How much lower would a sea level compression result of 150 psi be at 5500 ft, e.g. Denver?
02-14-2008, 07:12 AM
#21
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interesting point, didn't think about it that way, but
while I agree that makes since for the calculated portion of it, it doesn't make sense for the gauge reading. I'm willing to wager that most compression test gauges out there do not have a reference for current atmospheric pressure built into them, EG its just a relative pressure gauge just like a standard boost gauge, tire gauge, etc.
02-14-2008, 07:14 AM
#22
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Completely stolen from another racing site..
Enjoy...
This how we calculate the elevation differences. I took this from our Excel sheet we use to design motor configurations.
GP = (CRE^1.2 × AP) - AP
Where:
GP = gauge pressure in PSI
CRE = effective compression ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Now let's correct for elevation.
ECRE = CRE - [(altitude/1000)x 0.2]
CGP = (ECRE^1.2 × AP) - AP
Where:
CGP = Corrected gauge pressure in PSI
ECRE = Elevation Corrected Effective Compression Ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Example:
Assume a near stock Compression of 8.395 to 1 using a 133042 cam with ABDC of 65 degrees. That's a CRE of 6.731
GP = (CRE ^ 1.2 x AP) - AP
130.18 psi = (6.73^1.2 x 14.7) -14.7
ECRE = CRE - [(altitude/1000)x 0.2]
5.431 = 6.731-((6500/1000) * .2)
CGP = (ECRE ^ 1.2 x AP) - AP
97.29 psi = (5.431^1.2 x 14.7) -14.7
So the difference between sea level and 6500 is 130.18 - 97.26 = 32.98 psi
This how we calculate the elevation differences. I took this from our Excel sheet we use to design motor configurations.
GP = (CRE^1.2 × AP) - AP
Where:
GP = gauge pressure in PSI
CRE = effective compression ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Now let's correct for elevation.
ECRE = CRE - [(altitude/1000)x 0.2]
CGP = (ECRE^1.2 × AP) - AP
Where:
CGP = Corrected gauge pressure in PSI
ECRE = Elevation Corrected Effective Compression Ratio
AP = atmospheric pressure 14.7psi @ SeaLevel
Example:
Assume a near stock Compression of 8.395 to 1 using a 133042 cam with ABDC of 65 degrees. That's a CRE of 6.731
GP = (CRE ^ 1.2 x AP) - AP
130.18 psi = (6.73^1.2 x 14.7) -14.7
ECRE = CRE - [(altitude/1000)x 0.2]
5.431 = 6.731-((6500/1000) * .2)
CGP = (ECRE ^ 1.2 x AP) - AP
97.29 psi = (5.431^1.2 x 14.7) -14.7
So the difference between sea level and 6500 is 130.18 - 97.26 = 32.98 psi
02-14-2008, 07:36 AM
#29
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irrelevant for the point you're trying to make though
if I get the point you're trying to make anyway...
heat generated by compression doesn't really have anything to do with altitude.
heat generated by compression doesn't really have anything to do with altitude.
02-14-2008, 07:38 AM
#30
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You need to know the atmospheric pressure at both locations to calculate the difference
If you assume that the test was performed at the standard pressure of 14.7 psi you still need to know the atmosphereic pressure at 5500 ft.