1 (this indicates that separation speed of the particles is the same as the approach speed) which is probably the easiest place to start. By changing the value you will see different behaviour between particles after collisions.
Sorry not to be able to write this all down in formula for you, but this is not really the appropriate place for it. Living in the UK I have no idea what "9th grade" is so I can't assess if the above is too advanced for your current level of education. Here in the UK this type of problem would typically be covered at A-level mathematics education level.
Hopefully though it will give you an indication of the terms and topics that you can teach yourself/ research in order to achieve your goal.

Three.js lookat seems to be flipped

I have a demo of what I mean here: Test Site or (Backup)
For some reason, even though the mouse vector is correct my object is rotated by 90 degrees always in favor of the positive Y axis. The only call that this could be going wrong, as far as I can tell, in is the call: ship.mesh.lookAt(mouse);, I call this every time the screen is animated.
Can anyone tell me what to do to fix this and why it is doing it?

object.lookAt( position ) orients the object so that the object's local positive z-axis points toward the desired position.
Your "ship's" front points in the direction of the local positive y-axis.
EDIT:
To re-orient your geometry, apply a matrix right after the geometry is created, like so:
geometry.applyMatrix( new THREE.Matrix4().makeRotationX( Math.PI / 2 ) );

How to do this coordinate system operation more efficiently?

I'm making a 3D game, where the player's back should always be facing the camera and he should move in that direction. I didn't come to the "back facing the camera" part yet, but I believe that it will be simple once I figure out how to move the player in the right direction...
Though it is a 3D coordinate system, height can be ignored (z-axis) because no matter how high the camera is, the player should always be going in the same speed (the camera system is planned to function much like in the game World of Warcraft).
Now, I have summarized my problem to this...
Point (0, 0) is the players position.
Point (x, y) is the camera's position.
The camera is (dx, dy) units away from the player (and because player is at (0, 0), it is also (x, y) units away, although this is a position vector, not a translation one)
Problem: how do I get a point (a, b) in this 2D space that lies on a circle r = 1 but is on the same line as (0, 0) and (x, y)?
Visualization:
By doing this, I should have a 2D vector (a, b), which would, when multiplied by -30, act as the speed for the player.
I know how to do this, but in a very expensive and inefficient way, using the Pythagora's theorem, square roots, and all those out-of-the-question tools (working in Javascript).
Basically, something like this:
c = sqrt(dx*dx + dy*dy); //Get the length of the line
rat = 1/c; //How many times is the desired length (1) bigger than the actual length
a = x*rat;
b = y*rat;
There must be something better!
For reference, I'm making the game in Javascript, using the Three.js engine.

There is nothing to make more efficient here, these calculations are standard stuff for 3D scenes.
Don't optimize prematurely. There is no way this stuff is a bottleneck in your app.
Remember, even if these calculations happen on each render(), they still only happen once every several milliseconds - 17ms assuming 60 FPS, which is a lot. Math.sin() / Math.cos() / Math.sqrt() are plenty efficient, and lots of other calculations happen on each render() that are much more complex.
You'll be just fine with what you have now.