[vina]

Originally Posted by pranavt

Peak pressure at 16 degrees atdc is regarded as optimal. Less shock on the bearings and piston pin, more power transferred to the crank. Having peak pressure at TDC would destroy bearings like nobody's business.

bearing may or may not be destroyed, but it doesn't help in getting the direction of rotation set (I mean unless the momentum of the shaft keeps it going, at TDC the crank can go either way clockise or anti clockwise).

[Born2Slow]

IMHO big bores are a necessary evil i.e heavier piston, inefficient combustion etc, But it has to be done if you want to produce max power from a given displacement by virtue of its slow piston speed and larger valves that can be fitted.
If everything else is same then both long and short stroke should have similar power and torque curve. But it will be stupid to make a large head and fit small valves and heavy pistons if the same can be done with a smaller bore and lighter piston by increasing the stroke.

[vina]

Originally Posted by Born2Slow

IMHO big bores are a necessary evil i.e heavier piston, inefficient combustion etc, But it has to be done if you want to produce max power from a given displacement by virtue of its slow piston speed and larger valves that can be fitted.


I think you may be right;

If everything else is same then both long and short stroke should have similar power and torque curve. But it will be stupid to make a large head and fit small valves and heavy pistons if the same can be done with a smaller bore and lighter piston by increasing the stroke.

I don't know about that

Sutripta, Pranav any comments on the first part? Higher rpm will also mean higher piston speed, and there is a limit to that too.

[Mpower]

Let us assume they have similar frictional losses over a cycle, similar thermal losses (as a percentage of heat lost)

we're starting with an erroneous assumption but OK..

Originally Posted by vina

If that is the case then for 4 rotations (or 2 for a 2-stroke or 6 for a 6-stroke, etc. however let's limit ourselves to 4stroke here, others can be extended similarly) of the shaft - both have to do similar amount of work (efficiency is similar and same displacement, fuel-air ratio and compression ratio means you put same amount fuel per 4-stroke cycle into each engine).

Now over one complete rotation:

Work = Avg-torque * 2*pi where pi = 3.1416 (roughly).

For four complete rotations it'll be:

work = 4* Avg-torque * 2*pi


Now if work over one 4-stroke cycle is W1 for engine 1 and W2 for engine 2, and torque generate by the two engines are T1 for engine 1 and T2 for engine 2 then using the above work-torque equations and the fact that W1=W2 (deduced above) we get:

T1=T2

you don't even have to look for bore, stroke etc.

My head is spinning

How did 'average torque' become same for both engines.

Originally Posted by vina

the piston area compensates the stroke exactly (if displacement is equal) - so once displacement is same stroke should be irrelevant.

Kindly explain how?

Originally Posted by pranavt

Shouldn't it be the other way round? As in the spring height being determined by the stem length? Sounds counter-intuitive to do it the other way.

In a way they influence each other so you have to find a happy medium.

Originally Posted by Shan2nu

BTW, I've read that an engine achieves peak torque when the exhaust velocity is between 240-260ft/s. So could the stroke length have an effect on this?

where??
In what way is peak torque is related to exhaust velocity.
Exhaust velocity depends on BMEP, EVO and port goemetry.

Originally Posted by Shan2nu

For example, 1kg of force acting on a 1mtr lever creates 1kgm of turning force. But 1kg of force acting on a 3mtr lever will produce 3kgm of turning force.

you/re on the right track butttttt........ there's no free lunch. The Force has to travel a longer distance (perimeter) to make the circle.

[vina]

Originally Posted by Mpower

we're starting with an erroneous assumption but OK..

Why do say that?

My head is spinning.

I thought it was obvious. anyway,

W1=4*T1*2*pi, W2=4*T2*2*pi

if W1=W2 then put the above in here and you get T2=T1

[pranavt]

Originally Posted by vina

bearing may or may not be destroyed, but it doesn't help in getting the direction of rotation set (I mean unless the momentum of the shaft keeps it going, at TDC the crank can go either way clockise or anti clockwise).

Correct. But the same stands even for peak pressures just after TDC, say upto 10 degrees ATDC. Basically the same thing happens during pre-ignition or detonation which is why it kills an engine.

Originally Posted by Mpower

where??
In what way is peak torque is related to exhaust velocity.
Exhaust velocity depends on BMEP, EVO and port goemetry.

It's the other way round. If the exhaust velocity is exceeding those numbers, there's going to be friction and backpressure. Used to calculate exhaust pipe diameters basically. At those numbers, you have the best compromise (diameter not small enough to cause back pressure and not big enough to slow down gases and hurt scavenging)

[Mpower]

I dont see how W1 became = W2.

Before using, lets understand where the formula came from...

W= Force x dist = F_piston x dist travelled
dist travelled = perimeter = 2pi R where R is stroke
so W=F.2.pi.R (for one cycle)

So if stroke or R is larger then Work done (positive work) per cycle is higher and for a given rpm, the power (rate of work) is also higher. F_piston comes form mass of the charge combusted which is same because displacement+VE is same for both engines.

