Problem 8.9

In the textbook consider equations 8.7, 8.10 and 8.11, now substituting in equation 8.13 we have

TLL =

[ 1, 0, L2

0, 1, -L3

0, 0, 1 ] ,

For case a) substituting for L2 and L3 we have TLL =

TLL =

[ 1, 0, 1

0, 1, -2

0, 0, 1 ]

Also,

[ X Y 1]_ee = TLL [x₁ y₁ 1 ] (A)

Since the co-ordinate system is located at J1 we have [x₁ y₁ 1 ] = [0 0 1], multiplying with TLL we have EE coordinates to be [ 1 -2 1 ];

For case b) we have the EE coordinates to be [ 1 2 1 ], cross multiplying both sides of equation (A) by the inverse of TLL and solving we get L2 = 1 and L3 = -2 (or the direction of the joint is on the other side now.

Problem 8.10)

Again consider equation 8.22 from the textbook we have ,

[ 1, 0, l2*cos(the)+l3*cos(beta)]

[ 0, 1, l2*sin(the)+l3*sin(beta) ]

[ 0, 0, 1 ]

Where beta = theta – alpha

Case a) Solving for link lengths l2 = 0.5 and l3 = 1.0 and alpha = beta = 45 deg and given that the origin is at J1

We have

X_ee = l2*cos(the)+l3*cos(-alp+the) = 1.3536

Y_ee = l2*sin(the)-l3*sin(-alp+the) =  0.3536

Case b) given [ X Y 1]_ee = [1.414 , 0.5] , again multiplying both sides by inverse of T_rr

707/500+l2*cos(the)+l3*cos(beta) = 0

1/2+l2*sin(the)-l3*sin(beta) = 0

The above given simultaneous equation can then be solved to obtain values for theta and beta.

There will be two possible solutions

Beta = -.328 or -.325

Theta = .363 or .316