How to calculate odometry give two poses?
So I have the pose at times t - 1 and time t. We are given: that the the new pose, at time t, is
x_{t - 1} = [0,0,pi/6] x_{t} = [0.2,0.1,(11*pi)/60]
In my head, I have the two poses. I just need to calculate the difference of the two and that'll be odometry, right? So, subtracting the two vectors: [0.2,0.1,(11*pi)/60] - [0,0,pi/6] = [.2, .1, pi/60]
However, according to the answer checker that doesn't seem to be the right approach. Can someone suggest a strategy to go about this? Any help is very much appreciated.
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