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Compression Ratios and Forced Induction

Old 09-24-2008, 02:35 PM
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I dont see the logic in that? The higher CR is going to have a more linear power curve and the Low CR high boost is going to have more of a bell curve. That would make a high CR low boost application better for a daily driver/track car and the low CR high boost application better for a drag car.

And Jason the 4% rule as far as ive seen only applies to cars running on atm pressure (14.7). I dont know where you get the 3psi reference but that certainly isnt the case on a honda.

Once again the only definitive way to test the theory is to put it into practice.

So we'll see what the dyno has to say in the next few months.
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Old 09-24-2008, 04:17 PM
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jc rotor, did you see my post 33? Give me your opinion on my math. Is it correct? Or is it close? Do you agree with the 1/3, 1/3, 1/3 rule? Is an 11:1 motor more than 43% efficient at making chemical energy into mechanical energy?
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Old 09-24-2008, 04:27 PM
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Originally Posted by patsmx5
jc rotor, did you see my post 33? Give me your opinion on my math. Is it correct? Or is it close? Do you agree with the 1/3, 1/3, 1/3 rule? Is an 11:1 motor more than 43% efficient at making chemical energy into mechanical energy?
No way. Even very efficient power plants work hard to get that kind of efficiency from coal-fired steam turbines, with very sophisticated heat cycle controls and schemes. They are much more efficient overall at extracting energy than piston engines are.

Think more like 20% for a gas piston engine.
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Old 09-24-2008, 04:30 PM
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Originally Posted by jc_rotor
And Jason the 4% rule as far as ive seen only applies to cars running on atm pressure (14.7).
Thermodynamics. It will be true whether or not the car is FI'ed. Think about it.

If the thermal efficiency of an Otto cycle is say, 20% at 9:1, at 10:1 the output will increase by 4%, for a new thermal efficiency of 20.8%. I mentioned "Otto cycle" in my post. As an M.E. you should have recognized the term.

I dont know where you get the 3psi reference
I pointed out that 8:1 -> 11:1 is a 12% gain. 12% gain on a 15 psi setup is like a 3 psi gain.

Last edited by JasonC SBB; 09-24-2008 at 05:08 PM.
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Old 09-24-2008, 04:33 PM
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Originally Posted by Ajb
I think lower CR is better for the daily driver. You may have less bottom end ...
I pointed out above that my car did NOT lose bottom end going from 9.5:1 to 8.4:1 - dyno measured. And spoolup IMPROVED.
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Old 09-24-2008, 05:09 PM
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jc_rotor,

You seem to think that equivalent pressure at the top of the stroke is all that matters. It isn't. JasonC has pointed out the dramatic difference in temperatures (and hence the likelihood of detonation).

To that, I'll add that the lower compression engine makes more power at the same peak cylinder pressure. Why: The volume/quantity of high pressure gas in the LCHB (low compression, high boost) engine is larger than with the HCLB (high compression, low boost) engine. So, while the chamber pressure may be the same at the top of the stroke, the LCHB engine has more gas to distribute through the cylinder during the power stroke than does the HCLB engine. Using 8:1 vs 11:1, the 8:1 engine will retain roughly 15% more pressure/torque/power mid stroke and 30% more pressure by the time the piston reaches BDC.
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Old 09-24-2008, 05:10 PM
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Originally Posted by patsmx5
I would assume we hold other things constant. IE, you don't run a bigger turbo, i don't run nitrous, etc. I know you can do X, Y, and Z to your high comp motor and make more power. That's not the point and that works both ways anyways.

So if all other things are constant, which is makes more power? That's my question. If you are saying, "Well see, I can run a bigger turbo, better manifold, and a less restrictive turbine and still have the same spool, so I'll make more HP", then that's not a fair comparison. I want to look at the physics of it. I seriously don't see the low comp/ high boost motor moving 30% more air/fuel and falling short on power to a high comp/low boost motor when they both have the same dynamic compression ratio. I have some reasoning, but I need to put it to paper to show the numbers. To me, for your argument to be valid, you would have to be a whole lot more efficient, and I don't see that happening.

