Cartesian Coordinates, Points, and Transformations

Transcription

1 Cartesian Coordinates, Points, and Transformations CIS Russell Taylor

2 Femur Planned hole Pins CT image Tool path Femur Planned hole Pins F CT CT image Tool path COMMON NOTATION: Use the notation F obj to represent a coordinate system or the position and orientation of an object (relative to some unspecified coordinate system). Use F,y to mean position and orientation of y relative to. 2

3 Tool path Pin Pin 2 Pin 3 Planned hole Femur Assume equal CT image Want these to be equal Can control Tool holder Tool tip Can calibrate (assume known for now) Tool path Base of robot Pin Pin 2 Pin 3 Planned hole Femur Assume equal CT image 3

4 More correctly, this would be F Base,Wrist Tool holder F WT Tool tip FTip = FWrist FWT Target I F Wrist F Target Base of robot Question: What value of F will make F = F? Answer: Tip Target F Wrist = F Target I F WT = F Target F WT Wrist Tool holder F WT Tool tip FTip = FWrist FWT Target I F Wrist F Target Notational Note We Base use of the robot notation A B to represent composition or transformation. Where the contet is clear, we may also use AB for the same thing. Question: What value of F will make F = F? Answer: Tip F Wrist = F Target I F WT = F Target F WT Target Wrist 4

5 Tool path FCP = FHole FHP F HP Femur Pin 2 Pin 3 Pin b b 2 b 3 CT image Assume equal Planned hole F Hole Question: What value of F Wrist will make these equal? Tool tip Tool holder F WT Tool path F Wrist Base of robot F CT Pin 2 Pin 3 Pin b b 2 b 3 CT image FCP = FHole FHP 5

6 But: We must find F CT Let s review some math Tool tip Tool holder F Wrist F WT F = F F I F Wrist CT CP WT = F F F CT CP WT Tool path Base of robot F CT Pin 2 Pin 3 Pin b b 2 b 3 CT image FCP = FHole FHP Coordinate Frame Transformation F = [ R, p] 0 F y z y 0 z 0 Slide acknowledgment: Sarah Graham and Andy Bzostek 6

7 b F = [R,p] 7

8 b F = [R,p] b F = [ I,0] 8

9 b F = [R,0] v = R b R b v = R b R p F = [R,p] b p v = vr + p = R b + p 9

10 Coordinate Frames v = F b = [R, p] b = R b + p b F = [R,p] Forward and Inverse Frame Transformations Forward F = [ R, p] v = F b = [ R, p] b = R b + p Inverse F F v = b b = R = R ( v p) v R = [ R, R p] p 0

11 Composition Assume F = [R, p ], F 2 = [R 2, p 2 ] Then F F 2 b = F (F 2 b) = F (R 2 b + p 2 ) = [R, p ] (R 2 b + p 2 ) = R (R 2 b + p 2 ) + p = R R 2 b + R p 2 + p = [R R 2,R p 2 + p ] b So F F 2 = [R, p ] [R 2, p 2 ] = [R R 2, R p2 + p ] v v col row v = v y v z = length : [ v v v ] y 2 z 2 y v = v + v + v Vectors 2 z u = vw z v w v w y dot product : a = v w = ( v w + v w + v w ) = y y z z v w cosθ cross product : u = v w v yw z v zw y = v zw v w z v w y v yw, u = v w sinθ Slide acknowledgment: Sarah Graham and Andy Bzostek

12 Matri representation of cross product operator Define 0 az a y Δ Δ ˆ a = skew( a ) = az 0 a ay a 0 Then a v = skew( a ) v Ais-angle Representations of Rotations a α c = Rot( a,α) b ( ) = bcosα + a bsinα + a( a b) cosα b Rodrigues' rotation formula (named after Olinde Rodrigues) Rotation of a vector b by angle α about ais a (Assumes that a is a unit vector, a = ) 2

