I would like to see a dyno run that showed 400 lb. ft. as well. Heard about it, but nobody's ever produced one. On a NA 300.

 

? Col. Flashman would be one example...but then we'd debate the accuracy of the dyno used.

 

Probably not. Most of the telltale signs of a manipulated dyno chart are right on the chart itself, you just have to know what to look for.

 

What I mean is a scanned in dyno sheet, reproduced for us to see as original. With the power output curves, etc. etc. so we can see the behavior of the motor. I've seen SS's so it's been done, I'd just like to see these 400+ lb. ft. 300's.

I'd think if you shelled out the bucks for a dyno run you have a right to expect a printed graph of the motor's output.

 

Most junk dyno sheets aren't junk because someone doctored them up after they were printed. Most are junk right off the printer. I have seen a sheet on a 98 TA that showed 58 rwhp and over 1200 ft-lbs of torque. That was done by clipping the inductive pickup on a plug wire and telling the dyno that it fired every 90 degrees instead of every 720.

 

I know this is pulling up an old thread but I wanted to comment on some of the earlier posts about number of cylinders vs displacement vs area swept by the rings.

The area swept by the rings is not determined solely by engine displacement.

Imagine that you have a 100 cubic inch engine with 2 cylinders and a 1" stroke. This means the bore is 7.97" (ridiculous, but this is just for illustration). The math:

Cylinder volume = (pi) * (cylinder radius squared) * (stroke)

which is: 3.14159 * ( 7.98/2 ) * (7.98/2) * 1 = 50.0

The circumference of the piston (pi * bore) determines the area swept by the rings. With a 7.98" bore, the circumference per piston is 25.06". This means there is a total of 50.12" of ring length in contact with cylinder walls.

If we now imagine a 1 cylinder 100 cubic inch engine with a 1" stroke (everything is the same as the first engine except we have one cylinder instead of two), the bore must increase to 11.28". Now the circumference is 35.44" per piston, and since there is only one piston, total ring length is 35.44".

Conclusion: same displacement with fewer cylinders means less area swept by rings.

 

I agree with what was stated.

However, if your comparing a 300 I-6 to a 302/351w v-8, the bore doesn't increase with less cylinders, the stroke does, longer stroke = more area swept by the rings.

 

So would you rather pull 600 lbs 20 feet or pull 800 lbs 20 feet? It's as simple as that.

 

Yes, the stroke does matter. What's the bore/stroke of the I6, and what's the bore/stroke of the v8? It's easy to calculate the total swept area:

bore * pi * stroke * cylinders

To get an exact result you would need to factor in ring thickness and tension, but those being equal you can compare swept area directly.

 

whats this have to do with the engine? im puzzled...

 

We're trying to understand how the 300 I6 may be more or less efficient in various ways than a 302/351w V8.

Which might ultimately support or undermine claims of a certain power output. Or might just be nitpicking that won't answer any questions.

 

After reading this thread, all I can say is I'm not feeling so "superior" about having a 460 in my truck anymore!

 

300 actual displacement is 300.08563

302 actual displacement is 301.5936

swept area for (300) 300.08563ci is 299.9328

swept area for (302) 301.5936ci is 301.44

now take swept area of the 300 aka 299.9328 divide by cubic inches 300.08563 = .9994907 per ci

now take .9994907 X (ci of 302) or 301.5936 = 301.43999 which is very close to the original swept area for the 302, actually a difference of .00001

Lets say they are near identical.






Silver streak

I am still trying to grasp your concept, to draw in X cubic inches with two identical bore engines, regardless of the cylinders, the rings have to move Y distance regardless.

Picture this, all cylinders are at TDC at the same time and BDC at the same time, for my example engines. First example with 8 cylinders, obviously 8 cylinders would require a shorter stroke, than one with 6 to move 300ci given each share a simillar bore. Given the 10lbs force per cylinder rule, the 8 may take 800 lbs to move 300ci, however it moves over a shorter distance, as compared to a 6 cylinder engine that takes 600lbs of force to move 300ci but it moves over a longer distance. Wouldn't force in relation to distance play some role in resistance?


Its interesting though, I have an open mind, one day it will become clear....yeah right.



No more bumping this thread, it gives me a headache.

 

Motorhead: I didn't realize they were actually that close! You're right, that is essentially identical. And the per-cubic-inch number is important too. Thanks for actually running the numbers there, I would have expected the V8 to have a bigger swept area.

 

Is there a formula for force / distance / resistance?