[Samurai]

Originally Posted by DirtyDan

Sammy, Sutripta has eloquently addressed what I had in mind. He just said it much better than I did.

Actually, he has done more than that. Yesterday he made me understand [offline] how torque can be visualized when there is no obvious displacement or rotation. So there will be distribution in a locked MLD, it can be seen only mathematically.

[mallumowgli]

Originally Posted by Samurai

Then tell me how I should have interpreted the following statement:

Torque available at the propeller rotates the slipping wheel at a higher rpm - Guess I should have used the term force -but that became clear after reading amit's post!!

It won't, I don't know where you got this from.

In an open differential, there is no transfer of torque in case of slip. If the wheel with traction requires 40Nm and the slipping wheels require just 10Nm, then both wheels will get just 10Nm each. The propeller can only deliver 20Nm in total, which will be spilt equally to each half axle. This example is the same one I gave earlier in post#73 and it was validated by Amit and Sutripta.

I think I need to re-read the whole thing again - I though that was where we disagreed. I always said if a differential is designed to split available torque in the ratio 50-50, it will split it in the ratio 50-50.

My confusion is here - Consider the following statement from Amit's

So in a 4WD vehicle the Front axle differential will transmit more torque to the front wheel which has more resistance in the same proportion as the ratio of the resistances of the two front wheels.

Same is for the rear axle differential.

And the third differential ( if it exists) will transfer more torque to either the front or rear propeller shaft depending on where the resistance is more.

If the vehicle is rear heavy and all four tyres are on nearly the same type of ground surface more torque might be transferred to the rear axle as it will have higher traction values.
For eg. In a 4WD Tractor doing a loader operation (filling the bucket with the gravel) more torque (as high as 90%) is transfered to the front Propeller shafts rather than the rear because sometimes the rear wheels get lifted off the ground! I say this from my personal design experience

Will more torque get transmitted to the front wheels if rear wheels are losing traction(Unless you designed something in-between specifically designed to transfer torque that way)? Am considering an open central differential here

[Samurai]

Originally Posted by mallumowgli

Torque available at the propeller rotates the slipping wheel at a higher rpm

Torque available at propeller is a misnomer. Consider open differential. Torque at the propeller depends on the load from the slipping wheel. If the slipping wheel demands only 10Nm, the propeller can only provide 20Nm.

Originally Posted by mallumowgli

I think I need to re-read the whole thing again - I though that was where we disagreed. I always said if a differential is designed to split available torque in the ratio 50-50, it will split it in the ratio 50-50.

We were originally discussing in the MLD thread. So it is possible some of us were talking MLD, and you thought otherwise?

Originally Posted by mallumowgli

My confusion is here - Consider the following statement from Amit's

I don't know what kind of differential he is talking about. It certainly doesn't sound like open differential.

[Sutripta]

Originally Posted by lucifer1881

Let us examine the torque vs. rpm curve above. Torque and rpm have a straight-forward relationship. At any given rpm, the torque is the same. I cannot explain any further than this. So I will rest my case.

If graphs are sacrosanct, rather than the workings of an engine, then:-
Implicit in these (printed) curves is that these are for WOT. For each throttle position you will get a different curve. Thus a family of curves. Thus three variables:- rpm, torque, throttle position. On a two dimensional graph, one will be held steady as a parameter. For the normal torque, power/ rpm, throttle position (WOT) is the parameter. We can replot these as throttle position - torque, keeping rpm as a constant. When the varying nature of the torque as a function of throttle position for any given rpm will become clear.

How about explaining the missing power in
http://www.team-bhp.com/forum/4x4-te...ml#post3329713

Regards
Sutripta

[amit_purohit20]

Originally Posted by DKG

Shom in your second diagram assume one vice applies a force of 20 lbs and the second vice applies a force of 40 lbs ...

First of all this is not possible as what you said. For a single shaft (get the differential out now for some time) which has two different set of resisting torques at different ends the shaft will first try to overcome the least resistance then it can not still rotate because the other higher resisting torque is there so it will try to overcome the higher one too and then only it can rotate. So the torque in the shaft will be only one torque that is the maximum one once it starts rotating.

Originally Posted by DKG

Since an assertion has to be explainable....

I didn't understand a single thing out of what you have written. Could you please use some diagrams to illustrate and elaborate your point.

Originally Posted by Sutripta

@amit
I disagree. I say in locked mode the two halfshafts can carry different torques, so torque at each wheel can be different. (Just the sum has to be equal to the driving torque).

No a single shaft (as is the case of a locked differential) can carry only one torque at a time. If a shaft carries different torques at different places on the same shaft then the law of equilibrium will fail.

