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# Gimbal Lock. Trying to describe mathematically.

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### #1nikitablack  Members   -  Reputation: 1015

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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.

### #2clb  Members   -  Reputation: 2143

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Posted 05 September 2012 - 09:12 AM

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### #3Álvaro  Crossbones+   -  Reputation: 19099

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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.

### #4Catmull Dog  Members   -  Reputation: 136

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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.

### #5nikitablack  Members   -  Reputation: 1015

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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!

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).

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.

### #6Álvaro  Crossbones+   -  Reputation: 19099

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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.

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