1. Cincinnati Milacron T3-776 Reverse Analysis

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

5. Problem Statement

6. 1st Step
Obtain and then close-the-loop

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

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

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

11. square and add these two equations to get

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

14. 7 6 5 4 3 2 1
Solve for 5

15. 7 6 5 4 3 2 1
Solve for 4

17. 7 6 5 4 3 2 1
Solve for 6

19. Section 11.4.8 – Geometric Solution