Formula934

iTrader: (4)

Can someone explain to me how engine torque is derived. I know torque is force multiplied by distance but my question is how does this relate to overall engine torque. It is my understanding that in a cylinder torque cannot be made when the piston is at the top of the stroke because its a vertical movement. Hopefully, someone can help me understand this and maybe chime in to help me understand why strokers (396 preferably) build more torque across the boards. Thanks alot, any advice is appreciated.

3.4camaro

iTrader: (1)

WRONG.

Am I explaining this right?

Old SStroker

Nope. But you have fallen into a common pothole. Don't feel alone.

Generally a larger displacement engine makes more torque than smaller one simply because it is larger. This is especially true if the BMEP (Brake Mean Effective Pressure) or torque per cubic inch is the same for both, which would not be uncommon.

Hint: Read up on BMEP.

Formula934, this might help:
http://science.howstuffworks.com/fpte4.htm

Alvin

iTrader: (2)

Bingo!

tee-boy

iTrader: (8)

Torque is directly proportional, everything else constant, to the total amount of air/fuel mixture entering the cylinder. The bigger the displacement, the more air/fuel that will enter. Air and Fuel is good for Torque.

3.4camaro

iTrader: (1)

Wait wait, so the fact that the throw of the crank is longer, and creating a longer lever has nothing to do with it?

Seems counter-intuitive....

Old SStroker

Think about this:

For equal cylinder displacements, the long stroke will have a necessarily smaller bore, so the cylinder pressure is acting on a smaller area, and givng less "push" to the crank.

Does that turn on the light?

If the bores are the same and the stroke longer, there will be more displacement in the "stroker" and we are back to square one: "No replacement for displacement" (except perhaps lots of money to make a better BMEP)

tee-boy

iTrader: (8)

Same displacement but diff strokes... The shorter stroke will produce more total force and over a shorter distance. Longer stroke will produce less force over a longer distance.

The torque curves for the 2 motors will be different.

Old SStroker

Please unconfuse me.

topend

strokers make more torque because there is more displacement. When u increase stroke and/or increase bore a motor u add displacement which in turn increases the torque of the engine.

It does not matter what the stroke or bore of the engine , total engine displacement is what matters when speaking of torque.

tee-boy

iTrader: (8)

The torque curves won't look the same. The stroker will accomplish peak tq at a lower rpm.

Old SStroker

Will the peak torque (BMEP) be the same? Why will the longer stroke engine of the same displacement make peak torque at a lower rpm?

So if peak torque of the stroker is at a lower rpm, does that mean that high rpm torque of the stroker will be less than that of the "bore" engine? If yes, you know what that means, right? Help unconfuse me some more.

MrBill97396

iTrader: (10)

Explain

tee-boy

iTrader: (8)

Torque is force times distance, correct? The charge or force has a lower surf area to act on in the stroked motor; however, the pistion will travel a longer distance. The non stroked motor has a larger surface area for force to act upon and will travel a shorter distance.

MrBill97396

iTrader: (10)

At the end of the 2006 season, these F1 engines made in the vicinity of 750 HP at an astonishing 19,000 RPM. Assuming peak power is around 18,500, the torque at peak power would be 213 lb-ft and peak-power BMEP would be 219 psi. Peak torque BMEP would likely be at least 10 psi greater. There can be no argument that 219 psi at 18,500 RPM is truly amazing.

However, let's look at some astounding domestic technology. The 2006 Nextel Cup engine is a severely-restricted powerplant, being derived from production components. It is based on a production cast-iron 90° V8 block and 90° crankshaft, with a maximum displacement of 358 CID (5.87 liters). A typical configuration has a 4.185" bore with a 3.25" stroke and a 6.20" conrod (R/S = 1.91). Cylinder heads are similarly production-based, limited to two valves per cylinder. The valves are operated by a single, engineblock-mounted, flat-tappet camshaft (that's right, still no rollers as of 2007) and a pushrod / rocker-arm / coil-spring valvetrain. It is further hobbled by the requirement for a single four-barrel carburetor. Electronically-controlled ignition is not allowed, and there are minimum weight requirements for the conrods and pistons.

How does it perform? At the end of the 2006 season, the engines were producing in the neighborhood of 840 HP at 9000 RPM (and could produce more at 10,000 RPM, but engine RPM has been restricted by means of a rule limiting the final drive ratio at each venue). 840 HP at 9000 RPM requires 490 lb-ft of torque, for a peak-power BMEP exceeding 206 PSI. Estimating peak torque to be 550 lb-ft (probably in the neighborhood of 7800 RPM) yields a peak BMEP of nearly 232 PSI.

courtesy of www.epi-eng.com

3.4camaro

iTrader: (1)

These 2 things offset each other, if I'm getting SStroker correctly.

On a side note, I got a 96% on my engineering test this week. WOOT!

tee-boy

iTrader: (8)

Yes they do. But I still say the curves will be different.

THE_PROFESSOR

integrating pressure- dee- volume leads to an equation like:

Power = (BMEP * Piston area * Volumetric Eff * Mean piston speed)/4

for a steady-state condition

Torque = P/(2 * pi * Rev/sec)

combine 'em, if you wanna solve for torque.

analyze 'em if you wanna discover something about engine characteristics............
what factors are stronger for undersquare/oversquare?? what about V.E. wrt piston speed??

flame on!

Formula934

iTrader: (4)

thanks guys this stuff is all making sense to me. However, if were measuring torque in ft. pounds, lets say hypothetically a 383 makes 500 ft pounds, is the calculations of engine torque using the same principles as lets say a person putting 200 pounds of force on a 1 ft breaker bar turning off a lugnut=200ft/pounds. So the motor is only putting out a little more than double the amount of a torque the average person could exert? I guess what im asking is, Is it possible for torque to be calculated on other variables? I fully understand what OldStroker is saying and many others about having more force driving the piston down because the displacement is larger, allowing more air/fuel to enter the cylinder. Im just trying to rationalize these claims with torque being derived from force X distance=torque.

3.4camaro

iTrader: (1)

Let's try and get into some math now. If I keep the BMEP constant(same psi, but perhaps not the same area to which the pressure is exerted), I think we can show that the stroke and bore changing will offset, won't it?

So a BMEP for a standard LS1, given factory specs at 350 ft/lbs on an engine dyno, comes out to:

BMEP = 150.8 x TORQUE (lb-ft) / DISPLACEMENT (ci)

150.8*350/346 ~ 153 psi, assuming the 150.8 is a constant which keeps all my units in check.

so if you have a 3.910" bore(stock LS1 bore), the piston face area would be:

area = pi*r^2

3.14159*(3.910/2)^2 ~ 12 square inches of piston face.

12" * 153psi = 1837 pounds of force down on the crank. COOL!

So, we have 1837 pounds of force pushing on a lever of length 3.6"(stock crank throw IIRC).

At this point, I call on SStroker to check my work, and to lead the way, because I'm pretty sure the force isn't perpendicular to the lever in the case of our engines. This probly has alot to do with ignition timing and whatnot.