Cincinnati Milacron T3-776 Reverse Analysis

a12 = 0 a34 = 0 a45 = 0 a56 = 0 12 = 90° 23 = 0 34 = 90° S2 = 0 S3 = 0 S5 = 0

Problem Statement

1st Step Obtain and then close-the-loop

Solve for 1 (and thereby 1) vector loop equation: from set 14 of the table of direction cosines X32 -X*32 Z3

The Z component equation is written as Z3 = c34c23 – s34s23c3 Since 23 = 0 and 34 = 90°, Z3 = 0 Equation (11.99) reduces to

Solve for 3 substituting s12=1 and c12=0 into the X and Y components of set 14 gives

square and add these two equations to get

Solve for 2 These two equations represent 2 equations in s2 and c2.

7 6 5 4 3 2 1 Solve for 5

7 6 5 4 3 2 1 Solve for 4

7 6 5 4 3 2 1 Solve for 6

Section 11.4.8 – Geometric Solution