I have the pose at times t - 1 and time t. We are given: that the the new pose, at time t, is
$$x_{t - 1} = [0,0,pi/6]$$ $$x_{t} = [0.2,0.1,(11*pi)/60]$$
(Where the vectors are [x, y, theta] with respect to some world frame.)
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?