TWO-LINK MANIPULATOR q=[102] is the joint vector • r = [x_y] is the Position of the End Effector • Forward Kinematics Equations: x=a+b=l₁cos0₁ + 12 cos(01 +02) y = c + d = l₁sinė₁ + l2sin(0₁ +02) Trigonometry for beginners - https://www.youtube.com/watch?v=PUBOT aZ7bhA Y-axis y Link 2 Joint 2 Link 1-1 Joint 1 End Effector r= X-axis X For the given Two Link Manipulator (Fig 1), If 11-1m and 12=1m 01-40° and 02-10° Calculate the coordinates of the end effector (x,y)
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- Evaluate the 3-DOF wrist as shown in Figure 2, use the conventional method to determine 1. Linear velocity and 2. Angular velocity NOTE: for JOINT 3 ( 03 ) only Connected to robot Figure 2: Wrist assembly The known position and orientation of the end of the arm point is. [-C,S2C3 + S1S3 C;S2S3 +S1C3 |-S;S2C3 – C,S3 S,S2S3 + C,C3 -C2S3 C,C2 S,C2 S2 °T3=°T;'T2?T3= C2C3 [G 0 S, 0 S, 0 -G 0 °T 1 0 0 1 -S2 0 C, 0° C2 0 S, 0 'T2 1 1 [C3 -S3 0 07 S3 C3 0 0 2T3= 1 0 0 0 1 00010 IIA slider-crank mechanism is required to do the following crank angle-slider position combinations. Crank Angle (Q) Slider Position(S) Position 1 0° 12 cm Position 2 90° 10 cm Position 3 120° 8 cm Once the mechanism is completed, ie r, L, e are found. Draw the mechanism at 0°,90° and 120° crank angles and measure x coordinate of slider, which is 5 , and check if the measured values are 12 cm, 10 cm , 8 cm. (Hint:Square root operation makes this problem difficult.You have to get rid of the square root sign. this we can do by leaving the square root term alone on one side of the equation and then squarring both sides.The figure shows a linkage in one position. What is the instantaneous acceleration of point Pif the link O₂A is rotating CW at 40 rad/sec? 4/5 44 201 0₂ Show Transcribed Text MULTIPLE CHOICE a. 8000 in/sec^2 b. 23,073 in/sec^2 c. 15,282 in/sec^2 d. 9,005.8 in/sec^2
- For the 3-DOF Industrial manipulator arm as shown in Figure 3, determine the joint displacements for known position and orientation of the end of the arm point. 0. 0. Y1A Y2 A Y3 X2 X3 21 22 23 Yo xo Figure 3. 3 DOF RRR Industrial manipulator arm The link transformation matrices are given by C3 -S3 0 a3C3 S3 C3 1 C2 -S2 0 a,C2 C1 S1 -S1 -C1 d1 S2 C2 2T3 %3D T2 1 1 1 1 1 S1 C\C2a3+C1a2 C1C2C3 – C1S2S3 -C¡C2S3 – C1S2S3 S,C2C3 – S1S2S3 -SĄC2S3 – S1 S2C3 -C1 SịC2a3+ Sja2 Szaz + d1 1 OT3 = = °T, 'T2 ?T3 S½C3 + C2S3 - S2S3 + C2C3What is the equivalent root of the system of the figure using the displacement of the block as a generalized coordinateThe figure below shows an offset slider crank linkage. The links lengths are: link2 = a= 100 mm and link3 = b = 600 mm. The offset is c = 190 mm. we need to : 1. determine the maximum horizontal position of the slider B (dmax) and the corresponding input angle 02 2. determine the minimum horizontal position of the slider B (dmin) and the corresponding input angle 02 Y 03 y В R3 R4 R, 04 R2 02 d ► X R1 The maximum horizontal position of the slider is dmax = Choose... + The input angle 02 corresponding to dmax , in degree and measured CCW from X axis, is = Choose... + The minimum horizontal position of the slider is dmin = Choose... + The input angle 02 corresponding to dmin , in degree and measured CcW from X axis, is = Choose.. +
- 1. Given a four-bar linkage with the link lengths L1 = d = 100 mm, L2 = a = 40 mm, L3 = b = 120 mm, L4 =c= 80 mm. For 0 = 60°, m2 = 5 rad/s (CCW), RCA=110 mm and 83 = 30°, find 03, 04, VĄ, V B, V c for open circuit by a) Graphical method b) Instant center method c) Vector loop method. Then compare the obtained results.Evaluate the 3-DOF wrist as shown in Figure 2, use the conventional method to determine 1. Linear velocity and 2. Angular velocity NOTE: for JOINT 3 ( 03 ) only Connected to robot arm Pitch Roll Yaw Figure 2: Wrist assembly The known position and orientation of the end of the arm point is. C, C2 [-C,S2C3 + S¼S3 -S;S2C3 – C1S3 C2C3 C¡S2S3 + S1C3 S4S2S3 + C,C3 -C2S3 01 S,C2 S2 1 °T3=°T¡'T2²T3Problem 3: The RPH robot of Figure 3 is shown in its zero position. Determine the end- effector zero position configuration M, and the screw axes S; in {s}. n. F -inta. Sorowo- Zs {s} Xs L₁ ŷs Lo 0₁5 L2 02 {b} zb 103 pitch h = 0.1 m/rad Ấb ŷb L3 Figure 3: An RPH open chain shown at its zero position. All arrows along/about the joint axes are drawn in the positive direction (i.e., in the direction of increasing joint value). The pitch of the screw joint is 0.1 m/rad, i.e., it advances linearly by 0.1 m for every radian rotated. The link lengths are Lo = 4, L₁= 3, L2= 2, and L3= 1 (figure not drawn to scale).
- Given: N1 = 10, N2 = 20, N3 = 30, N4 = 40 J1 = 1, J2 = 2, J3 = 3, J4 = 4, J5 = 5, D1 = 1, and D2 = 2 Determine: 01(s)/Tl(s) T₁(1) 0₁ (1) CC N₂ D2, J2 N₁ J₁. Di N3 J3 N4 N₂0 JS J41. Find a combination of link lengths where motion of a point on output link is one quarter of a circle. 2. Find the value of all 0, 0, 0, and y in open and close configuration Read the value of link lengths and the input angle 8., then use the formulae given below to calculate the value of unknowns 03, 0, and y K₁ = = K₂= d K2 K3 = a²-b²+c²+d² 2ac A = cos 0₂ - K₁ - K₂ cos 0₂ + K3 B = -2 sin 0₂ C = K₁ (K₂ + 1) cos 02 + K3 -B± √B²-4AC 2A 0412 = 2tan-1 d K₁ = — K5 = c²d²a²-6² 2ab D = cos 0₂ - K₁ - K4 cos 0₂ + K5 E = -2 sin 0₂ FK₁+ (K₁ - 1) cos 02 +K5 0312 2 tan-1 (-E± -E± √E²4DF 2D Y = 04-03finishing the program of the part by using SKY CNC system format R4 2x45° 28 50 60 75 040 032 024