13 Eponential representation Consider a rotation about ais n by angle θ. Then e skew( n)θ = I + θskew( n) + θ 2 2 skew( n) 2 + By doing some manipulation, you can show Rot( n,θ) = e skew( n)θ = I + skew( n)sinθ + skew( n) 2 ( cosθ) = I + skew( n)sinθ + ( n n T I)( cosθ) = Icosθ + skew( n)sinθ + n n T ( cosθ) Note that for small θ, this reduces to Rot( n,θ) I + skew(θ n) Rotations: Some Notation Rot( a,α ) = Rotation by angle α about ais a R a (α ) = Rotation by angle α about ais a R( a) = Rot( a, a ) R yz (α,β,γ ) = R(,α ) R( y,β) R( z,γ ) R zyz (α,β,γ ) = R( z,α ) R( y,β) R( z,γ ) 3

14 Rotations: A few useful facts Rot(s a,α ) a = a and Rot( a,α ) b = b Rot( a a,α ) = Rot(â,α ) where â = a Rot( a,α ) Rot( a,β) = Rot( a,α + β) Rot( a,α ) = Rot( a, α ) Rot( a,0) b = b i.e., Rot( a,0) = I Rot = the identity rotation ( ) Rot(â,α ) b = ( â b )â + Rot(â,α ) b ( â b )â Rot(â,α ) Rot( ˆb,β) = Rot( ˆb,β) Rot(Rot( ˆb, β) â,α ) Rot(â,α ) R β = R β Rot(R β â,α ) R α Rot( ˆb,β) = Rot(R α ˆb,β) R α Rotations: more facts If v = [v,v y,v z ] T then a rotation R v may be described in terms of the effects of R on orthogonal unit vectors, e = [,0,0] T, e y = [0,,0] T, e z = [0,0,] T R v = v r + v y ry + v z rz where r = R e r y = R e y r z = R e z Note that rotation doesn't affect inner products ( R b ) ( R c ) = b c 4

15 Rotations in the plane R v cosθ y sinθ R y = sinθ y cosθ + cosθ sinθ = sinθ cosθ y θ v = [, y] T Rotations in the plane R e e y = cosθ sinθ sinθ cosθ = R e R e y = r ry 0 0 θ 5

16 3D Rotation Matrices R e ey ez = R e R ey R ez = r ry rz T rˆ T T R R = rˆ y r ry rz ˆ rz T T T r r r ry r rz 0 0 T T T = y y y y z = 0 0 r r r r r r T T T rz r rz ry rz rz 0 0 Properties of Rotation Matrices Inverse of a Rotation Matri equals its transpose: R - = R T R T R=R R T = I The Determinant of a Rotation matri is equal to +: det(r)= + Any Rotation can be described by consecutive rotations about the three primary aes,, y, and z: R = R z,θ R y,ϕ R,ψ 6

17 Canonical 3D Rotation Matrices Note: Right-Handed Coordinate System R (θ) = Rot(,θ) 0 0 = 0 cos(θ) sin(θ) 0 sin(θ) cos(θ) R y (θ) = Rot( y,θ) cos(θ) 0 sin(θ) = 0 0 sin(θ) 0 cos(θ) R z (θ) = Rot( cos(θ) sin(θ) 0 z,θ) = sin(θ) cos(θ) Homogeneous Coordinates Widely used in graphics, geometric calculations Represent 3D vector as 4D quantity For our current purposes, we will keep the scale s = v s ys zs s y z 7

18 Representing Frame Transformations as Matrices 0 0 p v 0 0 p y v y v + p = P v 0 0 pz v z R 0 v R v 0 I p R 0 R p P R = = [ R, p] = F R p v ( R v) + p F v = 0 F CH F CA F BE F BD F C F D F E F DU F GC F G F U F EH F BG F UA F H F B 8

19 F CH F CA F BE F BD F C F D F E F DU F GC F G F EH F U F BG F UA F H F B F GH Give a formula for computing the pose F GH of the surgical tool coordinate system relative to the patient rigid body coordinate system F G F CH F CA F BE F BD F C F D F E F DU F GC F G F EH F GH = F BG F BE F EH F U = F BG F BH F BG F UA F B F GH F H F BH Give a formula for computing the pose F GH of the surgical tool coordinate system relative to the patient rigid body coordinate system F G? What are the components F GH = [R GH, p GH ]? R GH = R BG R BE R EH p GH = F BG p BH = F BG F peh BE p GH = F BG (R peh BE + p BE ) p GH = R BG (R BE peh + p BE p BG ) 9

20 Tool tip v = F Wrist, p WT = F CT b Tool holder F WT F Wrist, Base of robot v F CT Pin 2 Pin 3 Pin b b 2 b 3 CT image 20

