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Old 2nd February 2014, 07:33   #196
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Default Re: Torque generation and distribution

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Originally Posted by amit_purohit20 View Post
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.
The Strain Gauge Torque Sensors ! Been ages since I heard that term. Last heard in college! (Really relived those time momentarily thanks to you!)

But I thought in his problem Sutripta said the beam is rigid. No twisting. So all the torque is either countered or applied. That's why I was thinking why the drop to zero.
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Old 2nd February 2014, 16:27   #197
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Default Re: Torque generation and distribution

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The Strain Gauge Torque Sensors ! Been ages since I heard that term. Last heard in college! (Really relived those time momentarily thanks to you!)

But I thought in his problem Sutripta said the beam is rigid. No twisting. So all the torque is either countered or applied. That's why I was thinking why the drop to zero.
I too have been away from Data Acquisition group who work with these stress/strain Sensors on a everyday basis. But yes do understand bare minimum basics to understand how they work.

Regarding rigid beam - Practically all shafts twist, also I assumed the twist to make others understand the effect of torque otherwise I would have neglected it.
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Old 2nd February 2014, 21:08   #198
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Default Re: Torque generation and distribution

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Originally Posted by amit_purohit20 View Post
Sir, do you agree with what I said about torque, acceleration and other things or you have a different opinion?
We start from the same point, but reach different conclusions!

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Originally Posted by Sutripta View Post
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.)
And that I think needs to be resolved.

Tomorrow morning onwards, will probably be off the grid. Think will continue discussion after I'm back. Or maybe it'll be resolved by then.

Regards
Sutripta
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Old 3rd February 2014, 01:00   #199
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Default Re: Torque generation and distribution

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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.)
The introduction of constant torque drive had me confused for a while. Once I found a corollary with constant current source, it started making sense.

But we still need to explain constant torque drive. Otherwise it might give some of the early contributors of this thread an Aha moment. Didn't they say 247Nm rated engine will deliver 247Nm all the time?

Nope, constant torque drive isn't that.

A regular car engine is an open loop system. The torque depends on the load, but not the other way around, load is not depended on torque. So Thar engine will not deliver 247Nm torque unless at full load.

But constant torque drive will mostly be a electric motor, which takes negative feedback. That means a closed loop system. Let's say the motor is delivering 100Nm. Say at load X, the rpm is 1000. If the load suddenly drops, the rpm will pickup instantly to bring the torque back to 100Nm. If the load increases, the rpms will drop instantly to bring the torque back to 100Nm.
The motor circuit is designed to sense the load and adjust the torque output accordingly, hence it is a closed loop. The power (HP) or work done will vary with the rpm as torque is constant.

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Therefore, if the wheel was stationary and then released, the torque will remain same, even if the speed goes from 0 to 1000rpm.

I have no practical experience of constant torque motors. Won't it be dangerous to have these motors at no load? I guess the motor circuit will have breakers (Zener diodes) to avoid such situations. May be those motors always have load.
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Old 3rd February 2014, 03:39   #200
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Uptill now I was not thinking of the torque source. So all the explanations in my last 3-4 posts were assuming constant torque and constant rpm source and not the constant torque source which will accelerate the shaft till the earlier torque levels are reached.
Say 50 Nm was the torque before clamps were released, after releasing, the torque source accelerated the shaft to achieve 50 Nm. Fine but remember here its the acceleration not the angular velocity.

So what happens next after 50 Nm is reached? It will fall down right?

If not because its a constant source torque then the shaft should maintain constant acceleration to match 50Nm. That means shaft would reach very high rpms till the whole thing breaks down.

So the notion of a source which maintains constant torque even after release of clamps is not right.

Thats why I assumed a source which will be able to apply a torque of 50 Nm considering resisting torque is there and wont go beyond a certain fixed rpm say 500
In such a case whatever I have written in previous posts should hold true.
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Old 3rd February 2014, 23:27   #201
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Default Re: Torque generation and distribution

When the shaft is clamped, the torque is 50Nm. That is in equilibrium.

When the clamp is released, the shaft starts rotating fast enough to counter the 50Nm of applied torque. Slowly at first, for need of breakaway torque, then faster until a steady speed is reached. Again we have equilibrium.

Keep in mind we are not talking about frictionless systems.

Since the constant torque source will maintain 50Nm torque thanks to closed loop system, it also ensures that it is fighting against 50Nm of opposing force all the time. Releasing the clamp, didn't release the load.

let's look at the time lapse.

