Compression at Elevation
- Thread starter BeeDoo
- Start date
At higher elevation, the air is less dense; therefore, your compression (and horsepower) will drop some as you climb.....
B
BadAftertaste
Well-known member
Don't know of a chart, but it's basically 3% loss of atmosphere for every 1000ft. in elevation.
fwiw-
fwiw-
as a general rule you can figure 0.5psi decrease with 1000ft elevation change (up)... the only problem is the varying temperature/humidity can increase and/or decrease that even more then the actual elevation change.
http://www.grow.arizona.edu/Grow--GrowResources.php?ResourceId=214
Compression doesn't drop, but the total true pressure you get will. Compression in a motor is a ratio, 14:1 or 10:1 or whatever. That doesn't change. But at altitude you start with a lower starting pressure so when you compress it, say 10:1 the final pressure is less.
sled_guy
sled_guy
as a general rule you can figure 0.5psi decrease with 1000ft elevation change (up)... the only problem is the varying temperature/humidity can increase and/or decrease that even more then the actual elevation change.
http://www.grow.arizona.edu/Grow--GrowResources.php?ResourceId=214
You should have said .5psi of atmospheric pressure, dont want people more confused. It might sound like 125lbs of cranking compression and for every 1000 ft of altitude you loose 1/2 of 1 lb, example 122.5 cranking compressio (via a compression tester gauge) at 5000 feet of elevation, which isnt the case!
Same thing SG, You are correct about the ratio's cause they are just that... ratio's, it sounds like BeeDoo is talking about cranking compression.
So, hmm interesting.
Using that calculator.
at 10,000 feet theres ~10 lbs/in^2
at 0 feet theres 14.7 lbs/in^2
Does that mean, if you would compress your intake air pressure to ~15 lbs/in^2 at 10k ft you'd have the same HP as if you were @ 0 feet?
Meaning, 5 lbs of boost @ 10k ft is equal to 0 lbs of boost @ 0 feet, in reference to HP?
Or is it not directly translated that way?
Assuming no deficiency in the 'compression' system.
Example: My 800 Rev has 150 HP @ 0 feet at WOT. My Rev has 110 HP @ 10k feet at WOT. My 800 Rev has 150 HP @ 10k feet at WOT with 5 lbs of boost.
Or am I wrong, and boost numbers are not the same as air pressure numbers?
Just got me thinking some.
On top of that, how do you calc "air pressure" on the fly or in real time... Considering you have barometer and temp variables available?
Using that calculator.
at 10,000 feet theres ~10 lbs/in^2
at 0 feet theres 14.7 lbs/in^2
Does that mean, if you would compress your intake air pressure to ~15 lbs/in^2 at 10k ft you'd have the same HP as if you were @ 0 feet?
Meaning, 5 lbs of boost @ 10k ft is equal to 0 lbs of boost @ 0 feet, in reference to HP?
Or is it not directly translated that way?
Assuming no deficiency in the 'compression' system.
Example: My 800 Rev has 150 HP @ 0 feet at WOT. My Rev has 110 HP @ 10k feet at WOT. My 800 Rev has 150 HP @ 10k feet at WOT with 5 lbs of boost.
Or am I wrong, and boost numbers are not the same as air pressure numbers?
Just got me thinking some.
On top of that, how do you calc "air pressure" on the fly or in real time... Considering you have barometer and temp variables available?
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Here's a chart.http://www.2-stroke-porting.com/altiden.htm