21 Tool tip v = F Wrist, p WT = F CT b v 2 = F Wrist,2 p WT = F CT b 2 Tool holder F WT F Wrist,2 Base of robot F CT v 2 Pin 2 Pin 3 Pin b b 2 b 3 CT image Tool holder F Wrist,3 F WT v 3 Tool tip v = F Wrist, p WT = F CT b v 2 = F Wrist,2 p WT = F CT b 2 v 3 = F Wrist,3 p WT = F CT b 3 Base of robot F CT Pin 2 Pin 3 Pin b b 2 b 3 CT image 2

22 Frame transformation from 3 point pairs b 3 F CT b b 2 v 3 v v 2 F rob Frame transformation from 3 point pairs F CT b b 2 F = F F rc rob CT b 3 v 3 v Frob vk = FCT bk vk = Frob FCT bk v = F b k rc k F rob v 2 22

23 Frame transformation from 3 point pairs v = F b = R b + p k rc k rc k rc Define 3 3 v = v b = b u = v v a = b b m k m k 3 3 k k m k k m FrC a k = RrCa k + p rc RrCak + prc = RrC ( bk bm) + prc R a = R b + p R b p RrCa k = v k v m = u k p rc = v m R bm rc rc k rc k rc rc m rc a b m b a 2 Solve These v v 3 v m u u u 3 2 a b 3 3 b 2 v 2 Rotation from multiple vector pairs Given a system Ra k = u k for k =, ", n the problem is to estimate R. This will require at least three such point pairs. Later in the course we will cover some good ways to solve this system. Here is a not-so-good way that will produce roughly correct answers: Step : Form matrices U= u u and A a a [ " ] = [ " ] Step 2: Solve the system RA = U for R. E.g., by R = UA T Step 3: Renormalize R to guarantee R R = I. n n 23

24 Renormalizing Rotation Matri T Given "rotation" matri R = r ry rz, modify it so R R = I. Step : Step 2: Step 3: a = ry rz b = r a R z a b r z normalized = a b rz Calibrating a pointer But what is b tip?? F ptr b tip vtip = Fptr btip F Russell lab H. Taylor ; lecture notes 24

25 Calibrating a pointer b post = Fk btip = R b + p k tip k F ptr F k b tip b post F Russell lab H. Taylor ; lecture notes Calibrating a pointer For each measurement k, we have b = R b + p post k tip k I. e., R b b = p k tip post k Set up a least squares problem " " b " tip Rk I b post p k " " " F ptr b tip b tip b tip F ptr F ptr 25

26 Registration Transformations F CH F CA F BE F BD F C F D F E F DU F GC F G F EH F U F BG F UA F H F B Given a coordinate system F C and another coordinate system F G (e.g., a CT scan and a tracked "rigid body" attached to the patient, and points c i in the coordinate system F C and points g i in the coordinate system F G,then the "registration transformation" F GC between F G and F C one in which for F GC ci = g i if and only if c i and g i refer to the same or corresponding points. Use in surgical navigation F CH F CA F BE F BD F C F D F E F GD F DU F GC F G F EH F BA = F BD F DU F UA F BA = F BG F GC F CA F U F BG F UA F H F BG F GC F CA = F BD F DU F UA F B F CA = ( F BG F GC ) F BD F DU F UA If an anatomic structure is identified at pose F UA in ultrasound image coordinates give the formula for computing the corresponding pose F CA in CT coordinates F GC = F BG F BD F DU F UA F CA F GC = F GD F DU F UA F CA where F GD = F BG F BD 26

27 Kinematic Links End of link k- Fk, k End of link k F k F k Base of robot R k, p k F k = F k F k,k = R k,p k R k,k,p k,k Kinematic Links F k F k- L k Θ k F k = F k F k,k R k, p k = R k,p k = R k,p k = R k,p k R k,k,p k,k Rot( r k,θ k ), 0]i[I, L k Rot( r k,θ k ), L k Rot( r k,θ k ) 27