T0 second - 50Nm of torque is fighting the clamp
T+ second - 50Nm of torque is fighting the inertia of the just released shaft
T1 second - 50Nm of torque is accelerating [not just turning] the shaft at 50rpm/sec
T10 second - 50Nm of torque is accelerating the shaft at 25rpm/sec [accleration is reducing every second now]
T20 second - 50Nm of torque is turning the shaft at 1000rpm, acceleration is 0. Equilibrium is reached.

Why should the torque drop? The 50Nm is required to maintain the shaft at 1000rpm.
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Old 6th February 2014, 00:35   #202
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Default Re: Torque generation and distribution

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T10 second - 50Nm of torque is accelerating the shaft at 25rpm/sec [accleration is reducing every second now]
T20 second - 50Nm of torque is turning the shaft at 1000rpm, acceleration is 0. Equilibrium is reached.

Why should the torque drop? The 50Nm is required to maintain the shaft at 1000rpm.
Okay, that's a closed loop constant torque system.

In an open loop constant torque system acceleration continues indefinitely.
We had one of these in the lab here that we were tinkering with. But it broke down in early in our very first run. That was 8 months ago and we are still awaiting spares from M&M, the manufacturer.
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Old 9th February 2014, 00:54   #203
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But constant torque drive will mostly be a electric motor, which takes negative feedback. That means a closed loop system. Let's say the motor is delivering 100Nm. Say at load X, the rpm is 1000. If the load suddenly drops, the rpm will pickup instantly to bring the torque back to 100Nm.
Can it really do so? No it can not. By picking up rpm will not bring torque back.

T=I*Angular acceleration

Torque is dependent on acceleration and not angular velocity (rpm) so that means to get the torque of 100 Nm the angular acceleration will have to be constant and not the angular velocity (rpm).

So in this example, lets say the angular acceleration needed is 10 rpm/second square to reach a torque reading of 100 Nm.

Once 10 rpm/s2 is reached the torque levels should fall, but as per the definition its a constant torque source so to maintain constant torque on the shaft the motor will make the shaft to accelerate constantly at 10 rpm/s2.

So after a period of time the shaft would have reached speeds (rpm) crazy enough to break the whole apparatus.

If you see the graphs, I always say that once the rpms come into picture the torque domain ends and Power starts.

And we should use the Power formula of:

P= 2*Pi*N*T/60 where N is rpm and T is torque so simplified equation is

P= K*N*T where K is constant. This formula can be used for the steady state condition.

So with more rpms means more power.

Quote:
Therefore, if the wheel was stationary and then released, the torque will remain same, even if the speed goes from 0 to 1000rpm.
It cannot be the same. When the wheel was stationary the torque on the shaft was the max. torque what the source can apply say 100 Nm

Once you unclamp the wheel (say remove resistance) the torque level should fall down. And once a steady-state is reached the torque will be at 10 Nm (say this is the torque required to just keep the wheel rotating read system resistance). Please note here we are not using a constant torque source which will itself ensure that 100 Nm is maintained on the shaft. I am talking about a source which was applying 100Nm when the clamps were applied.

Quote:
I have no practical experience of constant torque motors. Won't it be dangerous to have these motors at no load? I guess the motor circuit will have breakers (Zener diodes) to avoid such situations. May be those motors always have load.
Same for me about closed loop and open loop. But what I understand is that if you have a constant torque source (which somehow tries to maintain constant torque on the shaft irrespective of the load) it has to have a safety feature, otherwise it will blowup itself to pieces.

In my earlier examples (previous to 4-5 posts) I was trying to consider a theoretical source which will apply a max torque of 100 Nm and try to stay at a constant rpm of say 500 rpm.

I had explained the same thing earlier...

Quote:
Originally Posted by amit_purohit20 View Post
Uptill now I was not thinking of the torque source. So all the explanations in my last 3-4 posts were assuming constant torque and constant rpm source and not the constant torque source which will accelerate the shaft till the earlier torque levels are reached.
Say 50 Nm was the torque before clamps were released, after releasing, the torque source accelerated the shaft to achieve 50 Nm. Fine but remember here its the acceleration not the angular velocity.

So what happens next after 50 Nm is reached? It will fall down right?

If not because its a constant source torque then the shaft should maintain constant acceleration to match 50Nm. That means shaft would reach very high rpms till the whole thing breaks down.

So the notion of a source which maintains constant torque even after release of clamps is not right.

Thats why I assumed a source which will be able to apply a torque of 50 Nm considering resisting torque is there and wont go beyond a certain fixed rpm say 500
In such a case whatever I have written in previous posts should hold true.
Quote:
Originally Posted by Samurai View Post
When the shaft is clamped, the torque is 50Nm. That is in equilibrium.