Here's an overly simplified example, but perhaps it will explain my reasoning.

Generally speaking, at best, 1/3 the air/fuel mixture is converted from chemical to mechanical energy. So out of 100HP worth of air/fuel, the motor makes 33HP to the crank.

But say you are high compression, and your motor is 12% more efficient. Then you will make 12% more of 33. So 1.12*33=37

But I accept the standard 1/3 rule, but move 30% more air/fuel instead. so 33*1.3=43.

So I just made more power. Feel free to tear my argument apart. I know it's oversimplified, but I think yours is too.

EDIT: continuing...

So for you to make more power than me, you will have to be a lot more efficient. Specifically, you will have to be greater than 30% more efficient than my 8:1. So you are saying that your 11:1 motor is more than 30% more efficient at turning chemical energy into mechanical energy? Ie, your motor can turn more than 43% of the chemical energy into mechanical energy by use or increased compression?

"So if all other things are constant, which is makes more power? That's my question."

If all other things are constant, and both are tuned perfectly and they have the same effective CR, they should make the same similar power, the lower CR from the more air/fuel, the higher CR from higher efficiency.

You are looking at it differently than I am, but you also have to consider that a 24 psi charge is goinng to be hotter than a 14 psi charge, and the heat from compression should be the same according to pressure in the CC b/c they have the same dynamic effective CR. I cant say for sure because I havent done extensive research but they should be very similar in power.
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Old 09-24-2008, 05:14 PM
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Originally Posted by MeOughtta
jc_rotor,

You seem to think that equivalent pressure at the top of the stroke is all that matters. It isn't. JasonC has pointed out the dramatic difference in temperatures (and hence the likelihood of detonation).

To that, I'll add that the lower compression engine makes more power at the same peak cylinder pressure. Why: The volume/quantity of high pressure gas in the LCHB (low compression, high boost) engine is larger than with the HCLB (high compression, low boost) engine. So, while the chamber pressure may be the same at the top of the stroke, the LCHB engine has more gas to distribute through the cylinder during the power stroke than does the HCLB engine. Using 8:1 vs 11:1, the 8:1 engine will retain roughly 15% more pressure/torque/power mid stroke and 30% more pressure by the time the piston reaches BDC.
Exactly. There is more "stuff" in the LC/HB motor. When combustion begins, it's going to release more energy. That simple.
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Old 09-24-2008, 05:16 PM
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ok i was hoping to copy a proven set up but i cnt find the specs on stp's egt or the 616 hp miata. is there a simple chart to figure out estimated hp in relation to cr, boost, octane, and what other veriables im missing? the question sounds like a noob question sorry, but ill leave the details to the engineers and experts i just want to know what piston cr i should get? i know your going to say well how dependable and how much do you want to spen? well i want the car to be some what dependable it will mainly be a weekend warrior and maybe once a year i may drive from chicago to ohio. i dont care about running low octane i always run 93 or better.
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Old 09-24-2008, 05:21 PM
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Originally Posted by MeOughtta
jc_rotor,

You seem to think that equivalent pressure at the top of the stroke is all that matters. It isn't. JasonC has pointed out the dramatic difference in temperatures (and hence the likelihood of detonation).

To that, I'll add that the lower compression engine makes more power at the same peak cylinder pressure. Why: The volume/quantity of high pressure gas in the LCHB (low compression, high boost) engine is larger than with the HCLB (high compression, low boost) engine. So, while the chamber pressure may be the same at the top of the stroke, the LCHB engine has more gas to distribute through the cylinder during the power stroke than does the HCLB engine. Using 8:1 vs 11:1, the 8:1 engine will retain roughly 15% more pressure/torque/power mid stroke and 30% more pressure by the time the piston reaches BDC.
Yes but its also a less complete burn of the volume that is there.

