Has anybody used the piston as the combustion chamber? Make a bath tub sort of cut into it and get real close to the head on a flat head?

 

The 348 and 409 Chevy "W" engines had the deck at an angled at something other than 90* to the bore centerline. The combustion chambers in the head were almost nonexistent. There were just the valves and the spark plug. The combustion chamber was made up by the deck of the head, a wedge shaped cylinder (wedge, due to the angle cut of the block) and the piston top. It wasn't all in the piston, but it's the closest I've seen.

RACE ON!!!

 

OK Guys, here’s some numbers to ponder.

I’ve looked at the distance from TDC vs crank angle achieved with a 5.7 inch rod and a 6.0 inch rod in a standard 350 CI chevy with a 3.48 inch stroke. With the crank 10 degrees ATDC, the 5.7” rod has the piston -.0344 in the hole, a 6.0” rod is -.0340” in the hole, a difference of .0004”. At 20 degrees, the 5.7” rod is -.1361 in the hole, the 6.0” rod is -.1345” in the hole, a difference of .0016”. At 45 degrees, the 5.7” rod is -.6440 in the hole, the 6.0” rod is -.6371” in the hole, a difference of .0069”. At 90 degrees, the 5.7” rod is -2.0120 in the hole, the 6.0” rod is -1.9978” in the hole, a difference of .0142”.

From a perspective of a longer rod giving you more time at the top, or a shorter rod pulling the piston down faster for more resistance to detonation, the difference between a 5.7” rod and a 6” rod seems pretty small.

There are, however, other issues involved in the long rod vs short rod debate. One that was mentioned previously is side loading. The piston would not see this side loading if the piston center axis was always aligned with the rod (as it is at TDC and BDC). However, as the crank rotates, the angle of the rod to piston increases and peaks when the crank is at 90 or 270 degrees. When the piston pushes on the angled rod, it is forced against the cylinder wall. This angle between the piston and rod is easy to calculate. Let’s assume 0 degrees defines when the piston and rod are in alignment. As one case, with the crank at 90 degrees, a 5.7” rod has an angle of 17.77 degrees, a 6.0” rod is 16.86 degrees, a difference of .91 degrees. As CFI noted, this might be a bigger issue than piston position when you actually calculate vector forces.

Finally, there is a final issue that I have not heard much talk about in my hot rod books and that is the angle between the rod and the crank throw. The best analogy I can think of for this is a merry-go-round. If you were standing outside a merry-go-round and someone hands you a ten foot pole and asks you to spin the merry-go-round. To get the best push, you would align yourself so the pole is tangent to the outside of the merry-go-round (or perpendicular to the radius). The crank and rod share the same situation. The best push is when the crank throw and rod are at 90 degrees. It is interesting that this occurs sooner from TDC in a short rod than a long rod. In other words, when the combustion is pushing harder on the rod, the shorter rod may actually be in a better position to take advantage of it. But again, as in the above example, the difference between a 5.7” rod and 6.0” rod is pretty small.

There are a million other factors to consider in all this. A 5.7” rod piston and 6.0” rod piston would have different pin heights if the deck was held at the standard 9.025”. The weight of the rods would be different. There’s cylinder pressure vs crank position, cam events, etc, etc, etc. I guess this is why Smokey Yunick was so good, he wouldn’t be writing about it right now, he’d be slapping together a test motor.

I’ll finish this with: these are the ramblings of an amateur. I am not an automotive engineer, so don’t take anything above as gospel. There could be mistakes and could be several factors beyond what I’ve brought up that could make a difference in the issue of rod length. I’m just hacking around with this stuff.

 

Cris,

About the side loading... In one of Smokeys books (The 3 book series) he talked about using an offset pin in the piston. To change the angles that were most effiecient when it came to the rotation.

I wish I had taken some notes when I read through the books... with page numbers!

 

Good, info, Cris. As you can see, when you crunch the numbers the difference between a 5.7" and 6.0" rod is pretty much noise level. That's why I think all the hype about rod length is just that - hype!

Many early SBs have offset pins, but I'm not sure if this design practice carries through to today. Same basic deal. Offsetting the pin can reduce piston side load, which can reduce the tendency for piston/wall scuffing, but it doesn't seem to make much, if any, difference. The key is skirt design/clearance and oil quality.

Duke

 

There is a bit of a problem with some of those numbers. OK, one number. I didn't check any of the calculations, but common sense tells you that the 90* number is off. At 90* the crank is at half stroke. In a 350 engine with a 3.48" stroke, the piston will have traveled (3.48/2) 1.74" down from TDC. It doesn't matter if it's a 5.7", 6.00" or a 10" rod. Halfway is halfway. If you are using a 4.00" stroke for your calculations, they still, both, should be 2.00", exactly.


A few posts back, I mentioned that when the crank arm and the con rod are at a 90* angle, that is the point of the greatest efficiency in power transfer, and the most piston travel per degree of crank rotation. I best visualize it by comparing myself on a bicycle, to the engine. My thigh provides the power. From my knee to my foot is the con rod, and the arm from the pedal to the crank hanger bearings is the crank throw. When the pedal is directly above the crank hanger, it doesn't matter how much you push. The leverage isn't there, and you cannot apply any power to the crank. When your shin and the pedal arm reach 90*, you can really pour it on.

