15" tire and 10.6lbs wheel

Is it good idea to replace the stock 175/65-14 tire with 195/50-15 tire and 10.6lbs alloy wheel?

why wouldnt it be, your speedo wil be a litle off but that will happen unless you get a tire low profile enough to make the whole wheele teh same size

aka: if you go from 14in to 15 in rims then make the tires 1in shorter to make the rotaton the same

195/ 55 /-15 are virtually the same size as the escorts stock wheel+tire (175/65/14 or 185/65/14 dependign on year)

195/ 50 /-15 are about 2-3 inches smaller than the 175/65/14

whats the largest tire you can fit on the front without modifing the wells? the rear?

whats the largest tire you can fit on the front without modifing the wells? the rear?

Ive seen a couple examples online with 18's if you went with a 215/35-18 they would probably fit, they might look funny though (theyre pretty wide). 205/40-17 is probably as big as youd wanna go, same deal for the rear.

wides good.. more traction.. the higher they are the more use they make of the rpms.. so biggers better even if you gotta replace the speedometer or wich ever.. thanks.

I'd keep the diameter around stock or smaller. The taller tires just slow you down, it makes for taller gearing. It would be like trying to race some one on a 10 speed huffy, but starting it 10th gear, you wont be winning any races like that. The heavery rim doesnt help much either.

Light weight 15" wheel's with 205/50 or 195/55 would be perfect for exceleration.

the zx2's have either:
185/65-14 (maybe its 60 ???)
185/55-15
205/55-15 S/R only

as stock sizes

i suppose you have to balence what you want.. high max speed or high acceleraton.. hm.. im runnin 15s on my car at the moment wiht the alloy rims that come origenal with the car (by the way is there a better bearing for a 93 escort then fords?)

im also running 15s 195/50/15 are the tires that came with the rims, but 55 is closes to stock diameter, i like 195/55/15 best it terms of price and looks otherwise id be running 16s or 17s the tires are just alot more expensive.