[mallumowgli]

Originally Posted by Jeroen

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

Last few weeks I've not been able to access Tbhp and I was thinking that there would be a lot of reading to do since I last left to catch up. But this statement confirms things have not moved forward much - still stuck at Torque or no torque. Not sure how much more torque will be needed to get this stuck thread to move forward

[Samurai]

Originally Posted by amit_purohit20

The torque generated cannot exceed the resisting torque.

This is how I came up with my least force definition. I used to say torque generated is the least force required to turn the wheel [resisting torque]. Somewhere along the line, I dropped the "generated", and started saying torque is the least force required to turn the wheel. That is how my statement became screwed up.

Originally Posted by Jeroen

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.

Originally Posted by mallumowgli

But this statement confirms things have not moved forward much - still stuck at Torque or no torque. Not sure how much more torque will be needed to get this stuck thread to move forward

After page 9 (Jan 4th), nobody has tried to make a case for torque without resistance. From the poll that was added afterwards, it was clear that the opinions are still divided. After 3 days the poll was removed because it would reflect badly on the people who chose wrongly. However, I would say the thread has moved on since there is no more open disagreement on that point. Hopefully, they will do their own research and change their mind someday.

[amit_purohit20]

Infact I was supporting the poll to understand how many people do agree to this theory of torque depending on resisting torque and for those who do not agree we could try to give some explanation. Anyway my intent was good not to pin point some people.

I again reiterate it has been proved again and again that torque depends on resisting torque. I feel we should park the case here for it.

The next googly for me was when people asked me to explain the same thing by using the angular concept of torque. I tried it in my previous post.

But both the equations T= F*R or T= I* Alpha do not prove the resistance theory.

The resistance theory can be proved only by the Varignons law or the Law of static equilibrium.

Other than the resisting theory of torque do we have any other questions on torque or differentials?

@Jeroen I agree to your point.
Consider this case a wheel is in the air and for one throttle position say the wheel in air is at 50 rpm. Say the torque is T1 in this case.

Now if you press the accelerator hard the rpms increased from 50 to 100. During this transition period where the wheel accelerates from 50 rpm and reaches 100 rpm we should experience this increase in torque (which I again feel is not substantial) say T2 (max.)
As soon as the wheel reaches 100 rpm this extra torque (T2-T1) will vanish and it will again become T1.
In most practical cases this momentary extra torque is not sufficient to pull the stuck vehicle out.

[Sutripta]

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.

Originally Posted by Jeroen

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.

What happens if we consider the drive to be a constant torque (for the rpms we are concerned with:- ie 0 to say 1000 rpm) drive? Let us concentrate on the time from just before to just after wheel leaves the ground. (Assume the wheel leaves the ground cleanly. Think of brakes operating on a free shaft if wheels cause conceptual problems.)

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

What happens if we consider the drive to be a constant torque (for the rpms we are concerned with:- ie 0 to say 1000 rpm) drive? Let us concentrate on the time from just before to just after wheel leaves the ground. (Assume the wheel leaves the ground cleanly. Think of brakes operating on a free shaft if wheels cause conceptual problems.)

Regards
Sutripta

I am not sure what you want to say here.

But assuming a shaft carrying 50 Nm of torque at 1000 rpm (Constant torque).

Now if the rpms of the shaft go up by say 500 rpm. As the shaft is accelerating from 1000 to 1500 rpm the torque will increase and as soon as the rpm reaches 1500 rpm and stabilises ie. no more acceleration the torque caused by acceleration will cease to exist.

Say the angular acceleration of the shaft is 1 radian/second square.
And Moment of Inertia is say 20 kg-metre square.

Then T= I* alpha
or T= 20*1 = 20 N-m.

If I understand it right this torque of 20 N-m is over and above the 50 Nm of torque which it was previously carrying. This torque is because of the MI and change in acceleration and not because of the resistance on the shaft.

So the total torque on the shaft will be 50+20= 70 Nm till the shaft is accelerating with angular acceleration of 1 radian/second square.

Point to be noted is we are here mixing two states static (50 Nm) and dynamic (20 Nm).

PS: In the above example I have taken random values of MI and Alpha. In real world if we do with the real values the addition of torque caused by acceleration will not be so much substantial.

[Sutripta]

Originally Posted by amit_purohit20

I am not sure what you want to say here.

You say the torque (measured at the halfshafts) will increase.
Jeroen says it will decrease.
I was trying to act as an (honest) broker to see if middle ground could be reached. ie. torque will remain the same.

