Mechanical Engineering Matlab Exercises Part 2

Matlab Exercise 5 а) solution, parts 1. thro11gh 5. Joint rates Joint angles Towards the end of the simulated time, this roЬot is approaching the elЬow-straight singularity. If the simulation time goes Ьеуопd 5 sec, the Jacobian matrix determinant will approach zero, and the joint rates will approach infinity. For this exercise, this occurs after t==5.4 sec, to the nearest one-tenth second. The elЬow-straight singularity corresponds to а loss in опе translational freedom, in the direction of links 1 and 2 aligned in а straight line. Ь) Check your JacoЬian matrix results for the initial and fiлal joint angle sets. % Exercise 5 using the Corke Matlab Robotics Toolbox % Or. ВоЬ Williams, Ohio University, 2/2002 clc; сlеаг; % Constants OR = pi/180; L 1 = 4; L2 = З; LЗ = 2; % ОН parameters alp(1) = О; а(1) = О: d(1) = О; th(1) = О; alp(2) = О; а(2) = L 1; d{2) = О; th{2) = О; аlр(З) = О; а{З) = L2; d(З) = О; th{З) = О; Ц1} = link([alp(1 ),а{1 ),th{1 ),d{1 ),0],'mod'); Ц2} = link{[alp(2),a{2),th{2),d{2),0],'mod'); ЦЗ} = link([alp(З),a{З),th(З),d(З),0],'mod'); % R joints and Craig ОН convention ThreeR = robot{L, 'PlanЗR'); % Сгеаtе ЭR гоЬоt object q0 = [10 20 ЗО]'ОR; qf = (12.8415.84 -27.12]"0R; % lnitial angles % Final angles (from my Matlab solution) % Jacoblan {3} wr1 {О}, expressed in {О] J_00 = jacob0{ТhreeR,q0); J00 = J_00(1 :2,:); JOO = [JOO;J_00(6,:)]; J_fO = jacobO{ThreeR,qf); JfO = J_f0(1 :2,:); JfO = [JfO;J_f0(6,:)]; lnitial Extract ЗхЭ matrix Final Extract ЭхЗ matrix Using this Matlab program, the following JacoЬian matrices res11lt: -2.19 -1.so о Initial: 0 J 3 = [ 6.54 2.60 О] 1 1 1 о] -2.33 -1.44 Final: 0 J 3 = [ 6.�3 2.63 О 1 1 Note the Corke toolЬox results iл the JacoЬian matrix relating the motion of {3} with respect to {О}, expressed in {О} coordinates. For the Jacoblan matrix expressed in {3} coordinates, use function jacobn() instead of jacobO(). The аЬоvе matrices do not correspond to the symЬolic Jacoblan matrix given in the proЫem statement since that \Vas for motion of {Н} with respect to {О}, expressed in {О} coordinates; the rate equations for the Corke results are (motion of {3}, not {Н}): So, when using the Corke toolЬox JacoЬian matrices, one must first numerically transform Cartesian velocities and force/moment vectors from {Н} to {3} using the rigid-Ьody velocity and static force transformations given in the Ьооk (Section 5.11). Matlab Exercise 6А solution, parts 1. through 5. Joint Angles time (sec) Witfюttt Gravity Effect Towards the end of the simulated time, this гоЬоt is approaching the elЬow-straight singularity. If the simulation time goes Ьeyond 1 sec, the joint rates, accelerations, and torques will approach infinity. For this exercise, this occurs after t= 1.09 sec, to the nearest one-hundredth second. The аЬоvе joint torque results are for гоЬоt dynamics ,vithout gravity (i.e. the plane of nюtion is normal to gravity). The next plot shows the joint torque results for roЬot dynamics considering gravity (i.e. g is in the negative У direction). AII of the other plots are identical. Joint Torques 250 200 Е 150 100 50 0.2 0.4 0.6 time (sec) With Gravity Effect 1 Matlab Exercise 6В а) For the given roЬot, parameters, and motion condition, the recursive Newton-Euler inverse dynamics solutio11 yields the following required drivi11g joint torques: j т1 853 