[fitzwell]

so last night we were in the garage, over a beer or two, discussing the relative merits of ford vs chevrolet, long rod / short rod and stroker motors when the following ensued.....if you increase the stroke of the motor, all other things being equal, does the compression ratio change ?
Theory a......increased stroke takes in more air, chamber & piston configuration remain the same, more air, higher compression
Theory b.......even though you increase the amount of air to be compressed, if the chamber & piston are identical, the ratio does not change

discuss...........................

BTW, there is beer wagered on this

[DoubleVision]

All else equal, Increasing the stroke lenth raises the compression ratio, since it`s starting point is from futher down in the bore than the crank before it. It`s compressing a greater distance.

[mstngjoe]

By increasing the stroke, you also increase the swept volume.

So yes....... more stroke = higher static compression ratio (all else being equal).

[predator carb guru]

i agree. if you leave all things equal, but make a stroke increase, you increase displacement. more displacement compressed into the same space as the previous lesser displacement (again, keeping piston dome volume, combustion chamber volume, and cylinder bore size the same, only changing the stroke) will result in increased compression. here's a simple way to look at it: let's say you have a cylinder with a 1" bore, and a 3" stroke. total cylinder length is 4". let's also say there is 1" above the piston at tdc. when the piston travels from bdc to tdc, it will compress the volume above it to 1/3rd of it's original volume. this would mean that the compression ratio would be 3:1. ( i think i got that example right. it's early in the morn for me, i can't think right...heheheh)

[71C10]

Try it yourself: http://kb-silvolite.com/calc.php?action=comp

355 SBC
64cc combustion chamber
13cc piston dish
.040 head gasket
4.166 gasket bore
4.030 cylinder bore
.000 deck
3.480 stroke
Compression = 9.468

383 SBC
64cc combustion chamber
13cc piston dish
.040 head gasket
4.166 gasket bore
4.030 cylinder bore
.000 deck
3.750 stroke
Compression = 10.125

So who won the beer?

[ap72]

Um... if you increase the stroke a significant amount ( a stroker engine) and leave the rod and the piston the same. You will end up with a bent rod from the piston smashing into the chamber. Otherwise it will raise the compression ratio. that's why high compression on short stroke engines is harder to achieve than on long stroke engines.

[71C10]

Originally Posted by ap72

Um... if you increase the stroke a significant amount ( a stroker engine) and leave the rod and the piston the same. You will end up with a bent rod from the piston smashing into the chamber. Otherwise it will raise the compression ratio. that's why high compression on short stroke engines is harder to achieve than on long stroke engines.

Yes the piston pin height must be changed if you stroke or de-stroke an engine. The theoretical question posed was what is the effect on compression if you used pistons that were the same with of course the exception of necessary pin height changes to accommodate the change in stroke.

Thanks for pointing that fact out though for our less informed readers

[curtis73]

Short answer: compression ratio doesn't change with the amount of air brought in. That is dynamic compression, or just cylinder pressure that is affected. Static compression ratio is the total volume of the cylinder at BDC divided by the total volume at TDC. So, increasing the stroke but keeping the same compressed volume will increase the compression ratio.

Long answer: If you have an engine cylinder which displaces 1000 cc (one liter) at BDC and 100 cc at TDC, you have a 10:1 compression ratio. If you increase that BDC volume to 1200 cc with stroke but keep the same compressed volume of 100 cc, you have increased its static compression ratio to 12:1 In order to retain the same compression ratio, you need to increase the combustion space at TDC by 20 cc in this example.

The amount of air that the engine actually ingests depends on RPM, cam event timing, intake design, throttle position, and dozens of other factors. Which means the actual pressure inside that cylinder varies widely based on those factors. Now we're talking about VE or volumetric efficiency, which is basically a percentage of mass. If an engine is operating at 100% efficiency, that means at BDC of the intake stroke, the mass of air inside the cylinder would weigh the same as the mass of an equal volume in ambient air. That is to say, the cylinder has completely filled based on surrounding conditions. When the intake valve closes, there is neither negative nor positive pressure in the cylinder. In truth, most street engines operate between 75-90% VE and naturally aspirated race engines usually operate at a peak VE of 80-100%. Forced induction engines actually operate sometimes well over 100% VE since they are receiving pressurized air. Its also important to point out that when I speak of those VE numbers, I'm speaking of their peaks at WOT. On a dyno at WOT, those peak VEs occur at an RPM where airflow is maximized by the wonderful selection of matching parts you've chosen for your application. This is why mismatching parts hurts engine performance so much. Using an intake that tunes peak VE for 3500 rpms, but a cam that tunes it for 2500 rpms means they will never work together to make power at the same time... but I digress.

