Torque

[ChrisPy]

On a graph, the maximum torque USUALLY starts to taper off well before the maximum BHP.

Unless the torque reaches a peak and then just drops off to nothing, the torque curve has to taper off before the bhp curve tapers off; for any normal torque curve it is a mathematical certainty. BHP is simply an arithmetical derivative of torque (with the rpm multiplied in).

 

For those that are graphically inclined, it would be interesting to get the power/torque curves from your engine specification sheet, and note the maximum duty point for your boat (the rpm when the throttle is fully open). From that point you can draw a curve y = x-squared back to the origin (0,0). that curve approximates to the load from the boat and the propellor, it shows the power and torque developed at any intermediate rpm up to the maximum.

 

If the maximum duty point is not in between the point of maximum torque and the point of maximum power, then you probably have the wrong propellor to achieve maximum performance. Of course that probably doesn't matter for canal boats in normal cruising mode.

[Guest]

mi hors cant tork burav herd o mr ed he can, he got no brakes on im eever so no bhp

Common fault with that model it would seem.

[Timleech]

For those that are graphically inclined, it would be interesting to get the power/torque curves from your engine specification sheet, and note the maximum duty point for your boat (the rpm when the throttle is fully open). From that point you can draw a curve y = x-squared back to the origin (0,0). that curve approximates to the load from the boat and the propellor, it shows the power and torque developed at any intermediate rpm up to the maximum.

 

But remember that the Power/rpm curve for the prop will be approximately y = x cubed

 

Tim

[ChrisPy]

But remember that the Power/rpm curve for the prop will be approximately y = x cubed

 

Tim

maybe, but we both understand the idea, Tim, unlike some other (s?) I could mention

[RobinJ]

I've been doing some calculations (compression test checking valves/rings) and come up with an interesting conclusion.

Power is directly dependant on the amount of air burned in a given time.

Torque is directly dependant on the bmep developed in the cylinder.

So when an engine is bored out (+40) the air and size of piston is slightly increased, the compression ratio is slightly reduced (if head not skimmed), but it would appear that this makes it more efficient because:-

i) it needs less heat to start (compressed air reaches slightly higher temperature!)

ii) developes more power (acts on a slightly larger piston!)

So in theory produces slightly more torque!

I always assumed that a big compression ratio produced lots of power, but here combustion should be better with slightly more air available, the same amount of fuel is burnt but acting on sightly larger area, this presumably means slightly more power?

Does it also means that this alters the torque/power graph and so produces highest torque at higher revs.?

I conclude this probably also makes it run slightly warmer?

Edited September 28, 2007 by RobinJ

[Supermalc]

Back to basics ..........

 

An air fuel mixture is ignited, which causes it to expand.

 

The air fuel mixture is in a confined space (the cylinder) which can expand (the piston moves)

 

The larger the cylinder diameter, and/or the longer the stroke of the piston, the more air/fuel mixture is drawn into the engine. (hence the larger the engine, the greater the expansion of mixture, so the more power)

 

These are just the basics, however as many factors have to be taken into consideration, it now become rather more complicated.

 

But from this simple observation, it should become obvious if you want more power you increase the size of the engine, or repeat the cycle more times (run the engine faster).

 

The easiest way of making a given engine more powerful it to increase it's size. However as the early engine makers discovered, soon the mechanical efficiency of the engine decreases, so any increases are lost, if not decreased.

 

Suck - Squeeze - Bang - Blow.

 

Suck:- suck as much air/fuel mixture into the cylinder as you can.

 

Squeeze:- squeeze it as hard as possible before it reaches a temperature to ignite the mixture. (diesels obviously have the fuel squirted into air, hot enough to cause ignition)

 

Bang:- get the most energy out of the expanding air/fuel mixture as you can.

 

Blow:- clear the cylinder of the spent gasses, ready for the next cycle.

 

If you increase any of the above you will increase the power of the engine.

[Gibbo]

Back to basics ..........

 

An air fuel mixture is ignited, which causes it to expand.

 

The air fuel mixture is in a confined space (the cylinder) which can expand (the piston moves)

 

The larger the cylinder diameter, and/or the longer the stroke of the piston, the more air/fuel mixture is drawn into the engine. (hence the larger the engine, the greater the expansion of mixture, so the more power)

 

These are just the basics, however as many factors have to be taken into consideration, it now become rather more complicated.

