[dhanushs]

Originally Posted by ex670c

My Question to him was -
Q1) How much torque will go to one wheel, with 3 locked differentials, with 3 wheels in the air?
Answer - 100% (excluding windage)

Q2) Why will the torque go to the one wheel.
Answer - For any WORK to be done there has to be a RESISTANCE the reason for the work being done, and to do the work you have to apply POWER

A lot of discussion happened when I was away. Which I will be going through.

However, I got the above answer from here:

[Samurai]

Originally Posted by ex670c

If Torque goes to where there is resistance?

Why Don't Open Differentials do Work, when one wheel is Lifted in the air?

In a locked differential, the torque at the propeller is the sum of torque at both wheels. Let's ignore the axle ratio since it doesn't change the work done.

But the construction of open differential is very different. It is designed to take the path of least resistance. Think of how the spider gear gives way and start turning other way when one of the side gears offers more resistance. Whichever side offers lesser resistance, the open differential sends only sends the torque required by the lesser resistance, but to both sides. Since both sides get same amount of torque, the wheel that gets enough torque will turn and other wheel that needs more torque won't turn. If the car is merely turning, the inside wheel faces more resistance and will turn less, and outer wheel faces less resistance and will turn more.

Originally Posted by ex670c

How does the Differential allow the wheels to spin and different RPMs?

How does the Differential transfer power to the Wheels?What happens inside the differential?

Arka, most of us know the construction of open differential as well as the behavior. But the dynamics of power transfer is not easy to understand. In fact, I don't understand fully either. Trust me, I have tried hard enough. That knowledge transfer to my brain hasn't happened yet.

Originally Posted by DKG

So what makes you think that there is no load in an IC engine prior to the flywheel? Without your initial compression load you have no engine !!

And what makes you think that is what I think? When I say no load, I mean real conditions like a car in idle, but gear in neutral.

I have described my understanding of load in the very first page:

Originally Posted by Samurai

Engine has the potential to generate X amount of torque. But only when required. Consider an engine that can generate 240Nm at 2000rpm. If the gear is in neutral, then the engine will only generate enough torque to spin the flywheel at 2000rpm. So you will never see it generate 240Nm of torque at neutral. It won't generate 240Nm while driving downhill or straight road either. That may only happen when the vehicle is climbing uphill with 10 people on-board.

Consider a human example. You may have the strength to lift a 50kg bag of sand. But if you are asked to lift a 100gram bag of sand, you can't apply the same force as required by the 50Kg bag.

Originally Posted by dhanushs

However, I got the above answer from here:

Hey, I said the same thing in post#8 in reply to that. Damn, nobody trusts software guys anymore.

[DKG]

Sharath I just explained that the very principle of an IC engine requires load to function. Idling and neutral is of no consequence. Given load is intrinsic to the design and functioning of an IC engine your question whether it develops torque without load doesn't pan out. IC engines are designed to bear load on the combusting mixture and so torque is a result at the crank.

The initial load is created by the starter at the flywheel and subsequently once the engine fires up the angular momentum of the flywheel and crank bear load on the combusting mixture.

[Samurai]

Ok, I am very confused now. What do you think I was saying?

[DKG]

Originally Posted by Samurai

whether engine can generate torque without load (resistance)?

Lol no perhaps I am confused. What do you imply by the above question? Is there not an assumption that it's possible for an IC engine to generate torque without load?

I am not assuming you postulate that, just that no one can say an IC. Engine can develop torque without load. There are internal loads intrinsic to the design.

[Samurai]

It was a rhetorical question. From the beginning of this thread, I am fighting against the notion that engine can generate torque without load/resistance. There are many who disagree. You can see their arguments in the first few pages of the thread. Although no one is arguing for that camp anymore, you can see from the poll that many people still believe engine torque does not depend on load.

For example, the following statement is wholehearted supported by many (I find it shocking):

Originally Posted by lucifer1881

Similarly, if an engine is rated for 240Nm at 2000 rpm it will generate that torque irrespective of whether the transmission is engaged or in neutral.

[DKG]

An engine spinning at 2000 rpm will have considerable angular momentum at the flywheel. What if we brake the flywheel suddenly with a weighing scale at a distance of 1 meter from the flywheel axis? Would it register a force of 240 newtons?

[Samurai]

Angular momentum is a much more complex phenomena, too much for my brain to process.

I'll consider myself lucky if I get to understand the principle behind open differential, I mean the dynamics of power transfer within it.

[dhanushs]

Originally Posted by Samurai

Hey, I said the same thing in post#8 in reply to that. Damn, nobody trusts software guys anymore.

Hey, got confused with your answer. Sutripta answered "Yes".

[alpha1]

Originally Posted by DKG

An engine spinning at 2000 rpm will have considerable angular momentum at the flywheel. What if we brake the flywheel suddenly with a weighing scale at a distance of 1 meter from the flywheel axis? Would it register a force of 240 newtons?

How suddenly is very important.
The change in angular momentum is caused by torque.
Analogy is change in linear momentum is caused by force.

Now if you say that you have system angular momentum of 100 kg m2 s-1 (N m s).
You brake this down in 0.1 s.
It means you had exerted a torque of 100/0.1 kg m2 s-2 = 1000 N m
If you had braked this down in 1 s, the torque you need to exert = 100 N m

The force registered is dependent on the lever arm too. So lets us fix the lever arm as 0.1 m then for 1 N m of torque, we will register 1/0.1 = 10 N force.

[DirtyDan]

Torque without load?......some of modern physics says there exist objects with no mass. If I spin one?.....anybody know where I can get an adapter plate for photons, neutrinos, gluons?