My whole point the whole time of this, with or without arguing the extremes of either case or the (not likely) chance that everything but CR is different. Is that there is absolutely no reason to lower your CR lower than 9:1 unless you are going to run some ridiculous amount of boost.

The best CR for turbo cars IMO is 9.5-10.5:1

Im doing 11:1 because I want to see what happens to the miata motor specifically when running this high of a CR. All engines react differently to CR changes because of the cams, bore to stroke ratio and so forth. I think that 11:1 is going to be the sweet spot for boost on my motor because of the way im doing everything. Ive already reached 140HP dyno proven naturally aspirated, more than 8% power I "should have gained" from efficiency. So im very interested in how its going to run at 14psi and im anticipating 400+ to the wheels.
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Old 09-24-2008, 05:26 PM
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Originally Posted by jc_rotor
"So if all other things are constant, which is makes more power? That's my question."

If all other things are constant, and both are tuned perfectly and they have the same effective CR, they should make the same similar power, the lower CR from the more air/fuel, the higher CR from higher efficiency.

You are looking at it differently than I am, but you also have to consider that a 24 psi charge is goinng to be hotter than a 14 psi charge, and the heat from compression should be the same according to pressure in the CC b/c they have the same dynamic effective CR. I cant say for sure because I havent done extensive research but they should be very similar in power.
I'm definitely looking at it differently. So what is right? I used the 1/3, 1/3, 1/3 rule. Even if it was 20% as ZX-Tec suggest, the idea that you will have to be MORE THAN 30% MORE EFFICIENT from changing compression ratio's alone to make more power than LC/HB still stands. So is that reasonable to you?

I think my case is clear. I say that will never happen.

Hotwheels, go make yourself a thread and ask for opinions. We are talking physics right now, not answering phone calls.
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Old 09-24-2008, 05:28 PM
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Originally Posted by Hot_Wheels
ok i was hoping to copy a proven set up but i cnt find the specs on stp's egt or the 616 hp miata. is there a simple chart to figure out estimated hp in relation to cr, boost, octane, and what other veriables im missing? the question sounds like a noob question sorry, but ill leave the details to the engineers and experts i just want to know what piston cr i should get? i know your going to say well how dependable and how much do you want to spen? well i want the car to be some what dependable it will mainly be a weekend warrior and maybe once a year i may drive from chicago to ohio. i dont care about running low octane i always run 93 or better.
Leave the stock CR of 9:1 if its an early 1.8. Get 1mm overbore pistons if you are going to build the bottom end with A-beam rods. Run 14-18psi. Should get you considerable power and be very reliable. Make sure you get fully programable engine management and get it properly tuned. Should get you north of 275 HP easily.
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Old 09-24-2008, 05:51 PM
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Originally Posted by patsmx5
I'm definitely looking at it differently. So what is right? I used the 1/3, 1/3, 1/3 rule. Even if it was 20% as ZX-Tec suggest, the idea that you will have to be MORE THAN 30% MORE EFFICIENT from changing compression ratio's alone to make more power than LC/HB still stands. So is that reasonable to you?

I think my case is clear. I say that will never happen.

Hotwheels, go make yourself a thread and ask for opinions. We are talking physics right now, not answering phone calls.
If you are using the 1/3, rule, then (assuming) you have 30% more power from the added volume, you take that 100HP at 30 and THEN do the efficiency. so instead of 100HP worth of air fuel, its 130*33.33%=43.329 HP from it, and the 11:1 is 12% more effcient its 100HP*45.33%=45.33 HP. Very similar as you can see.

These calculations are not accurate though, you are assuming too much.
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Old 09-24-2008, 06:31 PM
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Originally Posted by jc_rotor
If you are using the 1/3, rule, then (assuming) you have 30% more power from the added volume, you take that 100HP at 30 and THEN do the efficiency. so instead of 100HP worth of air fuel, its 130*33.33%=43.329 HP from it, and the 11:1 is 12% more effcient its 100HP*45.33%=45.33 HP. Very similar as you can see.

