Thinking in print and out loud.

Well hmm...if I visualize 6 10 lbs rocks and I move these six rocks with 60 lbs of force for 3 miles...no doubt it would be easier than moving 8 10 lbs rocks for 3 miles but in this case...to make it equal (displacement)...you would have to move the 60 lbs rocks .9 more miles...should balance out to equal work...resistance... No? We can agree that resistance has been applied as 10 lbs per cylinders.

Still trying to grasp it all.

 

OK:

6 pistons times 4 inches per stroke = 2 piston feet / stroke
8 pistons times 3 inches/stroke = 2 piston feet / stroke

piston feet / stroke means "number of feet moved by all pistons in the engine per stroke (i.e., half a crank rotation)"

10lbs times 6 pistons = 60 piston lbs
10lbs times 8 pistons = 80 piston lbs

piston lbs means "(linear) pounds of force required to move all pistons in the engine". All along we've assumed this is constant even if velocity changes.

60 piston lbs times 2 piston feet / stroke = 120 piston squared lb-feet per stroke
80 piston lbs times 2 piston feet / stroke = 160 piston squared lb-feet per stroke

What is a piston squared? We need to end up with piston lb-feet per stroke, which would represent "lb-feet of work required to move all pistons one stroke (half crank rotation)". But there is an extra piston factor in the result!

 

to answer motorheads question, NO. I have not recieved my motor yet. It has been 2 1/2 months since Larry first told me they would get to it in about a week and a half. I have paid for it though and do expect it eventually, I would recomend finding your own machine shop. I will update if and when I ever get the thing and get it running.nuf said? Also I am not a math wizard, but I think there may be a factor being misinterperated. the phrase"wrench on the crank" I think is more of a factor than any resistance created by ring friction. does any one know leverage rates? I mean I figure that If I had to move a big load by hand, such as a crank I would chose the longest wrench. since the displacement is the same in the motors(apx.), in a full cycle of firing, wouldn't the same amount of displacement equal the same amount of overall force acting on the piston tops.lets forget about ring and bearing friction because I beleave it to be a moot point. Now if you weigh three hundred lbs, and you stand on a 2" long wrench, you will move more leverage but the extra two shorter wrenches do throw a wrench in things. All I know is that I have diven 302's and 300's, and the 300 will out pull a 302 anyday. I'm not saying beat it in a race, I'm saying, you throw 2000lbs in an Identical truck, same gearing everything but the two motors. which wins then? I'd put all my money on the i6.please tell me if I'm wrong about this. any one actually have a math book or are we all just pulling formulas out of thin air?

 

Ok so I looked it up on a math website. the force of a lever is distance from the fulcrum, which would be the center of the crank shaft, times the effort acting on it. So the leverage on the crank shaft would be 2" X 50ci of explosive force times six cylinders divided by four revolutions of the crankshaft per revolution of the camshaft.
in a 302, you have an apx. 1.5 inch lever with only 37.75 cubiic inches of displacement working on it given there are eight instead of six if you figure it all out the 300 out torques the 302 by 32 percent at any given revolution. this is actually pretty dumbed down and not scientific but again someone tell me if I'm way off or what?

 

all things equal, the motor with the most average hp between shift points should win.


right?

 

I screwed up and said four revolutions of the crankshaft. I believe actually the crankshaft goes around twice for every cylinder to fire. I don't know much about all this stuff I just wanted to join in this conversation. All I know is that in town my 300 will almost keep up with my 460. with the same load behind it. that is saying alot considering how healthy the 460 is and throwing in the old addage "there is no substitute for cubic inches". now get out on the open road and it's a whole different story. I don't know really.

 

I thought when calculating friction forces surface area had no part of the equation? The only time surface area comes into play is when you are trying to determine the force needed to move an object from a stand still.

And.. while I often enjoy discussions like this, I think you guys are nitpicking on one tiny aspect of a huge (HUGE!) pile of factors that go into engine efficiency. There's all kinds of losses:

- Friction from piston skirt and rings (as we're discussing). There's also the side-loading friction forces to consider here too.
- Recipricating motion losses, mostly determined by the weight of each recipricating assembly.
- Valve train losses.
- Pumping losses.
- Timing.
- Compression ratio.
- Then you get into thermodynamic issues with heat being lost to coolant, expansion ratios, flame temperature bla bla bla.

It's huge.. I think back in the 70s it was mostly fluke how well an engine turned out in the efficiency dept. They're learning more tricks but I'm sure a lot of it is still trial and error.

 

That's the basic concept, but not exactly right.

mdmbkr
I never said the 300 was 33% more efficient than the 300. There you go putting words in my mouth again. At best it might be 10% or so. The rod angles of the 300 don't help it any. The difference we are talking about is probably about equivalent to the difference between regular rings vs. low tension rings.

You won't have piston squared because piston isn't a unit of measure. You're either just making it up or don't have a clue. The number 6 is a coefficient and is dimensionless in the force equation. If you have three rocks and add their weight together do you come up with rocks cubed?

Motorhead, you've got the right idea. The only problem is that you would be moving the rocks the same distance. If you look at is at one cylinder at a time you would be moving each rock further, but since we all agree that the swept area is the same for the sum of all cylinders you are actually doing less work with the 300. Work=(sum of forces) x (sum of distance).

 

I think we are trying to determine ring resistance, with all things being equal for now, once we all come to an agreement, we would explore the main bearings/rod bearings etc., maybe thats just me.




Think I am sorry I asked??



silver streak,

Think I need to think on that for a while...

