roll cage

[forked horn]

will a yj cage fit in to a CJ??

[OJTV8CJ]

Depends on the type of cage.  Stock the bars to the windshield frame may be a different connection.

[Mo_Corley]

Stock YJ and CJ roll bars didn't go to the windshield.

I would do some measuring but I think most CJ and YJ stuff is supposed to interchange.

[OJTV8CJ]

Uhhh okay.  All those bars that I have taken out aren't really there?

Disclaimer:  That is not a Jeep that I have worked on or anyone that is affilliated (sp?) with OJT.  Just a picture to illustrate what you don't see going from the windsheild back to the B hoop of the stock cage.  

My name is Brad and I approve this message.  '07

[TanYJ]

Actually, i think the spreader bars have trunkated holes for the fasteners at the windshield.  The mounting plate for the bar actually fits down into the windshield, not just sandwiches to it.  I can take a few shots of my '88 AMC Wrangler (yes, it has AMC all over it) if you need to see how to do it.  I have no idea about the Chrysler Wranglers, though.



You can see the spreader bar in this one.  But to answer your question, yes, it can be done, but it would look like hammered $h..T.  You'd have to drill through the windshield frame of the CJ, or INSERT IGNORE nut-serts, etc.

[Mo_Corley]

Was that on all years of YJ?  I thought they just had a hoop like the CJs.

[TanYJ]

All years of YJ had the spreader bars.

[OldYJ]

yj bars will fit in a cj the spreaders will not unless you drill mountnting hole in the windshield frame. all yj and tj have the spreader(stock that is) cj do not. I have put two yj family cages in cj before.



good one brad

[GURU]

Actually, i think the spreader bars have trunkated holes for the fasteners at the windshield.    I can take a few shots of my '88 AMC Wrangler (yes, it has AMC all over it) if you need to see how to do it.  I have no idea about the Chrysler Wranglers, though.

WORD OF THE DAY BROUGHT TO YOU BY DICK: TRUNKATED

CLOSEST DEFINITION I COULD FIND;

The truncated icosahedron is an Archimedean solid. It comprises 12 regular pentagonal faces, 20 regular hexagonal faces, 60 vertices and 90 edges.

This polyhedron can be constructed from an icosahedron with the 12 vertices truncated (cut off) such that one third of each edge is cut off at each of both ends. This creates 12 new pentagon faces, and leaves the original 20 triangle faces as regular hexagons. Thus the length of the edges is one third of that of the original edges.
 :bowdown:
 :bowdown:
 :bowdown:
 :bowdown:
 :bowdown:
 :bowdown:

[forked horn]

the family stly is what i am going for. It is looking like just building one maybe easier. Db

[TanYJ]

I wasn't aware that i actually knew such a formidable word....... :eyemo: