Question

RX3-Y 4Yw=20-Yı 1999 Xe 02 EE Problem 2: (4.5 points) Consider the PRP manipulator as shown in Fig 1. The angle and link length variables are 9, x. 92, and 93. a. Draw all the co-ordinate frames and write out the DH parameter table. b. Find the transformations for each row of the DH table, and compute the overall transformation. c. Derive the full 6 by n Jacobian matrix J9). 1191 - Figure 1

Answer 1

Solution

Joint 3 (P) share 4Yw=20=Y1 ZX to 93 Joint 2 (R) F EE JU 19, Figure 1 Joint 1 (P) X=Xo toko

Frame Representation

93 13 72 ELL you {2} LE? - 90 --- -1 X 203 - 12 mg 1B - Samo -- Same origin trto >10 -K {o}

Using the frame assignment in the diagram above, we can obtain DH parameters as shown in the below,

Oil dil dil dil o a o 9001 9,790 olqz 1010 olo

Forward Kinematics

Using general transformation matrix,

cos 8; – sin8,cosa; sin 8, sind; al cos, Tri - 1) = sino, cos,cos; -cos,sino, a,sino, O sinal cosa, di 10 001

- 1 o o o 7 | o o -1 o 1 оч, ороо ороо оноо тя оо — ооо - 1t= |с, | о #1, 2 Г1 o o o | 0 1 o o По о 143

And final transformation matrix will be,

°, - °, , , [-ѕ, oc, 93 | “Т, - о -1 o o la o Sz Qq+5293 o o o 1

Jacobian Matrix

Using general Jacobian matrix,

Prismatic Revolute Linear x (du - d Rotational Ri-1

o o o о он E) ( рас ер (3) o o o . о о

। 0 0 0 0 ( 0 01 : S. 0 97 = 0 0 - 1 1 0 -1 0 | 0 1 0 | la o Sz] Ko | -5 - W 3

rol J- OOOOO gorooo

3 0 - 0 0 0 -S243 C2+z 0 0 - 0 0 0 4 0 | 0 0 0 1

Therefore this is the 6 by 3 Jacobian matrix for the given robotic manipulator.