Lecture 4: Exponentials & Twists

Transcription

1 ME Introduction Lecture 4: Exponentials & Twists Noah J. Cowan Fall Skew symmetric matrices Let so(n) :={S R n n : S = S T } denote the skew symmetric matrices. Note that given any matrix, M R n n,thats = 1(M part. Proof: M T )istheskew-symmetric S T = 1 (M M T ) T = 1 (M T M)= 1 (M M T )= S Let b: R! so(), and _ : so()! R be given by! 1 b: 4! 5 7! 4! 0!!! 0! 1 5,!! 1 0 _ : 4 0!!! 1! 0! 1 5 7! 4! 5,!! 1 0! Note that a b =âb, since a b a b a b = 4a b 1 a 1 b 5 = 4 a 1 b a b 1 0 a a b 1 a 0 a 1 5 4b 5 a a 1 0 b Two useful facts (Lemma.1, MLS). See HW1. RŵR T = d Rw R(a b) =(Ra) (Rb) 1

2 Matrix Exponentials The exponential of a matrix is given by its power series expansion, i.e. exp{a} := I + A + 1! A + 1! A +... MLS Exercise 8, p74. Suppose = (t), and note that d dt ea = d dt I + A + 1! A + 1! A +... =(A + 1! A + 1! A +... =(A )(I + 1! A +1! A +... =(I + 1! A +1! A +... ( A) =(A )e A = e At (A ) Note that this does not imply that, in general, d dt ea(t) = ȦeAt. Exponential representation of rotations Motivation: Suppose a point q(t) isrotatingaroundanaxiswithdirectiongivenbya unit vector! R.Then q(t) =! q(t) =ˆ!q(t) and thus, if we assume a solution we see that it works, namely q(t) =eˆ!t q(0) q(t) = d dt eˆ!t q(0) = ˆ!eˆ!t q(0) = ŵq(t) and thus, if we rotate for units of time, we have MLS (.9) R(!, ) =exp{ŵ } Obviously, ˆ! so() that is, we are exponentiating a skew symmetric matrix! Exponentials of skew symmetric matrices are rotation matrices. This is because we can show that R 1 = R T and R =1. SeeMLS,Proposition.4. Let R =exp{ ŵ}

3 Then, the Rodrigues formula is given by R = I +sin ŵ +(1 cos )ŵ. This can be seen as follows. From the definition of the exponential mapping, we have that R = I + ŵ + ŵ +! ŵ + (1) It turns out, we are able to write ŵ n,forn>intermsofŵ and ŵ,andthenwe can group terms to get the desired result. Note that for an arbitrary vector v R, we have ŵ n v = w (w (w v) )(ntimes). The first cross product w v results in a vector orthogonal to w. Sincew is a unit vector, each successive product is a rotation of / aroundw, and the sequence of four vectors w v, w (w v), w v, w (w v) isrepeateduntilthelastcrossproduct. Sincethisistruefor all v, thenwehaveŵ n = ±ŵ for odd powers of n and ŵ n = ±ŵ for even powers. Another way of seeing this is that ŵ = (I ww T ), which is a projection onto the plane orthogonal to w, andfurtherthat(i ww T ) n =(I ww T ). Thus for n =1,,,... ŵ n 1 =( 1) n+1 ŵ, and ŵ n =( 1) n+1 ŵ. We can therefore rewrite R by simplifying the power series expansion (1), namely R = I +(! + 5 5! )ŵ +( which is the so-called Rodrigues formula. 4 4! + 6 6! )ŵ = I +sin ŵ +(1 cos )ŵ,

4 Quick review of Euler angles. We can now write Euler angles in terms of matrix exponentials, e.g. ZYX Euler angles are given by R(,, )=e be e be e be 1 where e 1 =[1, 0, 0] T, e =[0, 1, 0] T and e =[0, 0, 1] T,asusual. 4 Generalizing exp to rigid motions: twists This is just like rotations, only slightly more complicated. Turn to MLS, Figure.5, p40. Let p(t) beapointinspace,rotatingaboutanaxisw R, kwk =1,andletq R be a point on the axis. Assume the tip rotates with a unit velocity, then Let ˆ be given by with v = ṗ(t) =w (p(t) q) apple ŵ v ˆ = w q. Then, we can write this all in homogeneous coordinates: or simply apple ṗ 0 Just as before, we can solve this to get = ˆ ṗ = ˆ p apple p 1 p(t) =e tˆ p(0) Similarly, if we have a pure translation, then we have ṗ(t) =v, and,asbefore p(t) =e tˆ p(0) but this time, ˆ = apple 0 v This motivates the generalization of so(), namely the set of twists se() := {(v, ŵ) :v R, ŵ so()}. Note that so() and se() are linear vector spaces. 4

5 Let =(v, w) R 6. The mapping b : R 6! se(), and its inverse _ : se()! R 6 are given in homogeneous coordinates by ˆ = apple v w b apple bw = v and apple bw v _ = apple v w = The vector R 6 are called twist coordinates for the twist ˆ se(). Mathematical fact: g = eˆ is a rigid transformation. Moreover, given g SE(), it is possible to find a twist ˆ se(), such that eˆ = g (surjectivity). 5 Computing the exponential of an element of se() Let =(v, w) R R,representarotationalaxisand slide axis.defineatwist apple ŵ v ˆ = which is an element of se(). Let g =exp{ ˆ } SE(), namely apple R d g = 0 T 1 = I + ˆ + ˆ +! ˆ + Note that ˆ n = appleŵn ŵ n 1 v There are two cases. First, assume that kwk =0,inwhichcase ˆ n =0,forn thus apple I v exp{ } = I + = 0 T 1, corresponding to a pure translation., and Second, assume without loss of generality (WLOG) that kwk =1andrecallŵ n = ±ŵ for odd powers of n and ŵ n = ±ŵ for even powers. 1 We can therefore rewrite the terms R and d of g by simplifying the power series expansion, namely 1X k R = k! wk = I +sin ŵ +(1 cos )ŵ, k=0 which is Rodrigues formula derived before, and, similarly, by brute force 4 d =0+ v +(! 4! + 6 )ŵv +( 6!! = I +(1 cos )ŵ +( sin )ŵ v 5 5! + 7 7! 1 If kwk 6= 1, just scale everything by kwk, namely exp{ kwkˆ /kwk} =exp{ ˆ}. )ŵ v 5

6 The MLS way of doing this on pp4-44 is a bit more elegant, but we arrive at the same place. Note that d =(I eŵ )w v + ww T v from MLS (.6) =(I eŵ )ŵv + ww T v = I (I +sin ŵ +(1 cos )ŵ ) ŵv + ww T v = sin bw (1 cos )ŵ v + ww T v = sin bw +(1 cos )ŵ v +(ŵ + I)v MLS, Lemma. = I +(1 cos )ŵ +( sin )ŵ v 6