On my '97 I measured from the outboard edge of the leaf spring to the
edge of the fender lip (hang string with plumb and measure to spring)
and if I rememeber correctly is was 12-13 inches. Figure half inch
for clearance to the spring and you've got room for a 12" section
width tire!!! The problem is the brake drums are so far outboard you
need a front drive type wheel to make it all work.
As for the front....a local Chevy dealer had a '97 on the used car lot
on one of those 45 degree incline racks. I stopped and poked my head
under there for a look. With the stock 15x7 cast alum. wheels there
is no addtional room to the inboard side because of the spindle. The
factory opt. 15x8 cast wheels add the addtional inch to the outside.
I read that the '98 Dakota R/T will have 255/65/15's on (I'm guessing)
15x8 rims. This I'm sure will fit no problem as the 255's are only
.40 inches wider per side than the currently optional 235/70/15's and
this additional width is above the area of the spindle. The backspace
on the 15x8 inch wheel is approx. 5.5 inches. I say approx. because
the engineering dwg. locates the mounting surface by offset from the
rim centerline. The offset is a positive 1.0 inch to the outboard.
Remember a wheel actual width is approx. 1 inch greater than it's
advertised width due to the measurement is from tire bead to bead.
sean
'97 318, 5spd, 3.55SG
______________________________ Reply Separator _________________________________
Subject: Re: Re[2]: shortening axle housing
Date: 2/13/97 1:19 PM
Quick question on the Wheel widths.
I realize that the 7" offset will bring the outside edge of the Wheel
inside of the Fender edge, but how wide is the inside of the Fender?
What is the biggest size tire (wheel) that the fender will support?
I also assume that these are only for the rear wheels, so my question
also goes for the front, What is the larget tire the fender will allow?
-rICK
'90 4x4 3.9L Auto(A500)
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