Originally Posted by 350zDCalb

how many times the ambient compression was compressed to get the given number at 5500ft altitude:

we'll do the following arithmitic in inches of mercury conversions (because Gurgen didn't know the psi numbers off hand) normal atmospheric pressure..drop of one inch of mercury for every thousand feet elevation...5.5 inches dropped....minus 29.92 (14.7psi)

29.92-5.5= 24.42 atmospheric pressure in inches of mercury at 5500ft

24.42/29.92 = percent of air at this altitude compared to sea level: .82%

given 82%... convert to psi

.82*14.7 = 12.0psi (this is the amount of atmpsoheric pressure at 5500ft altitude)

compression test here was in 130psi ballpark...to see what these numbers would have been at sea level: 130/12= 10.83 (compression ratio for compression testing purposes--not CR of piston)

14.7 * 10.83= 159.2 ---> that would be MY compression at sea level

leave the other variances to difference in piston to wall clearences, accuracy of the compression gauges, etc..

*** this entire math lesson was brought to you by a conversation with physics/math master Gurgen***

owe, my head hurts!

Originally Posted by GurgenPB

Or, more simply, next time jsut add 22.5% to your number (by multiplying your compression number by 1.225). Above is the long 'derivation' version.

Originally Posted by 350zDCalb

sure, but i wanted to explain how the figure was derived!