Jump to content

  • Log In with Google      Sign In   
  • Create Account

Gimbal Lock. Trying to describe mathematically.


Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

  • You cannot reply to this topic
5 replies to this topic

#1 nikitablack   Members   -  Reputation: 610

Like
0Likes
Like

Posted 05 September 2012 - 08:25 AM

Greetings! I'm trying to understand what is Gimbal Lock. After some digging I can explain it with words like this - with some combinations of cardinal axes and Euler Angles we lost one degree of freedom because two axes became parallel. Now I want to prove it mathematicaly. This what we have:

.....|1.....0.......0|
Rx = |0....cosA..sinA|
.....|0...-sinA..cosA|


.....|cosB..0...-sinB|
Ry = |0.....1.......0|
.....|sinB..0....cosB|


.....|cosC...sinC...0|
Rz = |-sinC..cosC...0|
.....|0.......0.....1|


Now we can get Gimbal Lock multiplyig matrices in that order (from left to right): Ry * Rx * Rz, where A = 90 deg. We have:

|1 0 0|
Rx = |0 0 1|
|0 -1 0|

..........|cosB..0...-sinB|...|1...0...0|...|cosB...sinB...0|
Ry * Rx = |0.....1.......0| * |0...0...1| = |0.......0.....1|
..........|sinB..0....cosB|...|0..-1...0|...|sinB..-cosB...0|


...............|cosB...sinB...0|...|cosC...sinC...0|
Ry * Rx * Rz = |0.......0.....1| * |-sinC..cosC...0| =
...............|sinB..-cosB...0|...|0.......0.....1|


..|cosB * cosC - sinB * sinC......cosB * sinC + sinB * cosC.....0|
= |0..........................................0.................1| =
..|sinB * cosC - cosB * sinC......sinB * sinC - cosB * cosC.....0|


..|cos(B + C)...sin(B + C)...0|
= |0................0........1|
..|sin(B + C)..-cos(B + C)...0|

So, our final matrix is very strange. I thought it should be same as Ry since rotations are the same. Why it difers?

Edited by Volgogradetzzz, 05 September 2012 - 08:33 AM.


Sponsor:

#2 clb   Members   -  Reputation: 1789

Like
1Likes
Like

Posted 05 September 2012 - 09:12 AM

I remember writing about gimbal lock in a thread some time ago, does this thread help?
Me+PC=clb.demon.fi | C++ Math and Geometry library: MathGeoLib, test it live! | C++ Game Networking: kNet | 2D Bin Packing: RectangleBinPack | Use gcc/clang/emcc from VS: vs-tool | Resume+Portfolio | gfxapi, test it live!

#3 Álvaro   Crossbones+   -  Reputation: 13912

Like
1Likes
Like

Posted 05 September 2012 - 10:04 AM

[Note: I didn't read what clb wrote.]

[Disclaimer: If you have never studied Differential Geometry, this description might not be very helpful to you.]

The way I think about it, the space of 3D rotations is the Lie group SO(3) (being a Lie group just means that on top of the group structure given by composition of rotations, it is also a smooth manifold). Euler angles are a particular chart (a "well-behaved" map from an open set in R^3 to some part of the manifold) that covers most of the manifold, but not all of it. If you push the parametrization to the border of the open set where the chart is defined, you'll cover all the manifold, but the parametrization is degenerate, which is what we call gimbal lock.

Gimbal lock is described in similar terms in this Wikipedia page.

#4 Catmull Dog   Members   -  Reputation: 136

Like
1Likes
Like

Posted 05 September 2012 - 01:14 PM

Basically when you rotate, you rotate the very axes of rotation themselves, so it quickly becomes non-intuitive.

Trackball style essentially counters this by applying the transpose of the camera matrix.

#5 nikitablack   Members   -  Reputation: 610

Like
0Likes
Like

Posted 06 September 2012 - 01:22 AM

Thanks everyone. Alvaro, you're very smart. But I can't understand any word from this theory (. Where did you learn such staff?
And my question is bit another. I understand why GL occurs. I can't understand why my matrices multiplication product with GL set not matches with single rotation matrix about single axis. They should match - they are the same rotations!

CLB, this is a quote from your linked post:

Mathematically we can see this as follows. Let's represent rotation using Euler angles, R = Rx(a) * Ry(b) * Rz©, where Ri(v) is a rotation matrix about axis i by angle v, and use the Matrix*vector convention. Fix b to -90 degrees (I think, or, b=90 if I messed up the sign). Then it can be seen that Ry(-90deg) * Rz© == Rx©. Therefore when b=-90, the rotation equation for the remaining two axes is R=Rx(a) * Rx© = Rx(a+c).

How did you get Ry(-90deg) * Rz© == Rx©??

Also consider a quote from wikipedia http://en.wikipedia.org/wiki/Gimbal_lock:

Loss of a degree of freedom with Euler angles
...
Let's examine for example what happens when B = 0. Knowing that cos0 = 1 and sin0 = 0, the above expression becomes equal to:

Why angle B is zero???? GL occurs when angle is 90 degrees. Please explain.

#6 Álvaro   Crossbones+   -  Reputation: 13912

Like
0Likes
Like

Posted 06 September 2012 - 08:05 AM

Also consider a quote from wikipedia http://en.wikipedia....ki/Gimbal_lock:

Quote
Loss of a degree of freedom with Euler angles
...
Let's examine for example what happens when B = 0. Knowing that cos0 = 1 and sin0 = 0, the above expression becomes equal to:
Why angle B is zero???? GL occurs when angle is 90 degrees. Please explain.


There are many conventions that people call "Euler angles". I believe proper Euler angles are what the Wikipedia page is using, which is a rotation about the z axis followed by a rotation about the x axis, followed by another rotation about the z axis (yes, the z axis appears twice). In those circumstances, gimbal lock happens when the rotation about the x axis is 0, as they explain.




Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.



PARTNERS