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I'm having trouble finding any good information on this topic. Basically I want to find the component of a quaternion rotation, that is around a given axis (not necessarily X, Y or Z - any arbitrary unit vector). Sort of like projecting a quaternion onto a vector. So if I was to ask for the rotation around some axis parallel to the quaternion's axis, I'd get the same quaternion back out. If I was to ask for the rotation around an axis orthogonal to the quaternion's axis, I'd get out an identity quaternion. And in-between... well, that's what I'd like to know how to work out :)

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    Orthogonal and perpendicular is the same for vectors. You probably meant parallel for the identity quaternion case. Commented Sep 29, 2011 at 13:57
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    Wow; well noticed! I've read this several times (and have had colleagues independently search for the same thing and find this question) and have never noticed that before. I meant parallel for the no-change case (e.g. the component of a rotation around the Y axis, around the Y axis, will remain unchanged), and have updated the question to reflect that. Thanks! Commented Oct 1, 2011 at 9:39

5 Answers 5

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There is an elegant solution for this problem, specially suited for quaternions. It is known as the "swing twist decomposition":

in pseudocode

/**
   Decompose the rotation on to 2 parts.
   1. Twist - rotation around the "direction" vector
   2. Swing - rotation around axis that is perpendicular to "direction" vector
   The rotation can be composed back by 
   rotation = swing * twist

   has singularity in case of swing_rotation close to 180 degrees rotation.
   if the input quaternion is of non-unit length, the outputs are non-unit as well
   otherwise, outputs are both unit
*/
inline void swing_twist_decomposition( const xxquaternion& rotation,
                                       const vector3&      direction,
                                       xxquaternion&       swing,
                                       xxquaternion&       twist)
{
    vector3 ra( rotation.x, rotation.y, rotation.z ); // rotation axis
    vector3 p = projection( ra, direction ); // return projection v1 on to v2  (parallel component)
    twist.set( p.x, p.y, p.z, rotation.w );
    twist.normalize();
    swing = rotation * twist.conjugated();
}

And the long answer and derivation of this code can be found here http://www.euclideanspace.com/maths/geometry/rotations/for/decomposition/

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11 Comments

Please note paper "SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA" with general derivation
@minorlogic this looks great, but how do we deal with the singularity? I probably misunderstand something, but if this solution breaks some small percentage of the time, then surely it's not very useful?
singularity can be easy catched checking "twist " length is not zero. it is not present in code above (to simplify code).
How would I check for the singularity and where would it appear?
This is the best answer, however there is one step missing: if the dot product of ra and p is negative, then you should negate all four components of twist, so that the resulting rotation axis points in the same direction as direction. Otherwise if you are trying to measure the angle of rotation (ignoring the axis), you'll confusingly get a mix of rotations and their inverses, depending on whether or not the rotation axis direction flipped during projection(ra, direction). Note that projection(ra, direction) computes the dot product, so you should reuse it and not compute it twice.
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I spent the other day trying to find the exact same thing for an animation editor; here is how I did it:

  1. Take the axis you want to find the rotation around, and find an orthogonal vector to it.

  2. Rotate this new vector using your quaternion.

  3. Project this rotated vector onto the plane the normal of which is your axis

  4. The acos of the dot product of this projected vector and the original orthogonal is your angle.

     public static float FindQuaternionTwist(Quaternion q, Vector3 axis)
 {
     axis.Normalize();

     // Get the plane the axis is a normal of
     Vector3 orthonormal1, orthonormal2;
     ExMath.FindOrthonormals(axis, out orthonormal1, out orthonormal2);

     Vector3 transformed = Vector3.Transform(orthonormal1, q);

     // Project transformed vector onto plane
     Vector3 flattened = transformed - (Vector3.Dot(transformed, axis) * axis);
     flattened.Normalize();

     // Get angle between original vector and projected transform to get angle around normal
     float a = (float)Math.Acos((double)Vector3.Dot(orthonormal1, flattened));

     return a;
 }

Here is the code to find the orthonormals however you can probably do much better if you only want the one for the above method:

private static Matrix OrthoX = Matrix.CreateRotationX(MathHelper.ToRadians(90));
private static Matrix OrthoY = Matrix.CreateRotationY(MathHelper.ToRadians(90));

public static void FindOrthonormals(Vector3 normal, out Vector3 orthonormal1, out Vector3 orthonormal2)
{
    Vector3 w = Vector3.Transform(normal, OrthoX);
    float dot = Vector3.Dot(normal, w);
    if (Math.Abs(dot) > 0.6)
    {
        w = Vector3.Transform(normal, OrthoY);
    }
    w.Normalize();

    orthonormal1 = Vector3.Cross(normal, w);
    orthonormal1.Normalize();
    orthonormal2 = Vector3.Cross(normal, orthonormal1);
    orthonormal2.Normalize();
}

Though the above works you may find it doesn't behave as you'd expect. For example, if your quaternion rotates a vector 90 deg. around X and 90 deg. around Y you'll find if you decompose the rotation around Z it will be 90 deg. as well. If you imagine a vector making these rotations then this makes perfect sense but depending on your application it may not be desired behaviour. For my application - constraining skeleton joints - I ended up with a hybrid system. Matrices/Quats used throughout but when it came to the method to constrain the joints I used euler angles internally, decomposing the rotation quat to rotations around X, Y, Z each time.

