THE_PROFESSOR

Anybody can type, mzoo, give explanation!

Old SStroker

Do you really want to go there?

You offer a target-rich environment.

mzoomora

iTrader: (36)

You do understand that given the same displacement an engine with a longer stroke will also have a longer distance for the piston rings to travel. The more they move the more friction they make, which usually equals wear. Not to mention the increase in piston SPEED since it is travelling a longer distance in the same period of time at a given rpm, which also equals more wear. So even at idle, low rpm cruising, highway speeds, etc., the piston rings would see more wear.
3.48 stroke x 2000 rpm = 6960" of travel and travel per minute per minute
3.82 stroke x 2000 rpm = 7640" of travel per minute(10% increase)
So not only will it be going 10% faster, but also 10% further. The increase in distance is pretty easy to figure out how it will affect wear, but the increase in speed Im sure would require a formula I do not know. Plus the extra heat it would generate in the rings.
So increased distance also at an increased speed equals increased wear. Not really that hard to understand, and I didnt think it would need to be explained. Also not hard to understand why GM wouldnt do it when warranty would be an issue.


No how about YOU explain how it makes no difference.

Old SStroker

Let's look at two engines of the same displacement, one a 3.000" stroke x 4.607" bore (call it the "Bore" engine) and the other a 4.000" stroke x 3.989" bore (call it the "Stroke" engine). Assume they are both 400 cubic inch V8 and that they each make maximum torque of 531 lb-ft, which is 1.327 lb-ft per cubic inch. Check the math if you wish.

BMEP (Brake Mean Effective Pressure) is the mean or average pressure in the cylinder during a power stroke. It can be determined from a cylinder Pressure-Volume (PV) diagram. a BMEP of 200 lb/sq-in (psi) is equivalent to 1.3263 lb-ft per cubic inch. Let's use 200 psi as the pressure pushing down the pistons
in both the "Bore" engine and the "Stroke" engine. FWIW the "Brake" part means measured at the flywheel by a dyno or "Brake", the old name for a dyno...or a nickname often used by Brits.

How much instantaneous torque (lb-ft) does each engine produce when has a BMEP of 200 psi and the crank and rod are at a 90° angle?

"Bore" engine has a piston area of (4.607^2 x .7854) or 16.670 sq-in (in^2). So 200 lb/sq-in x 16.670 sq-in = 3334 lbs of force on the piston. At 90° crank/rod angle the moment arm is stroke/2 or 3/2=1.5 inches or 1.5/12 = .125 ft. Now 3334 lb x .125 ft = 416.75 lb-ft.

"Stroke" engine has a piston area of (3.989^2 x.7854) or 12.497 sq-in. So 200 lb/sq-in x 12.497 sq-in = 2499.4 lbs of force on the piston. At 90° crank/rod angle the moment arm is stroke/2 or 4/2 = 2.0 inches or 2.0/12 = .1667 ft. Now 2499.4 lb x .16667 ft. = 416.65 or, except for rounding of the numbers, the same instantaneous torque as the "Bore" engine.

You might see that the longer "arm" has less force pushing on it because the piston area on which the combustion forces act is smaller for the "Stroke" engine than for the "Bore" engine.

If we keep the 4.607 bore of the "Bore" engine but increase the stroke to the 4.000 of the "Stroke" engine, the instantaneous torque is 200 lb-sq-in x 16.670 sq-in x .1667 ft. or 3334 lb x .1667 ft = 555.8 lb-ft. Yep it's a lot more! Actuall i's 4/3 times more, the ratio of the strokes, and also the ratio of the displacement which is now 533.3 cubic inches.

All of this example assumes the same frictional losses as well as the same efficiency of pumping air for both engines. Others have pointed out that the "Stroke" engine will have more friction losses due to the higher piston speed and more area rubbed by the rings, and also that the "Bore" engine might be able to breathe better beacause it could support larger valves with less shrouding. Neglect all those things in this example for they would make the "Bore" engine produce more BMEP.

