COMPRESSION RATIO info..
- Thread starter 67staaang
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65ShelbyClone
Founding Member
Its a complex subject, but I'll try to keep it as simple as I can for now. The compression ratio is simply the volume of the cylinder at TDC to the volume at BDC. If the cylinder has a volume of 600cc when the piston is at the bottom of the stroke and zero when the piston is at the top of the stroke, but you also have a 60cc combustion chamber, that makes a 10:1 compression ratio. Higher compression is more thermally efficient, but is harder on parts and requires higher octane gas.
jerry S
New Member
65ShelbyClone said:
Your "simple" explanation reminds me of when my hs math teacher was saying how simple it was to get an explicit solution to a differential equation by taking the natural log of x of both sides of the equation. Uhhh, ok.
Route666
Active Member
65shelby, isn't what you explained an 11:1 ratio? If the cylinder has a volume of 600 at bottom, and 0 at top, I assume you mean the stroked volume. You also have to add in the combustion chamber, which in your case was 60, making it 660 / 60 or 11:1.
Basically the compression ratio is what the words mean, in relation to the pistons and the place they live. The pistons compress the volume of space (filled with air/fuel) into a smaller space, and the ratio of the compression is the: compression ratio.
You can modify the ratio by using thinner and thicker gaskets (although thicker ones may reduce your ratio, they may increase the chance of detonation, by reducing the effectiveness of the squish / quench area)
You can also put a bigger or smaller crank in, which compresses a bigger amount (if a bigger crank) into the same space and therefore increases the ratio, and the opposite if de-stroking.
You can change your heads, different heads have different combustion chambers, which means if you get a smaller chamber, you'll compress slightly less air fuel into even less space, so the ratio will increase.
You can use pistons with different compression heights (the distance from the centre of the wrist pin bore to the top of the piston) which will change the size of the combustion chamber, as the top of the piston forms the floor of the compressed combustion chamber. Less height means less ratio.
You could put in pistons with a flat top, dome, or dish, which can reduce or increase the ratio. Domes will increase the ratio, dishes decrease, and flat tops are usually used to increase as well.
You could use shorter or longer rods, effectively does the same thing as compression height on pistons.
All the above will do what I have said if everything else stays the same. You could put in dome pistons in a hope to increase CR, but they actually have a shorter compression height, which more than cancels out the dome and you end up reducing your ratio.
From what I've read, for best performance and efficiency, on an NA motor, try to get the best CR for the fuel you plan to use, with a squish height of less than .040", and use only flat-tops or dishes. Domes get in the way of the quench effect created by the squish area.
Basically the compression ratio is what the words mean, in relation to the pistons and the place they live. The pistons compress the volume of space (filled with air/fuel) into a smaller space, and the ratio of the compression is the: compression ratio.
You can modify the ratio by using thinner and thicker gaskets (although thicker ones may reduce your ratio, they may increase the chance of detonation, by reducing the effectiveness of the squish / quench area)
You can also put a bigger or smaller crank in, which compresses a bigger amount (if a bigger crank) into the same space and therefore increases the ratio, and the opposite if de-stroking.
You can change your heads, different heads have different combustion chambers, which means if you get a smaller chamber, you'll compress slightly less air fuel into even less space, so the ratio will increase.
You can use pistons with different compression heights (the distance from the centre of the wrist pin bore to the top of the piston) which will change the size of the combustion chamber, as the top of the piston forms the floor of the compressed combustion chamber. Less height means less ratio.
You could put in pistons with a flat top, dome, or dish, which can reduce or increase the ratio. Domes will increase the ratio, dishes decrease, and flat tops are usually used to increase as well.
You could use shorter or longer rods, effectively does the same thing as compression height on pistons.
All the above will do what I have said if everything else stays the same. You could put in dome pistons in a hope to increase CR, but they actually have a shorter compression height, which more than cancels out the dome and you end up reducing your ratio.
From what I've read, for best performance and efficiency, on an NA motor, try to get the best CR for the fuel you plan to use, with a squish height of less than .040", and use only flat-tops or dishes. Domes get in the way of the quench effect created by the squish area.
I also caught 65shelbyclone's mistake. 11 to 1 here too.
65ShelbyClone
Founding Member
Route666 said:
70_Nitrous_Eater said:
The combustion chamber does not account for any swept volume; the cylinder does. The chamber volume remains constant. Since the piston is compressing 600cc into a 60cc chamber(1/10 the space), you get a simplified explanation of a 10:1 compression ratio.
It all depends on what you want. High compression usually means high performance. As stated before it is more efficent. But also as stated these higher pressures and stresses on the engine take a toll. Nobody wants to blow up there engine. Too high and you will get pre-detonation and your engine will ping, unless you retard the timing or use a higher octane. If you use a lower compression performance will drop off but the eninge will be more tolerant. You could run 87 pump gas. Usually 10:1 is a good street engine compression ratio.
65ShelbyClone
Founding Member
70_Nitrous_Eater said:
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.
65ShelbyClone said:
I understand what you are saying... only 600cc of fuel/air is entering the system. But that still does not take away from the fact that there is 660cc of vapor that is being compressed. 60cc of vapor is left over from the pervious powerstroke. A ratio is a before and after comparison. Before there was 660cc of vapor, afterwords there is 60cc of vapor.
Here's some food for thought for you. We all know that changing the piston can increase/decrease the compression ratio. Now... how does a change is piston work in your line of thought? Hint: The bore and stroke stay the same..so the total volume of air/fuel brought into the cylinder is also the same.
krash kendall
Active Member
Boyles law:
K = constant
V = Volume
P = Pressure
K is the litres of air you are beggining with and remains constant throughout the equation regardless of pressure. Volume dictates the actual volume that K takes up and is directly proportional to P.
P x K = V or V/P=K
You are starting with 660cc which remains constant and is equal to V at this point as it is at 1 atmoshere pressure. After P (pressure) is applied K (660cc) then takes up V (volume) of 60cc. The fact that stroke is involved is inconsequential the only thing that matters is that 660cc is now compressed to 60cc. The only way for the chambers volume to not be included would be if the chamber (and chamber only) were a total vacuum.
K = constant
V = Volume
P = Pressure
K is the litres of air you are beggining with and remains constant throughout the equation regardless of pressure. Volume dictates the actual volume that K takes up and is directly proportional to P.
P x K = V or V/P=K
You are starting with 660cc which remains constant and is equal to V at this point as it is at 1 atmoshere pressure. After P (pressure) is applied K (660cc) then takes up V (volume) of 60cc. The fact that stroke is involved is inconsequential the only thing that matters is that 660cc is now compressed to 60cc. The only way for the chambers volume to not be included would be if the chamber (and chamber only) were a total vacuum.
krash kendall
Active Member
To finish this up, in the compressed state:
K/V=P
660cc/60cc=11atmosphere
Since in the beginning we started with 1 atmosphere it becomes clear that the compression ratio is 11:1.
K/V=P
660cc/60cc=11atmosphere
Since in the beginning we started with 1 atmosphere it becomes clear that the compression ratio is 11:1.
67coupe351w
New Member
jerry S said:
I for one am impressed that you were learning about differential equations in high school.
And I also 5th that its 660 to 60
65ShelbyClone
Founding Member
D.Hearne said:Face it shelbyclone, you've lost.
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.
Mitch, I never said swept volume was the only factor.
Some sources to back up my apparent insanity:
Chevy High Performance, paragraph 2
One of many online compression calculators
Hot Rod, 2nd half of second paragraph and the best, simplest description IMO
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