These calculations are not accurate though, you are assuming too much.
Your math is wrong. Increasing the compression ratio will increase power to the crank by a max of 12%. That means the change in power to the crank will increase by 12%. You can't just add 12% to 1/3. 1/3 is an arbitrary value anyways as it is a constant in both scenarios.

Assume P=Power to the crankshaft, Q = efficiency added in %, R = increase in air/fuel, * represents multiplication.

We will let my 8:1 motor be the datum for which yours is measured off of.

So for me, Q is 1.00. For an 11:1 motor under ideal conditions, Q=1.12.

For me, R=1.3 as I will move 30% more air. For an 11:1 motor, R=1.00.

P = (R)*(1/3)*(Q)

For an 8:1 motor:

P = (1.3)*(1/3)*(1)= .433

For an 11:1 motor:

P= (1.0)*(1/3)*(1.12)=.373

Again the 1/3 is a constant in either case and is therefore arbitrary. Delete it from the equation and you will still get the same ratios.

All I'm assuming is that everything is held constant and we only change compression ratios. Assume whatever makes you happy, as long as it is the same for both motors. Only reason for this is to calculate the change in power from changing compression ratios.
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Old 09-24-2008, 07:29 PM
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Originally Posted by JasonC SBB
I pointed out above that my car did NOT lose bottom end going from 9.5:1 to 8.4:1 - dyno measured. And spoolup IMPROVED.

What else did you change?

There is NO WAY that is possible. HC engines are more efficient, which creates more heat and greater pressure. The turbo spools from the heat and pressure, so it's IMPOSSIBLE to spool faster with a LC engine UNLESS OTHER THINGS HAVE BEEN CHANGED.

I can name off lots of other changes you can make that will make a LC motor spool a turbo faster, but that would mean OTHER CHANGES besides just a lower CR.
-- 14yr engine builder/tuner who owns an import turbo performance shop

Even if you changed the tune to compensate for the new lower CR, you wouldn't be able to spool faster unless you tuned it BETTER for the LC vs the HC. All things in the tune being equal, meaning they are as close to the line to knock as you can get, and tuned to 11.5:1 perfect AF all the way across, the HC will spool faster and make more power on less boost.
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Old 09-24-2008, 11:56 PM
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Originally Posted by jc_rotor
"So if all other things are constant, which is makes more power? That's my question."

If all other things are constant, and both are tuned perfectly and they have the same effective CR, they should make the same similar power, the lower CR from the more air/fuel, the higher CR from higher efficiency.
Nooooooooo... Get that "effective CR" out of your head, it's next to useless!!! (Dammit y8s!! )


Try and think of this. Higher CR gives higher thermodynamic efficiency. END OF STORY.

Power output is proportional to efficiency and manifold absolute pressure, all else being equal.

Each point of additional compression adds 4% of output. END OF STORY. That piece of **** term "effective compression ratio" doesn't figure in this at all!

JESUS CHRIST I THOUGHT YOU WERE A MECH. ENGINEER????!!!
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Old 09-24-2008, 11:59 PM
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Originally Posted by ray_sir_6
What else did you change?
Minor DIY valve seat cleanup and 3-angle valve job.
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Old 09-25-2008, 12:52 AM
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Originally Posted by jc_rotor
30% more air in the combustion chamber does not equate to 30% more power.
Power is a function of BMEP. BMEP is a function of volume. It can be said that Power is a function of volume. If there's any doubt I'll type of the long fancy version tomorrow.
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Old 09-25-2008, 01:31 AM
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Power = BMEP * RPM.

BMEP is proportional to air consumed per cycle, and thermal efficiency.
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Old 09-25-2008, 01:35 AM
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jc_rotor,

AG Bell's book "4-stroke performance tuning" warns against the common mistake of raising compression ratio too high for the octane (whether FI or n/a). The temptation is to raise c/r, then retard timing to prevent knock. The latter will lose you more power than the raised c/r will gain you. High c/r is over-rated.
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