When the clamp is released, the shaft starts rotating fast enough to counter the 50Nm of applied torque. Slowly at first, for need of breakaway torque, then faster until a steady speed is reached. Again we have equilibrium.
With decreased load and steady speed can we have the same torque? No right, because its not the speed its the steady acceleration we need for the same torque.

Quote:
let's look at the time lapse.

T0 second - 50Nm of torque is fighting the clamp
T+ second - 50Nm of torque is fighting the inertia of the just released shaft
T1 second - 50Nm of torque is accelerating [not just turning] the shaft at 50rpm/sec
T10 second - 50Nm of torque is accelerating the shaft at 25rpm/sec [accleration is reducing every second now]
T20 second - 50Nm of torque is turning the shaft at 1000rpm, acceleration is 0. Equilibrium is reached.
My version (Considering a source which can apply 50Nm of torque and not constant torque and is say limited to a finite rpm due to practicality:
T0 second - 50Nm of torque is fighting the clamp.
T+ second - 50Nm of torque is fighting the inertia of the just released shaft
T1 second - Applied 50Nm of torque is accelerating [not just turning] the shaft at 50rpm/sec, but the reading of torque will go down because the load has decreased so 50Nm can not be seen on the torque measurement device.

T10 second - We see a torque spike as the shaft accelerates to a certain level and then starts deaccelerating. The torque spike is surely below the 50 Nm torque and above the 10 Nm torque (which is the bare minimum torque required to rotate the wheel read resisting torque without the clamp)

T20 second - Torque level is now 10 Nm and the rpms stabilised to say 500/1000/1500 any rpm based on the nature of the source and the whole system and its inertia.

Quote:
Why should the torque drop? The 50Nm is required to maintain the shaft at 1000rpm.
No the torque required to maintain 500 or 1000 or 1500 rpm is the same and is dependent on the resistance of the system.(Steady State)
Only when the system is accelerating (unsteady state) the torque required will be different.

Quote:
Originally Posted by DirtyDan View Post
In an open loop constant torque system acceleration continues indefinitely.
Totally agree.

Last edited by amit_purohit20 : 9th February 2014 at 00:55.
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Old 9th February 2014, 06:20   #204
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Default Re: Torque generation and distribution

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T=I*Angular acceleration

According to this formula when we are at

T + 0 seconds the shaft is not rotating, the clamp is on, angular acceleration must be zero...right? So the right side of the equation is also zero......torque would then also be zero at T + 0. This formula is not what you want, I think?
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Old 9th February 2014, 08:37   #205
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In an open loop constant torque system acceleration continues indefinitely.
I didn't respond to this earlier assuming it was one of your zen riddle. Now that Amit has agreed with you, I am confused. I thought any constant torque or constant current system has to be essentially a closed loop system. I can't imagine how you could have open loop constant torque system. If it exists, it is beyond my understanding.

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Originally Posted by amit_purohit20 View Post
Can it really do so? No it can not. By picking up rpm will not bring torque back.
Then we have fundamental problem right here. Don't you need more torque to turn something faster than before?

Last edited by Samurai : 9th February 2014 at 08:38.
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Old 9th February 2014, 09:11   #206
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I didn't respond to this earlier assuming it was one of your zen riddle. Now that Amit has agreed with you, I am confused. I thought any constant torque or constant current system has to be essentially a closed loop system. I can't imagine how you could have open loop constant torque system. If it exists, it is beyond my understanding.
Now that Amit has agreed with me I am confused, too....and highly suspicious of my results.

By the way, Amit, are you accusing me of using sub-standard rubber in the production of my shafts? I warn you, this is a serious charge.

Samurai, what you mean to say is that such a system cannot in principle exist? I may agree with you but we can still plug it in hypothetically as a mathematical object....and it is zen master Sutripta who stuck us with this zen koan, not me. I think he was trying to teach us something....and he may have succeeded...sneaky b*stard!....that's bustard, a type of large bird.

Last edited by DirtyDan : 9th February 2014 at 09:20.
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Old 9th February 2014, 23:07   #207
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According to this formula when we are at

T + 0 seconds the shaft is not rotating, the clamp is on, angular acceleration must be zero...right? So the right side of the equation is also zero......torque would then also be zero at T + 0. This formula is not what you want, I think?
Yes when the shaft is not rotating the formula is T1= F*r
When the shaft starts rotating the formula should be T1= F*r + T2= I*alpha
Total torque on shaft should be T1+T2
ie.
T1= 10 Nm (system resistance as mentioned in my previous post)
T2 exists because of the acceleration of the shaft.

Thats why I mentioned about the torque spike of greater than 10 Nm.

