Problem 2:

Part (a)

dh = {{0, 0, 0, θ1, 0}, {π/2, 0, 0, θ2, 0}, {0, L3, 0, θ3, 0}} ; dh // TraditionalForm

( ) 0 0 ... 2 0 0 θ2 0 0 L3 0 θ3 0

T01 = T[dh, 0, 1] ; T01 // TraditionalForm

( cos(θ1) -sin(θ1) 0 0 ) s ... 0 1 0 0 0 0 1

T12 = T[dh, 1, 2] ; T12 // TraditionalForm

( cos(θ2) -sin(θ2) 0 0 ) 0 ... cos(θ2) 0 0 0 0 0 1

T23 = T[dh, 2, 3] ; T23 // TraditionalForm

( cos(θ3) -sin(θ3) 0 L3 ) s ... 0 1 0 0 0 0 1

Part (b)

T03 = T01 . T12 . T23 // FullSimplify ; T03 // TraditionalForm

( cos(θ1) cos(θ2 + θ3) -cos(θ1) sin(θ2 + θ3) sin(&# ... 0 0 1

For remainder of the inverse kinematics, see solutions text.

Created by Mathematica  (October 21, 2003)