I was doing some numbers today and think I found something cool. Frist, assumptions: - The stock s/c lasts 100,000 miles - The 15% reduction s/c pulley is 15% smaller than a 100% (stock) s/c pulley, so 85% of the stocker's size. - A 15% diameter reduction would cause a 15% increase in rotational speed - The 15% reduction pulley causes the s/c to fail 15% faster than stock. Case 1: new car with stock s/c pulley, 100% of stock size. 0 miles, 100% s/c life left, new 50,000 miles, 50% s/c life left, half 100,000 miles, 0% s/c life left, failure Case 2: new car has 15% reduction s/c pulley added immediately, 85% of stock size. 0 miles, 100% s/c life left, new 42,500 miles, 50% s/c life left, half 85,000 miles, 0% s/c life left, failure supercharger life is 85% of the stock life due to the 15% higher rotational speeds causing failure 15% faster. Case 3, my case: Car had 15% pulley added at 27,924 miles, so at that point it had 72% life remaining, based on the 100,000 mile stock life span with the stock size pulley. Working with 72% life remaining, I then found 85% of of that life remaining to account for the accelerated declination effects of the smaller pulley on the 72,000 miles that were theoretically still in the s/c. So that is 72,000 miles*(1-.15). That yielded 61,200 miles remaining if I use the 15% reduction until failure. My car is at about 61,940 miles right now, 34,016 have been on the 15% reduction pulley. (61,940 right now - 27,924 installation = 34,016 mileage on 15% reduction pulley) So that means that I have 61,200(remaining s/c life with 15% reduction pulley) - 34,016(mileage on 15% reduction pulley) = 27,184 miles left until failure. Added to my current odometer, that means my s/c will fail at 89,124 miles. I know that the real world failures on these s/c is far from predictable, but I wanted a basic idea, and now, in true Star Trek fashion, I have the "exact time to failure." Any mathematicians or physicists out there please feel free to pick this apart.