Anyone buy into this? Isaac was a good teacher.

Cascazilla

[QUOTE=THE_PROFESSOR]This goes back to leverage. When the crank is at the horizontal positon, it is easy to see that the larger stroke crank will develop more torque from the conn rod. However, places more demands on crank, rods, block, due to more piston accelerations and more extreme angles.

All things being the same, a stroker is a torque machine, and a bore is a rpm/head flow engine. This is not conjecture. More bore means more effective valve area, which increases valve area flow by a squared factor.


For a 10% increase in displacement, plan on a more than 10% increase in torque (346 --> 383). Unless head/intake flow become a restriction.

QUOTE]

Wow, you are pretty smart with this stuff. Where can I learn all this? Do you have a PhD in Physics?

RoDan

Sunset'01Z

iTrader: (10)

just get a 347 and turbo it

THE_PROFESSOR

We have some explanations here, with reasonable thinking. Let me offer you my side view.

mzoo, you offer a compelling argument for wear. Further stroke means further distance travelled, and a reasonable and easy path for this would be that more wear would occur. I agree. However, I am having problems with the velocity thing. I think it has to come down to thin-film lubrication analysis, and I am not a Tribology engineer. We know that plain babbit bearings lose contact with the crankshaft during hydrodynamic lubrication thus no wear, unless overloading (high rpm) I have my doubts that this applies to ring-cyl wall interface. However, I will say this. In my opinion, the increased angle when taken too far, could very well be the dominating factor in ring wear due to the side loading.

Let's take a look! For a slider-crank mechanism, x, displacement is defined as: (no equation editor available)

x= R cos(theta) + L* {1-(R/L)^2 [sin(theta)]^2}^.5

where R is crankshaft throw (1/2 stroke), L is rod length (effective), theta is crank angle, measured from a line through center of crankshaft main bearing and thru center of piston. Crank throw on right, piston on left. Any position where there is no mathematical symbol denotes multiplication. We'll keep convention of ccw+, displacement = 0 at TDC.

Let's derive for velocity!! chain rule applies.

v = -R sin(theta)(omega) – L/2 {(1-(R/L)^2 [sin(theta)]^2}^ ½ (R/L)^2 2sin(theta) cos (theta * omega)

simplifying, using some creative algebra:

v = -R sin(theta)(omega) {1+ {1-(R/L)^2 [sin(theta)]^2}^.5 R / L cos(theta)

where omega is angular velocity, or thetadot.

Comparing velocity of the different strokes, 4.125 vs. 3.622, and rpm = 6000, or 36000 deg/sec, i came up with a ratio of 1.14, or an increase in velocity of 14%. I am using a rod of 6.125 eff. length for the stroker. mzoo has assumed that wear is proportional, I am again not a Tribologist.

Let's compare side loading, to see percentage of increase with the longer stroke. Using similar triangles and placing the crank at 90*, ratio of side loading is 15%. I got a push here, little difference to the proportion of velocity increase.

Of course this analysis is with the assumption that there is some kind of friction here, in lieu of hydrodynamic lubrication. Friction again is independent of speed. There is static and dynamic friction changes but once underway, frictional loading opposite of the velocity is the same at any speed. We could take a look at thin-film shear of the lubricant, undoubtedly a property of viscosity, etc.. But since is a minor effect and occurs only when assuming thin film lubrication, it is minimum. If we are assuming thin-film lubrication at the rings, throw wear out the window. Isn't that what the cross hatch is for..???

I am having a tough time finding any loads on the side of the piston, however, unless the result of the side-loading described before as a result of the crank at 90*. Conventional physics says that for two objects rubbing against each other, friction is independent of velocity. MY GUESS IS THAT IT IS A COMBINATION OF THE TWO , side scrub and a minor component due to the increased velocity. Again, I am not a tribologist.

