COMPRESSION RATIO info..

[Route666]

65shelby, no need to get steamed, even if certain others are poking with the proverbial stick.

The way it works is that yes the combustion chamber never "changes", but it's always there. There is the swept volume of 600, but there isn't a void left in the combustion chamber, that 60cc still has stuff in it, whether it be left over exhaust byproduct from the last stroke or fresh air/fuel from the new stroke as overlap cleans out the cylinder.

Your example works for 10:1 if the 60cc combustion chamber is included in your cylinder volume, but cylinder volume includes the stroked volume, gasket volume, piston deck clearance volume, piston releif volume, and them combustion chamber. In the simple version it includes the basics - stroked volume and combustion chamber. The way you explained it with 600 at bottom of stroke and 0 at top equates to just the stroked volume. 600 at bottom and 60 at top is the 10:1 you are talking about.

 

[Route666]

Heck here's a calculator (from a shop I know in my home town, for a bit of trivia) that allows you to work out what combustion chamber volume you need for a given CR and swept volume.

http://www.turbofast.com.au/Tfcomp.html

So 600cc is a 3.942 bore with a 3" stroke.

(BORE/2)^2 x PI x STROKE x CI_TO_CC = CC swept by piston. (600 in this case)

CI_TO_CC conversion factor is 2.54^3

Put those values into the calculator, with zero for gasket and deck clearance, and with 11 in the compression ratio and you get 59.974 and with 10 you get 66.638.

 

[D.Hearne]

No, I have not. What I HAVE done is explained this three times already. I may not have any Perfect Gas laws down, but I still correctly explained what compression ratio is. Let me dumb it down even further.....If I can make it simpler.

Round 4:
Compression ratio = the ratio of the cylinder volume at bottom dead center to the remaining total volume at top dead center. Once again, my example used a 600cc cylinder and a 60cc chamber.

You STILL have to include the chamber volume along with the swept volume at BDC. There isn't a slide valve covering it up till the piston gets back to TDC. The chamber is there all the time. It doesn't just magically appear when the piston gets back up at the top of it's stroke. And there's an explanation in the back of Ford Racing's catalog that also explains this. "Compression ratio is defined as the ratio between the total volume ( cylinder volume PLUS clearance volume )above the piston at BDC, and the clearance volume above it at TDC"

 

[krash kendall]

But thats just it, the 60cc chamber is dead volume that never changes. Only the cylinder's swept volume, in this case 600cc, is drawn in. You are then sticking 600cc into a 60cc space. 600/60 = 10, or 10:1. You would need a 660cc cylinder and a 60cc chamber to get 11:1.

The ratio that you are describing is between two constants. You are only pointing out that the cylinder volume is 10 times larger than the chamber volume. This only gives us the tools for determining the values in our compression equation.

(K+K)/V = P

(600cc+60cc) / 60cc = 11 atm

 

[Soaring1]

If you start out with 60, then you add 600.....that's 660cc. Not 600. 660cc to 60 cc equals 11:1 ratio. That's a no brainer.

 

[ForceFed70]

No, I have not. What I HAVE done is explained this three times already. I may not have any Perfect Gas laws down, but I still correctly explained what compression ratio is. Let me dumb it down even further.....If I can make it simpler.

Is this going to be a case of "The rest of the world is wrong and I'm right?"
This many people arn't wrong 65Shelby..

If you play with the online compression calculator you posed a link to, it'll show you for yourself. I think your getting confused by the definitions that you posted links so. When they say "cylinder volume at BDC" they mean the volume of the cylinder PLUS the volume of the combustion chamber.

But by all means.. plug the numbers into the calculator and find out for yourself. For 600cc of displacement by the piston, use a bore of 3.942 and stroke of 3.0 ( equals 600cc). For simplicity use 0 for all other numbers except combustion chamber volume. It'll show you right there that the net compression ratio is 11:1. Unless you want to include the online calculator as also being wrong...

 

[65ShelbyClone]

This many people arn't wrong 65Shelby..

They certainly can be and I've seen/had it happen. However, I'm the ass this time.

I found my error; all this time I havent realized that I was calculating all the possible variables like deck height, piston notches, and gasket thickness rather than only figuring with cylinder and chamber volume. I calculated the hell out of my engine with a 4" bore, 3" stroke, .010" in the hole, 61cc chamber, and .040" gasket. I knew I didnt have 11:1 or even 10:1, so how could anyone else? Duh....I guess I had the right definition with less understanding of it than I thought. I was so sure I was right and was too close to the problem too see it.