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5 Comments

I'm at work at the moment, I'll try this later and if it works I'll be somewhat chuffed :) (meant to put that on a separate line in the previous comment but apparently 'enter' means 'submit' in this particular text box)
there is a small problem, the returned value is always positive. which means that if I have a rotation of -PI/4 (or 7*PI/4) I get PI/4 as a result. Am I missing something here?
@João, you are not missing anything but maybe misunderstand slightly how this works :) The core of this method is the acos of the dot product of the resulting vectors - to put it another way - this method does not operate 'directly' on the quaternion but rather 'observes the results' of the application of that quaternion. Therefore, the angle returned will always be the smallest between the two vectors, and will be limited to 0-360 deg. You can recover whether the angle of transformation was negative or positive however using the cross product. Good luck!
@sebf Could you elaborate? Which vectors do you need to apply cross product to in order to find out?
ah, found it. You have to do the cross product between orthonormal1 and flattened
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minorlogic's answer pointing out the swing-twist decomposition is the best answer so far, but it is missing an important step. The issue (using minorlogic's pseudocode symbols) is that if the dot product of ra and p is negative, then you need to negate all four components of twist, so that the resulting rotation axis points in the same direction as direction. Otherwise if you are trying to measure the angle of rotation (ignoring the axis), you'll confusingly get a mix of correct rotations and the reverse of the correct rotations, depending on whether or not the rotation axis direction happened to flip when calling projection(ra, direction). Note that projection(ra, direction) computes the dot product, so you should reuse it and not compute it twice.

Here's my own version of the swing-twist projection (using different variable names in some cases instead of minorlogic's variable names), with the dot product correction in place. Code is for the JOML JDK library, e.g. v.mul(a, new Vector3d()) computes a * v, and stores it in a new vector, which is then returned.

/**
 * Use the swing-twist decomposition to get the component of a rotation
 * around the given axis.
 *
 * N.B. assumes direction is normalized (to save work in calculating projection).
 * 
 * @param rotation  The rotation.
 * @param direction The axis.
 * @return The component of rotation about the axis.
 */
private static Quaterniond getRotationComponentAboutAxis(
            Quaterniond rotation, Vector3d direction) {
    Vector3d rotationAxis = new Vector3d(rotation.x, rotation.y, rotation.z);
    double dotProd = direction.dot(rotationAxis);
    // Shortcut calculation of `projection` requires `direction` to be normalized
    Vector3d projection = direction.mul(dotProd, new Vector3d());
    Quaterniond twist = new Quaterniond(
            projection.x, projection.y, projection.z, rotation.w).normalize();
    if (dotProd < 0.0) {
        // Ensure `twist` points towards `direction`
        twist.x = -twist.x;
        twist.y = -twist.y;
        twist.z = -twist.z;
        twist.w = -twist.w;
        // Rotation angle `twist.angle()` is now reliable
    }
    return twist;
}
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3 Comments

If you negate all components of a quaternion you end up with the exact same rotation, so the extra steps you added achieve absolutely nothing.
@maff You're right. Then what is a reasonable stability heuristic to stop the axis flipping unpredictably?
Why are you worried about the axis flipping? There are cases like interpolation where you want dot(a, b) >= 0, but that's out of scope here.
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I tried to implement sebf's answer, it seems good, except that the choice of the choice of vector in step 1:

  1. Take the axis you want to find the rotation around, and find an orthogonal vector to it.

is not sufficient for repeatable results. I have developed this on paper, and I suggest the following course of action for the choice of the vector orthogonal to the "axis you want to find the rotation around", i.e. axis of observation. There is a plane orthogonal to the axis of observation. You have to project the axis of rotation of your quaternion onto this plane. Using this resulting vector as the vector orthogonal to the axis of observation will give good results.

Thanks to sebf for setting me down the right course.

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2 Comments

If axis of observation equals axis of rotation then that would produce a zero-length vector, so this case would have to be checked.
See my answer, it may solve your problem about result repeatability.
-1

Code for Unity3d

// We have some given data
Quaternion rotation = ...;
Vector3 directionAxis = ...;

// Transform quaternion to angle-axis form
rotation.ToAngleAxis(out float angle, out Vector3 rotationAxis);

// Projection magnitude is what we found - a component of a quaternion rotation around an axis to some direction axis
float proj = Vector3.Project(rotationAxis.normalized, directionAxis.normalized).magnitude;
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This answer is wrong on many levels: (1) The output is a float, but it should be a quaternion (2) The value of proj is always 1, because you're passing a normalized axis to Vector3.Project (3) rotationAxis can be constructed directly from x, y, and z of rotation. ToAngleAxis performs additional unnecessary computation

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