Does anyone know if after using the TRD Springs and getting the 1" drop if that throws the camber of the tires off at all??? Will the tires wear unevenly with the 1" drop. Please let me know if anyone knows??? thanks!!!

 

It should be highly doubted that this would happen

If its from TRD it shouldn't do anythign other than drop the car...

 

Anytime that you change the stock suspension set up you are going to change camber. The springs are shorter therefore your camber will be off. The real question is how much the camber will be off...is it enough to worry about or not...well that is kinda up to you. Yes your tires will wear out faster than if you had left the car alone but no I don't believe that the camber is off enough to be buying tires every two months. So the choice is up to you..whether or not to buy a camber kit. Just because something is from TRD doesn't mean squat...eiback manufactures them and they technically are for off road (racing) use anyways.

 

Cmon guys we all know this goes in the corner carvers section and not in the general section

 

moved...

 

I have had mine for a few months and have not noticed any uneven wear patterns occurring.

 

Got to an alignment shop, they'll tell you whats up. It's always a good idea to get an alignment after suspension work or modifications.

 

When you have them installed at a dealer they do an alignment. To me that says it should be done. But like PolyFM_TC said anytime you lower a car you're gonna change the camber. So I say get the alignment that way you're sure. I mean whats and alignment cost $60 bucks or so and whats a set of tires gonna run you $250+. You ca weigh the options.

 

More like $400 - $600 a set for decent tires! Definitely worth the $60 alignment if you are unsure

 

Thanks for the feedback!!

 

OH you're right! I meant a cheapies for sure!!

 

The TRD Springs do not greatly change suspension alignment. I had my TRD Springs on my car for 13k miles, rotated them front to back once at 7k and they wore flat acrossed the tire patch area.