View Full Version : Q - Compression Ratio to lb/in2 conversion?

05-12-2006, 08:58 AM
Does anyone know a way to convert compression readings in lb/in2 to a compression ratio? I'm sure there must be a simple conversion table, or rule-of-thumb to use, but I've never seen it.

Thanks in advance,


05-12-2006, 10:14 AM
static - don't you just divide by atmospheric (14.7psi)?
or am i missing something obvious.

05-12-2006, 11:22 AM
static - don't you just divide by atmospheric (14.7psi)?
or am i missing something obvious.

[/ QUOTE ]

That would be the logical assumption, but cam timing comes into play (overlap, duration and so forth).
The "trapped time" can differ with different cams.
Also, combustion chamber volume affects this.
For example, my race Spridget (with about 10:1 compression ratio) reads 205 psi on a compression tester.

05-12-2006, 12:24 PM
static - don't you just divide by atmospheric (14.7psi)?
or am i missing something obvious.

[/ QUOTE ]

At 195lb/in2 that wiuld put my TR at 13:1 - seems a bit high. Would be getting pre-ignition at that ratio.

I suspect something closer to 10:1.

Dave Russell
05-12-2006, 12:51 PM
Hi Rob,
It's not possible to accurately convert CR to PSI. Nial is correct. To carry
it a bit further, the following explanation may help.

Question: “If an engine has compression readings around 165 PSI, and
that is taken at face value; that means the compression ratio is
165/14.7 ~ 11.2:1. This is much higher than the published numbers.
That air must get really warm when you ‘squish’ it.”

Answer: That calculation is correct for "isothermal" compression which
assumes that no heat is gained or lost during the compression cycle, and
if the actual compression ratio is the same as the rated ratio.

To complicate the calculation, at low RPM, such as at cranking speeds, the
actual engine compression ratio is never the design calculated ratio
because in reality the piston cannot begin to compress anything until
the intake valve gets closed. If the intake valve were to be completely
closed at BDC -- which by design it never is -- the actual and rated
compression ratios would be the same. At higher RPMs and thus higher
intake gas velocities, the late closing of the intake valve is offset by the
gas inertia in the intake tract and the engine can approach or exceed its
design compression ratio.

The formula for "adiabatic" compression takes heat loss or gain
into consideration. For example, as the air is compressed it is heated
by the compression and the final pressure is higher than would be
predicted by isothermal compression. The variables are rather large and
the difference between the two calculations can vary by a factor of
almost two to one in the final pressure. The pressure rise due to
heating is an "exponential" function (i.e., more heat = more pressure =
more heat = more pressure). The "exponent" for air will vary from one
with complete heat loss to about 1.4 with no heat loss.

For a hypothetical engine with 9:1 compression ratio, assume that the
actual mechanical compression ratio is 6:1 due to late intake valve
closing. We can calculate that the pressure for isothermal compression
would be 14.7 x 6 = 88 PSI. With an adiabatic exponent of 1.14 this
would give a compression pressure of 165 PSI. (88 raised to the 1.14 power)

Whether heat is gained or lost during compression depends on several
things such as how quickly the compression takes place (starter and
engine cranking speed), temperature of the engine during the test (is
the air in the cylinder hotter or cooler than the air compressed in
it?), and a few other variables. The end result is usually that the
measured compression pressure is higher than would be calculated by
isothermal compression.

The compression pressures given at "face value" are empirically
determined from measurements and experience. There are just too many
variables to make accurate calculations.

05-12-2006, 01:17 PM
Wow - Thanks Dave for the very detailed explination. My college thermodynamics is starting to come back to me. I see your point - without knowing the heat transfer rate of the compression chamber you cannot know where between isothermal the and adiababatic compression you are. All I know is that entropy has to be increasing. At least mine is!


05-12-2006, 02:40 PM
Good thread, gents!
How would these numbers compare to measuring the ratio of cylinder/combustion chamber volume at TDC and BDC? Would that ratio be closer to the nominal compression ratio?

05-12-2006, 03:52 PM
Perhaps the easiest way to "solve" for this is if folks who know both their compression ratio and pressure could post them, together with the engine type.

We have 1 entry so far - aeronca65t's Spriget - 10:1 = 205lb

This is the "brute force" approach I know, but might prove quite informative.


Dave Russell
05-12-2006, 05:03 PM
Good thread, gents!
How would these numbers compare to measuring the ratio of cylinder/combustion chamber volume at TDC and BDC? Would that ratio be closer to the nominal compression ratio?

[/ QUOTE ]
The standard calculation for "design" compression ratio is combustion chamber volume at TDC plus cylinder volume at BDC, divided by combustion chamber volume.

The "real, effective" compression ratio at cranking & moderate rpm would be the cylinder volume at the point of intake valve closing plus the chamber volume divided by the chamber volume. Not sure what you mean by "nominal" ratio.

The cam timing is such that the intake valve doesn't actually close until the piston is already traveling back up on the compression stroke. Typical cams will have the intake valve closing somewhere between 30 & 70 degrees after BDC. This is the point where actual compression begins at low engine speeds. The cylinder is actually pumping some of the intake charge back out past the intake valve until it closes.

At higher rpm, the intake charge's inertia keeps the piston from pumping the charge back out even though the valve is still open & the piston is rising. This raises the effective compression ratio to somewhere around the design ratio. The more radical the cam timing, the more rpm is required to prevent this reversal. Larger ports & valves will also raise the rpm at which inertial cylinder filling takes place.

In other words for an engine designed for 9 to 1, the "real - effective - actual" compression ratio may be something like 5 to 1 at cranking speeds & lower rpm. It will approach 9 to 1 at higher rpm as the intake charge inertia gives more complete cylinder filling. The "real or effective" compression ratio for a given cam timing, is a constantly changing value depending on engine rpm, throttle opening, & intake inertia. As said, it also varies greatly with different cam designs.

Hope I haven't confused things more.
PS; Air density decreases about 3.4% for each 1,000 ft of elevation increase, the compression readings would be 10% lower at 3,000 ft than at sea level.

05-13-2006, 12:53 PM
Well, to help the rule of thumb method, mine on 9.3:1 gives 175psi. On 10.8:1 it gives 210psi.

05-13-2006, 05:16 PM
Wow, I like this thread... Should be in ALL the car forums. /ubbthreads/images/graemlins/thumbsup.gif