28 Kinematic Chains F0 = [ I, 0] R3 = R0,R,2R 2,3 = Rot( r, θ) Rot( r2, θ2) Rot( r3, θ3) p3 = p0, + R0, ( p,2 + R,2p2,3) F 2 = L Rot( r, θ) + L2Rot( r, θ) Rot( r2, θ2) + L Rot( r, θ ) Rot( r, θ ) Rot( r, θ ) L 2 θ 2 θ 3 L 3 F 3 F F 0 L θ Kinematic Chains If r = r2 = r3 = z, R3 = Rot( z, θ) Rot( z, θ2) Rot( z, θ3) = Rot( z, θ + θ2 + θ3) p3 = p0, + R0, ( p,2 + R,2p2,3) = L Rot( z, θ) + L2Rot( z, θ) Rot( z, θ2) + L3Rot( z, θ) Rot( z, θ2) Rot( z, θ3) = L Rot( z, θ) + L2Rot( z, θ + θ2) + L Rot( z, θ + θ + θ ) F 0 L θ L 2 θ 2 θ 3 L 3 F 3 28

29 Kinematic Chains If r = r 2 = r 3 = z, cos(θ + θ 2 + θ 3 ) sin(θ + θ 2 + θ 3 ) 0 R 3 = sin(θ + θ 2 + θ 3 ) cos(θ + θ 2 + θ 3 ) L cos(θ ) + L 2 cos(θ + θ 2 ) + L 3 cos(θ + θ 2 + θ 3 ) p 3 = L sin(θ ) + L 2 sin(θ + θ 2 ) + L 3 sin(θ + θ 2 + θ 3 ) 0 Small Transformations A great deal of CIS is concerned with computing and using geometric information based on imprecise knowledge Similarly, one is often concerned with the effects of relatively small rotations and displacements Essentially, we will be using fairly straightforward linearizations to model these situations, but a specialized notation is often useful 29

30 Small Frame Transformations Represent a "small" pose shift consisting of a small rotation ΔR followed by a small displacement Δp as Δ F = [ ΔR, Δp] Then Δ F v = Δ R v + Δp Small Rotations Δ R = a small rotation R a( Δ α) = a rotation by a small angle Δα about ais a Rot( a, a ) b a b + b for a sufficiently small Δ R( a) = a rotation that is small enough so that any error introduced by this approimation is negligible Δ R( λ a) ΔR( µ b) Δ R( λ a + µ b) (Linearity for small rotations) Eercise: Work out the linearity proposition by substitution 30

31 Approimations to Small Frames ΔF( a,δ p) " [ΔR( a),δ p] ΔF( a,δ p) v = ΔR( a) v + Δ p v + a v + Δ p a v = skew( a) v 0 a z a y = a z 0 a a y a 0 skew( a) a = a a = 0 v v y v z ΔR( a) I + skew( a) ΔR( a) I skew( a) = I + skew( a) Approimations to Small Frames Notational NOTE: We often use α to represent a vector of small angles and ε to represent a vector of small displacements In using these approimations, we typically ignore second order terms. I.e., α A αb 0, α A εb 0, ε A εb 0, etc. 3

32 Errors & sensitivity Often, we do not have an accurate value for a transformation, so we need to model the error. We model this as a composition of a "nominal" frame and a small displacement F actual = F nominal ΔF Often, we will use the notation F * for F actual and will just use F for F nominal. Thus we may write something like F * = F ΔF or (less often) F * = ΔF F. We also use v * = v + Δ v,etc. Thus, if we use the former form (error on the right), and have nominal relationship v = F b, we get v * = F * b * = F ΔF ( b + Δ b) = F (ΔR b + ΔR Δ b + Δ p) R ((I+ sk( α)) ( b + Δ b )+ Δ p)+ p = R ( b + α b + Δ b + Δ p)+ p R ( α b + Δ b + Δ p)+r b + p = R ( α b + Δ b + Δ p)+ v Δ v R ( α b + Δ b + Δ p) Small Errors F CH F CA F BE F BD F C F D F DU F F G GC F U F E F EH Suppose that there is a small systematic error in the tracking system so that * F B = ΔF B F B F B F BG F UA F H for F BG,F BD, F BE. How does this affect the calculation of F GH? * F GH * = (F BG ) * F BE F EH F GH ΔF GH = ( ΔF B F BG ) ΔF B F BE F EH ΔF GH = F BG ΔF B ΔF B F BE F EH F GH = F BG F BE F EH F GH = F GH F GH = I 32