I apologize for being a foaming-at-the-mouth jerk and I hope there will be no ill feelings in the long run.

 

[ForceFed70]

No need to apologize to me. I'm just glad we got it figured out.

It shows character to stick to your guns... but it shows honor and nobility to step back, rethink things, and admit you made an error.

 

[whwright]

It would have been a lot simpler if somebody pointed out that the "compression ratio" that is so often talked about is more technically correct when called "STATIC compression ratio."

Static compression ratio is simply the total uncompressed volume divided by the total compressed volume. It's an easy number to calculate to compare one engine configuration to another. Both the uncompressed and compressed volumes include the stroke, bore, combustion chamber, piston volume, head gasket, and so forth.

If you start considering "swept areas" and stuff like that, you're getting into the much more difficult to calculate "DYNAMIC compression ratio" and "EFFECTIVE compression ratio." To get those numbers you need differential equations and lots of emperical data.

I for one am not anxious to get back into differential equations.

 

[ForceFed70]

It would have been a lot simpler if somebody pointed out that the "compression ratio" that is so often talked about is more technically correct when called "STATIC compression ratio."

Static compression ratio is simply the total uncompressed volume divided by the total compressed volume. It's an easy number to calculate to compare one engine configuration to another. Both the uncompressed and compressed volumes include the stroke, bore, combustion chamber, piston volume, head gasket, and so forth.

If you start considering "swept areas" and stuff like that, you're getting into the much more difficult to calculate "DYNAMIC compression ratio" and "EFFECTIVE compression ratio." To get those numbers you need differential equations and lots of emperical data.

I for one am not anxious to get back into differential equations.

True... but in automotive purposes we always look at static compression when talking about compression ratio's. To calculate dynamic compression you'd almost have to have an engineering degree...not the mention the fact that dynamic compression changes with throttle and RPM.

Thus, You'd think that it would be assumed that we are looking at static compression in this discussion.

 

[whwright]

True... but in automotive purposes we always look at static compression when talking about compression ratio's. To calculate dynamic compression you'd almost have to have an engineering degree...not the mention the fact that dynamic compression changes with throttle and RPM.

Thus, You'd think that it would be assumed that we are looking at static compression in this discussion.

You are correct, and that is a good assumption.

But once the term "swept area" entered the discussion it became apparent that everyone's assumptions weren't based upon the same foundation.

It would have saved a lot of grief if someone had clarified that you were talking about STATIC compression ratio.

And by the way ... I am an engineer. At one point, I couldn't even spell engineer -- now I are one.

And I still can't figure out my engine's dynamic compression ratio, or its cylinder pressures for that matter.

 

[D.Hearne]

To calculate dynamic compression, that would take a computer. No-one, has the time to sit and figure all the variables needed to do that. If they do, then they need to get a job.

 

[65ShelbyClone]

No need to apologize to me.

Even so, I feel it needed to be said to everyone. I'm going to be more cautious and open minded on the topic of dynamic compression, though...

Wouldnt it be possible to roughly calculate dynamic compression by taking into account intake vacuum, A/F ratio, valve timing events, and static compression? I imagine it would need dyno testing and maybe some specialized equipment to measure actual volumetric efficiency.

 

[whwright]

When I was in college (longer ago than I like to admit) one of the projects in a senior level class was to write a program to calculate the air flow through a hair dryer. It took into account the RPM of the fan, the venturi in the dryer, the air resistance across the heating coils, basically anything we could apply a formula to.

And if we couldn't find a formula, we experimented until we figured one out. We basically used numeric methods to solve a set of complicated differential equations. (If I recall correctly, the name of the class was "numerical methods.")

We wrote the program in Fortran (does that give away how long ago it was?).

Yuck. I hope to never do anything like that again. Probably couldn't if I tried.

But someone (other than me!) could more than likely do the same sort of thing with variable cylinder volumes, valve events, manifold designs, induction components, static compression (!!), and lots of other engine variables. With enough number-crunching power, you could get a good feel for the dynamic compression (directly related to volumetric efficiency) of an engine assembly. I wouldn't be surprised to find that the major auto makers already have this sort of software that they use during hardware development. They might even use it to develop the look-up tables that modern engine management systems use.

