That would make my 2.0L Triumph TR7 engine a 240 cube engine.

In reality, it's only 122

One liter is around 61 or 62 cubic inches.

 

No, it's 1 liter = 61 cubic inches.

 

For all you geeks (like me) 1L = 61.024 cubic inches.

 

True it's easier to wrap your mind around, but if someone is going to go as far as learning about displacement they may as well wrap their mind around the rest of it.

 

Well, I don't know anybody who considers the volume of a bare cylinder when talking about engines. I'd suppose it's important to a builder who wants to know just how far he can push the displacement limits of a block. But everyone I know who builds race engines are only concerned with the area from the top of the piston when it's at the very bottom of it's stroke to the top of the piston when it's at the very top of it's stroke.

 

Geez, and all this time I thought my 6.0 was basically a 351 only diesel flavored.

Joe

 

Amazing how complicated it can get ain't it!

 

Fivonut, please don't take this as being critical. I was just trying to make it clear to the OP that when you stroke an engine you do indeed add displacement which as you describe above seems to be cylinder volume(which is not what you said earlier), because the piston will be lower in the bore at BDC and higher in the bore at TDC. You seem to have corrected that above.

Maybe it's just a case of misunderstanding which terminology is being used to describe which aspect of the cylinder/piston relationship.

Quote:
<TABLE cellSpacing=0 cellPadding=6 width="100%" border=0><TBODY><TR><TD class=alt2 style="BORDER-RIGHT: 1px inset; BORDER-TOP: 1px inset; BORDER-LEFT: 1px inset; BORDER-BOTTOM: 1px inset">Originally Posted by fivonut
I see your point and you're correct. The formula for displacement is based (partially) on the stroke of the engine not volume of the cylinder. You can lengthen the stroke, adding displacement, and never change the cylinder volume in the block itself.





</TD></TR></TBODY></TABLE>

 

No harm no foul!! It was indeed a misunderstanding of terminology. I couldn't think of a better way to say what I was thinking. When I said you wouldn't change the cylinder volume what I meant was the volume of an empty cylinder if you measured the height from the bottom of the empty bore to the deck (oops!! there's another technical term!!)

 

Since the Op used the 6 litre PSD I will complicate things by saying don't forget the turbo. At 15lbs of boost(14.7 to be exact) a 6 litre engine would now displace 12 liters and at 30lbs of boost (29.4 to be exact) it would displace 18 liters. More boost equals more air which equals more power. The can of worms is now open.

 

No, it is still displacing 6 liters of air, but it is displacing far more MASS AIR than at atmospheric. Roughly it is discplacing an equivalent about of 12 liters of air at 30lbs boost. None the less, good use of the can opener!

 

Here is a question I asked a while back:Hi all, I am trying to figure out how to calculate how boost translates into cubic inches or liters. For example, Kenne Bell at one time made a supercharger for the Ford v10. It was, if I recall, 2.2 liters and I would guess that would be at full boost which I think was 8lbs. So at 8lbs of boost I would think that would make the engine 9.0 liters (6.8 plus 2.2). Am I correct in this assumption? So now what about a 6.0 Diesel at 28lbs of boost, how much air would the engine be getting. I hope you understand what I am trying to say, it is not always easy to write what I am thinking.











Not quite. The 2.2l quoted is (roughly) the volume of air moved by the supercharger's screws in one revolution. The KB's superchargers spin at several times the motor's speed to achieve a volume is excess of the motor’s displacement.

Here's how you want to think about it in gross terms: At sea level atmospheric pressure is about 14.7 psi. Boost is relative to atmospheric pressure and is typically measured in psi, but for this exercise we'll start by using a unit of measure called Atmospheres or ATM. 14.7 psi is one ATM, 29.4 is 2 ATM, and 7.35 is 0.5 ATM. A theoretically perfect normally aspirated engine will move its displacement (6.8l in your example) with each revolution. (In actuality it will move slightly more or less depending on the elasticity of air molecules, restrictions, etc. and is why tuned intakes and headers improve power). Anyhow, if your blower is capable of 1 ATM you are moving approximately twice as much air (like running a 13.6l normally aspirated motor) of 2 ATM is three times as much air (like running a 20.4l normally aspirated motor).

Now if we go back to 8psi, which is 8/14.7 or 0.544 (ATM), which is like running a 6.8l * 1.544 = 10.5l normally aspirated motor. Make sense? A 6.0l diesel at 28 psi is like a 28/14.7 = 1.90 ATM which is running like a 6.0l * 2.90 = 17.48l normally aspirated motor.


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Last edited by FTE Herman : 02-06-2008 at 09:42 PM.

 

Right, but a 4-cycle engine will only draw in that much air/fuel mixture over the course of TWO revolutions.

 

You're right!!! Brain malfunction on my part!! sorry and thanks!! ( I knew better...function of my age)

 

nope, the 6 liter diesel is 365 cubic inches.
the 7.3 diesel is 444 cubic inches.
in gas engines,
the 4.6 is 281 cubic inches,
the 5.4 is 330 cubic inches.
and the 6.8 V10 is 415 cubic inches.