33 Small Errors F CH F CA F BE F BD F C F D F DU F E Suppose that there are additional errors in the tracking of each tracker body so that F GC F BG F G F U F UA F H F EH * F B = ΔF B F B ΔF B for F BG,F BD, F BE. How does this affect the calculation of F GH? F B F GH * = F GH ΔF GH = (F BG * ) F BE * F EH ΔF GH = F GH ( ΔF B F BG ΔF BG ) ( ΔF B F BE ΔF BE )F EH ΔF GH = ( F BG F BE F EH ) ΔF BG F BG ΔF B ΔF B F BE ΔF BE F EH = F EH F BE F BG ΔF BG F BG F BE ΔF BE F EH F = [R,p] b v 33

34 F * = F ΔF v v v * = + Δ b b b * = + Δ F * b * = v * F ΔF ( b + Δ b) = v + Δ v F * = F ΔF v v v * = + Δ b b b * = + Δ 34

35 Errors & Sensitivity Suppose that we know nominal values for F, b, and v and that T T -ε,-ε,-ε Δ v ε,ε,ε (i.e., Δ v ε) What does this tell us about ΔF = [ΔR,Δ p]? F * = F ΔF * v = v + Δv * b = b + Δb ; Copyright 999, 2000 rht+sg Errors & Sensitivity so v * = F * b * = F ΔF ( b + Δ b) ( ( ) + Δ p ) + p ( ) + p ( ) ( ) = R ΔR( α) b + Δ b R b + Δ b + α b + α Δ b + Δ p = R b + p + R Δ b + α b + α Δ b + Δ p v + R Δ b + α b + Δ p if α Δ b α Δ v = v * v R Δ b + α b + Δ p Δ b is negligible (it usually is) ( ) = R Δ b + R α b + R Δ p 35

36 Digression: rotation triple product Epressions like R a b are linear in a, but are not always convenient to work with. Often we would prefer something like M( R, b) a. R a b = R b a = R skew( b) a = skew( ) T R b a Digression: rotation triple product Here are a few more useful facts: ( ) = ( R a ) ( R b ) ( ) = R R a R a b a R b Consequently (( ) b ) skew( a) R = R skew(r a) R skew( a) R = skew(r a) 36

37 Previous epression was ( ) Δ v R Δ b + α b + Δ p Substituting triple product and rearranging gives Δ v R R R skew( b) Δb Δ p α So ε ε ε Errors & Sensitivity R R R skew( b) Δ b Δ p α ε ε ε Now, suppose we know that Δ b β, this will give us a system of linear constraints ε ε ε β β β Errors & Sensitivity R R R skew( b) I 0 0 Δ b Δ p α ε ε ε β β β 37

38 Error from frame composition Consider R * R * 2 = R * 3 where R * = R ΔR,R * 2 = R 2 ΔR 2, R * 3 = R 3 ΔR 3 and ΔR I + sk ( α ), ΔR 2 I + sk ( α 2 ), estimate ΔR 3 I + sk α 3 R ΔR R 2 ΔR 2 = R R 2 ΔR 3 R (I + sk( α ))R 2 (I + sk( α 2 )) R R 2 (I + sk( α 3 )) ( R R 2 ) R (I + sk( α ))R 2 (I + sk( α 2 )) I + sk( α 3 ) R 2 R R (I + sk( α ))R 2 (I + sk( α 2 )) I + sk( α 3 ) I + R 2 sk( α )R 2 + sk( α 2 ) + R 2 sk( α )R 2 sk( α 2 ) I + sk( α 3 ) R 2 sk( α )R 2 + sk( α 2 ) sk( α 3 ) ( ) Since R i( a R b) = (R a) b for all R, a, b we get R 2 sk( α )R 2 = sk(r 2 α ) sk( α 3 ) sk(r 2 α ) + sk( α 2 ) = sk(r 2 α + α 2 ) α 3 R 2 α + α 2 Error from frame composition Consider F * F * 2 = F * 3 where F * = F ΔF,F * 2 = F 2 ΔF 2, F * 3 = F 3 ΔF 3 and ΔF I + sk ( α ), ε, ΔF I + sk α 2 ( 2 ), ε 3, estimate ΔF 3 I + sk ( α 3 ), ε 3 From before, we have α 3 R 2 α + α 2. So now we just need ε 3. p * 3 = R (ΔR ( p 2 + R 2 ε 2 ) + ε ) + p p 3 + ε 3 R ( I + sk( α )) p 2 + R 2 ε 2 = R p2 + R R 2 ε 2 + R i = p 3 + R R 2 ε 2 + R i ε 3 R R 2 ε 2 + R i α p 2 + R ε = R R 2 ε 2 R i p 2 α + R ε = R R 2 ε 2 R sk( p 2 ) α + R ε ( ) + R ε + p ( α p 2 + α ε 2 ) + p + R ε α p 2 + α ( ε 2 ) + R ε 38

