SOLVED PROBLEMS SET 3

Transcription

1 SOLVED PROBLEMS SET PROBLEM During tsk performed by robotic mnipultor, its end-effector is required to be oriented with respect to the bse frme s described below The pproch vector is oriented by two ngles s u = u cosα cos β + u sinαcos β + u sin β, where α is the zimuth ngle nd β is the elevtion ngle The side vector is required to remin horizontl so tht u = u sinα + u cosα s ) Show tht the norml vector ( u n ) is expressed s follows in this tsk u = u cosαsin β + u sinα sin β u cos β It is strightforwrd n b) Express the ngulr velocity ω of the end-effector in the bse frme in terms of α, β, nd their rtes du 0 s / dt= u αcosα u αsin α, du 0 s / dt= ω us = [ ωu + ωu + ωu ] [ u sinα + u cos α], du 0 s / dt= u ( ωcos α) u ( ωsin α) + u ( ωsinα + ωcos α) These equtions imply tht ω = α, ω sinα + ω cosα = 0 See below tht the second eqution is utomticlly stisfied by ω nd ω Similrly, du 0 / dt= u ( αsαcβ βcαsβ) + u ( αcαcβ βsαsβ) + u ( βcβ), du 0 / dt= ω u = [ ωu + ωu + ωu ] [ u cαcβ + u sαcβ + u sβ], du 0 / dt= u ( ωsβ ωsαcβ) + u ( ωcαcβ ωsβ) + u ( ωsα ωcα) cβ Reclling tht ω = α, these equtions imply tht ω = βsinα, ω = βcosα; u n u u s M K Özgören

2 ω sα ω cα = β This third eqution is lso utomticlly stisfied by ω nd ω Hence, the ngulr velocity is obtined s ω = u βsinα u βcosα + u α c) Using the results of prt (b), express lso the ngulr ccelertion α of the end-effector in the bse frme in terms of α, β, nd their first nd second rtes α = d0ω / dt = d0[ u βsinα u βcos α + u α]/ dt, α = u ( βsinα + αβ cos α) u ( βcosα αβ sin α) + u α PROBLEM () u () u O 0 = O () O O () u () u s 4 u O 4 = O 5 u 5 u O 6 u u () 5 () () u () u For the R PR mnipultor shown in the figure, it hs lredy been shown tht the orienttion of its hnd nd the loction of its wrist point cn be expressed s follows with respect to the bse frme: ˆ u u u u C e e e e = 5 6; = +, 5 = 5 + π, 6 = 6 + π / u u r = e e u + u s + [ ( cos ) ( sin )] 4 Show tht the ngulr velocity of the hnd (ω ) nd the liner velocity of the wrist point ( w ) cn be expressed by the following formuls: ω u u ω = e e ; M K Özgören

3 ω = u ( sin + ) u ( + sin ) + u ( cos + cos ) u u = ; w e e w w = u ( cos + cos + s sin ) 4 + u ( cos + s ) + u ( sin + s ) 4 4 SOLUTION Strting from the given Ĉ, ω cn be written directly s u u = e e Let ω u u u u u u 5 u e u 5e e u 6e e e u ω = + + ω Then, 5 u u e u u 5u 6e u ω = + + ω = ( u cos + u sin ) u + u + ( u cos u sin ) ω = u( sin + 5) u( + 6 sin 5 ) + u( cos + 6 cos 5 ) Differentition of the given r leds to u u w e u e = u + u s + [ ( cos ) ( sin )] 4 u u e e u u u s [ ( + cos ) ( + sin )] 4 u u e e u u s + [ ( sin ) ( + cos )] 4 u u u u w e e e u e = u + u s + [ ( cos ) ( sin )] 4 u u e e u u s + [ ( + cos ) + ( + sin )] u Agin, let u w e e w Then, = 4 u u e e u u s [ ( sin ) + ( + cos )] 4 w = ( u cos + u sin )[ u( + cos ) u( s4 + sin )] + u [( + cos ) s cos ] + u [( s + sin ) sin ] 4 4 w = u ( + cos )cos + u ( s4 + sin )sin + u [( + cos ) s cos ] + u [( s + sin ) sin ] 4 4 w = u( cos + cos + s4sin ) + u ( cos + s ) + u ( sin + s ) 4 4 M K Özgören

