Excuse me if this has been covered already, but I searched and only found this thread: https://honda-tech.com/forums/tech-misc-15/itr-lsd-will-fit-into-99-00-si-trans-1292666/ which still leaves me with some questions. I'm trying to swap an OEM ITR LSD into my 99 Si transmission.

I understand that the bearings need to be out of a GSR/ITR because the inner race is bigger than on the bearings for the Si's diff, which is ok because I'd like to replace the bearings anyway. But what's this about the ring gear? I can't be sure if I need a new ring gear, or new bolts for the ring gear, or both. Is the attachment surface on the ITR ring gear thinner than the Si's, and is that where the clearance is made up? Or is it made up by the bolt head being shorter? Or is it a combination?

I'm trying to make this as OEM and clean and tidy as possibly and certainly don't mind spending a couple extra bucks to avoid having to cut anything up and to make sure that everything works perfectly and reliably.

Thanks!

 

I believe it is the based upon the final drive bolt assembly. B16 (Si) and LS share one family, while ITR /GSR share another family of direct fitment. At least that's what I did when putting in OEM and aftermarket LSDs in my SI and GS-Rs

 

You'll need a B18C final drive (countershaft and ring gear, matching and non 98+ JDM).

 

Cool, thank you. I see a lot of places selling ITR final drives and saying it is a direct fit for a GSR final drive, and I have a GSR trans with a cracked case, would it be possible to use the GSR final drive instead of the ITR? Not second guessing you guys, just wondering.

 

Yes. non 98+ JDM ITR FD is the same as the non-LSD GSR FD.

 

Perfect, thanks so much!