39 Inverse of frame transformation with errors F i = F = [R, R p] F * i = ( FΔF) F i ΔF i = RΔR,RΔ p + p = [(RΔR), (RΔR) ( RΔp + p )] = [ΔR R, ΔR R ( RΔp + p )] = [ΔR R, ΔR Δ p ΔR R p] ΔF i = F ( ) [ΔR R, ΔR Δ p ΔR R p] = [R, p]i[δr R, ΔR Δ p ΔR R p] = [RΔR R, RΔR Δ p RΔR R p + p] Inverse of frame transformation with errors Suppose we know that ΔR is "small", i.e.,δr I+ sk( α), and for notational convenience we write Δ p = ε, we get ( ) R ΔR i = RΔR R R I+ sk( α) R( I sk( α) )R = RR Rsk( α)r = I Rsk( α)r = I sk ( R ) α = I sk(r α) Δ p i = RΔR Δ p RΔR R p + p ( ) ε ( I sk(r α) ) p + p R I sk( α) = R ε +R ( α ε ) p + (R α) p + p R ε + (R α) p = R ε p (R α) = R ε sk( p)r α ( ) = R ε R ((R p) α ) ( ) = R ε Rsk(R p) α (also) = R ε (RR p) (R α) = R i ε + (R p) α 39

40 Error Propagation in Chains F k F k- L k θ k F * k = F* k F* k,k F k ΔF k = F k ΔF k F k,k ΔF k,k ΔF k = ( F k F k )ΔF k F k,k ΔF k,k = ( F k,k ΔF k F k,k )ΔF k,k Error Propagation in Chains F k F k- L k θ k ΔF k = ( F k F k )ΔF k F k,k ΔF k,k = ( F k,k ΔF k F k,k )ΔF k,k Note: This is same as what we could have obtained by substituting in formulas from the error from frame composition slides given earlier. ΔR k = ( R k,k ΔR k R k,k )ΔR k,k ( R k,k ( I+ skew( α k ))R k,k )( I+ skew( α k,k )) I+ ( R k,k skew( α k )R k,k )+ skew( α k,k ) = I+ skew(r αk k,k + α k,k ) 40

41 Eercise Suppose that you have Δ Rk, k = ΔR( ak ) I + skew( ak ) Δ p = e k, k k θ 3 F 3 Work out approimate formulas for [ ΔR 3, Δp in terms of L, r θ a and e. You should k k k k, k, k k come up with a formula that is linear in L, a, and e F k. F 2 3 ] L 2 θ 2 L 3 F 0 L θ Eercise Suppose we want to know error in F = F F 0,3 0 3 F = F F F F F 0, ,,2 2,3 F = F F F * * * * 0,3 0,,2 2,3 F Δ F = F F F * * * 0,3 0,3 0,,2 2,3 Δ F3 = F0,3 F0, ΔF0, F,2 ΔF,2 F2,3 ΔF2,3 = F2,3 F,2 F0, F0, ΔF0, F,2 ΔF,2 F2,3 ΔF2,3 = F2,3 F,2 ΔF0, F,2 ΔF,2 F2,3 ΔF2,3 F F 2 L 2 θ 2 θ 3 L 3 F 3 Now substitute and simplify F 0 L θ 4

42 Another Eample F B F TU F BU p Tf p Uf p Bf Another Eample p = F p Tf TU Uf FTU = FB FBU = [ RB RBU, RB pbu + pb ] p = R R p + R p + p Tf B BU Uf B BU B Also p Tf = F B p Bf p Bf = F BU p Uf F TU p Tf F B p Uf F BU p Bf = R BU p Uf + p BU p Tf = R B R BU p Uf + R B p BU + p B 42

