compression ratio of modern saws

[Plasmech]

What is usually the compression ratio of a modern chainsaw? If we need specifics, Stihls. I know less about 2-cycle than 4-cycle for sure. So 2-cycles usually have a (relatively) very short stroke?

 

[Crofter]

What is usually the compression ratio of a modern chainsaw? If we need specifics, Stihls. I know less about 2-cycle than 4-cycle for sure. So 2-cycles usually have a (relatively) very short stroke?

If you consider the compression starting only after the piston has covered the exhaust port it is very short indeed, so other things have to be considered.

Not that so simple; corrected or calculated from displacement and combustion chamber volume? Static as in cranking or dynamic as under load? However you describe it the operating pressures at ignition will be similar to four strokes. The function of the piston being the valves and such things as port shape and intake efficiency having an effect on dynamic compression make compression ratios a bit vague in importance.

How the engine is ported has a bigger influence on actual combustion pressures than simply knowing the bore and stroke and combustion chamber volume!

 

[J.W Younger]

so if you raise the exh port you could be lowering cyl pressure at one rpm and possibly raising it at anouther or maybe breaking even or not. what if you lower ex port will that lower crankcase pressure and less dense intake charge or what.I t hink baby steps are in order for me, maybe not so much for the O.P

 

[Crofter]

I dont see lowering the exhaust port as directly lowering the crankcase pressure but certainly the density of the charge in the crankcase is one thing that will influence the combustion pressures. Intake port timing would be more directly related to base compression than exhaust port timing I think.
Lots of trade offs can occur in tuning and they vary depending on what operating rpm you consider.

Consider what the effect a tuned pipe will have on dynamic compression.

 

[SilverBox]

The simplest answer is to take compression reading in psi and divide by 14.7.

ie. 150 psi = ~ 10 to 1 compression ratio.

 

[J.W Younger]

THANKS, THAT WILL GIVE SOMETHING TO THINK ABOUT. i KNEW IT WAS COMPLICATED SO JUST FOCUSING ON EXH SIDE . sorry for the caps

 

[J.W Younger]

didn't mean to hi-jack your thread plastech

 

[Plasmech]

didn't mean to hi-jack your thread plastech

No problem. OK well I don't have a compression tester, has anybody tested any Stihl models?

What I'm getting as is trying to figure out if super high octane gas is *really* needed in these saws. Octane does ONLY one thing, and that is lower the chance of pre-det. I've actualy had people tell me it cleans the cylinder and combustion chamber out :monkey:

 

[Lakeside53]

you need 89 or above.... Stihl are around 160-175 new.

 

[Crofter]

No super high octane needed but there has been oodles of threads discussing the pro and con of premium fuels. I think some models are around 160 psi.

What is pre det?

 

[fourfivefour]

What is pre det?

I would assume he means pre ignition. Pre ignition is detonation.
Detonation is the noise you hear in a car when it's pinging or rattling, sometimes heard when lugging down an engine while running cheap gas.
Your probably not going to hear a saw pinging.

If the formula above is true 160/14.7=10.88 to 1 and 175\14.7= 11.9 to 1
In my book that's to much compression for 89 octane especially in an air cooled engine. Unless, they are more efficient at cooling than a liquid cooled engine?
Is it even an accurate comparison, is an air cooled chainsaw less prone to detonation than a liquid cooled auto engine?
A liquid cooled auto engine wouldn't last very long burning 89 octane with that much compression. Running 93 octane and it would be pinging and eventually eating it's self apart (scoring) unless you backed the ignition timing way down. But then it wouldn't run worth a :censored:.

My Jred dealer recommended 91 or above for my new 2165. I was running 93 octane but switched to 110 Sunoco (overkill, I know). I didn't notice any loss or gain of power (didn't check it with a stop watch or tach). If it sits for a while and looses some octane I'll still have enough to be safe.
The best part is that it smells like I'm at the racetrack:greenchainsaw:
That's my .02............454

 

[Mad Professor]

The simplest answer is to take compression reading in psi and divide by 14.7.

ie. 150 psi = ~ 10 to 1 compression ratio.

My 1969 LT1 motor gave 190 psi +/- 3 psi, after decking the block and milling the heads, 13:1. Good thing I ran AV gas.

 

[timberwolf]

The simplest answer is to take compression reading in psi and divide by 14.7.

ie. 150 psi = ~ 10 to 1 compression ratio.

Compression ratio is a measured and calculated number, actual compression does reflect the compression ratio, but there are aspects of trapping efficiency, charge density and ring sealing brought into it by trying to aproximate it this way so it does not really give a reliable reference number.