So what I'm saying is basically, increasing the static compression ratio makes a general, across-the-board shift upward in peak cylinder pressures (all other things equal). In your example, the stroked engine would ingest more air and compress it into the same size combustion space. This is the exact premise behind why hotter cams require more static compression ratio; Hotter cams allow bleeding of cylinder pressure at lower RPMs. Increasing the static compression raises the actual cylinder pressures back up to recover that lost pressure. Without that cylinder pressure, large cams make wimpy torque down low in the RPMs.

[xntrik]

Originally Posted by curtis73

Short answer:

So, increasing the stroke but keeping the same compressed volume will increase the compression ratio.

Long answer: If you have an engine cylinder which displaces 1000 cc (one liter) at BDC and 100 cc at TDC, you have a 10:1 compression ratio.

.

If I may intergect something, please. Using this illustration given.

In the illustration given the displacement/ swept volume (bore x stroke) would be only 900 cc. Yes it would be 10/1. But the engine is only a 7.2 litre not an 8.0 litre.

There might be a mixing of terms here concerning "displacement".

A cylinder displacement of 1000 cc (swept volume) and a compressed "chamber" volume of 100 cc when the piston is at TDC = 1100 cc total....

when the piston gets to TDC, the total 1100 is compressed into 100. Therefore you have an 11/1 static compression.

This is a commonly made error. Hey, I have done it myself...... ping ping

[Rick WI]

The swept volume/displacement of a cylinder, which is calculated as (3.1417 / 4) x Bore x Bore x Stroke is not added to the volume at TDC. The swept volume is the displacement, period. That's what makes it a "ratio" equation. If that were true then displacement would grow as compression ratio dropped.

Curtis, your definition is correct.

[xntrik]

I apologize if I was ineffective in my explanation.

Here's a article exlaining compression ratio that agrees with what I said....

Read paragraph 2, "What is compression ratio."

But it is a magazine.. you know how those magazine guys are.

http://hotrod.com/techarticles/54258/

http://e30m3performance.com/myths/mo...comp_ratio.htm

[Rhansen]

Originally Posted by ap72

bent rod from the piston smashing into the chamber.

lol - that's when it's time to pick up more beer

[Rick WI]

What that says and you you are saying are not the same thing at all.

Think of it this way X. The swept volume is the displacement of the cylinder. That is a fixed value based on the bore diameter and the stroke. The volume, or space within the cylinder when the piston is at TDC is composed of the gasket thickness, piston cc valve reliefs or dish, combustion chamber size, valve selection and deck height clearance. That volume does nothing to contribute to displacement. All those variables comprise the 100cc's of area that were used in curtis's example, or as the article explains and defines as clearance volume. So......clearance volume has no affect on displacement whatsoever. Therefore if a cylinder displaces 45ci (based on bore and stroke and the clearance volume at TDC is 4.5CI the compression ratio is 10:1. Since clearance volumes are measured in CC's when blueprinting a motor we will typically convert all the units, including displacement to CC's.

Does that make sense?

[xntrik]

Originally Posted by Rick WI

Therefore if a cylinder displaces 45ci (based on bore and stroke and the clearance volume at TDC is 4.5CI the compression ratio is 10:1. Since clearance volumes are measured in CC's when blueprinting a motor we will typically convert all the units, including displacement to CC's.

The article says the same thing. I'm missing where your getting something else out of it.

Does that make sense?

Please click here for picture:
http://e30m3performance.com/myths/mo...comp_ratio.htm

Hot Rod paragraph 2 says: "If the volume of the cylinder with the piston at BDC is 10 times greater than the volume of the combustion area with the piston at TDC, then 10 units of volume get squeezed into 1 unit of space, and the compression ratio is 10.0:1"

Your illustration indicates that 45 (bore x stroke swept volume) + 4.5 (TDC volume) = 49.5
49.5 divided by 4.5 = 11/1


Thanks for your input,
x

[Rick WI]

Gotcha, I see what your saying. Talk to ya later.