Introduction to Robotics Kinematics

Link Description

Kinematics Function of a link Link length Link twist

What are the kinematics functions of this link? a = 7  = 45 0

Describe the connection between two links Link offset d Joint angle 

Summary of the link parameters in terms of link frames. ai = the distance from Zi to Zi+1 measured along Xi  i = the angle between Zi and Zi+1 measured about Xi di = the distance from Xi-1 to Xi measured along Zi  i = the angle between Xi-1 and Xi measured about Zi We usually choose ai > 0 since it corresponds to a distance; However,  i, di,  i are signed quantities.

There is no unique attachment of frames to links: 1. When we align Zi axis with joint axis i, two choices of the Zi direction. 2. When we have intersecting joint axes (ai=0), two choices of the Xi direction, corresponding to choice of signs for the normal to the plane containing Zi and Zi When axes i and i+1 are parallel, the choice of origin location for {i} is arbitrary (generally chosen in order to cause di to be zero).

Three link Arm (RRR) Schematic Parallel axes Find coordinate systems and a, , d,  of all the three accesses

z is overlapping the joint’s axis x is perpendicular to the two joint’s axis y is …?  0 =  1 =  2 = 0 a 0 = 0; a 1 = L1; a 2 = L2 d 1 = d 2 = d 3 = 0  i =  i

Three link Arm : RPR mechanism “Cylindrical” robot – 2 joints analogous to polar coordinates when viewed from above. Schematic: point – axes intersection; prismatic joint at minimal extension Find coordinate systems and a, , d,  (i=3)

 0 = 0;  1 = 90;  2 = 0 a 0 = 0; a 1 = 0; a 2 = 0 d 1 = 0; d 2 = d 2 ; d 3 = L 2 ;  1 =  1;  2 = 0 ;  3 =  3;

Schematic RRR Parallel / Intersect (orthogonal) axes Find coordinate systems and a, , d,  of all joints Two possible frame assignments and corresponding parameters for the two possible choices of Z and X directions.

 1 = -90;  2 = 0  1 = 90;  2 = 0 a 1 = 0; a 2 = L 2 d 1 = 0; d 2 = L 1 d 1 = 0; d 2 = -L 1  ­ 1 =  1 ;  2 = -90+  2  ­ 1 =  1 ;  2 = 90+  2 Option 1

 1 = 90;  2 = 0  1 = -90;  2 = 0 a 1 = 0; a 2 = L 2 d 1 = 0; d 2 = L 1 ;d 1 = 0; d 2 = -L 1  ­ 1 =  1 ;  2 = 90+  2  ­ 1 =  1 ;  2 = -90+  2 Option 2

 0 = 0;  1 = 90;  2 = 0 a 0 = 0; a 1 = 0; a 2 = 0 d 1 = 0; d 2 = d 2 ; d 3 = L 2 ;  1 =  1;  2 = 0 ;  3 =  3;