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I was figuring 1,020 hp at the crank with 15% loss through the trans (typical for a th400). If the drivetrain loss was even higher, as you say 25%, then the rwhp would be even lower... (1020 x 0.75) = 765 rwhp. To be at 1,100 rwhp would require 1,300 at the crank with a typical 15% loss through...

That's wild dude!

Thanks for the correction. I'll re-do my math... So a stock coyote at 460 hp is about 920 hp on 14.7 psi of boost. So 18 psi would be somewhere around 1,020 hp at the crank right? So then that would be something like 870 rwhp (going through a th400). So if all that is reasonably accurate, then...

I know, right? Someone help me with this math. I always thought you double the manifold pressure you roughly double the horsepower? So a stock coyote at 460 hp is about 920 hp on 14.7 psi of boost. So 16 psi would be somewhere around 960 hp at the crank right? So then that would be something...

Reason I asked the question is I was curious about the guy in this post https://www.mustang6g.com/forums/threads/new-all-time-best-stock-motor-coyote-record.134190/#post-2763771 who is running 8.20's on all stock motor. I guess he has $35K in the turbo, fuel system, electronics, trans, etc. but...

Just curious, how much rwhp would it take to run 8.20's at 3,800 lbs?