 

[rbohm]

ok here is a SIMPLFIED mathmatical model of how compression ratios work, and why more is better, to a point. let us take two engines that are otherwise identical, but one has 8:1 compression, and the other has 10:1 compression. let us also assume for simplicity's sake that standard atmospheric pressure is 15lbs(yes i know it is 14.7 but it make the math easier). at this point we are only looking at a known compression ratio, and NOT trying to calculate it. as the piston moves to tdc, we are compressing the a/f mixture(15psi as it come into the cylinder) by the compression ratio, thus we get;

15x8=120psi for the 8:1 engine, and

15x10=150psi for the 10:1 engine.

when the spark ignites the fuel mixture, cylinder pressure goes up by 10 times at its peak pressure, thus

120x10=1200psi for the 8:1 engine and

150x10=1500psi for the 10:1 engine.

to get the NET cylinder pressure pushing on the piston, we have to subtract the initial compression pressure, thus

1200-120=1080 net psi for the 8:1 engine and

1500-150=1350 for the 10:1 engine.

as you can see the 10:1 engine has 270 more net psi pushing on the piston than the 8:1 engine has, and this translates into more working capacity. as compression ratio increases, you get still more net pressure, BUT you also get to a point of diminishing returns, meaning that the extra pressure doesnt add enough to the bottom line(so to speak) to where it makes sense to raise compression much higher than 10:1 for a street engine, and 13:1 for a race engine. as was stated, the higher compression ratios load piston rings and head gaskets much more(not quite the square of the increase) so you need to control detonation as best as you can. remember this is a SIMPLIFIED mathmatical model to make things easy to understand for the layperson who has only a passing understanding of how an engine works. for you engineers out there i know there is a far more complexe mathmatical model you use that is far more accurate, but the results are the same, just the number are different.

 

[Route666]

They certainly can be and I've seen/had it happen. However, I'm the ass this time.

I found my error; all this time I havent realized that I was calculating all the possible variables like deck height, piston notches, and gasket thickness rather than only figuring with cylinder and chamber volume. I calculated the hell out of my engine with a 4" bore, 3" stroke, .010" in the hole, 61cc chamber, and .040" gasket. I knew I didnt have 11:1 or even 10:1, so how could anyone else? Duh....I guess I had the right definition with less understanding of it than I thought. I was so sure I was right and was too close to the problem too see it.

I apologize for being a foaming-at-the-mouth jerk and I hope there will be no ill feelings in the long run.

I'm just glad you've got it right and you weren't too stubborn to listen to others to work it out... in the end.

Oh I'm also glad that you're one of the few willing to admit when they're wrong. It takes solid character to do that.

 

[Randy'65]

Boy I'm glad you guys settled this before I had to dig out my old Thermodynamics books.

 

[ForceFed70]

Wouldnt it be possible to roughly calculate dynamic compression by taking into account intake vacuum, A/F ratio, valve timing events, and static compression? I imagine it would need dyno testing and maybe some specialized equipment to measure actual volumetric efficiency.

It's possible to calculate that sort of thing... but it'll take a long time and a crapload of math.

Also keep in mind that there are leaks in the system so we need calculate how much compression is lost due to blowby. This also gets complicated because the amount of blowby changes with engine RPM. We'd also have to look at temperatures..both of the air/fuel comming in as well as cylinder.

Not easy by any means...

 

[whwright]

I think you guys are right on the money when you say that 9:1 or so is optimum for a street engine. FWIW: my 460 runs 9.5:1 and I haven't noticed it detonating at all.

Just for grins, I punched a bone-stock 351W into a horsepower estimating program. I used "stock" heads and "manifolds with performance mufflers." The cam was "stock," and the induction was "600 cfm with dual plane intake."

Then I asked it to calculate HP with different static compression ratios. I didn't change anything else, just the compression. The software modeled the different ratios by changing the combustion chamber size, which would be analogous to changing heads without making any other changes.

Here's what I got:

Ratio HP ft-lbs
==== === =====
6:1 151 265
7:1 170 293
8:1 186 316
9:1 199 335
10:1 211 351
11:1 220 365
12:1 229 377
13:1 238 388
14:1 246 397
15:1 253 406
16:1 259 414
17:1 265 421
18:1 270 427

The software wouldn't accept a compression ratio below 6:1, nor above 18:1 (surprise, surprise!).

From those numbers, it looks like the point of decreasing returns occurs right at 10:1, above which the increase becomes less than 10 HP per point of compression increase.

Interesting? Or did I waste my time?

Wayne.

 

[Route666]

Definitely fascinating mate, although they're increasing in power like this

19
16
13
12
09
09
09
08
07
06
06
05

Definitely an exponential dropoff, it's kinda hard to say where the real loss of benefit for increase in CR is, because 5 horsepower might mean more to someone than someone else.

There isn't a great deal of efficiency to be gained in increasing CR, but if you go from 8.5 to 10, that's definitely worth it. IMO