Originally Posted by Sutripta

Please do make up your mind.

Regards
Sutripta

You got me wrong. I agree that a propeller shaft feeding the open differential will carry 10Nm + 10Nm = 20 Nm when the differential is not locked.

I disagree with your theory of same shaft carrying different torques at different ends.

Originally Posted by SS-Traveller

The degree of angular deflection is not important. What matters is that some twisting happens.

What would be the T then?

Not unless there is any slippage between the square bar and the jaws of the vice - reasonable to expect that (slippage) in a round bar, hence I took the example of a square bar here.

Sorry Sir the angular deflection is important in Case II without which nobody can comment at what you are trying to arrive. If you place the spanner in the middle twisting will be equal on both sides of the shaft as length of shaft on both sides of the spanner is the same. Please refer the equation which I have put forth for proper understanding of how the length of shaft affects the torque required to create a unit deflection.

Originally Posted by Samurai

Actually, he has done more than that. Yesterday he made me understand [offline] how torque can be visualized when there is no obvious displacement or rotation. So there will be distribution in a locked MLD, it can be seen only mathematically.

I again strongly disagree here and would be really interested to know what offline talk happened. Just dont use mathematics, also support it with imagination of what happens between the teeth of the differential planet pinions and differential sun gears. Till the time one is not clear about the force distribution on the pair of teeth in contact between the planet pinion and sun gear it is difficult to understand.

Originally Posted by mallumowgli

My confusion is here - Consider the following statement from Amit's

Will more torque get transmitted to the front wheels if rear wheels are losing traction(Unless you designed something in-between specifically designed to transfer torque that way)? Am considering an open central differential here

My statement was:
So in a 4WD vehicle the Front axle differential will transmit more torque to the front wheel which has more resistance in the same proportion as the ratio of the resistances of the two front wheels.

Same is for the rear axle differential.

And the third differential ( if it exists) will transfer more torque to either the front or rear propeller shaft depending on where the resistance is more.


Oops I stand corrected here. I had written this post in the night @ 2pm.
Sorry for confusing you and others. I have written exactly opposite of what I wanted to tell. I apologize.

The corrected statement is:

Case I
In a 4WD (actually what I mean here is AWD) vehicle having Front Unlocked Differential, Rear Unlocked Differential and a Centre Unlocked Differential (which powers the front and the rear propeller shafts)...

The unlocked front and rear differentials will transfer torque/ power to the wheels having the least resistance.

The Centre Unlocked Differential will transfer torque to either the front or the rear propeller shaft which ever has less resistance.

Case II
In a 4WD (AWD) vehicle having Front Locked Differential, Rear Llocked Differential and a Centre Locked Differential (which powers the front and the rear propeller shafts) and one of both the front and rear wheel is in air....

The Front and Rear Locked Differentials will transfer torque to the wheel on the hard ground.

If you try to measure torque now at the wheel (or half shaft) which is in air you will get the same torque value as you get on the other half shaft (connected to the wheel on the hard ground) because both the half shafts have been joined mechanically to form a single shaft.

And as I stated earlier a single shaft will carry only one torque value.

The Centre locked differential will transmit torque which ever is the highest between the two half shafts (the front and rear propeller shafts)
Both these half shafts are now again joined mechanically here so the Front and the Rear propeller shaft will carry the same torque.

If the 4WD vehicle is a rear heavy vehicle that means we have more traction on the rear wheels. The Front and Rear propeller shafts (In case of Locked Centre Differential) will carry the torque which can be transferred to the Rear Differential (as this is higher than the front).

So In a Unlocked Differential the torque is transferred to the wheel having the least resistance.

In a Locked Differential the torque carried by the half shafts is same and is equal to the highest resisting torque (or highest resistance offered by any one set of wheel on an axle).

I confused the two conditions of a Locked and Unlocked Differential in my previous post and I had said exactly the opposite.

I hope the above explanation makes things clear.

This is only for Sutripta Sir, others dont read the same and get confused:

I again repeat for a single shaft we cannot have multiple torques on the same shaft at different places unless the shaft is constrained at different places in different manner.

If you fix a shaft at say 4 places in its entire length and apply different torques at different locations on the same shaft then only we can have different torques on the same shaft. But practically it is no more a single shaft but as good as 4 different shafts.

[DKG]

Amit what I meant by an assertion being explainable is that it should fall within the elements of the equation that defines torque. That is, the force applied and the distance at which it is applied from the pivoting point.

