Maths - Conversion Quaternion to Euler - Forum

By: Andy Goldstein - andygoldstein
Corrections for Euler angles page et al  
2004-01-18 01:43

Martin, I've worked through the Euler angle and quaternion pages, and the related transformations in detail, and I have some major corrections for the Euler angles page. I also have detailed derivations for some of the Euler/quaternion transformations. All this is written up as a Word document with liberal use of the equation editor. What's the best way to get this to you?

Cheers - Andy

By: Martin Baker - martinbaker
RE: Corrections for Euler angles page et al  
2004-01-18 09:18

Andy,

Thank you very much.

I will update the pages when I get it, is it also OK to include your document on the site?

The latest version of word that I have is word2000, if you are using a later version I might not be able to read it, in this case would it be possible to export to HTML for me (this converts equations to gifs).

Thanks,

Martin

By: Michaele Norel - minorlogic
RE: Corrections for Euler angles page et al  
2004-03-31 15:53

Hi Martin !

on this page
https://www.euclideanspace.com/maths/geometry/rotations/conversions/quaternionToEuler/index.htm

When you convert quat to euler
heading = Math.atan2(2.0 * (q1.x*q1.y + q1.z*q1.w),(sqx - sqy - sqz + sqw));
bank = Math.atan2(2.0 * (q1.y*q1.z + q1.x*q1.w),(-sqx - sqy + sqz + sqw));

the "heading" and "bank" will be found correctly for for unit and nonunit quaternions too.

The use asin - is not good choice at all. You can swith the "attitude" in to atan2 , if you place there the
atan2( sin_attitude, cos_attitude );

just you need to find cos_attitude ( as i remember this can be found using one sqrt call)

And than your code will take a nonunit quaternions too, with comparable speed.

By: Martin Baker - martinbaker
RE: Corrections for Euler angles page et al  
2004-04-01 01:28

Hi minorlogic,

Yes, it would be very good to remove any requirement for a unit quaternion as input.

How do we check whether this is true? I guess that if we have:

k*x , k*y , k*z , k*w

where k=constant scaling factor.

Then if k cancels out in the equations then there is no requirement for unit quaternion? I can see that this applies to the expressions for heading and bank but not the expression for attitude.

But if we are using tan(a) = sin(a)/cos(a) how can we cancel out any constant scaling factor? I cant work out how to do this?

Martin

By: Andy Goldstein - andygoldstein
RE: Corrections for Euler angles page et al  
2004-04-02 08:30

Well, there's a fundamental problem here. For the two angles that are computed as ATANs, k cancels out because we're dividing one set of terms from the quaternion by another. That's a happy coincidence in the rotation matrix you get from Euler angles (in https://www.euclideanspace.com/maths/geometry/rotations/euler/index.htm .

However, for the remaining angle (theta in our current discussion) there are no terms in the matrix that lend themselves to solving for theta using a division. The only tractable term is the sin (theta) term; I don't see how you can solve for theta using some of the more complex terms that contain both sin (theta) and cosine (theta).

Normalizing a quaternion is straightforward enough: you simply divide each component of the quaternion by sqrt (w**2 + x**2 + y**2 + z**2),

- Andy

By: Andy Goldstein - andygoldstein
RE: Corrections for Euler angles page et al  
2004-04-02 10:22

After another 10 minutes thought...

If we combine the normalization of the quaternion into the expression for theta, we can save a sqrt operation, since the numerator term contains products of two quaternion components. So if for a normalized quaternion, theta = asin (2wy - 2xz), then for an unnormalized quaternion,

theta = asin ((2wy - 2xz) / (ww + xx + yy + zz))

In terms of complexity, this is consistent with the atan expressions for the other two angles.

- Andy

By: Martin Baker - martinbaker
RE: Corrections for Euler angles page et al  
2004-04-02 23:06

Andy,

That's brilliant; a well spent 10 minutes in my opinion.

I've included this on the webpage.

Martin