"Doubt is not a pleasant mental state, but certainty is a ridiculous one." -- Voltaire

[DKG]

Originally Posted by alpha1

The change in angular momentum is caused by torque.

So when an increase in angular momentum is caused by torque can we say a decrease in angular momentum generates torque? I am inclined to think so.

Applying a reverse torque to brake, the angular momentum then applies an equal and opposite torque?

In a drivetrain it's like varying loads at the wheel scrub angular momentum off the flywheel to utilise the torque it then generates.

This same principle is being used by hybrids to utilise deceleration of vehicle momentum to run generators to recharge batteries.

[alpha1]

Originally Posted by DKG

So when an increase in angular momentum is caused by torque can we say a decrease in angular momentum generates torque? I am inclined to think so.

Applying a reverse torque to brake, the angular momentum then applies an equal and opposite torque?

In a drivetrain it's like varying loads at the wheel scrub angular momentum off the flywheel to utilise the torque it then generates.

This same principle is being used by hybrids to utilise deceleration of vehicle momentum to run generators to recharge batteries.

Decrease in angular momentum is also caused by torque (exerted by external agent - like brake shoe) - but in opposite direction.

Yes this is equal to the torque being generated by the rotating flywheel on the brake shoe while decelerating.
In fact we should talk more in terms of energy/work/power/momentum since that is potentially less confusing (and also confirm with the law of conservation - not applicable to torque and force).

-x-

Also, does anyone here agrees with me when I say that measuring engine torque figures is an absolutely hogwash for comparison with another vehicle harboring a different engine. What we are really interested is the motive force at wheel-surface contact, which is dictated very strongly by the gearing and the wheel diameter.

In other words, it is bhp or kW (at the particular RPM) that we should use to compare - since you cannot reduce or increase power via gearing or wheel size. Ideally it should be wheel HP, but since manufacturers don't disclose this figure BHP is also fine for comparison.

eg a vehicle having higher BHP @1500 RPM will always pull stronger and faster than a vehicle having lower BHP @1500.

(Note I am least concerned about peak BHP, which happens at different RPM for different engines)

[amit_purohit20]

Originally Posted by DKG

The above equation does not factor in the load either without which torque does not apply. RPM and the resultant angular momentum in my opinion play a role in affecting the resultant load a shaft experiences. That in turn determines how much torque is applied.

I see torque and angular momentum as one "generating" the other.

To analyse anything engineers are taught the principle of static equilibrium. So the shafts are considered immovable (like being fixed in a wall) and the whole calculations done.
Here in our case the infinite resistance offered by the wall is not there and so the traction sets the limit for the torque to get generated and transferred.

Originally Posted by DKG

Amit your example of a block appears illogical.

You have said that it's a frictionless system

I didn't say its a frictionless system. What I said was neglect friction losses for a while. Once you think with this thought you will understand what I have to say.

Now coming the new definition or the other definition of torque in the angular domain.

T= I*Alpha
Where T is torque in Nm
I is moment of Inertia or the resistance provided by the shaft to change its position of rest. I is constant for a given shaft whose geometrical dimensions do not change. So in all our discussions we can safely say this is a constant.
Alpha is the angular acceleration, not the angular velocity. So if the shaft is accelerating there is increased torque but as soon as the shaft has accelerated and reached a certain rpm the increase in torque will no more be there.

So when a wheel spin occurs say in case of a wheel in air. The shaft will experience a slight increase in torque till because of it angular acceleration. Once the shaft (attached to the wheel spin) reaches a certain rpm with respect to the throttle position the extra torque caused by acceleration will vanish as it is no more accelerating.

I wonder whether this little extra torque generated due to the shaft acceleration causes a little extra torque jerk to be transmitted to the other wheel which in some cases (where there is only a slight more torque required to move the vehicle) will make the vehicle to start moving again.

The above case is a very very theoretical case.

Bottom line still remains, For a shaft -

Torque cannot be generated if there is no resisting torque. The torque generated cannot exceed the resisting torque.
At most there will be be some torque increase till the shaft is accelerating. Once the acceleration stops the shaft will no more carry this extra torque (torque difference above the resisting torque).

Guys need your opinion on this. Based on the physics and equations.

[Jeroen]

Originally Posted by amit_purohit20

So when a wheel spin occurs say in case of a wheel in air. The shaft will experience a slight increase in torque till because of it angular acceleration. Once the shaft (attached to the wheel spin) reaches a certain rpm with respect to the throttle position the extra torque caused by acceleration will vanish as it is no more accelerating.

I wonder whether this little extra torque generated due to the shaft acceleration causes a little extra torque jerk to be transmitted to the other wheel which in some cases (where there is only a slight more torque required to move the vehicle) will make the vehicle to start moving again.

The above case is a very very theoretical case.

Bottom line still remains, For a shaft -

Torque cannot be generated if there is no resisting torque. The torque generated cannot exceed the resisting torque.
At most there will be be some torque increase till the shaft is accelerating. Once the acceleration stops the shaft will no more carry this extra torque (torque difference above the resisting torque).

Guys need your opinion on this. Based on the physics and equations.

As you say it's a bit theoretical.
I would say the minute the wheel starts leaveing the ground the amount of total torque decreases sharply as there is no resistance. So the question comes down whether the additional torque as you mention causing the acceleration comes on top of a rapidly decreasing torque.

In theory I would say it will drop, because in theory the torque in the shaft drops as the wheel leaves the ground, so one replaces the other so to speak. The latter being smaller than the original torgue so to speak. But I can see you can built a case for it going up as well.

Anyway, glad to see some of us agree that torgue needs a resisting force to begin with. Not sure some of our members on this thread agree, but there you go.

Jeroen