My undergraduate and master's degrees are Mechanical Engineering, My PhD work is in Civil, Structural. This doesn't mean jack squat until legitimate technical argument is utilized. If you have a point, OFFER AN ARGUMENT.

sstroke, you have put together an interesting point of view. I would like to address on this thread later on tonight!

THE_PROFESSOR

SSTroker, you have offered a sound and technical argument. Not to directly dispute you here or to defend a position, but let's take a look at a different point of view. I think we might possibly be looking at differing RPM bands.

You have your stroker and your bore version. One has a bore of 4.61 inches, the other has a bore of 3.99 inches.

Let's see what we can fit into this effective valve area. If we hold an intake/exhaust diameter ratio of around 1.3, a 2.5 diameter intake valve combined with a 1.94 exhaust valve utilizes 4.44 inches of the total valve distance of 4.61. We could go slightly larger. With the 3.99 bore, we could use an intake of 2.14 dia. and an exhaust of 1.66 inches, 3.80 inches utilized. Intake flow curtain is gonna be of circumference of 13.94 inches and 11.93 inches for the bore combo and stroker combo, respectively

For HP, which valve combination do you think will fill the same cubic inches of these two engines more effectively? I am of the opinion that this points to a result that the bored engine is a high-rpm performer. Perhaps the big bore will develop it's own high torque, developing the fore-mentioned BMEP at a higher RPM.

For a stroke engine, other than the argument that I have already presented regarding leverage, I will offer this for you. First, with a smaller bore and smaller intake valve, a smaller runner volume will certainly accompany. This smaller intake runner as well as a smaller exhaust runner lends to high incoming air charge velocity, aiding low end torque. Again, opposite from the bored engine in that the effective RPM is lower.

Now, here is some conjecture. With a longer stroke, we are agreed that increased piston velocity will result, in particular, at that 90* crank angle. Incoming air will start at near zero at TDC (piston vel = 0), albeit aided by scavenging. As the piston gains velocity the incoming air will struggle to meet that increase in volume, gaining velocity. Perhaps this increase in piston speed will aid low-end flow as well, at the detriment of high end, because of limited valve area.

Friction again is independent of speed. There is static and dynamic friction changes but once underway, frictional loading opposite of the velocity is the same at any speed. From above post: “We could take a look at thin-film shear of the lubricant, undoubtedly a property of viscosity, etc.. But since is a minor effect and occurs only when assuming thin film lubrication, it is minimum. If we are assuming thin-film lubrication at the rings, throw wear out the window”. Also, friction is independent of surface area. There are situations where this is not true, notably the interlocking of a rough asphalt and a flexible tire that molds itself to these imperfections. I am of the opinion that this is not the case, we can stick to the rule. Either way, take note that the stroker has higher velocity, and the bore engine has more friction surface. Much like your argument regarding torque leverage vs. bore area, these may cancel out.

I also offer some other points of view:

http://autospeed.com/cms/A_108647/article.html
So, to increase torque, increase the stroke, or increase the capacity(volume), or both.............Increasing stroke will increase torque, so theoretically it would be good to have very long stroke engines. The problem is, if the stroke is too long, the volumetric efficiency decreases, particularly with increasing revs (which is why long stroke engines don't like a big rev, apart from the rotating friction and harmonics....

http://www.hotrod.com/howto/69883_stroke_any_engine/
The first step is usually deciding what the crank stroke will be, and since the priority of this story is maximum torque, we’ll focus on gaining the longest possible stroke.

http://en.wikipedia.org/wiki/Undersquare#undersquare
An undersquare engine usually has a lower redline than an oversquare one, but it may generate more low-end torque.

