Conjugate of quaternion doesn’t give expected result

orientationquaternionsrotations

I'm using this site to play with quaternions.
All of my quaternions are unit quaternions.

I find quaternion of some Euler Angles(x, y, z) by using the website -inputs are degree and ZYX order Euler- and then by inputting the conjugate of the founded quaternion, I expect to see Conjugate of my Euler Angles (-x, -y, -z)

My flow is:

Euler(Input) => Quaternion(Result) => Conjugate of the Quaternion(Input) 
=> Euler Conjugate (Which is original Euler multiplied by -1)(Result)

In Degree and ZYX format, I input values

x = 70
y = 30
z = 0

And resulting quaternion is

[x, y, z, w(scalar)]
[ 0.5540323, 0.2120121, -0.1484525, 0.7912401 ]

When I input conjugate of this quaternion, which is vector parts multiplied by -1:

[ -0.5540323, -0.2120121, 0.1484525, 0.7912401 ]

Resulting Euler angles as (Z Y X) are:

[ x: -72.5047593, y: -9.8465479, z: 28.4812339 ]

Which is not related to my first angles (70, 30, 0). Shouldn't the conjugate of a quaternion give results of Euler angles multiplied by -1, (-70, -30, 0)

I tested the result of (-70, -30, 0) degrees and the resulting quaternion is

[ -0.5540323, -0.2120121, -0.1484525, 0.7912401 ]

Which has x and y components multiplied by -1, but z component is preserved. What is the point I'm missing in this problem?

Best Answer

You're missing two things. First, the conjugate quaternion should give you the inverse rotation, that is, a rotation that undoes your original Euler-angle rotation. The inverse rotation is not obtained just by changing the signs of the Euler angles. You can invert a rotation by reversing the rotation around each individual axis, but only if you apply it to the axes in reverse order. Merely changing the three angles in your set of Euler angles, you are claiming you can apply the rotations to the axes in the same order as before, which does not work in general.

Second, Euler-angle representations are not unique. You could input a set of angles, get a quaternion, and then try converting that quaternion back to Euler angles; you are not guaranteed to get the same Euler angles back even in that case, where you are dealing with only one rotation. For example, this answer deals with one particular interpretation of Euler angles and one particular set of conversion functions to and from quaternions, where the second set of Euler angles will be different from the first if the $y$-axis rotation in the original Euler angles was greater than a right angle.

Related Solutions

Euler Angles – Getting Euler (Tait-Bryan) Angles from Quaternion Representation

After a lot of searching, I have finally found a reference that has what I needed. Although it doesn't appear in a 'Quaternion to Euler' or 'Quaternion to Tait-Bryan' google search, when I gave up and started looking for 'Quaternion to Axis-Angle' with the intention of going through that representation as an intermediate step, I came across the following wonderful document:

Technical Concepts
Orientation, Rotation, Velocity and Acceleration, and the SRM
Version 2.0, 20 June 2008
Author: Paul Berner, PhD
Contributors: Ralph Toms, PhD, Kevin Trott, Farid Mamaghani, David Shen, Craig Rollins, Edward Powell, PhD
http://www.sedris.org/wg8home/Documents/WG80485.pdf

It covers a lot of the formalisms, but most importantly, shows derivations and solutions for 3-1-3 and 3-2-1 Euler angle representation. It also seems to cover inter-conversion between pretty much every other rotation representation I'm aware of, and so I would also recommend it as a good general reference.

Oh, and the actual solution for a 3-2-1 ($z-y-x$) Tait-Bryan rotation convention from that reference: $$ \phi = \operatorname{arctan2}\left(q_2 q_3 + q_0 q_1,\frac{1}{2}-(q_1^2 + q_2^2)\right) \\ \theta = \arcsin(-2(q_1 q_3 - q_0 q_2)) \\ \psi = \operatorname{arctan2}\left(q_1 q_2 + q_0 q_3,\frac{1}{2}-(q_2^2 + q_3^2)\right) $$

Note that the gimbal-lock situation occurs when $2(q_1 q_3 + q_0 q_2) = \pm1$ (which gives a $\theta$ of $\pm \frac{\pi}{2} $), so it can be clearly identified before you attempt to evaluate $\phi$ and $\psi$.

(Convention for arctan2 is $\operatorname{arctan2}(y, x)$, as hinted on page 3.)