 

But from this simple observation, it should become obvious if you want more power you increase the size of the engine, or repeat the cycle more times (run the engine faster).

 

The easiest way of making a given engine more powerful it to increase it's size. However as the early engine makers discovered, soon the mechanical efficiency of the engine decreases, so any increases are lost, if not decreased.

 

Suck - Squeeze - Bang - Blow.

 

Suck:- suck as much air/fuel mixture into the cylinder as you can.

 

Squeeze:- squeeze it as hard as possible before it reaches a temperature to ignite the mixture. (diesels obviously have the fuel squirted into air, hot enough to cause ignition)

 

Bang:- get the most energy out of the expanding air/fuel mixture as you can.

 

Blow:- clear the cylinder of the spent gasses, ready for the next cycle.

 

If you increase any of the above you will increase the power of the engine.

 

And if you increase the RPM you can increase the available power without increasing the available torque.

 

 

Gibbo

Edited September 28, 2007 by Gibbo

[Supermalc]

And if you increase the RPM you can increase the available power without increasing the available torque.

 

 

Gibbo

 

Sometimes - it all depends on the design of the engine ......... hence the confusion.

 

A good example of torque. Anyone remember the 998cc version of the Ford Anglia, the one with the rake back window. Well there were 3 sizes of engines, 998cc, 1198cc and 1498cc, all having the same cylinder size. The increase in cc was achieved by a longer stroke. The car was manufactured with the 2 smaller sized engines, and the 'boy racers' swapped them for the larger one, fitted to some Mk1 Cortinas.

 

If you were in the smaller engined car, you had to rev the engine to pull away at traffic lights, otherwise they often stalled. Fords got the reputation of having engines which were 'all wind and piss'. In other words you had to increase the torque to stop the engine stalling.

 

Yet if you had the 998 BMC in a Morris 1000, they didn't suffer from this problem. But the Anglia was faster, and had better acceleration.

[RLWP]

So when an engine is bored out (+40) the air and size of piston is slightly increased, the compression ratio is slightly reduced (if head not skimmed)

 

I don't think that is right. The bore is bigger, so there is more air in the cylinder with the piston at the bottom of the stroke. At the top of the stroke, the increased bore is now full of piston, so the increase in bore diameter makes very little difference. The combustion chamber hasn't changed size, so when the piston is at the top of the stroke, more air is squashed into what was the same space. Consequently, the compression ratio is actually raised.

 

Richard

[Gibbo]

Sometimes - it all depends on the design of the engine ......... hence the confusion.

 

A good example of torque. Anyone remember the 998cc version of the Ford Anglia, the one with the rake back window. Well there were 3 sizes of engines, 998cc, 1198cc and 1498cc, all having the same cylinder size. The increase in cc was achieved by a longer stroke. The car was manufactured with the 2 smaller sized engines, and the 'boy racers' swapped them for the larger one, fitted to some Mk1 Cortinas.

 

If you were in the smaller engined car, you had to rev the engine to pull away at traffic lights, otherwise they often stalled. Fords got the reputation of having engines which were 'all wind and piss'. In other words you had to increase the torque to stop the engine stalling.

 

Yet if you had the 998 BMC in a Morris 1000, they didn't suffer from this problem. But the Anglia was faster, and had better acceleration.

 

Revving the engine increases the power, not necessarily the torque.

 

Gibbo

[RLWP]

Revving the engine increases the power, not necessarily the torque.

 

Gibbo

 

Also, by revving the engine and dropping the clutch you are using the rotational energy stored in the flywheel and crankshaft to accelerate the car as well as the power produced directly by the engine.

 

Richard

[Supermalc]

I think you are both loosing sight of what torque actually is - the force, or turning moment of the engine.

 

The faster the engine goes, the greater the torque ........ until the mechanical losses, or other inefficiencies of the engine increases to a point the torque decreases.

 

And it is irrespective of how this is achieved.

 

As has been said, usually the greatest torque is achieved at maximum BHP.