Which IMO is what it should be. Same as in exactly the same. Which is different from practically the same which actually means there is some difference which we choose to ignore. (Talking strictly of the diff, not characteristics of the engine. Thus the simplification of replacing the engine with a constant torque source. Another assumption I should have added is that only the wheel has nonzero MoI.)

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

You say the torque (measured at the halfshafts) will increase.
Jeroen says it will decrease.
I was trying to act as an (honest) broker to see if middle ground could be reached. ie. torque will remain the same.

Which IMO is what it should be. Same as in exactly the same. Which is different from practically the same which actually means there is some difference which we choose to ignore. (Talking strictly of the diff, not characteristics of the engine. Thus the simplification of replacing the engine with a constant torque source. Another assumption I should have added is that only the wheel has nonzero MoI.)

Regards
Sutripta

I was neither answering you nor jeroen so no question of difference. I was not even talking of differential half shafts, I was just giving an example of a simple shaft to put my point of torque.

I agree to Jeroen that if you lift a wheel off ground, one is reducing the resisting torque and so the torque on that shaft is going down.
My point here was that because the shafts rpm is increasing or accelerating it will be subjected to a temporary increase in minuscule torque due to this momentary acceleration.
Now coming to your point like having a constant rpm and constant torque as power source the answer will be the same as mentioned above.
Note that the acceleration of the half shaft is caused here because of differential property (2N rpm) and not because propeller shaft rpm has increased.

Again I repeat that I have brought the point of temprorary increase in torque because someone mentioned in previous posts about Torques other formula having angular acceleration in it. Otherwise this temporary increase in torque is negligible for all practical purposes and should not be considered.
Amit

[Sutripta]

Originally Posted by amit_purohit20

My point here was that because the shafts rpm is increasing or accelerating it will be subjected to a temporary increase in minuscule torque due to this momentary acceleration.
Now coming to your point like having a constant rpm and constant torque as power source the answer will be the same as mentioned above.

Just so that I've properly understood what you are saying:
Other things remaining constant, if a wheel which was previously locked (by whatever means) is suddenly unlocked and starts turning, the shaft turning the wheel will see an increase in torque.

Have I got it right?

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

Just so that I've properly understood what you are saying:
Other things remaining constant, if a wheel which was previously locked (by whatever means) is suddenly unlocked and starts turning, the shaft turning the wheel will see an increase in torque.

Have I got it right?

Regards
Sutripta

No the torque will decrease as the resisting torque will decrease.

Forget about what I said earlier if you dont get my earlier point.

[ampere]

Originally Posted by amit_purohit20

But assuming a shaft carrying 50 Nm of torque at 1000 rpm (Constant torque).

If you are applying a torque you should see acceleration of the shaft. 50nm @ 1000 rpm is an instantaneous notion.

Now if the rpms of the shaft go up by say 500 rpm. As the shaft is accelerating from 1000 to 1500 rpm the torque will increase and as soon as the rpm reaches 1500 rpm and stabilises ie. no more acceleration the torque caused by acceleration will cease to exist.

I think its this applied torque which is causing the (angular) acceleration

Say the angular acceleration of the shaft is 1 radian/second square.
And Moment of Inertia is say 20 kg-metre square.

Then T= I* alpha
or T= 20*1 = 20 N-m.

The I Alpha that you compute should a resultant action of the torque applied.

If I understand it right this torque of 20 N-m is over and above the 50 Nm of torque which it was previously carrying. This torque is because of the MI and change in acceleration and not because of the resistance on the shaft.

So the total torque on the shaft will be 50+20= 70 Nm till the shaft is accelerating with angular acceleration of 1 radian/second square.

Point to be noted is we are here mixing two states static (50 Nm) and dynamic (20 Nm).

I have no idea of differentials. But from a pure physics point of view:

- From Newton's first law, object would be in the same state (constant rest or motion) until acted upon by an external force (or torque).
- Its this force/torque which should cause change of state of the object, by giving it an acceleration (angular).

- As the topic of the thread says (generation and distribution of torque) I think both have nothing to do with the object (to whom) the torque is applied. Torque is an external agency which would change the object's state.

- Now how the torque is actually generated and distributed I think that should be a different debate. (Fuel generating energy which is turning the shaft etc).

[Sutripta]

Originally Posted by amit_purohit20

Forget about what I said earlier

OK. Lets start afresh!

if you dont get my earlier point.

Which was?

Originally Posted by ampere

I have no idea of differentials.

So lets get rid of it. Not important in the immediate context.

