Tippyman

I have an EF I am making a custom traction bar (radius rods) for.

An EG/EK chassis has compliance bushings and no radius rod, so a traction bar is just made to be a little snug off of the control arm and you're good to go.

On an EF, the "traction bar" (aka radius rod) is actually locating the arm where it needs to be for caster and whatnot.

How do I determine how long to make this rod? Measure from the lower control arm of a stock car to some static point in the front suspension and duplicate it?

hondamark35

most that are commercially available (traction bars) have adjustable length radius rods by way of a threaded rod-end and jam nut at both ends. this allows the caster to be set properly at the alignment shop.

Tippyman

Mine will have that on one end. However one heim joint only gives you so much adjustment. How should I get it as close as possible?

spAdam

iTrader: (1)

If you use a rod end at both ends (one threaded left-handed) you can effectively double the amount of thread length and ease adjustment. Add some wrench flats to the arms and they will turn just like a steering tie rod. You should be able to mock up before you commit to your arm length to get yourself measured up for arm length. The down side would be doubling the slop by doubling the rod ends (as they wear).

Actually now that I think about how the EF arms are setup, I'm not sure that's the best solution, but you could still use threads at both ends of the arm to ease adjustment.

Tippyman

The problem with EF's is the lca has no locating arm. The radius rod (what some would call a traction bar) IS the locating arm. So I need to keep the LCA side of the bar solidly bolted to avoid twisting. I can mock it all up to look good in the wheel well, but how can I know it's good before cutting/welding?

I didn't know if there was some way to tell where the lower control arm was supposed to be. I have a stock EF parts car. I guess I'll just measure the lca to a point on the subframe/rack thats equal to both cars and emulate that. That's all I can think.