at 2:50 why can we reduce R'RT=-RR'T to R'=-R'T? This is true only in case when R=I, which is not true in general case? especially later you write R'=S(w)R -- so it suggests that R can be non-trivial
at 3:40, honestly I don't get what is being proved here. R'RT is skew-symmetric => there exists such vector w, that R'RT=S(w). Next we multiply both sides by R on the right and get that R'=S(w)R
at 4:20, I don't get why you say it is a sum of two rotations? R(t) is rotation, but is dt*R'(t) also a rotation? First of all, multiplication by a small value dt is suspicious: rotation saves volume, but if you multiply a matrix by a very small number, it has a very low chance to have the determinant of 1. Even more, as we remember, R'=S(w)*R. The R is a rotation, that's fine. But S(w) is a 3x3 skew-symmetric matrix, and since it is of an odd dimension, it's determinant is always zero: det(S(w)) = 0. This means that det(R')=0. So dt*R'(t) cannot be a rotation matrix.
I have a confusion: at 2:00, the relation of the skew-symmetric matrix and its transpose has been established. Now at 2:37 how can we get v = S(w) r to v = - S(r) w? (w=> omega)
@@Woolfrey Understood v = w x r = -r x w, but what is the role of S( ) here? I am getting confused with the skew-symmetric notation: how can I prove S(w) r = - S(r) w?
The time-derivative should be the same no matter what type of Euler angles you use to formulate the rotation matrix. Even when you rotate about Z, then Y, then X, when you execute it all together, you'll end up with some equivalent rotation about X, Y and Z
at 3:54, this w is s frame not b frame, is that correct?
which one is more often used?
At 7:12 how can you sub S(w) = S(a)q ?
at 2:50 why can we reduce R'RT=-RR'T to R'=-R'T? This is true only in case when R=I, which is not true in general case?
especially later you write R'=S(w)R -- so it suggests that R can be non-trivial
at 3:40, honestly I don't get what is being proved here. R'RT is skew-symmetric => there exists such vector w, that R'RT=S(w). Next we multiply both sides by R on the right and get that R'=S(w)R
at 4:20, I don't get why you say it is a sum of two rotations? R(t) is rotation, but is dt*R'(t) also a rotation? First of all, multiplication by a small value dt is suspicious: rotation saves volume, but if you multiply a matrix by a very small number, it has a very low chance to have the determinant of 1. Even more, as we remember, R'=S(w)*R. The R is a rotation, that's fine. But S(w) is a 3x3 skew-symmetric matrix, and since it is of an odd dimension, it's determinant is always zero: det(S(w)) = 0. This means that det(R')=0. So dt*R'(t) cannot be a rotation matrix.
I have a confusion: at 2:00, the relation of the skew-symmetric matrix and its transpose has been established. Now at 2:37 how can we get v = S(w) r to v = - S(r) w? (w=> omega)
This is taken from properties of the cross product: v = w x r = -r x w . If you expand the calculations then you will see this property holds
@@Woolfrey Understood v = w x r = -r x w, but what is the role of S( ) here? I am getting confused with the skew-symmetric notation: how can I prove S(w) r = - S(r) w?
@@sourkarm2 v = w x r = S(w)r. The skew symmetric matrix does the operation as the cross-product. Then, v = -r x w = -S(r)w.
@@Woolfrey Thank you.
But how do you calculate the time derivative of a rotation matrix of Z-Y-Z Euler angles?
The time-derivative should be the same no matter what type of Euler angles you use to formulate the rotation matrix. Even when you rotate about Z, then Y, then X, when you execute it all together, you'll end up with some equivalent rotation about X, Y and Z
Woolfrey Thanks for the quick response
dislike for too low voice.