Now many here seem to suggest that the torque in a locked differential is same regardless of the load at either wheel. I am inclined to disagree as it appears to me that the pivoting point is not the same for the wheel on the ground vis a vis the wheel in the air. If the resultant distance at which torque is applied is different at either end of a locked differential then the resultant torque at either wheel will be different as well. In summary I am suggesting that the pivot point varies for either end of a locked diff with one wheel on the ground and another in the air and hence torque experienced cannot be the same for both ends.

The force applied is the same for both ends of the diff but the torque will be different. The fact that the grounded wheel is exerting resistance at its circumference whereas the wheel in air exerts it's resistance of inertial mass at a point between centre to circumference means the torque applied has to be different.

[Sutripta]

Hi amit,
Lets take a simple scenario which everyone here can relate to. Only addition is we instrument the halfshafts and the prop shaft so that we can get a torque reading for all three shafts.
Assume the standard theoretical conditions of perfect mechanisms, no friction/ drag/ power loss etc.
Assume 1:1 diff ratio. (Not really necessary, introduced for simplicity).
Vehicle with automatic GB/ torque converter.
Chuck the non driving wheels so that the vehicle cannot move. (Actually nothing changes if the vehicle is moving, but is far simpler to visualise if the vehicle is stationary. Which necessitates introduction of the torque converter. Better than a slipping clutch!)

Open differential.
Put the vehicle in D and gently rev the engine.
What expected (steady state) observation be for each of the wheel-halfshaft combination?
Now jack up one of the driving wheels so that it is in the air.
How will the observation change?

Now repeat the same procedure with the diff locked.
What expected observations now?

By observation I mean torque and rpm. Qualitative, not quantitative

Regards
Sutripta

[SS-Traveller]

Originally Posted by amit_purohit20

Sorry Sir the angular deflection is important in Case II without which nobody can comment at what you are trying to arrive. If you place the spanner in the middle twisting will be equal on both sides of the shaft as length of shaft on both sides of the spanner is the same.

Okay, then assume that the square bar in Case II is 2x the length of the bar in Case I, and the spanner is applied exactly in the centre. To produce the same deflection as in Case I with torque T(x), will torque T(x) be sufficient in Case II, or would the value of (x) change?

[Samurai]

Originally Posted by amit_purohit20

Oops I stand corrected here. I had written this post in the night @ 2pm.
Sorry for confusing you and others. I have written exactly opposite of what I wanted to tell. I apologize.

Oh, thank you. That is why I expressed my doubt in both post#73 and post#93. I couldn't understand what kind of differential you were talking about. Not need for apologies, mistakes happen, remember my least-force definition?

Hopefully, the proponents of "constant torque at all load conditions" hypothesis will also admit their mistake and come out in open. We all make mistakes, there is no shame in admitting it.

Originally Posted by amit_purohit20

First of all this is not possible as what you said. For a single shaft (get the differential out now for some time) which has two different set of resisting torques at different ends the shaft will first try to overcome the least resistance then it can not still rotate because the other higher resisting torque is there so it will try to overcome the higher one too and then only it can rotate. So the torque in the shaft will be only one torque that is the maximum one once it starts rotating.
.
.
.
No a single shaft (as is the case of a locked differential) can carry only one torque at a time. If a shaft carries different torques at different places on the same shaft then the law of equilibrium will fail.
.
.
I disagree with your theory of same shaft carrying different torques at different ends.
.
.
I again strongly disagree here and would be really interested to know what offline talk happened. Just dont use mathematics, also support it with imagination of what happens between the teeth of the differential planet pinions and differential sun gears. Till the time one is not clear about the force distribution on the pair of teeth in contact between the planet pinion and sun gear it is difficult to understand.

Originally Posted by amit_purohit20

In a Locked Differential the torque carried by the half shafts is same and is equal to the highest resisting torque (or highest resistance offered by any one set of wheel on an axle).

Let's take a human example.

Hold a 6ft long metal pipe at the center, horizontally to the ground. Have a strong guy hold one end tightly. And there is a small child holding the other end. Let's say we need 50Nm to overcome the strong guy and only 1Nm to overcome the child.

How much torque is required to turn the pipe? According to you it is 50Nm, according to Sutripta it is 51Nm.

Now let us replace the child with a adult, whose grip needs 49Nm to overcome.

How much torque is required to turn the pipe now? According to you it is still 50Nm, according to Sutripta it is 99Nm.

I believe Sutripta is right here.

[mallumowgli]

Originally Posted by amit_purohit20

Case I
In a 4WD (actually what I mean here is AWD) vehicle having Front Unlocked Differential, Rear Unlocked Differential and a Centre Unlocked Differential (which powers the front and the rear propeller shafts)...

The unlocked front and rear differentials will transfer torque/ power to the wheels having the least resistance.