43 Another Eample Suppose that the track body to US calibration is not perfect F BU * = F BU ΔF BU = [R BU ΔR BU,R BU Δ p BU + p BU ] F B p Tf + Δ p Tf F BU ΔF BU p Bf * = F BU * p Uf I.e., p Bf + Δ p BF = F BU ΔF BU puf p Uf Δ p Bf = F BU ΔF BU puf p Bf ( ) = F BU (ΔR BU puf + Δ p BU ) R BU puf + p BU = R BU ΔR puf BU + R BU Δ p BU + p BU R puf BU p BU = R BU ΔR puf BU + R BU Δ p BU R puf BU Another Eample Continuing Δ p Bf = R BU ΔR BU puf + R BU Δ p BU R BU puf R BU ( I + skew( α BU )) p Uf + R BU Δ p BU R puf BU = R BU puf + R BU α BU p Uf + R BU Δ p BU R BU puf = R BU α BU p Uf + R BU Δ p BU = R BU p Uf α BU + R BU Δ p BU = R BU skew( p Uf ) α BU + R BU Δ p BU 43

44 ΔF B p Tf + Δ p Tf = F B ΔF B Another Eample pbf + Δ p Bf ( ) Δ p Tf = F B ΔF pbf B + Δ p ( Bf ) F pbf B pbf + Δ p ( Bf ) = ΔR pbf B ( + Δ p Bf ) + Δ p B ( I + skew( α B ))( p Bf + Δ p Bf ) + Δ p B = ( p Bf + Δ p Bf ) + α B p Bf + α B Δ p Bf + Δ p B p Bf + Δ p Bf + α B p Bf + Δ p B Δ p Tf F pbf B + Δ p Bf + α B p Bf + Δ p B ( ) F pbf B ( ) + p B ( R pbf B + p B ) ( ) = R B pbf + Δ p Bf + α B p Bf + Δ p B = R B Δ p Bf + α B p Bf + Δ p B Δ p Bf R BU skew( p BU ) α BU + R BU Δ p BU Δ p Tf R B ( Δ p Bf + α B p Bf + Δ p B ) Another Eample Δ p Bf R BU skew( p BU ) α BU + R BU Δ p BU Δ p Tf R B ( R BU skew( p BU ) α BU + R BU Δ p BU + α B p Bf + Δ p B ) R = B R BU skew( p BU ) α BU + R B R BU Δ p BU +R B skew( p Bf ) α B + R B Δ p B = R B R BU skew( p BU ) R B R BU R B skew( p Bf ) R B α BU Δ p BU α B Δ p B 44

45 Parametric Sensitivity Suppose you have an eplicit formula like L cos( θ) + L2 cos( θ + θ2) + L3 cos( θ + θ2 + θ3) p3 = L sin( θ) + L2 sin( θ + θ2) + L3 sin( θ + θ2 + θ3) 0 and know that the only variation is in parameters like Lk and θ. Then you can estimate the variation in p as a function k of variation in and θ by remembering your calculus. L k k p p ΔL θ L Δθ 3 3 Δp3 3 Grinding this out gives: Parametric Sensitivity p3 p3 ΔL Δp3 L θ Δθ where T L = [ L, L2, L3 ] θ = [ θ, θ, θ ] T 2 3 cos( θ ) cos( θ + θ ) cos( θ + θ + θ ) p3 = sin( θ) sin( θ + θ2) sin( θ + θ2 + θ3) L L sin( θ) L2 sin( θ + θ2) L3 sin( θ + θ2 + θ3) L2 sin( θ + θ2) L3 sin( θ + θ2 + θ3) L3 sin( θ + θ2 + θ3) p3 = L cos( θ) L2 cos( θ θ2) L3 cos( θ θ2 θ3) L2 cos( θ θ2) L3 cos( θ θ2 θ3) L3 cos( θ θ2 θ3) θ

46 More generally Suppose that we have a vector function v = g( q) = [g ( q),...,g m ( q)] T of parameters q = [q,...,q n ]. Then we can estimate the value of v + Δ v = g( q + Δ q) by where v + Δ v g( q) + J G ( q) Δ q J G ( q) = g g g q q j q n " " " g i # g i # g i q q j q n " " " g m g m g m q q j q n 46