In this note, we consider the motion of a rigid body B with respect to an inertial frame W. We will show that the body acceleration BaWB is simply related to the spatial acceleration WaWB by:
BaWB=XBWWaWBThis identity is not true for any twist, i.e. in general BmWB=XBWWmWB for an arbitrary m∈se(3).
Proof
We use the same notations as the spatial vector algebra cheat sheet, with XAB the 6×6 Plücker transform matrix that maps twists from frame B to another frame A. We will write twists with linear coordinates first:
AξAB=[AvABAωAB]As a warmup, recall (e.g. from the cheat sheet) that:
XBA(AξBA×)=(BξBA×)XBAFrom there, we can double check that:
(AξAB×)=XAB(BξBA×)XBA=(XABBξBA×)Thus AξAB=XABBξAB.
From twists to accelerations
Let us now switch to our practical example with a mobile frame B in the inertial frame W:
WξWB=XWBBξWBSpatial acceleration is, by definition, the time derivative of spatial velocity:
WaWB=Wξ˙WB=[Wv˙WBWω˙WB]Derivating the above identity with respect to time, we get:
WaWB=X˙WBBξWB+XWBBξ˙WB=XWB(BξWB×)BξWB+XWBBξ˙WB=0+XWBBξ˙WB=XWBBaWBNote how the first term cancel out because of the cross product between BξWB and itself, the former from the time derivative X˙WB and the latter because the twist we are time-derivating is also ξWB.
Discussion
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