Originally Posted by pranavt

It's the other way round. If the exhaust velocity is exceeding those numbers, there's going to be friction and backpressure. Used to calculate exhaust pipe diameters basically. At those numbers, you have the best compromise (diameter not small enough to cause back pressure and not big enough to slow down gases and hurt scavenging)

Velocity is driven by delta P and cross sec area. If its too high its means EVO is too early or port is too small and you are wasting gas energy which could have worked the piston. Of course you cannot use the same number for all engines because FIs will have much higher BMEP and therefore delta P. Agreed too low is bad for scavenging
Backpressure OK but what kind of Friction?

[pranavt]

Originally Posted by vina

Sutripta, Pranav any comments on the first part? Higher rpm will also mean higher piston speed, and there is a limit to that too.

I'm honestly lost here. The thread has deviated into about 350 different directions, to the point where I'm not even sure what we're discussion anymore lol. And some of the technical stuff is making me remember my college days. I'm a CS guy, some of this is way beyond my current level of knowledge.

[vina]

Originally Posted by pranavt

Correct. But the same stands even for peak pressures just after TDC, say upto 10 degrees ATDC. Basically the same thing happens during pre-ignition or detonation which is why it kills an engine.

That's right; nothing to disagree - but I don't understand where did I state that peak pressure is at TDC (except may be in a rough mathematical model)?

[pranavt]

Originally Posted by vina

That's right; nothing to disagree - but I don't understand where did I state that peak pressure is at TDC (except may be in a rough mathematical model)?

Not you, that was in reference to one of Sutripta's posts which I quoted in the original post discussing peak pressures.

[vina]

Originally Posted by Mpower

I dont see how W1 became = W2.

Before using, lets understand where the formula came from...

W= Force x dist = F_piston x dist travelled
dist travelled = perimeter = 2pi R where R is stroke
so W=F.2.pi.R (for one cycle)

So if stroke or R is larger then Work done (positive work) per cycle is higher and for a given rpm, the power (rate of work) is also higher. F_piston comes form mass of the charge combusted which is same because displacement+VE is same for both engines.

Mpower

No offence here but I'm not sure you remember high school Physics now.

The moment you say R is larger, without saying Force is smaller, you implicitly assume torque is higher - proving torque is higher with that assumption is .....

Read my post again, without butting your head - may be you'll understand better.

Let me summarise once again:

For a given displacement, compression ratio, fuel-air ratio and efficiency, two engines with different bore/stroke (but the things above same - even if VVT, and a hundred other things are differenct) will still burn the same amount of fuel per 4-stroke cycle and generate the same amount of work (that is the definition of efficiency). No amount of computations of force/displacement are going to change this one.

Once W1=W2 over a cycle, torque has to be equal.


By the way, you wrote something about my assumptions earlier - care to explain that?


If you believe stroke matters, page 824 on Heywood's book states emphatically that it doesn't. It also mentions that peak power depends on bore.

Now I'm prepared to believe that Heywood's assertion may not hold given the practicalities of an engine, but that's the part where you challenge the assumptions I make - the burden should be on you exactly which ones are wrong and why?

[Mpower]

See it wouldn't matter if I remembered or not because they didn't teach us in our high school that piston force is not the same iin equal disp engines with different bxs.

As far as the rest of my physics formula, even Newton would agree that its correctly applied. No offence but I dont like using formulas without knowing how they are derived and sorry I dont have that book. IF that is indeed the case then how come 95% of the pass car engines are undersquare.

As far as the other assumption it should be obvious that friction losses are higher on longer stroke engine.

BTW, didn't know that Heywood is sitting next your keyboard all this while.

[Born2Slow]

Originally Posted by Mpower

I dont see how W1 became = W2.

Before using, lets understand where the formula came from...

W= Force x dist = F_piston x dist travelled
dist travelled = perimeter = 2pi R where R is stroke
so W=F.2.pi.R (for one cycle)

So if stroke or R is larger then Work done (positive work) per cycle is higher and for a given rpm, the power (rate of work) is also higher. F_piston comes form mass of the charge combusted which is same because displacement+VE is same for both engines.

Here your assumption is wrong because everything else same the pressure inside the cylinder is same but the force F on the the piston will be equal to pressure X Area of the piston i.e[pi x (bore/2)^2]. So that will compensate for the stroke and so w1 = w2.

[Sutripta]

Originally Posted by Born2Slow

But it will be stupid to make a large head and fit small valves and heavy pistons if the same can be done with a smaller bore and lighter piston by increasing the stroke.

Product rationalisation?

Originally Posted by Mpower

In what way is peak torque is related to exhaust velocity.

I think it should be seen the other way around: If everything is OK, these are the ballpark figures you should get. First iteration. Then finetune.

Originally Posted by pranavt

Basically the same thing happens during pre-ignition or detonation which is why it kills an engine.

Preignition destroys pistons essentially I think.

Originally Posted by pranavt

The thread has deviated into about 350 different directions, to the point where I'm not even sure what we're discussion anymore lol.

360. Why discriminate against the other 10 degs.

Regards
Sutripta

[Mpower]

Originally Posted by Born2Slow

Here your assumption is wrong because everything else same the pressure inside the cylinder is same but the force F on the the piston will be equal to pressure X Area

OK so how is pressure the same if if are igniting a charge in a smaller bore vs larger bore.

Originally Posted by Sutripta

360. Why discriminate against the other 10 degs.

Back to square 1 then? LOL