Two acepted methods are corected and un corrected compression, corrected being the most applicable to a 2 stroke motor.

Run under normal conditions you will not get pre-ignition, pre-detonation, detonation or any such thing what ever you want to call it running 89 oct instead of 91. I have run compression well up over 200psi and run pump gas no problem, might not be able to cut all day like that as HP is getting pretty high and heat output starts to surpass cooling capacity, but that is a thermal issue more than a detonation issue.

Octane requirements change with engine RPM, saws running up over 10,000 to 12,000 RPM where they should don't require near the octane a 2 stroke bike needs when it's asked to pull at 4000 RPM. Also the smaller the bore the the lower the octane requirment.

There are a lot of simmilarities between chainsaws motors and dirt bikes, atvs or sleds, but there are a lot of differences too and as a result a good part of 2 stroke info well accepted for bigger piped motors does not apply so well to smaller box muffler two strokes.

 

[PatrickIreland]

The simplest answer is to take compression reading in psi and divide by 14.7.

ie. 150 psi = ~ 10 to 1 compression ratio.

Not strictly true - My VWs have a ratio of 10:1 and produce around 180-200psi on the gauge. Remember that the air heats up as it is compressed, upping the pressure.
I was always taught for 4-strokes, a rule of thumb was that the reading in PSI should be 16-18 times the compression ratio - for a healthy, usable (not "perfect") engine.

 

[timberwolf]

Good point, also elevation will throw a big curve into it unless it was always checked at sea level and @ 101.325 Kpa.

 

[Crofter]

You cant project compression ratio effects from four stroke engines to two strokes because the piston ported two stroke will seldom trap more than about 75% of a theoretical full charge because of exhaust dilution etc. As TW mentioned the operating RPM of an engine has a big influence on octane demand. So too does the absence of hot exhaust valves as part of the combustion chamber as well as the relative size of the combustion chamber. It is quite right to suggest that those kind of ratios would cause detonation problems in a four inch bore four stroke with peak torque at half the rpm, but that is apples to oranges comparison to a piston ported two stroke saw engine.

 

[SilverBox]

I dug around a little and guage / 14.7 isn't correct. That assumes a constant temp and we don't have that, its an adiabatic process when using air as your gas for the test (it may be a bit different when the fuel/air mix is in the cylinder).

So the correct formula at sea level and standard pressure assuming an ideal gas adiabatic process is:

CR = ((compression gauge reading +14.7)/14.7)^.71

This number will be significantly lower then calculating the swept volume / TDC volume, as there are losses past the rings.

for 150 psi true CR = 5.5 (when testing with and ideal gas in the chamber, which we know an air/fuel mix is not)

The less ideal the gas is the higher the .71 exponent becomes.

This whole discussion I belive was started because Plasmech wanted to know what gas to run. Buy Premium and good mix and don't worry about it till you hear pinging, you won't unless your saw is HIGHLY modified.

 

[timberwolf]

There looks to be an order of ops problem with the formula?

CR = (compression gauge reading +14.7/14.7)^.71 does not work if you substitute 160 psi in for the gauge reeding the answer comes out to 33 and change?

Maybe, but I'm not sure what the 14.7 are intended to represent, I assume one is your see level pressure the other would be actual?

CR = ((compression gauge reading +14.7)/14.7)^.71 this returns 5.8 which would be more plausable. Though still its just a guesstimate as there is no acounting for ring sealing, thermal dynamics, pumping efficiency, charge density...

 

[SilverBox]

There looks to be an order of ops problem with the formula?

CR = (compression gauge reading +14.7/14.7)^.71 does not work if you substitute 160 psi in for the gauge reeding the answer comes out to 33 and change?

Maybe, but I'm not sure what the 14.7 are intended to represent, I assume one is your see level pressure the other would be actual?

CR = ((compression gauge reading +14.7)/14.7)^.71 this returns 5.8 which would be more plausable. Though still its just a guesstimate as there is no acounting for ring sealing, thermal dynamics, pumping efficiency, charge density...

Your right. Adding the 14.7 to the gauge pressure is just to get absolute pressure. You have to add the 14.7 to the gauge reading before you divide by 14.7

 

[maico490]

Timberwolf: 101.325 Kpa.

A number I've been able to recall ever since my engineering degree days.
People look a little confused when you drop it in instead of 1 atmosphere.
(Not of course to be confused with 1 bar which is 100 Kpa)