6 Т= т2] = 1637.2] (Nm) 296.8 Т3 As а matter of i11terest, the required joi11t torques to resist gravity only (i.e. static conditio11, 110 JOlllt velocities or acceleratio11s) are give11 Ьelo,v. These gravity joi11t torques may Ье easily verified using FВDs апd methods of statics. The аЬоvе answer gives the total joint torques, i.e. gravity plus dynamics torques. ! Т8 = т82 ''gl] g3 [1847.6] = 495.1 (Nm) 49.1 Ь) Check your а) results with the Corke Matlab RoЬotics ТооlЬох. % Exercise 6В using the Corke Matlab Robotics Toolbox % Dr. ВоЬ Williams, Ohio University, 2/2002 clc; clear; % Constants DR = pi/180; L1 =4; L2=3; L3=2; % DH parameters alp(1)=0; а(1)= О; d(1)=0; th(1)=0; alp(2)=О; а(2)= L1; d(2) = О; th(2)=О; аlр(З)= О; а(З)= L2; d(З) = О; th(З)=О; Ц1}= link([alp(1 ),а(1),th(1 ),d(1),0,20,L 1/2,0,0,О,О,О.5,О,О,О,О, 1,0,0,01); L{2}= link([alp(2),a(2),th(2),d(2),0,15,L2/2,0,0,0,0,0.2,0,0,0,0,1,0,0,0]); ЦЗ}= link([alp(З),a(З),th(З),d(З),0,10,LЗ/2,О,О,О,0,О.1,О,О,О,О, 1,0,0,0]); L{1 }.mdh = 1; L{2}.mdh= 1; L{З}.mdh = 1; % R joints and Craig DH convention % 'help DYN' to see what each term is % Specify modified (Craig) DH convention ThreeR = robot(L, 'PlanЗR'); % Create ЗR robot object grav =[0;9.81 ;О); % То simulate g down, must accelerate entire robot up Ьу same g mag Q = [1 О 20 зoJ•DR; Qd= [1 23); Qdd = [0.5 1 1.5); % Assigned motion conditions tau = me(ТhreeR, [Q Qd Qdd}, grav) % Newton-Euler recursion to calculate total dynamics joint torques tau_G = gravload(ТhreeR,Q,grav) % Calculate gravity joint torques only Using this Matlab program, the same joiлt torques result as given аЬоvе. Note the gravity vector must Ье specified as g in the +У direction as explained in the comment for grav аЬоvе алd also in Chapter 6. Matlab Exercise 6С Forward dynamics solution given initial conditions and constant joint torques: Joint Angles 100 50 • • о 3 -50 • • • • • • 2 time (sec) 4 3 Joint Rates 3 0.8 - - - · · · · · - - · - - - · · · · · .,. · · · · · · · · 0.6 0.2 • • • • • • • • 2 time (sec) 3 4 Exercise 6С Corke Matlab RoЬotics ТооlЬох Program % Exercise БС usiпg the Corke Matlab Robotics ТооlЬох % Dr. ВоЬ Williams, Ohio Uпiversity, 2/2002 clc; clear; % Coпstaпts DR= pi/180; L1 = 4; L2 = З; LЗ = 2; % DH parameters alp(1) = О; а(1) = О; d(1) = О; th(1) = О; alp(2) = О; а(2) = L1; d(2} = О; th(2) = О; al р(З) = О; а(З) = L2; d(З) = О; th(З) = О; L{1} = link([alp(1),a(1),th(1),d(1),0,20,L1/2,0,0,0,0,0.5,0,0,0,0,1,0,0,0]); % R joints апd Craig DH coпvention L{2] = link([alp(2),a(2),th(2),d(2),0,15,L2/2,0.0.00 , ,0.2,0,0,0,0, 1,0,0,0]); % 'help DYN' to see what each term is , , 0,О.1,О,О,О,О,1,0,0,0]); , 0 , 0 ЦЗ} = liпk([alp(З),a(З),th(З),d(З),0,1О,LЗ/20 L{1).mdh = 1; L(2}.mdh= 1; L{З).mdh= 1; ThreeR = roЬot(L, 'РlапЗR'); tO = О; tf = 4; N=40; d1 = (11-IO)/N; 1 = [IO:dt:tf]; ао = [-60 90З0]*DR; Qd0 = [ООО]; [tsim,Q,Qd] = fdyn(ThreeR, % Create ЗА roЬot object % Specify time array % lnitial joint angles and rates ю, tf, 1orqfun', Q0, Qd0); % Solve forward dynamics numerically % torqfun.m Function for ВоЬбс.m, torque profiles for forward dynamics of plaпar ЗR function tau = torqfun(t,Q,Qd) 1au = [20; 5; 1]; % Given constant joint torques This program \vas used 10 generate the reqLJired forward dynamics simLJlation results for the given planar ЗR roЬot. Function 'torqfun' is used to specify the given constant driving joint torques. The