6 replies to this topic

[Koga73]

Hi,
I'm new to quaternions and want to use the resulting rotation matrix to translate a point a towards the quaternion given a set distance.


Without using quaternions or matrices I would do something like this:

public static function calcVector(angle:Number, distance:Number = 1):Vector2D {
var vec:Vector2D = new Vector2D();
vec.x = Math.sin(90 - angle) * distance;
vec.y = Math.sin(angle) * distance;

return vec;
}


How do I mimic this functionality using the quaternions resulting rotation matrix as my "angle" parameter?
Assuming the starting point is at (0, 0, 0) and I have a rotation matrix from a quaternion how do I translate the starting point towards the quaternion with a set distance of say 10 units?
I would like to keep everything in the form of vectors, quaternions, and matrices.

Thanks for the help!

Edited by Koga73, 10 December 2012 - 03:28 PM.

[Paradigm Shifter]

If your starting point is at the origin, applying a quaternion won't do any good (the origin is a fixed point for rotations), I'm guessing you want to apply the quaternion to the forward facing axis of something... try (1, 0, 0) if your objects face in the +x direction...

[Koga73]

In my case I have an XZ plane with the normal facing +Y.

I am using the following for my quaternion:
var quat:Quaternion = new Quaternion();
quat.fromAxisAngle(new Vector3D(0, 1, 0), rot * MathConsts.DEGREES_TO_RADIANS);
var mat:Matrix3D = quat.toMatrix3D();

So now I have a rotation matrix... I am wanting to make the plane orbit around the origin... so for instance if my distance is 10 units and my origin is (0, 0, 0) and my angle is 0° then I would expect the plane's resulting position to be (10, 0, 0).

I'm just not sure how to actually apply the distance and quaternion to get the planes resulting position?

[Paradigm Shifter]

You apply a quaternion by either turning it back into a matrix and doing matrix multiplication in the usual way, or you do

v' = q * v * conjugate(q)

where v = 0 + xi + yj + kz [x,y,z is the (x, y, z) position vector for the point.], and v' is the result of applying the quaternion rotation.

Note: rotating 0 degrees doesn't do anything... and by distance it sounds like you mean applying a scale?

You're being a bit unclear and I've had beers so I'll probably bow out of this for now ;)

[Álvaro]

I think PS's first post was correct. The OP needs to take the point (1,0,0), apply the rotation to it (either through a matrix or through the conjugation formula) and multiply by the distance. Equivalently, one could apply the rotation to (distance, 0, 0).

[Koga73]

Yeah as for rotating zero in my example I know it doesn't do anything it was just easy math for me to show what I expect the resulting point to be. As you stated above applying the rotation to (distance, 0, 0) does work!

var quat:Quaternion = new Quaternion();
quat.fromAxisAngle(new Vector3D(0, 1, 0), rot * MathConsts.DEGREES_TO_RADIANS);
  
var mat:Matrix3D = new Matrix3D();
mat.position = new Vector3D(10, 0, 0);
mat.append(quat.toMatrix3D());

plane.position = mat.position;

This is good and all but now I want to abstract it out into a function that takes in an axis vector, an angle, and a distance. It doesn't seem like the same code above would work if the axis were not the Y due to the position being hard-coded to (distance, 0, 0). I guess at this point I'm just looking for what that vector should be? The magnitude should always be the distance but does it matter what value I start it at?

Edited by Koga73, 10 December 2012 - 05:00 PM.

[Paradigm Shifter]

I'm guessing you want to start with the current facing of the object? If your plane is in XZ you should apply the rotation quaternion to the normal (which is the y axis).

Edited by Paradigm Shifter, 10 December 2012 - 05:04 PM.