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Originally Posted by Samurai View Post
I didn't respond to this earlier assuming it was one of your zen riddle. Now that Amit has agreed with you, I am confused. I thought any constant torque or constant current system has to be essentially a closed loop system. I can't imagine how you could have open loop constant torque system. If it exists, it is beyond my understanding.
Let me be clear I have not agreed/disagreed to the talk on open or closed loop systems. I have limited knowledge on that.

What I have agreed is that if we have a source which irrespective of load will maintain the same torque on the shaft, then the system is going to blow itself to pieces.

As soon as the load is reduced the source will try to maintain the torque because of which it will induce steady acceleration in the shaft. To maintain the constant torque this acceleration has to be constant which means at a point of time the rpm of the shaft will be so much that the centrifugal forces will blow the system apart.

Quote:
Then we have fundamental problem right here. Don't you need more torque to turn something faster than before?
Well No, that is what I understand.
Torque is required to accelerate the shaft to a higher rpm, but not to maintain it. Its the Power which is required to maintain the rpms.

That can be proved by the equation Power, P= K*N*T (explained earlier in my posts)
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Old 10th February 2014, 21:44   #208
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Default Re: Torque generation and distribution

A certain amount of torque will be needed to overcome friction at any speed and to maintain rotation at any speed.

Physicists and engineers routinely talk of torque prior to rotation. Breakaway torque.
According to them there is torque, lots of it maybe, before there is angular acceleration. Amit, I know you want torque to conform to Newton's laws...maybe it does even before rotation commences....I am just not sure.

I haven't any other great wisdoms to cast forth and no zen moments to inflict on you all.
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Old 10th February 2014, 23:31   #209
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With the explanations put forth by me I feel we are able to use all the formulas of torque and power without conflicts and still able to explain the way shafts and differential behave.
....Till the time Sutripta or other bhpians contradict and I am unable to answer
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Old 11th February 2014, 01:25   #210
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A certain amount of torque will be needed to overcome friction at any speed and to maintain rotation at any speed.
Exactly! That is same reason why the rotating shaft will reach equilibrium at a point. That happens when acceleration will continuously decrease and stop when it can't overcome any addition friction and additional inertia.

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Originally Posted by DirtyDan View Post
Amit, I know you want torque to conform to Newton's laws...maybe it does even before rotation commences....I am just not sure.
You know, you reminded me of something. When I was a teenager, I read these books on Physics (The Book Thread). These books were written in 1913, and had numerous example of inventors trying to create perpetual motion machines. Because they too were thinking about Newton 1st law of motion. But they all overlooked something, newton's second law. For perpetual motion to work, you not only need initial breakaway torque, but you need sustaining torque to overcome external forces like friction and gravity. You have to take net force into consideration.

Quote:
Originally Posted by amit_purohit20 View Post
Let me be clear I have not agreed/disagreed to the talk on open or closed loop systems. I have limited knowledge on that.
In that case, let us assume we have a constant torque source, and not worry about the design of it. We all know constant torque drives exists, that should be good enough for us.

Quote:
Originally Posted by amit_purohit20 View Post
What I have agreed is that if we have a source which irrespective of load will maintain the same torque on the shaft, then the system is going to blow itself to pieces.
Not really, not unless you are assuming very theoretical situation like no friction, no gravity, etc.

Quote:
Originally Posted by amit_purohit20 View Post
As soon as the load is reduced the source will try to maintain the torque because of which it will induce steady acceleration in the shaft.
It will induce acceleration, but not steady, in fact the acceleration will be decreasing until the equilibrium is reached, where acceleration is zero.

Quote:
Originally Posted by amit_purohit20 View Post
To maintain the constant torque this acceleration has to be constant which means at a point of time the rpm of the shaft will be so much that the centrifugal forces will blow the system apart.
Noooo... ok, let me give another example.

Let me reuse the old graph.



And let me borrow your formula for power (P= K*N*T).

In the above graph T is constant, and K is already constant, so P is directly dependent on N (rpm). Hope we agree so far.

Quote:
Originally Posted by amit_purohit20 View Post
Well No, that is what I understand.
Torque is required to accelerate the shaft to a higher rpm, but not to maintain it. Its the Power which is required to maintain the rpms.
But power is not independent of torque, P is a function of T.

So P = K * 50Nm * 1000rpm

Now you have rightfully said power is required to maintain that 1000rpm. But if you reduce T, power will go down. To keep that power, you need the torque that got you there. There you go, I am using your formula.

Let's take your scenario. You say the rpm will go on increasing if we sustain the 50Nm of torque, that means P will also go on increasing... possible? Not possible, the drive can't possibly provide infinite power, while we know constant torque drives are real.

http://engineering.mit.edu/ask/it-po...motion-machine
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