initial joint angles and rates are clearly visiЫe in the resulting plots. Matlab Exercise 7 Тhird-orda polynomial. а) Resu\t: 0(t)= 1201 3 -180t 2 +120 0(t)= 3601 2 -3601 0(t)= 7201 - 360 (Note: deg units throughout) i i (t)=720 0.2 0.4 0.6 0.8 1 7210 0.4 time 0.6 0-8 Note: though the Jerk J looks fine (constant of 720, the slope of the А plot), there is an infinite spike (negative) at the start and also at the finish. This is due to the finite change in А with zero change in time at those time locations. То fix this potential proЫem, use а 5 th order polynornial (next page). Ь) Fiftlz-oгder polynomial. 0(1)= -3601 5 + 9001 4 -6001 3 + 120 0(1)= -18001 4 + 36001 3 Result: 0(1) = -72001 3 - 1800r 2 2 + 108001 - 36001 (Note: deg units throughout) 0.2 0.4 time 0.6 0.8 No\V there is а finite jump iл J at the start and finish (the slope of А is non-zero at the start and finish), but there are no longer infinite spikes in J at the start and finish. The cost is а higher-order polynornial. Note the slope of V starts and ends at zего (zero А at the start and finish) as required; this ,vas not the case with the third-order polynornial since it ,vas not required. с) Two third-order polynomials with via point. 02 (t) = 157.5t3 - 225t 2 -22.5t + 120 82 (t)= 472.512 -4501 -22.5 01(t)=-142.5t3 +202.512 +60 0 (r) = -427.512 + 4051 Result: 1 01 (t) = -8551 + 405 02 (t)=945t-450 Though it is difficult to see on the S plot, the slope of S is not flat at the via time; i.e., the velocity is not ° ° zero at this point since it is not required. S peaks at 120.57 (slightly higher than the required 0, 120 ) just Ьefore the via time, so that S is sloping downwards at the via time. The Jerk J has the same potential proЫem as with the з•• order polynomial: there is an infinite spike in J at the start and also at the finish. Since we matched А at the via point (in addition to У), the Jerk has а finite jump at the via point, but no infinite spike. = d) Check the а. and Ь. results with the Corke Matlab RoЬotics Тоо!Ьох. % Exercise 7 using the Corke Matlab Robotics ТооlЬох % Dr. ВоЬ Williams, Ohio University, 2/2002 clc; clear; tf = 1; n = 50; t = [O:tf/n:tf]; qO = 120; qf = 60; % User-defined time array % lnitial and final joint angles, single joint; degree units throughout [q qD qDD] = jtraj(q0, qf, t) ; % Calculate 7 '-order joint trajectory 11 % Plot results figure; subplot(311 ); plot(t,q,'k'); set(gca,'FontSize', 16); grid; axis([0 tf min(q) max(q)]); ylabel('S'); subplot(312); plot(t,qD,'k'); set(gca,'FontSize', 16); grid; axis([0 tf min(qD) max(qD)]); ylaЬel('V'); subplot(313); plot(t,qDD,'k'); set(gca,'FontSize', 16); grid; axis([0 tf min(qDD) max(qDD)]); ylaЬel('A'); Using this Matlab program, the following joint trajectory plots result ote that the toolЬox only gives us access to the angle (S), angular rate (V), and angular acceleration arrays(A) (i.e., no angular jerk). The tоо!Ьох uses а 7•h-order polynomial for joint trajectories, so .the results should compare "' closely, but not exactly, with the Ь) results; the results are different from the а) results due to the З -order polynomial's inability to provide zero angular acceleration at the start and end. The 7•h•order polynomial should alJow for zero jerk at the start and end; ho,vever, this does not appear to Ье the case from the plot since the slope of А appears to Ье non-zero at the start and end.