 

Torque is what causes acceleration, or movement. I have had lots of experience of driving cars and bikes (not in competition) where you try to do this the fastest. If you rev the engine over the limit of it's 'power band' you lose time i.e. forward motion, or acceleration. This is where the torque decreases, but it is because of the inefficiencies in the engine, be it air flow, timing, fuel starvation, mechanical efficiency, any one of a great number of things. And this is why torque can decrease with engine revs.

 

But as BHP is a measure of power over TIME this can actually increase, even though the torque, or effective power of the engine is actually reduced.

[chris w]

Revving the engine increases the power, not necessarily the torque.Gibbo

 

I agree; in the original equation where BHP = torque x RPM/5252 (with torque in ft.lbs) then imagine for the moment that the torque is constant right across the rpm band.

 

From the equation, then Power will increase simply because the revs are increasing with absolutlely no increase in torque.

 

Power is the RATE of doing work. If the work (ie: torque) is constant but you "do" that torque more often (ie: higher revs) then power will increase.

 

Chris

As has been said, usually the greatest torque is achieved at maximum BHP.

 

No it's not. It would be very unusual for this to happen in any engine.

 

Chris

Edited September 28, 2007 by chris w

[Supermalc]

in the original equation where BHP = torque x RPM/5252 (with torque in ft.lbs) then imagine for the moment that the torque is constant right across the rpm band.

 

 

 

Chris

 

Ah, but it's not. That's the crux of the problem, or misunderstanding.

 

A steam engine has maximum power at no revs. So it has maximum torque at no revs ...... agreed?

[RLWP]

No it's not. It would be very unusual for this to happen in any engine.

 

Chris

 

Example here: http://www.mgexperience.net/article/fuel-power.html

 

Peak torque 3000, peak BHP 5200.

 

Richard

[chris w]

Ah, but it's not. That's the crux of the problem, or misunderstanding.

 

A steam engine has maximum power at no revs. So it has maximum torque at no revs ...... agreed?

 

 

Keep off the wacky bakky

 

If you put zero for the RPM in the HP vs Torque equation you get zero!!! simple maffs

 

Chris

[Supermalc]

Keep off the wacky bakky

 

If you put zero for the RPM in the HP vs Torque equation you get zero!!! simple maffs

 

Chris

 

Exactly my point ...... the equation doesn't always work.

 

Totally off topic.

 

My washing machine is overfilling. Never had to fault find this before so didn't know it is done by air pressure. Tube to the drum seems clear, and blowing into the switch operates it. Anyone got any experience of this?

[RLWP]

Keep off the wacky bakky

 

If you put zero for the RPM in the HP vs Torque equation you get zero!!! simple maffs

 

Chris

 

Careful, that equation has nothing to do with steam engines. Torque comes from a product of pressure on the piston and the throw of the crankshaft. In an internal combustion engine the pressure is intermittent due to the combustion of the air in the cylinder. In a steam engine the pressure is supplied by the boiler and in comparison is constant. Therefore the peak torque will happen when the crankshaft is at 90 degrees to the axis of the cylinder even if the piston is stationary. Just as well really as otherwise a steam locomotive would never have been able to start a train moving.

 

Richard

[chris w]

Exactly my point ...... the equation doesn't always work.

 

Of course it works (always) because................... that's how Power is DEFINED !!! Doh

 

Chris

 

Therefore the peak torque will happen when the crankshaft is at 90 degrees to the axis of the cylinder even if the piston is stationary. Richard

 

You're confusing "torque" with "force". Torque is "force through a distance" (ie: force x distance). What you have with the steam engine in the scenario you describe is maximum "force". No distance, no torque and ergo no work done (ie: no power)

 

Chris

Edited September 28, 2007 by chris w

[Supermalc]

The point I am trying to make is that sometimes it helps to distance yourself from the theory, and just try to concentrate on the practice.

 

The words we have are to tell us what to expect with a certain machine (in this case an engine).

 

So to say a steam engine has no torque (true in the mathematical sense) is of no bearing to the train engine driver who wants to know if he can pull a load of many tons from stationary.

 

SOMETIMES we need to know the torque of an engine, for it's given purpose.

 

SOMETIMES we need to know the BHP of an engine, for it's given purpose.

