Inverse Kinematics From Position to Angles

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Transcription:

Invere Knemat From Poton to Ange

Invere Knemat Gven a dered poton P & orentaton R o the end-eetor Y,, z, O, A, T z q,, n Fnd the jont varabe whh an brng the robot to the dered onguraton.

Sovabt 0 Gven the numera vaue o we attempt to nd vaue o,,, n. The PUMA 560: T Gven 0 a 6 numera vaue, 6T ove or6 jont ange,,,, 6. equaton and 6 unnown 6 equaton and 6 unnown nonnear, tranendenta equaton

Invere Knemat More dut. The equaton to ove are nonnear thu temat oed-orm outon not awa avaabe. Souton not unque. Redundant robot. Ebow-up/ebow-down onguraton. Robot dependent. outon!

The Worpae Worpae: voume o pae whh an be reahed b the end eetor Detrou worpae: voume o pae where the end eetor an be arbtrar orented Reahabe worpae: voume o pae whh the robot an reah n at eat one orentaton

Two-n manpuator I = reahabe wor pae a d o radu. The detrou Worpae a pont: orgn I there no detrou worpae. The reahabe wor pae a rng o outer radu + and nner radu -

Etene o Souton A outon to the IKP et the target beong to the worpae. Worpae omputaton ma be hard. In prate t made ea b pea degn o the robot.

Method o Souton A manpuator ovabe the jont varabe an be determned b an agorthm. The agorthm houd nd a pobe outon. Souton oed orm outon numera outon

umera Souton Reut n a numera, teratve outon to tem o equaton, or eampe ewton/raphon tehnque. Unnown number o operaton to ove. On return a nge outon. Aura dtated b uer. Beaue o thee reaon, th muh e derabe than a oed-orm outon. Can be apped to a robot.

Coed-orm outon Anata outon to tem o equaton Can be oved n a ed number o operaton thereore, omputatona at/nown peed Reut n a pobe outon to the manpuator nemat Oten dut or mpobe to nd Mot derabe or rea-tme ontro Mot derabe overa

Invere Knemat Probem Gven: Poton & Orentaton Fnd: jont oordnate o ED-EFFECTOR 0 T q, q, q 3,, q eed to ove at mot ndependent equaton n unnown.

Invere Knemat Probem ISSUES Etene o outon Worpae Detrou Worpae Le than 6 jont Jont mt prata Mutpe outon Crtera Agebra Sovabt oed orm numera Geometr number o outon = 6 d, r 0 or pont

Souton To Invere Knemat 0 T = 0 T T T 3 - T = A A A 3 A Gven: 0 T n n nz o o o z a a a z p p A p z θ -θ θ d θ θ θ - θ d θ r Fnd: q = q, q, q 3,, q jont oordnate

Souton To Invere Knemat 3 z z z z...a A A A p a o n p a o n p a o n Equaton 6 ndependent 6 redundant unnown LHS,j = RHS,j row =,, 3 oumn j =,, 3, 4

Souton To Invere Knemat Genera Approah: Ioate one jont varabe at a tme A - 0 T = A A 3 A = T unton o q unton o q,, q Loo or ontant eement n T Equate LHS,j = RHS,j Sove or q

Souton To Invere Knemat A - A -0 T = A 3 A = T unton o q 3,, q unton o q, q on one unnown q ne q ha been oved or Loo or ontant eement o T Equate LHS,j = RHS,j Sove or q Mabe an nd equaton nvovng q on ote: There no agorthm approah that 00% eetve Geometr ntuton requred

A Smpe Eampe Revoute and Prmat Jont Combned, Fndng : θ artan More Spea: θ artan artan pee that t n the rt quadrant Y S Fndng S: X S

Invere Knemat o a Two Ln Manpuator, Gven:,,, Fnd:, Redundan: A unque outon to th probem doe not et. ote, that ung the gven two outon are pobe. Sometme no outon pobe.,

The Geometr Souton,

The Geometr Souton, Ung the Law o Cone: aro θ oθ oθ θ o80 θ o80 o C ab b a Ung the Law o Cone: artan α α θ θ nθ θ n80 nθ n n C b B artan nθ arn θ Redundant ne oud be n the rt or ourth quadrant. Redundan aued ne ha two pobe vaue

The Agebra Souton,

aro θ n n n The Agebra Souton, θ θ θ 3 n θ o θ o θ On Unnown n o o n n n n o o o : b a b a b a b a b a b a ote

n o n o n n n o o o : a b b a b a b a b a b a ote n We now what rom the prevou de. We need to ove or. ow we have two equaton and two unnown n and o Subttutng or and mpng man tme ote th the aw o one and an be repaed b + arn θ

Invere Knemat Anata outon on wor or a ar mpe truture umera/teratve outon needed or a ompe truture

umera Approahe Invere nemat an be ormuated a an optmzaton probem

Funton Optmzaton Fndng the mnmum or nonnear unton

Formuaton So how to onvert the IK proe n an optmzaton unton? arg mn F θ θ θ C=C,C Bae 0,0

Iteratve Approahe Fnd the jont ange θ that mnmze the dtane between the hpothezed harater poton and uer peed poton arg mn C hpothezed poton peed poton θ θ θ C=C,C Bae 0,0

Iteratve Approahe Fnd the jont ange θ that mnmze the dtane between the hpothezed harater poton and uer peed poton arg mn, o o n n θ θ θ C=, Bae 0,0

Iteratve Approahe Mathemata, we an ormuate th a an optmzaton probem: arg mn The above probem an be oved b man nonnear optmzaton agorthm: - Steepet deent - Gau-newton - Levenberg-marquardt, et

Gradent-baed Optmzaton

Gau-ewton Approah Step : ntaze the jont ange wth 0 Step : update the jont ange:

Gau-ewton Approah Step : ntaze the jont ange wth 0 Step : update the jont ange: How an we dede the amount o update?

Gau-ewton Approah Step : ntaze the jont ange wth 0 Step : update the jont ange: arg mn

Gau-ewton Approah Step : ntaze the jont ange wth 0 Step : update the jont ange: arg mn Known!

Gau-ewton Approah Step : ntaze the jont ange wth 0 Step : update the jont ange: arg mn Taor ere epanon

Gau-ewton Approah Step : ntaze the jont ange wth Step : update the jont ange: mn arg mn arg 0 Taor ere epanon rearrange

Gau-ewton Approah Step : ntaze the jont ange wth Step : update the jont ange: mn arg mn arg 0 Taor ere epanon rearrange Can ou ove th optmzaton probem?

Gau-ewton Approah Step : ntaze the jont ange wth Step : update the jont ange: mn arg mn arg mn arg 0 Taor ere epanon rearrange Th a quadrat unton o

Gau-ewton Approah Optmzng an quadrat unton ea It ha an optma vaue when the gradent zero,...,,...,,...,...... M M M mn arg

Gau-ewton Approah Optmzng an quadrat unton ea It ha an optma vaue when the gradent zero,...,,...,,...,...... M M M mn arg b J Δθ Lnear equaton!

Gau-ewton Approah Optmzng an quadrat unton ea It ha an optma vaue when the gradent zero,...,,...,,...,...... M M M mn arg b J Δθ

Gau-ewton Approah Optmzng an quadrat unton ea It ha an optma vaue when the gradent zero,...,,...,,...,...... M M M mn arg b J Δθ b J J J T T