It looks like we've got some guys thinking. Cool!

RACE ON!!!

 

The pin was offset .060" from center to help keep the engine quiet, but it resulted in a less direct push on the crank, when it counted. Back in the "day", stock class racers swapped pistons from one side of the engine to the other to gain an advantage.

RACE ON!!!

 

CFI-EFI:

At 90 degrees, it is not down half the stroke. Remember the connecting rod as been dragged not only down by half the stroke but OUT by half the stroke. This is why the net distance the piston gets pulled down is more than half the stroke. If you made the rod infinitely long, the piston would be down by half the stroke. Draw it out and it should make sense. I have hen-sketches of cranks and rods all over the place now. In the case where the rod length is half the stroke (now that you would ever do this), the piston is down by the FULL stroke at 90 degrees.

 

Moronic question...

If I have a circle with 360 degrees, where is 0 degrees? On a clock, 12 o'clock, 3 o'clock... etc?

Trying to get my visual of what you guys are talking about with 90 degrees. I see that as a quarter turn... but it depends on where you start! TDC a quarter turn clockwise from the snout of the crank would put your rod journal pointing straight down would it not?

 

Zero can be where ever you define. For engines, it's most common to define zero as TDC, which would put the crank throw at 12 o'clock. Of course, zero and 360 are the same thing; 90 ATDC places the crankthrow at 3 o'clock (for engines that rotate CW when viewed front the front, like Chevies), 90 BTDC places the throw at 9 o'clock, and 180 degrees is BDC, which places the throw at six o'clock.

Duke

 

Gotcha...

I had a momentary lapse of brain function... was thinking about the Chevy like it was a boxer!

 

Thanks for the info Cris,

This is one of the better reading threads I have seen in a long time....

 

Cris:

I ABSOLUTELY

What seemed "natural" when I wrote that, is flat out WRONG. I thought about this while watching TV last night and I have seen the error of my thinking. I was hoping to print a retraction, before I got caught, but...

Several things occur to me that prove me wrong. One, is with different rod lengths, when the crank is at 90* the angle between the crank and the rod will be different. With a different angle and the different rod length, the leg of the triangle from the crank centerline to the pin will be different. THAT blows my half stroke at 90* with any rod statement.

Another thing is that, as I said, the piston movement is the greatest, per degree of crank rotation, when the crank arm and the rod are at a 90* angle to one another. That 90* relationship occurs BEFORE the crank reaches 90*, regardless of rod length. Therefore the average speed (and total distance traveled) of the piston is greater in the first quadrant of rotation, than in the second. As your numbers show, the piston reaches half stroke before the crank gets to 90*.

As I said, I hadn't checked your actual figures, but I disputed their accuracy, especially at 90*. I was WRONG. I retract that former statement. I still, haven't verified your numbers, but I, NOW, have no reason to doubt them.

Good work, and thanks.

RACE ON!!!

 

Its ok CFI... I think it was above my head whatever you said that didnt make sense cause I didnt catch it the first time around!

 

You need to pay closer attention. Every once in a while, I get it right.

RACE ON!!!

 

I'm not sure if its the attention that needs correction or the level of understanding!

Its very interesting stuff, and I can wrap my brain around some of it... but golly jee whiz its a bit above my level of competency with math!

 

it is interesting to see the actual difference in piston position (long rod vs short rod) and the difference is quite small at times.
but it makes me wonder about how the cam timing would play out (verses piston position) on them both.

say the intake valve opens at 14 deg BTDC and close's at 50 deg ABDC. the postion of the piston in the bore would be slightly different for each length connecting rod.
it would be interesting to see what the cam timing would be to get the same piston position on the other lenght connecting rod.
hmmmmm.

don't have the time right now to figure it, maybe tonight.

 

Unlike others, better equipped, I haven't used any math. Only mental visualizations. Follow the pictures these posts paint, and it will all of a sudden make sense.



Cam timing and the rpm range is what makes rod length tuning work, but as it has been pointed out, the difference in piston position at a given degree of crank rotation is so minuscule, that it plays no part in the engines we build. For all out racing, where the slightest edge is worth big bucks, maybe, but in my opinion, not for 99.99% of us. That doesn't make this discussion uninteresting or moot, just one with little practical application.

RACE ON!!!

 

What it does for me is make me wonder about a real long rod on a short stroke crank... if the rod is the lever working the crank... maybe instead of gearing the two to match we could take advantage of the ratio!

 

The con rod isn't a lever. All it does is push. That is, transfer the pressure exerted on the piston to the crankshaft. That transfer is most efficient when the rod and crank form a 90* angle. Among other things, the length of the rod, can affect the crank position where this occurs, SLIGHTLY.

The LEVER is the crank arm. If you want more leverage twisting the crank, increase the length of the crank arm. Generally, we call that "stroking" the engine. Like making a 383 out of a 350.

I don't understand the last part, of your last sentence, at all. What are you referring to as being "geared"? What is being "matched"?

RACE ON!!!