If there is an overall message here, it is that there are rarely two different engines that have “all else equal”. But that being said, strokers lead to torque, bored lead to HP, if maximized for their respective dimensions, like GM would certainly have the means to do. Let's take a look at one of your recent posts: 20 Sept 2007, Thread “Any DeStrokers or Short Strokers out there with 4.125 bore”

That's one approach. Another is to build a strong flat torque curve (stroker) that will pull you along with fewer gear changes and put the HTC (human traction control) in the driver's right foot.

Sometimes you find that you are faster in a higher gear where the (stroker's) torque response to the tires is "damped" not by lack of torque but by the lack of gear. When you have the traction to put more down it's there. In the hands of a competent driver more torque makes it faster. No offense intended.

My thinking is dominated by the areas I am most involved in, Solid Mechanics notably. And I fully recognize it would be of poor character to discount others on their arguments based on my background. My time with WJ taught me that. I post so that, with an easily understandable explanation, others may better understand the answer to their question. It is a pleasure, it is what i enjoy. I appreciate your interest in searching for my posts, but I am afraid I am unconcerned if that provides a "target rich environment". Please feel free to argue a different point of view, though. I appreciate your ability to develop an argument here.

Old SStroker

Prof, your recent posts are along way from "This goes back to leverage. When the crank is at the horizontal positon, it is easy to see that the larger stroke crank will develop more torque from the conn rod."

I'm not saying I agree with all of your thoughts on what gets air into an engine, but that is another subject.

My "stroker" quotes about roadracing engines were in reference to larger displacement vs. smaller displacement when rules don't prohibit this. With an LS engine resleeving to get a large enough bore to substantially increase the displacement is more costly than stroking. In previous generations of the SBC where .040 might be a safe overbore, stroking is the practical way to get more displacement.

You may have missed this one from the same thread:

"One more try. It isn't the stroke that makes torque down low or up high, it's the size of the engine and how it is configured to shape the torque curve.

You could very well do a big-bore short stroke engine with long intake runners, proper heads and valve timing and make great torque per cubic inch in the low-medium rpm range. You also configure a long stroke engine to make the same torque per cubic inch in the same rpm range. The way the air is gotten into and out of the engine determines the shape of the torque curve and also the amount of torque and power the engine makes. The larger displacement engine could make more of everything."

Friction, or FMEP is important, but I'm not convinced piston ring to cylinder wall friction is independent of contact area and rubbing velocity (or speed as you said). Approximately 70% of an engine's friction comes from the rings, and friction losses increase with rpm. I'm not talking PMEP or pumping losses, just internal friction. This can be measured by motoring on a dyno, and in some good engine simulators it can be fairly accurately calculated.

Rings and cylinder bores wear with engine cycles. In my example the "Stroke" engine has about 15% more square inches of ring/bore contact area than the "Bore" engine of the same displacement. I find it difficult to believe that more heat and friction losses ar NOT associated with the extra sliding contact area.

Not to pick apart the magazine articles, but the Autospeed author suggested the same force on the con rod from the smaller bore stroker, but in his door analogy he had a smaller force and a longer moment arm. There's a name for this type of reasoning, but I can't recall it.

Perhaps folks confuse cause and effect here. It is intuitive to think the longer lever arm will make more torque. What isn't necessarily intuitive, with the same displacement engines, the force on the lever arm is less. That was my point in the tome above.

The Hot Rod article was all about getting more displacement from an engine with limited overbore. They got the larger displacement part right, but they did mention that the longer moment arm helps make the torque. ONLY IF THE FORCE IS THE SAME as it might be with the same bore. Once again, more displacement equals more torque. It's the "long arm" explanation that confuses...evidently both writers and readers.

Anyway, in your words, "Anybody can type...give explanation!"
It's good that you are doing that now.

Here's a quote from Newton that I like. He used it in physics class often.

"The latest authors, like the most ancient, strove to subordinate the phenomena of nature to the laws of mathematics." --Sir Isaac Newton

THE_PROFESSOR

Mission accomplished here.





Reality is just imagination. Imagination is true reality.
--It's the Don Quixote syndrome.

