O K. This is the skinny that I got from my roomate who frequents LS1 tech.com.That is where I saw the information originate from.You will need a little fabricating experience to make this happen or if you dont then have to pay a custom shop to do it. I guess basically the center sections are real close to being the same size between a 10 bolt and a 8.8. The guy who did this cut the axle tubes off of his 10 bolt.Cut the axle tubes off of the donor 8.8. The diameter of the two axle tubes is the same "he" said.You must index the 10 bolt axle tubes in the correct position on the 8.8 and then weld them on. That way you have your spring,shock,and panhard rod mounting points the same. I know out here in Phoenix you can pick up a 8.8 for @ $75 from the junk yard.It would be a good starting point. one part I left out was you have to make sure of the measurements between the 2 drums.The torque arm provision is something I don't know about.I don't remeber what the guy did for this.I think competition engineering makes a kit to eliminate the torque arm I have heard. I saw pictures of his 4th gen and it was definetly a 8.8. I have a Mustang and a Camaro,I recognized it as a 8.8.If I ever get the power numbers up on my Camaro where it will need a heavier rear end this is the way I will go. I have a spare 10 bolt laying around all I would need is a donor 8.8.It was also an Integral unit that they did this to.Not the one with a drop out center section. I HAVE NOT DONE THIS BEFORE. But I do believe it is possible for an average guy with some experience could pull off.Just my .02

James

 

For that trouble, I'd go to the 9 bolt or 9 in. Sounds like a PITA to use the 8.8 unless it is so much cheaper which I doubt. Personally if I wanted to make power, I'd go with the 9 in so I could easily change gear rations at the track and later on the way home. That or the 12 bolt. However, the difference in price is not that great from what I am told

 

Badassbowtie400 That combination you just stated is a 352 ??? Here is the math 4.00 x 4.00 = 16, 16 x 0.7854 = 12.5664, 12.5664 x 3.5 = 43.9824, 43.9824 x 8 = 351.8592.

Genereally when some on says 377 they are talking about a destroked 400. 4.125 x 4.125 = 17.015625, 17.015625 x 0.7854 = 13.364071875, 13.364071875 x 3.48 = 46.506970125, 46.596970125 x 8 = 372.055761 or 372 cid. Now take a .030 over 400 (4.155" bore) and you get 377.4871601424 or 377 cid.

Martin

 

Just waiting for a reply.

Martin

 

i have a 350 with a 4.000" bore and a 3.750 stroke and i am running 6.0 rods