Tires are considered bald when one or more of their grooves reaches 2/32 of an inch deep, compared with about 10/32 of an inch for new tires. So, you have 8/32 of usable tread or 1/4". (I found this on the internet concerning tire tread depth and life)
Doing some geometry (circumference = 2 x pi x radius)
Tire X = 225-50-18 with full tread (10/32)...circumference is 84.466" or 26.9" diameter
Tire Y = 225-50-18 with tread at 2/32 (- 1/4" as we are missing 8/32 of tread now)....What is the difference?
Larger tire diameter is 26.9" which is a radius of 13.45". We are subtracting .25" from radius to be 13.20. 13.20 x 2 = 26.4" radius. 26.4" * 3.14 = 82.896" circumference
What is the percentage?
82.896 / 84.466 = .981 x 100 = 98.1%.
So, a 225-50-18 tire new (10/32 tread life) has 1.9% more tread than the one that is worn to replacement (or tread has only 2/32 left)
If the above calculations are correct then how will adding a new tire on one side really throw off the AWD system? It's less than 2%. I'm not trying to say anyone is wrong but I simply can't believe that a car manufacturer would install a system with such strict tolerances from a tire perspective.
This is an interesting topic and why I'm digging up math I have not used since the 70's
