The MGA With An Attitude
COMPRESSION RATIO vs PRESSURE PP105
At 09:47 PM 10/11/2006 0400, someone who shall remain anonymous wrote:
"Malcolm Jeffcock wrote:
.... today the readings are 14: 40 psi; 30 psi; 190psi ; 50 psi."
"13:1 compression in #3. ....
.... if you take the pressure indicated on the compression tester and divide it by 14.7 which is atmospheric pressure you have actual compression ratio."
Sorry, but not even close. Gauge indication is not linear with compression. I can't let this bit of misinformation circulate in public. Since compression ratio and test pressure comes up periodically I will expand on the theory a little, hopefully enough to get the basics across. Have patience (and have mercy on me for eating bandwidth with small details).
"If you take volume at bdc and divide it by volume at tdc you get a theoretical compression ratio."
True. But that has not much to do with pressure gauge readings. Modern cams have considerable valve timing overlap, so the I and E valves are never both closed at bottom of stroke. To make a proper comparison between compression test pressure and theoretical compression ratio you have to know the cam profile.
"I'd expect the theoretical compression ratio would be higher than 13:1."
Maybe close to 13:1 (by sheer coincidence). But you can't tell directly from the pressure gauge reading, as the amount of air in the cylinder depends greatly on the cam profile and the amount of valve timing overlap. If you switch a stock cam to a high performance cam while doing nothing else to the engine, you end up with lower gauge reading for the compression test because of increased valve timing overlap.
"It all goes back to PV=nRT."
Correct. And for earth air nR=53.3 (with "normal" atmospheric conditions). But you have the wrong calculation method. You need to use absolute pressure (gauge reading +14.7psi). Also pressure change is not directly proportional to volume change. You need to use the formula for adiabatic expansion of air (which you have noted). When you manipulate the formula to make the temperature factor drop out you get this:
P2/P1 = (V1/V2)^1.41
or
V2/V1 = (P1/P2)^0.71
Refer to any science or engineering handbook.
The carrot "^" indicates exponent (read power of).
When the gauge pressure reading is 190psi, the absolute pressure is 204.7psi.
P1=204.7
P2=14.7
V2/V1 = effective compression ratio
V2/V1 = (204.7/14.7)^0.71 = 6.5/1
SURPRISE !!!
The "1" is the total volume above the piston at top of stroke.
The "6.5" is the effective piston displacement plus the final top volume
Effective displacement is then only 5.5 parts of that 6.5 total.
For 1800cc full displacement is 450cc for one cylinder.
For 13:1 theoretical compression ratio the top volume is 450/12 = 37.5cc
For the above calculated 6.5/1 effective compression ratio, the effective displacement would be 37.5x5.5 = 206cc, which is less than half of the total intake stroke. That would indeed be a radical race cam with lots of valve overlap. So perhaps the compression ratio is not that high, and the effective displacement (at cranking speed) is likely to be a little greater.
I worked these calculations for a 1600 engine (1588cc) with stock 8.3/1 CR and a very optimistic 140 psi gauge pressure at cranking speed. Here the head chamber volume is 38cc, and total top volume is 54.4cc. This works out to have 59% cylinder fill with the stock cam profile (assuming no compression leaks).
If I had a cam with zero valve overlap where both valves close exactly at bottom of stroke, and I could assume no compression leakage, then my 8.3/1 compression ratio might turn up a compression test gauge reading of 276psi with complete cylinder fill. WOW!!!
This is just to stress the fact that you can't assume any exact relationship between test pressure and compression ratio without knowing the cam profile. If you have some idea whether you have a stock cam or a race cam, then you can make a rough guess at the compression ratio from the pressure reading.
