High altitude compression tests
#1
#3
#4
17" at idle isn't bad at all! especially if the needle is pretty steady? i usually only go 4 pulses, 5 at the most. when i see the needle stop climbing i go 1 more pulse but i hit all cylinders the same amount. so there is a 10psi or less difference from the highest to the lowest reading?
#5
I am pretty sure there is no correction factor. Pressures are measured in "absolute" and "guage". If I carry my guage up your mountain in still reads "zero" when I get to the top. The reading on most guages is the difference between the atmospheric side and the pressure side of a diaphram or whatever. A barometer would be an example of an absolute guage, there is some calibrated reference inside that is constant. So you are pulling for 17" of vacuum relative to your atmospheric, if you bring it down to sea level, I am thinking you will still be pulling 17" relative to the higher atmospheric. Ditto for the compression. Hopefully, someone can confirm/deny this line of thinking.
Last edited by Ecuri; 10-01-2005 at 03:20 PM.
#6
i'm thinking there is a correction factor, otherwise racers would not change jetting for altitude. if there were no need for correction, the available air would compress the same at sea level or 6500' and the jetting requirement would remain the same for both. in the realm of compression testing i don't think it plays quite the same role as jetting for altitude though. i recall seeing some correction data but i didn't have any use for it at the time so it kept on going. at 6500' i'd expet the readings to be a little on the low side, how much low though? low readings would tend to "cover" for a borderline engine. with all readings being low, the variance would also be minimized. 10% at 100psi average is about the bottom line for a good/bad engine. could a 15% engine run for a good long time? sure, but you know where you stand, these things usually don't get better by themselves. that's my story and i'm stickin to it. until someone convinces me different?
#7
I think you're right on the compression side. Less air to compress at altitude would lower the guage reading. I would guess the correction could be fiqured with cross multipication, VLS would need to know the barometric pressure where he lives and whatever sea-level is normally.
I'm not sure about the vacuum side, as there is no 'multiplying' going on, but you got me wondering...hmmmm...... (where is Eric when you need him, that guy knows everything!)
I'm not sure about the vacuum side, as there is no 'multiplying' going on, but you got me wondering...hmmmm...... (where is Eric when you need him, that guy knows everything!)
Trending Topics
#8
my assistant has suggested we consider these,
http://www.anver.com/document/vacuum...20pressure.htm
http://www.dragsource.com/index.php?...s&calctoview=8
http://www.anver.com/document/vacuum...20pressure.htm
http://www.dragsource.com/index.php?...s&calctoview=8
#10
Sorry but I don't even come close to knowing everything. Just enuf to get me in trouble most of the time!
Most gages do read what we call PSIG or relative to the surrounding pressure not absolute. However AFAIK like the vacuum link above I would assume there would be a lowerd pressure shown for a compression test but I don't know. It has been too long since I worked any gas law equations.
Most gages do read what we call PSIG or relative to the surrounding pressure not absolute. However AFAIK like the vacuum link above I would assume there would be a lowerd pressure shown for a compression test but I don't know. It has been too long since I worked any gas law equations.
#11
Just my .02 so take it for what its worth...
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 lets 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 lets 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