Why we need a Skew-symmetric matrix to define acceleration?

Meossi91

Meossi91

Answered question

2022-08-10

I have a question about acceleration equation.
[ f x f y f z ] = [ u ˙ v ˙ w ˙ ] + [ 0 w v w 0 u v u 0 ] [ p q r ] + g [ sin θ cos θ sin ϕ cos θ cos ϕ ]
Here, why uvw should be skew-symmetric matrix and why skew-symmetric matrix of uvw should be multiplied with pqr?

Answer & Explanation

Aspen Beard

Aspen Beard

Beginner2022-08-11Added 9 answers

This is likely related to motion in a rotating frame. The antisymmetric matrix follows because this term is of the form R 1 d R, where R is a rotation matrix. In particular R is orthogonal so R 1 = R T and R T R = 1 ^
Take the differential of this:
0 = ( d R T ) R + R T d R = ( R T d R ) T + R T d R
showing that Ω = R T d R plus its transpose is nil, i.e. Ω T + Ω = 0, meaning Ω is antisymmetric.
This term is the rate of change of a rotating frame as seen from a lab frame.
So why should we need to consider R T d R? Take any fixed rotation matrix R 0 and consider R 0 R = r. The matrix r is simply the compound rotation of R 0 and the original R , i.e. we have done a rotational shift of R 0 to the coordinate system. Note then that r T d r = R T R 0 T R 0 d R since R 0 is constant. Since R 0 is a rotation, R 0 T R = 1 ^ so that r T d r = R T d R, independent of R 0 , and thus independent of the shift of origin in the rotational coordinates.

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