Lets take a shaft. A (fly)wheel at one end, a constant torque source at the other. For initial simplicity, let the shaft perfectly rigid, with no MoI. (the equivalent of the perfect string:- weightless and nonextensible).
Let us clamp the wheel so that it cannot turn. At time t=0, let the clamps be released.

What can we say about our simple system (torques, (angular) velocities and accelerations) at times 0 - delta t, and 0 + delta t?

Regards
Sutripta

[amit_purohit20]

Originally Posted by ampere

If you are applying a torque you should see acceleration of the shaft. 50nm @ 1000 rpm is an instantaneous notion.

Yes and No both. Yes if the shaft has overcome the resisting torque say 50 Nm. You try to apply a torque on this shaft above 50 Nm (now you go in the power domain) the extra torque which you have tried to apply will cause the acceleration of this shaft.

No- Take a shaft fixed at one end in a concrete wall. Now if you try to turn the other end with a 50 N-m torque which is very well less than the resisting torque, the torque of 50 Nm will be applied but without causing any acceleration in the shaft. It would only cause the shaft to twist on its own axis.

Originally Posted by ampere

I think its this applied torque which is causing the (angular) acceleration

Offcourse the torque has to be applied to get angular acceleration.

I prefer to use the word Power here because torque can be talked of in static sense but when things start moving better to use power.

Because, Torque need not be zero if there is no movement but Power has to be zero if there is no movement. Power is Work done/time. Work done is Force* Displacement. So if there is no displacement there is no work done. And if there is no work done there is no power.

Originally Posted by ampere

The I Alpha that you compute should a resultant action of the torque applied.

Agree. A shaft cannot accelerate on its own.

Originally Posted by ampere

I have no idea of differentials. But from a pure physics point of view:

- From Newton's first law, object would be in the same state (constant rest or motion) until acted upon by an external force (or torque).
- Its this force/torque which should cause change of state of the object, by giving it an acceleration (angular).

Agree.

Originally Posted by Sutripta

OK. Lets start afresh!

Originally Posted by Sutripta

Which was?

The point of extra torque (however miniscule it might be) due to acceleration. I asked you to forget this because when we talk about torque we hardly talk about torque spikes and instantaneous torque. (No offence was meant, When I read my words again I felt it to be rude Sorry)

Originally Posted by Sutripta

Lets take a shaft. A (fly)wheel at one end, a constant torque source at the other. For initial simplicity, let the shaft perfectly rigid, with no MoI. (the equivalent of the perfect string:- weightless and nonextensible).
Let us clamp the wheel so that it cannot turn. At time t=0, let the clamps be released.

What can we say about our simple system (torques, (angular) velocities and accelerations) at times 0 - delta t, and 0 + delta t?

Regards
Sutripta

Lets say when the wheel is hold (fixed) the constant torque source applies a 50 Nm of torque.
So at time t=0 Torque T= 50Nm
When the wheel is released suddenly and assuming there is no other resisting torque,
At t= 0 + delta t Torque T of 50 Nm should directly fall down to zero as there is no resisting torque now.

But this torque applied is going to cause the shaft to start rotating.
1) Rotate at what rpm's?
2) Accelerate at what acceleration values?

The above two points will be based on the nature of torque source.

Please note that here I am not talking about engine as a torque source. I have taken a ideal or a theoretical torque source.

Say the torque source will rotate at 500 rpm.

So now the shaft accelerates from 0rpm to 500 rpm. Say this takes 5 seconds.

During this period of 5 seconds we will see a torque spike more than zero as the shaft is accelerating.

What value of torque spike?
Well it should depend on T= I*Alpha equation.

If alpha is constant in the above equation we will observe a constant torque on the shaft. (Here I am assuming I is not a non-zero number)

[ampere]

Originally Posted by amit_purohit20

Yes and No both. Yes if the shaft has overcome the resisting torque say 50 Nm. You try to apply a torque on this shaft above 50 Nm (now you go in the power domain) the extra torque which you have tried to apply will cause the acceleration of this shaft.

No- Take a shaft fixed at one end in a concrete wall. Now if you try to turn the other end with a 50 N-m torque which is very well less than the resisting torque, the torque of 50 Nm will be applied but without causing any acceleration in the shaft. It would only cause the shaft to twist on its own axis.

Agreed. But when I said, the beam (shaft) should accelerate, I meant a beam which is not restrained. You analysed both scenarios.

Because, Torque need not be zero if there is no movement but Power has to be zero if there is no movement. Power is Work done/time. Work done is Force* Displacement. So if there is no displacement there is no work done. And if there is no work done there is no power.

