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Custom driveshafts

Mr_Roboto

Doing the jobs nobody wants to
TCG Premium
Feb 4, 2012
25,665
30,537
Nashotah, Wisconsin (AKA not Illinois)
ok, so this is what I came up with. First off, I took some measurements. I took the height difference from the trans yoke to the 9" yoke as compared to the output of the under drive unit. I got 4 1/8 inch difference. Putting a plum bob on the rear yoke and measuring to the front one I get 39.375. At this point, I'm going to throw out some info:

1)Assuming I get the yoke, I get 40.375 because the yoke being shorter thus increasing the spacing.

2)We know that the distance between the output and the axle yoke will stay pretty much the same. Even though it's live axle, it should be fairly well located.

3)I can tell the worse case scenario that the output yoke goes above the height of the pinion yoke, which means that one leg of the triangle becomes zero. What this means is the distance between the two yokes is ultimately going to be the distance between them. As it goes back up it forms a new triangle, and the distance between them will get progressively longer again the further higher than the output yoke the axle yoke goes.

4)Due to Mr. Pythagoras having a theorem that says X^2+Y^2=Z^2 referring to the lengths of the sides of a triangle, we can break this down into a mathematical problem.

My margin is 5/8 inch right now assuming I got a short yoke again.

40.375^2+4.125^2=Z^2 for the unloaded, fully extended length.
1630.140625+17.015625=1647.15625=Z^2
sqrt(1647.15625)=Z=40.585172785144083947072682640252
40.5851727-40.375=0.2101

Based on this math, I should have over 100% margin on 5/8 inch.
 
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