 

Actually, I would move the rocks a distance equivelent to 300 ci in both cases but with the six I would have to move (6 cylinder) 60 lbs 3.98 miles vs the (v-8) 80 lbs for 3...I kinda see what your saying though...and your probably right but as it sits...I can't convince myself. Your saying since its perpectual or each piston fired is staggered/ in sequence, this makes a difference but in my mind, whether they all fire at once or in sequence, they will be fired. Kinda like, you can move 6 rocks all at once each moving 4 miles, or you can move each rock a little until each move 4 miles, regardless, they are moving 4 miles.


Your saying since the 6 has two less cylinders, it requires less effort but you must travel further with less cylinders, to achieve the same swept area that we agree upon.

In my mind your confusing travel or stroke with work completed or displacement. I dunno, will think it over some more.



Picture this:


You move 8 cylinders 3 inches and you move 6 cylinders three inches, advantage 6 cylinders but wait, when you factor in equal displacement with equal bore, you need to move the 6 cylinders another .98", which would be added resistance.

 

I came up with another analogy, watch out, I love analogies, anything to avoid complex formulas.

If I pick up a 10 lbs ball 6 times, I have in effect lifted 60 lbs, at least this is how some weight lifters look at it.

Lets say for every inch of movement, your resistance is 10 lbs.

v-8, each bore of the eight cylinder engine has three inches of movement or 30 lbs per cylinder X 8 = 240 lbs combined effort.

I-6, each bore of the six cylinder engine has four inches of movement or 40 lbs per cylinder X 6 = 240 lbs combined effort.

Same. no?



In my mind, silver streak, when you state 10 lbs resistance per cylinder, this only means, thats the effort required to set the piston in motion. Resistance or drag comes into play when you move the piston over a certain distance. The analogy I provided I think is crude, and breaks it down into 1" distances, 10lbs being a measurement for drag/resistance, whereas in an actual cylinder the drag is constant.

 

I put a little more thought into this.



The formula:

Work=(sum of forces) x (sum of distance).

implies the cylinders creates energy, not resistance/drag of and that each cylinder creates the same amount of energy, which isn't true, as a simillar displacement/bore engine with less cylinders would create more force per cylinder given equal variables.

we won't consider port velocity or flow and air fuel mixtures per cylinder are perfect, only variable is displacement per cylinder to maintain equal displacement vs number of cylinders, bores are the same.

If each engine is 300 ci, the 8 would displace 37.5 ci per cylinder, the 6 would displace 50 ci per cylinder. Now we would have to assume, 37.5 ci displacement creates 10 lbs of force or .2666(infinite)) per ci.

for the eight we have 10 lbs per cylinder or the ability to make 80 lbs of force combined.

for the six we have 50 ci per cylinder X .2666 (force lbs per ci) = 13.5 lbs of force per cylinder X 6 cylinders = 79.98 (actaully near identical, only off due to the infinite numerals), so both engines have the ability or sum of forces to create 80 lbs of force X distance = the same for either.

This isn't a good example to represent drag or resistance.


With the above in mind

work = (sum of force) which would be the same for both engines X (distance) which has nothing to do with either, other than this is the distance they much travel as a whole engine.







On a side note, I sent an e-mail to an engineer, this was his reply, he kept things simple for my benefit.

To sum up what he stated regarding rotating mass friction:

*The v8 has 5 mains the inline six 7 = v8 wins

*the six has two less rod journals but it has a greater angle to spin through due to, stroke and rod length = probably equal

*he stated the greatest friction would come from the rings, while there are two less (sets) they move at a greater speed and have increased side loading with the six than the eight cylinder engine.

His opinion, the 8 may be a little better in this aspect if not equal.



I knew the first reaction would be but its an inline, shouldn't the force against the piston be better dispersed than with a v-8 configuration. The reply:

*The long 4" stroke gives the rod a high angularity to the bore on the way down. It is like putting a ladder up against the wall. If the base of the ladder is only a short distance form the wall then there is little force against the wall, but if you move the base of the ladder several feet away from the wall, then the (side) loading on the wall is greatly increased.



If someone knows an actual ford employee we can put this to rest easily, , ford does "friction horsepower loss" testing, so someone sitting at a desk right now has this info stashed in his files.

 

Silver Streak: I'm really not interested in a semantics debate or putting words in anyone's mouth. You said:

I said:

Were you not talking about ring friction when you said 120 ft-lbs vs 160 ft-lbs? 160 is 33% bigger than 120.

A piston IS a unit of measure .. one piston. You can hold it in your hand. We count them in all the calculations we've been doing. What I'm saying is you mistakenly counted them twice in your 120 vs 160 calculation.

If you have 3 rocks and add their weight together you get the total weight. But if you multiply them (just like you multiplied in your earlier work calculations) you would get pounds (or kg or whatever) cubed.

Just like if you have a 12" square piece of paper and you want to find its area. 12 inches times 12 inches is 144 inches squared ( = 1 foot squared). But 3 12" long strings add to 36 inches.

 

Whatever....We're just going to have to agree to disagree. I'm an engineer and an ASE Certified master tech. Everything mentioned above by the engineer can have holes poked in the logic all day long.

Exactly my point! I wasn't multiplying anything by pistons. I was adding 10lbs+10lbs+10lbs+10lbs+10lbs+10lbs. That gives you 60 lbs, not 60 piston lbs. This is the same as 6x10, is it not?

I've put all the effort into this that I'm going to. It's a lost cause.

 

Using your earlier math, if you calculate the work done by ONE cylinder in the 302 you would get:

3" of travel times 10lbs of force = 30in-lbs

Multiply by 8 cylinders for a total of 240 in-lbs (20 ft-lbs).

And ONE cylinder in the 300:

4" of travel time 10lbs of force = 40 in-lbs

Multiply by 6 cylinders for a total of 240 in-lbs (20 ft-lbs).

Why the discrepancy?

And we know from firsthand experience that it doesn't take 120 ft-lbs to turn the crank bolt on the 300, especially if all the valves are open.