It depends on the type of the movement. If the movement has acceleration, then there is a torque being applied.

Torque is the ability aspect while power is the rate at which that ability can be delivered. Power can also be seen as rate at which work is done. (Or energy is delivered) The way I think, that ability for work comes from Force (Torque)

At t= 0 + delta t Torque T of 50 Nm should directly fall down to zero as there is no resisting torque now.

This part I did not get. Why do you say the applied torque should go to zero? The instant the restraint on the joints are removed, the applied torque of 50 Nm should cause the beam to accelerate.

Thats what I thought should be the answer to Sutripta's question. In on other words, when the beam is restrained, all the applied torque should generate a reaction at the junction to keep the beam static. The instant the constraint is removed, all the applied torque should cause angular acceleration.

[Sutripta]

Originally Posted by amit_purohit20

Lets say when the wheel is hold (fixed) the constant torque source applies a 50 Nm of torque.
So at time t=0 Torque T= 50Nm
When the wheel is released suddenly and assuming there is no other resisting torque,
At t= 0 + delta t Torque T of 50 Nm should directly fall down to zero as there is no resisting torque now.

But this torque applied is going to cause the shaft to start rotating.
1) Rotate at what rpm's?
2) Accelerate at what acceleration values?

The above two points will be based on the nature of torque source.

Please note that here I am not talking about engine as a torque source. I have taken a ideal or a theoretical torque source.

Say the torque source will rotate at 500 rpm.

So now the shaft accelerates from 0rpm to 500 rpm. Say this takes 5 seconds.

During this period of 5 seconds we will see a torque spike more than zero as the shaft is accelerating.

What value of torque spike?
Well it should depend on T= I*Alpha equation.

Originally Posted by Sutripta

What happens if we consider the drive to be a constant torque (for the rpms we are concerned with:- ie 0 to say 1000 rpm) drive?

Originally Posted by Sutripta

Thus the simplification of replacing the engine with a constant torque source.

Originally Posted by amit_purohit20

If alpha is constant in the above equation we will observe a constant torque on the shaft. (Here I am assuming I is not a non-zero number)

Cause and effect.

Regards
Sutripta

[amit_purohit20]

Originally Posted by ampere

It depends on the type of the movement. If the movement has acceleration, then there is a torque being applied.

Not necessarily we can apply torque and still not have acceleration. It all depends upon what type of power/torque source you are having.

For eg. A torque source can apply say 50 Nm but still not cause much acceleration in the shaft.

I would give an example here of a long stroke engine and a short stroke engine. (Considering all other bigger parameters nearly equal).

A long stroke engine will be able to apply more torque than a short stroke engine but a short stroke engine would be able to accelerate the shaft more than the long stroke.

Thats why what matters is the nature of power/torque source.

Originally Posted by ampere

Torque is the ability aspect while power is the rate at which that ability can be delivered. Power can also be seen as rate at which work is done. (Or energy is delivered) The way I think, that ability for work comes from Force (Torque)

Refering your last line here that "ability for work comes from Force"
Might be might not be.
For example the force on a piston of a ships engine is far more greater than the force on the piston of a cars engine. But yet the Cars engine can accelerate faster than a Ships engine.

So the acceleration of a shaft depends upon what power/torque source its connected to.

Infact in real world once a preset value of torque required to overcome resistance is delivered by the engine any further acceleration is caused by the increase in power and not the torque.

Originally Posted by ampere

This part I did not get. Why do you say the applied torque should go to zero? The instant the restraint on the joints are removed, the applied torque of 50 Nm should cause the beam to accelerate.

Thats what I thought should be the answer to Sutripta's question. In on other words, when the beam is restrained, all the applied torque should generate a reaction at the junction to keep the beam static. The instant the constraint is removed, all the applied torque should cause angular acceleration.

Yes I agree that the torque of 50 Nm will be used to create acceleration but then its converted from torque to power now.

When you measure torque on a shaft using sensors. What they measure is the twist of the shaft. Before the clamp was released there was a twist in the shaft due to 50 Nm torque so we get measured reading of 50 Nm on measurement apparatus.

But as soon as we remove the clamps there is no resisting torque and the shaft is hardly twisted anymore. So you will not be able to get 50 Nm torque readings. The torque now has become power as the shaft has started rotating.

In the above case there will be some small torque reading because the shaft is still very very very slightly twisted due to its own MI and acceleration of the shaft.

Originally Posted by Sutripta

Cause and effect.

Regards
Sutripta

Sir, do you agree with what I said about torque, acceleration and other things or you have a different opinion?