[Math] Quaternion to Euler with some properties

It looks like the rotation order is $y x z$, but angles negated for some reason. In other words, the rotation matrix for Euler angles $(\chi,\gamma,\theta)$ used is most likely $$\mathbf{R}(\chi,\gamma,\theta) = \left[\begin{matrix} \cos \gamma \cos \theta - \sin \chi \sin \gamma \sin \theta & \cos \gamma \sin \theta + \sin \chi \sin \gamma \cos \theta & -\cos \chi \sin \gamma \\ -\cos \chi \sin \theta & \cos \chi \cos \theta & \sin \chi \\ \sin \gamma \cos \theta + \sin \chi \cos \gamma \sin \theta & \sin \gamma \sin \theta - \sin \chi \cos \gamma \cos \theta & \cos \chi \cos \gamma \end{matrix}\right]$$ The rotation matrix corresponding to versor aka unit quaternion $\mathbf{Q} = (w, x, y, z)$ is $$\mathbf{R}(w,x,y,z) = \left[\begin{matrix} 1 - 2 (y^2 + z^2) & 2 (x y - z w) & 2 (x z + y w) \\ 2 (x y + z w) & 1 - 2 (x^2 + z^2) & 2 (y z - x w) \\ 2 (x z - y w) & 2 (y z + x w) & 1 - 2 (x^2 + y^2) \end{matrix}\right]$$ The stated problem is to find the correspondence between the two.

Looking at $\mathbf{R}_{23}$ we can tell that $$\chi = \arcsin(2 y z - 2 x w)$$ Since $\mathbf{R}_{21}/\mathbf{R}_{22} = -\tan\theta$ if $\mathbf{R}_{22} \ne 0$, $$\theta = -\left\lbrace\begin{align} & -\arctan\left(\frac{x y + z w}{1/2 - x^2 - z^2}\right),\; & x^2 + z^2 \ne 1/2 \\ & -\frac{\pi}{2}, \; & x y + z w \ge 0 \\ & \frac{\pi}{2}, \; & x y + z w \lt 0 \end{align}\right.$$ Similarly, $\mathbf{R}_{13}/\mathbf{R}_{33} = -\tan\gamma$ if $\mathbf{R}_{33} \ne 0$, $$\gamma = \left\lbrace\begin{align} & -\arctan\left(\frac{x z + y w}{1/2 - x^2 - y^2}\right),\; & x^2 + y^2 \ne 1/2 \\ & -\frac{\pi}{2}, \; & x z + y w \ge 0 \\ & \frac{\pi}{2}, \; & x z + y w \lt 0 \end{align}\right.$$

As a result, I suggest trying the following pseudocode:

epsilon = 0.00001
halfpi = 0.5 * 3.14159265358979323846

function euler_angles(w, x, y, z):
    temp = 2*(y*z - x*w)
    if (temp >= 1 - epsilon):
        xa = halfpi
        ya = -atan2(y, w)
        za = -atan2(z, w)
    else if (-temp >= 1 - epsilon):
        xa = -halfpi
        ya = -atan2(y, w)
        za = -atan2(z, w)
    else:
        xa = asin(temp)
        ya = -atan2(x*z + y*w, 0.5 - x*x - y*y) 
        za = -atan2(x*y + z*w, 0.5 - x*x - z*z)

    return (xa, ya, za)

which utilizes the atan2(dy,dx) function available in most programming languages.

(Above pseudocode edited on 2017-09-30, to add the checks for gimbal lock, where xa = ±0.5 π. The epsilon represents the machine epsilon in this context, and depends on the precision of the variables; typically, it should be a compile-time constant that is easy to modify when porting or running the code on different architectures. The value shown should work well for visualization.)

The corresponding conversion from Euler angles to a quaternion is easy to compute, if one realizes that the three rotations as quaternions are easy to express: $$\begin{align} &\mathbf{R}_x(\chi) = \left( \cos\frac{\chi}{2} + \sin\frac{\chi}{2} \mathbf{i} \right )\\ &\mathbf{R}_y(\gamma) = \left( \cos\frac{\gamma}{2} + \sin\frac{\gamma}{2} \mathbf{j} \right )\\ &\mathbf{R}_z(\theta) = \left( \cos\frac{\theta}{2} + \sin\frac{\theta}{2} \mathbf{k} \right )\end{align}$$ Their Hamilton product $\mathbf{Q} = \mathbf{R}_y(\gamma)\mathbf{R}_x(\chi)\mathbf{R}_z(\theta)$ gives the desired quaternion here.

In pseudocode,

function quaternion(xa, ya, za):
    cx = cos(-0.5*xa)
    sx = sin(-0.5*xa)
    cy = cos(-0.5*ya)
    sy = sin(-0.5*ya)
    cz = cos(-0.5*za)
    sz = sin(-0.5*za)
    w = cx * cy * cz + sx * sy * sz
    i = cx * sy * sz + sx * cy * cz
    j = cx * sy * cz - sx * cy * sz
    k = cx * cy * sz - sx * sy * cz
    return (w, i, j, k)

Remember that both of these functions depend on the rotation order, and that the order of rotations here is $y x z$ with negated angles.

Best Answer

Related Solutions

Euler Angles – Getting Euler (Tait-Bryan) Angles from Quaternion Representation

[Math] Quaternion to Euler with some properties

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