Engine compression ratio is a volume ratio not a pressure ratio

I ORIGINALLY WROTE: A good place to start is just multiply the compression ratio time the atmospheric pressure to obtain an expected wet reading.

Roughly 6x14.7=88psi. That is the MAXIMUM theoretically possible. Expect it to be less.

*BUT LATER I got to thinking and that's not right. It ignores the heating of the charge as it is compressed. And here the heat causes more pressure. Aw shucks.... and we don't know real piston velocity and we also don't know how much heat is lost to the surroundings as the piston goes up. ugh.... it's gettng ugly. *

So I should have said that "6x14.7=88psi. That is the **MINIMUM** theoretically possible. Expect it to be** MORE**. I'm not sure by how much.

The rule of thumb is that whatever the reading, all of the cylinders shouldn't vary by more than one atmosphere. Even better, they should be within half an atmosphere. **THAT's ***STILL TRUE*

Good Luck,

rScotty

Engine compression ratio is actually a volume ratio not a pressure ratio. The equation I use to calculate the compression pressure is based on the isentropic equation pV^gamma=Constant (eqn 1) where p is the pressure, V is the volume and gamma is the ratio of specific heats of air.

Sparing you all the gory details, the equation for what you see on a compression gauge;

delta p = p1 x [(compression ratio)^gamma -1], (eqn 2)

where p1 is atmospheric pressure, gamma is 1.4 and delta p is the compression gauge pressure

Now the isentropic equation assumes constant mass in the cylinder but because of valve timing the mass in the real engine is not constant. so K is a constant that adjusts the theoretical value of compression gauge pressure to account for this.

Eqn 2 then becomes delta p = K x p1 x [(compression ratio)^gamma -1] (eqn 3)

Gamma =1.4. From experience K is between 0.70 and 0.75.

So a compression ratio 6.0 engine would have compression gauge pressure at an atmospheric pressure of 14.7 psia would be 124.6 psig (K=0.75). This is what you could expect as a maximum for a engine in good mechanical condition.

I hope this helps you.