 

It all depends on the application we are going to use it for.

[chris w]

The point I am trying to make is that sometimes it helps to distance yourself from the theory, and just try to concentrate on the practice.

 

The words we have are to tell us what to expect with a certain machine (in this case an engine).

 

So to say a steam engine has no torque (true in the mathematical sense) is of no bearing to the train engine driver who wants to know if he can pull a load of many tons from stationary.

 

SOMETIMES we need to know the torque of an engine, for it's given purpose.

 

SOMETIMES we need to know the BHP of an engine, for it's given purpose.

 

It all depends on the application we are going to use it for.

 

Sorry, but it's mathematics pure, plain and simple because that's how we as humans have chosen to define it.

 

If you want to define other parameters in your own way, then all you need to do is derive a set of equations to model the behaviour seen in practice and voilá.

 

Chris

[RLWP]

Of course it works (always) because................... that's how Power is DEFINED !!! Doh

 

Chris

You're confusing "torque" with "force". Torque is "force through a distance" (ie: force x distance). What you have with the steam engine in the scenario you describe is maximum "force". No distance, no torque and ergo no work done (ie: no power)

 

Chris

 

Torque is force times distance, hence foot-pounds and newton-metres. The steam engine exerts a force equal to the steam pressure times its piston area at a distance equal to the throw of the crank. The torque generated is the force from the piston times the crank throw. This will be at it's greatest when the engine is stationary and the crank is at 90 degrees because there will be no pressure drop between the boiler and the cylinder due to gas flow.

 

It's the same as a torque wrench or a windlass. It's why a weak and feeble person gets on better with a longer windlass. The exert a smaller force at a greater distance than someone using a regular windlass. Short windlass large force, long windlass smaller force, same torque generated to open the paddle.

 

BUT... As you have said, this has nothing to do with power. If my steam engine is stationary and generating it maximum torque but isn't moving, then no work is being done and therefore no power. The power that can be output by my windlass winder will depend on the speed at which they can wind the windlass. The work done will be the same (enough to lift the paddle) but the faster it is done the more power.

 

Chris, you're right. I got carried away and mixed up force - torque in this case, and power.

 

Richard

[RobinJ]

I don't think that is right. The bore is bigger, so there is more air in the cylinder with the piston at the bottom of the stroke. At the top of the stroke, the increased bore is now full of piston, so the increase in bore diameter makes very little difference. The combustion chamber hasn't changed size, so when the piston is at the top of the stroke, more air is squashed into what was the same space. Consequently, the compression ratio is actually raised.

Thats what I thought, but when you look at indirect injection, it appears the compression ratio actually goes down slightly, since the combustion chamber is bigger than whats left when the piston is up!

[John Orentas]

I've been doing some calculations (compression test checking valves/rings) and come up with an interesting conclusion.

 

So when an engine is bored out (+40) the air and size of piston is slightly increased, the compression ratio is slightly reduced (if head not skimmed), but it would appear that this makes it more efficient because:-

i) it needs less heat to start (compressed air reaches slightly higher temperature!)

ii) developes more power (acts on a slightly larger piston!)

So in theory produces slightly more torque!

I always assumed that a big compression ratio produced lots of power, but here combustion should be better with slightly more air available, the same amount of fuel is burnt but acting on sightly larger area, this presumably means slightly more power?

Does it also means that this alters the torque/power graph and so produces highest torque at higher revs.?

I conclude this probably also makes it run slightly warmer?

 

 

 

 

Much more complex than this, you should be talking about 'Swept volume'. Also the act of compression of the air in the cylinder grossly increases the air temperature and therefore the pressure to a level way above what the calculation would suggest. Thats why static calculation never gives you the 400+ psi figure you need for ignition.

 

The effect of re-boring makes very marginal differences. You are all getting very bogged down with this torque business.

Edited September 28, 2007 by John Orentas

[RLWP]

You are all getting very bogged down with this torque business.

 

I don't think we are alone in this either. I did a quick search to come up with a few facts and figures and the internet has many, many threads about torque, power, BHP and all that jolly stuff.

 

Mind you, I thought in off-roading it was torque that stopped you getting bogged down....

 

Richard

 

p.s adiabatic! now there's a good word!