it's called Physicsand Calculus ... just do the math
an over simplication sure... but it's true.
...what?
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Engine/motor geeks.. Question..
- Thread starter FREAKYSTLYEY
- Start date
...what?
I'm not going to say one way or the other(tho I'm leaning towards your point of view), but it's not just 75psi. It's about 75psi * 6.5 compression ratio, so the unignited cylinder pressure is 487.5psi. Of course, once the fuel is ignited this pressure goes WAY up. So, fuel around the edges of the cyl might be squeezed before they burn to the point where they are entering the range of solids. Then again, temp will also be extremely high, so that is also unlikely.
What equations are you referring to?
75 psi is a fair bit, but nothing insane, pressure-wise. Not sure what you mean when you say "volatile", as being especially volatile is going to make it more difficult to revert to solid phase.
I did a quick lookup of a PT phase diagram for nitro, and at 298k (which is a much lower temp than intake temp even with fuel cooling), the freezing point is at ~0.4GPa, or ~58000psi (and nitro is a stable superpressed liquid up to about 2GPa). At higher temps it would have to be even higher. Even in a slow-burning combustion scenario, the chamber pressures are not going to peak that high. I'm happy to be wrong, though, since I've never worked with nitro engines in an engineering context.
I'm curious what Lee H has to say. Nitro engines are really cool, though.
it can be explained with physics and mathematics...
This is what I say when someone ask a question an engineer can explain. I usually leave the room promptly after this comment. Someone always has a puzzled look on their face when I return.
Just being a smart arse... Carry on nothing to see here.
Well, read what I said again. I realize it gets compressed, but look at the numbers involved- 500psi is nothing. I stated that I don't think that the post-ignition pressures are also anywhere near the relevant values, because if it had such an intense pressure spike you'd be doing damage. Given how stable nitro is at high pressures, it is more likely that it would stay liquid anyway. This is ignoring the fact the whole issue of temperature. The pre-combustion and combustion pressures are just way too low to instigate a phase change.
I still don't know what the "it" you are talking about refers to.
It's all in the cams. It has nothing to do with the engine layout. V-twins have a heavy flywheel to keep the engine smooth, so the cams need to be tuned for high torque at low rpm.
Check out this video,
it isn't me but I do it all of the time to show off the smoothness of my Valkyrie:
it isn't me but I do it all of the time to show off the smoothness of my Valkyrie:
"It" refers to everything regarding the operation of a combustion engine.
math, physics and now that I think about it chemistry when you start talking about the fliuds within the system ...
So... what you are saying is that you really have nothing of value to add to this thread.
Is that correct?
I'd actually say that the answer to the OP question "Why does an inline 4 on a sport bike rev higher and make has it's power higher up than a v or parallel twin. BUT a inline rev revs higher than a v6 and that revs higher than a v8. "
is "because it can". Someone else stated that something about the relationship between torque and HP, but I'll give it to you mathematically: HP=(Torque*RPM)/5250.
Consider the simplest Otto cycle motor: single cylinder.
A large displacement motor, is capable of producing significant torque at low RPM. So you get decent HP w/o high revs. All other things being equal, a smaller displacement motor simply cannot make torque that the larger motor can. But there is much less stress of slinging a small piston around, and because it doesn't move as much air in one cycle, the valves are smaller and can open and close faster. So it can rev very high and produce significant HP.
So, the answer is "per cylinder, a smaller displacement CAN rev higher and a larger displacement CAN produce more torque"... "because it can"
is "because it can". Someone else stated that something about the relationship between torque and HP, but I'll give it to you mathematically: HP=(Torque*RPM)/5250.
Consider the simplest Otto cycle motor: single cylinder.
A large displacement motor, is capable of producing significant torque at low RPM. So you get decent HP w/o high revs. All other things being equal, a smaller displacement motor simply cannot make torque that the larger motor can. But there is much less stress of slinging a small piston around, and because it doesn't move as much air in one cycle, the valves are smaller and can open and close faster. So it can rev very high and produce significant HP.
So, the answer is "per cylinder, a smaller displacement CAN rev higher and a larger displacement CAN produce more torque"... "because it can"
I have been underestimated in some of these posts. I understand now.
Basically shorter travel of the piston (among other things) allows it to rev faster (hp) but decreases low end torque.
You can have a .6l inline sport bike rev faster than a .6l twin. Because the piston travels much less distance but it lacks low end power.
In theory a 2.5 v8 should out rev a 2.5 inline 4. But no manufacturerakes a v8 that small.
However there are 3.5l v8s and 3.5 v6s.
Basically shorter travel of the piston (among other things) allows it to rev faster (hp) but decreases low end torque.
You can have a .6l inline sport bike rev faster than a .6l twin. Because the piston travels much less distance but it lacks low end power.
In theory a 2.5 v8 should out rev a 2.5 inline 4. But no manufacturerakes a v8 that small.
However there are 3.5l v8s and 3.5 v6s.
Or, I did contribute, and most of the simplified answers have already been stated in the thread. I sighed, though, because in reality it's much more complicated than what you stated and so the thread really has nowhere to go. For example, revving faster and revving higher are unrelated issues, and the issue of powerband is another issue as well. There are a number of variables- some chosen, some inherent- that contribute to each of those issues, and boiling it down to overall displacement vs cylinder count/arrangement yields nothing more than some empty generalizations. Yes, a 2.4L V8 will probably have peak horsepower at a higher RPM than a 2.4L I4, but it's usually because it's been designed for a very different application, thus the other input parameters (ie, intake manifold and exhaust design, cylinder head and valvetrain design, bore/stroke, on and on and on) are going to be very different as well.
So from the very beginning, the answer was "well...depends." The theoretical reasons are going to be different from the cost, reliability and efficiency reasons, too.
So from the very beginning, the answer was "well...depends." The theoretical reasons are going to be different from the cost, reliability and efficiency reasons, too.
I think there might be confusion between "rev faster" and "rev higher". To alot of people, that's probably the same thing: the ability for a motor to produce power at a high RPM. But I think Angus is considering "rev faster" as a quicker change in RPM. Might redline at 5k, but the motor would get there in .01sec.
I'm not considering either specifically- I was saying the language was getting used interchangeably, but they are two separate issues each with a different set of contributing variables.