"against stupidity the gods themselves contend in vain" Friedrich von Schiller

Cascazilla

I don't want to make anyone mad because I actually just work at Lowe's, but I am saving up to by my cousin's Camaro.

I just remember from Physics class about torque and power. Proffesor, are you really Warren Johnson?

Anyhow, when you say that longer stroke engines have more torque, why does the torque matter if you don't talk about the rpm with it? Why does the leverage the rod has on the crank matter if you don't say how much area the piston has. Professor, I am starting to doubt you a little.

Oldstroker says that people say the longer stroke engine will make more torque even with the same displacement. I have heard that too. I dunno

BTW: Professor, you said something about "more torque makes it faster". Is this factoring rpm in? Why doesn't a power speed calculator tell you how much "torque" you need to run 10.0 @ 138? It gives you horsepower. Is power more important tthan torque or the same?

THE_PROFESSOR

I am not WJ.

To continue this discussion is fruitless. Best of luck on your conclusions.

rocksws6

iTrader: (4)

Horsepower = torque x rpm / 5252

Old SStroker

You have asked some very good questions. I'm not sure you are as "aw shucks" as you come across, but here is my take on what you asked:

1) Yes torque matters, but so does the RATE at which you apply the torque, which is called power. So don't we really want to make the most TORQUE we can at all rpms we use in our race, and especially make as much torque as possible at the highest rpm possible? That last part was paraphrased from a ProStock guy who races a two-car team with his son.

2) Why does torque matter? It goes to what actually accelerates a vehicle. Prof. Newton said that F=Ma. If you solve for a (acceleration), a=F/M. That says Force pushing on the vehicle chassis (in the direction you want it to go) makes the Mass of the vehicle accelerate. That's not power nor torque, but FORCE. The FORCE comes from TORQUE applied to the drive wheels.

If the drive tires are getting the power or torque to the track, we can figure how much FORCE is being applied to the track surface if we know the torque at the wheels and the tire radius. Let's look at a snapshop of one instant in a drag run: Let's say 400 lb-ft at the flywheel, a 3.00:1 first gear ratio and a 4.00:1 rear gear. To make the math easier, let's forget losses due to friction, etc. for this thought experiment. OK, so 400 lb-ft x 3.00 xs 4.00 = 4800 lb-ft at the tires. Let's say the tires have a rolling radius of 12 inches (1 ft) again to make the math easy. If TORQUE is FORCE x Distance, then FORCE = TORQUE/distance or 4800lb-ft/1ft or 4800 lbs. This force is acting at the tire/road contact patch trying to move the road back away from the car.

Prof. Newton proposed a Third Law which said "For every action there is an equal and opposite reaction." The reaction to the tire pushing on the track is an equal and oposite force pushing on the vehicle thru the axle centerline and then thru the suspension links to the chassis. This is the force that accelerates the vehicle (hopefully) forward.

Now look at this happening aver a period of time. The torque has to be applied more and more often at the tires as the vehicle accelerates to a higher velocity (mph). Torque applied at some RATE (# of times per second or minute or week) defines power. We usually use # of times per minute (rpm).

When we get the car to the end of the strip , let's say @ 138 mph you are obviously not in first (3.00) gear. You are still applying enough torque to the drive wheels to overcome the drag on the vehicle, which at 138 mph in a passenger car body is a significant amount. The vehicle is also accelerating at the finish line (hopefully) and the engine is making torque at some high rpm near it's hp peak. It has to be doing that to keep up the 138 mph.

Now we have torque and rpm so we can figure power. Actually the power/speed calculator backs you into torque. Let's say it says you need 600 hp to get your 138 mph and you know you are crossing the finish line at 7000 rpm. HP= Torque x RPM/5252, so Torque = HP x 5252/rpm or in this case about 450 lb-ft @ 7000.