4 PROBLEM Consider the sme mnipultor introduced in Problem ) Determine the joint velocities (the first derivtives of the joint vribles) corresponding to specified w nd ω t certin position of the mnipultor b) Identify the motion singulrities Illustrte them nd discuss their consequences Hint: As the first step, obtin ω nd w in terms of ω, w, nd the current position Then, proceed with ω nd w SOLUTION ) The given equtions for w nd ω led to the following sclr equtions: ( cos + cos + s4sin ) = w, () cos+ s 4= w, () sin+ s 4= w; () sin + 5 = ω, + 6 sin5 = ω, cos + cos = ω (6) 6 5 If cos + cos + s4sin 0, Eq () gives = w /( cos + cos + s4sin ) With the knowledge of, nd if cos 5 0, Eqs, (6), nd give ω, 5 = sin ω, 6 = ( cos )/cos 5 ( ω sin ) = With the knowledge of, nd if sin 0, Eqs () nd () give, = ( w s4 )/( sin ) s w 4 = cos + b) As noted bove, this mnipultor hs three kinds of motion singulrities: * The first kind of singulrity occurs if cos + cos + s 4 sin = 0 This eqution implies tht the wrist point lies on the xis of the first joint In this kind of singulrity, becomes rbitrry s n extr joint spce freedom but w vnishes s tsk spce restriction 4 M K Özgören

5 * The second kind of singulrity occurs if cos5 = cos( 5 + π) = 0, ie if 5 =± π / This implies tht the pproch vector becomes prllel to the prllel xes of the second nd third joints Consequently, nd 6 become rbitrry within the combintion ± 6 = ω s n extr joint spce freedom but ω, becoming ω = cos, loses its independence s tsk spce restriction * The third kind of singulrity occurs if sin = 0, ie if = 0 or =± π This implies tht OO 4 becomes orthogonl to ligned OO nd OO Thus, both nd s 4 drive the wrist point in the sme direction nd therefore they become undistinguishble In this kind of singulrity, Eq () implies tht w = s 4, ie w loses its independence s tsk spce restriction On the other hnd, Eq () implies tht s 4 ± = w, ie s 4 nd become rbitrry within the indicted combintion s n extr joint spce freedom PROBLEM 4 () u u 5 u () u u () 4 () s () u () u The figure shows the sketch of Unimte Robot The orienttion of its hnd nd the loction of its wrist point hs lredy been expressed s follows with respect to the bse frme ˆ ˆ(0,6) u u 4 u5 u6 u π / C = C = e e e e e, u u u 5 r = r = e e s u + u + d e 4u [ ( ) ]; = π /, 4 = + 4, 6 = 6 + π / ) Show tht the ngulr velocity of its hnd nd the liner velocity of its wrist point cn be expressed in the bse frme s follows: u u u 4 u u 4 u u e u e e u e e e 5u, ω = + 5 M K Özgören

6 u u u u u se + e u w= e [ u ( d c + s c s ) + e ( s u u ) + d e u ] It is strightforwrd b) Referring to the equtions given bove, identify the velocity influence coefficients tht constitute the wrist point Jcobin mtrix of the mnipultor u u u 4 u u 4 u 5 =, = 4 =, 5 =, 6 = J u J J e u J e e u J e e e u u u u u 4 r = r = + 5 u u u u J 4 r = e e u, Jr4 = d5e e u, Jr5 = Jr6 = 0 J e u ( d c s c s ), J e [ e ( s u u ) d e u ], PROBLEM 5 Consider the sme mnipultor introduced in Problem 4 ) Derive expressions to determine the rtes of the joint vribles corresponding to specified motion of the hnd b) Indicte nd discuss the motion singulrities s well ) Inverse Kinemtics : Let's re-write the wrist point velocity eqution s u e w= u ( d5c4 + sc s ) u u ( ) 4 u e s u u d e u s e u Let's pre-multiply both sides by t u : 5 4 u ue w wc ws ( dc sc s ) t = = Hence, if dc sc s 0, wc ws = dc sc s With known, let's write the ngulr velocity eqution s u u + u e 5u = ω, where ω is known s Let's write this eqution further s u u ( ) ω = e 4e ω u ( + c ) u + u u s = ω M K Özgören

7 Hence, if sin5 0, 5 = ω, 6 = ω /sin 5, 4 ( 6 cos 5 ) ω ω /tn 5 ω = + = With known nd 4, let's write the wrist point velocity eqution this time s Here, w is known s ( su u) + su = ( s ) u + ( s ) u = w u u u = ( ) 54 w e e w u d c s c s d e 4u Hence, noting tht s > 0 lwys, = w/ s, s = w + Finlly, 4 = 4 b) Motion Singulrities: First kind of motion singulrity occurs if dc + sc s = This eqution implies tht the wrist point becomes locted on the xis of the first joint In this kind of singulrity, becomes rbitrry while the wrist point motion becomes restricted so tht wc = or w = wtn ws 0 Second kind of motion singulrity occurs if sin 5 = 0 In such configurtion, the xis of joint 6 becomes prllel to the xes of joints nd 4 Then, 6 nd 4 become rbitrry However, their sum or difference cn still be determined s = ; σ 5 5 σ ω = cos =± The corresponding restriction in the hnd's ngulr motion is ω = 0 7 M K Özgören