Originally posted by: Zenmervolt
Originally posted by: TheSlammaLOL...
For every 2000~ft of altitude due to less air density you can lower the octaine by 1... being as Denver is 5280ft it's not quite 3.
Not so for turbocharged/supercharged engines. Because of the forced induction they will have exactly the same cylinder pressures as they have at sea level.
ZV
It's not exactly the same, since the boost is the manifold pressure over the ambient pressure, not absolute pressure. That's the difference between psia (ambient) and psig (gauge).
If it was exactly the same, turbocharged cars wouldn't lose power with altitude, but they do. Not as much as NA cars, though.
Here's a post I wrote about it on a car forum. The thread was on correction factors, and I was showing them how the same correction factor cannot be applied to a turbo car and a NA car. I used a car running 19 psi boost as an example.
NA car= (atmospheric)
Turbo = (atmospheric+boost)
So at sea level you have:
NA car= (14.7) total of 14.7 psi
Turbo= (14.7 + 19) total of 33.7 psi
We're using sea level as the baseline, so at sea level you both have 100% of your power.
Now in Denver, at 5280 feet, the atmospheric pressure is 12.3 psi. So we have:
NA car= (12.3) total of 12.3 psi, 83.6% of what he had at sea level.
Turbo = (12.3+19) total of 31.3 psi, 92.8% of what you had at sea level.
So if you both had 500 hp at sea level, in Denver the NA car now has 418 hp and you now have 464 hp.
Since the boost level is added on top of the atmospheric pressure and remains constant regardless of altitude, the more boost you run the less altitude will affect you.
Quick comparison at Denver's altitude with a NA car, car running 5 psi, a car running 10 psi, and a car running 20 psi:
NA car = (12.3) total of 12.3 psi, 83.6% of sea level
5 psi turbo =(12.3+5) total of 17.3 psi, 87.8% of sea level
10 psi turbo=(12.3+10) total of 22.3 psi, 90.2% of sea level
20 psi turbo=(12.3+20) total of 32.3 psi, 93% of sea level.