If your engine could make another 25 lb-ft of TORQUE@ 7000 it could make the vehicle faster. 475 lb-ft @7000 is about 633 HP, right? That extra 33 hp could result in more mph.

A thoulght experiment for you, Cas: How else could we get more torque to the rear tires at the finish line without a really changing the engine? Forget driveline efficiency, concentrate on finding more rear wheel torque at the finish line (and over the last third or so of the strip).

PROF= WJ? ROFL here.

Mike94ZLT1

iTrader: (3)

This thread makes my head hurt

3rdgentug

iTrader: (23)

Cascazilla

Does that mean you are really not very smart? I am disappointed.

fueledpassion

For the OP,

GM didn't make larger stroked motors for the F-Body and C5's AT THE TIME, because it was obviously not necessary to reach the performance goals desired by the GM engineering crew...more technology, more tuning, more money required to stroke the LS series for a reliable, fast, and completely drivable street motor..something they felt wasn't worth producing the work for, especially in Corvette's sensitive and edgy situation w/ sales.

When GM produced the LS1, they had C6 in mind more than anything else. Building a strong but simple and stock platform was the goal so that things like the LS7 were created in the future. They knew there would be awesome revisions which could have been put into action long before they did..but because GM and Corvette have a strict code of production w/ their generation to generation models...it was not time to put those huge changes into action UNTIL the next gen, which we call GEN4 now.

And since F-Bodies shared the same motor platforms w/ the C5's, they were subjects w/ the same outcomes.

Dude..back in 2000, my C5 was bad-***. Period. No one but Viper's messed w/ that car. Even in 2004, the LS1 and LS6 were bad to the bone motors, and even better when actually sitting in the car(s) designed for it. Just remember that a low 11's sport car was not necessary 4-10 years ago to reach a good sales pitch..which is what all of this came down to. Sales.

Gannet

Old SStroker is correct. An engine's torque output, and the RPM where that output occurs, is dependent on a number of factors in the engine's design, but stroke length itself is NOT one of them. Two engines with the same displacement, and the same volumetric efficiency at a given RPM, will produce the exact same torque at that RPM, no matter what their stroke length is. "Lever arm" doesn't enter into it, for the very reason Old SStroker explained: the surface area that is transferring the force from the expanding gas is proportionate to the stroke, if displacement is held constant.

As a practical matter, and just considering racing purposes, the effective ranges of bore to stroke were established long ago. Short stroke engines can have a lot of valve area, and hence can breathe well. Their primary disadvantage, putting valvetrain issues aside, is elevated piston speeds and the difficulty in preserving ring seal at high piston speeds. This is what sets the upper limit on what is possible. The only real advantage to a long stroke engine is displacement - but that's the most powerful advantage there is.

When considering GM's original design decisions other factors come into play, including fuel efficiency, emissions requirements, cost of manufacture, durability, and reliability. When comparing the LS series to earlier engines, they are relatively long stroke, and I suspect this is largely for fuel economy and emissions reasons. Note that when GM goes racing with the LS, they tend to use short-stroke variants where possible.

The reason a long-stroke design tends to be better for emissions and economy is because it has a relatively smaller combustion chamber and smaller piston area. Pistons that are smaller in diameter make detonation easier to control, and that in turn allows higher compression, and that yields better fuel economy. The reason detonation is easier to control with a smaller piston is that detonation tends to occur far from the spark plug, out at the "edges" of the combustion chamber. Smaller pistons reduce this distance, and therefore make it more likely that controlled combustion will occur prior to detonation setting in. Large combustion chambers tend to lead to higher emissions because they have more surface area, which leads to more localized cooling of the burning charge, which leads to less-complete combustion.

All just imo and, although I used to work as an automotive engineer, it wasn't in this area of expertise, and I didn't stay at a Holiday Inn Express last night. So fwiw.

Bottom-line: if power is the primary concern, run the largest displacement that is practical given other constraints. The stroke is what it is.