Isn't it "will transfer torque equivalent to that (or twice that) of the wheels having least resistance"?

Originally Posted by Samurai

Hopefully, the proponents of "constant torque at all load conditions" hypothesis will also admit their mistake and come out in open. We all make mistakes, there is no shame in admitting it.

If this includes me, I think I was the first to admit, immediately after Jeroen's post

Let's take a human example.

How much torque is required to turn the pipe now? According to you it is still 50Nm, according to Sutripta it is 99Nm.

I believe Sutripta is right here.

Well....Vow!!!

[amit_purohit20]

Oh Bacche ki Jaan loge kya ( How much more will you squeeze me? )

Originally Posted by Samurai

Let's take a human example.

Hold a 6ft long metal pipe at the center, horizontally to the ground. Have a strong guy hold one end tightly. And there is a small child holding the other end. Let's say we need 50Nm to overcome the strong guy and only 1Nm to overcome the child.

How much torque is required to turn the pipe? According to you it is 50Nm, according to Sutripta it is 51Nm.

Oh yes again the teacher comes to the rescue! This example will solve all questions arising in the minds of people here.

In a locked differential I was obsessed with the idea of one wheel in the air so I forgot to think about any resistance from it. For all practical terms I was taking it as zero and hence saying that the torque carried by the propeller shaft = torque transferred to the wheel on the ground.

But your example has enlightened me. I admit!

Yes in your example of resisting torques having 50Nm and 1 Nm on the opposite ends of the same shaft. The torque required to rotate this shaft will be 51 Nm :50+1, I was always treating this 1 as zero for all practical purposes.

Originally Posted by Samurai

Now let us replace the child with a adult, whose grip needs 49Nm to overcome.

How much torque is required to turn the pipe now? According to you it is still 50Nm, according to Sutripta it is 99Nm.

I believe Sutripta is right here.

Yes as you said in the second example it will be 50+49 = 99 Nm. The propeller shaft will carry 99 Nm torque considering differential ratio of 1:1.

(If in my above posts I have said the opposite of what I say here again I stand corrected! Apologies once again!)

First time I agree with Sutripta Sir . Thanks Sir

But still I disagree with his usage of words like torque distribution on a single shaft. It cannot have different torques at different places on the same shaft.

Lets again take the example of 50 Nm + 49 Nm = 99 Nm.
We cannot say here that the torque on one end of shaft will be 50 Nm and the other end will be 49 Nm. The max torque when measured on the shaft will be 99 Nm all throughout the shaft.

If you have a device which can measure torque placed at any random position on this rotating shaft you will get 99 Nm. Normally such devices are strain rosettes used for data measurement.

But practically in a locked differential the two half shafts that originate from differential are two different shafts.

So if you place strain rosette on one half shaft you will get 50 Nm.
If you place on the other side you will get reading of 49 Nm.
If you place on the top of the differential cage or propeller shaft (Diff. ratio being 1:1) you will get 99 Nm.

If this is what Sutripta Sir mentions as Distribution then I will agree because there are two half shafts here.

But If Sutripta Sir had to say that on a single shaft you will get torque distribution I will disagree. (As in the Case of Samurai's single 6ft shaft example)

So with this explanation I think we all can agree and are on the same table now. Right?

Originally Posted by SS-Traveller

Okay, then assume that the square bar in Case II is 2x the length of the bar in Case I, and the spanner is applied exactly in the centre. To produce the same deflection as in Case I with torque T(x), will torque T(x) be sufficient in Case II, or would the value of (x) change?

So Ultimately you caught me. Thanks I stand corrected I have goofed up in simple addition mathematics but not the thought process.

Case I - Shaft of Length L and spanner at the end. Torque required is say T.

Case II - Shaft of Length L hold at two places and spanner in between. Torque required is 4T and not 2T as stated by me earlier. Sorry!

Case III - Shaft of Length 2L hold at two places and spanner in between.
Torque required is 2T.

In all the above cases the angular deflection of the shaft is same.

Thanks indeed for pointing my shortcomings!

Originally Posted by DKG

Amit what I meant by an ....
applied has to be different.

Sir by all means I tried to understand what you want to say, but because of my limitations I really could not understand your point about the "pivots". I would really need you to draw something to understand further.

But I feel with the above explanations which I have given where in now I, Samurai and Sutripta all are on the same table. (Nearly!). I think the above explanations should satisfy your queries too.

Originally Posted by Sutripta

Hi amit,
Lets take a simple scenario which everyone here can relate to. Only addition is we instrument the halfshafts and the prop shaft so that we can get a torque reading for all three shafts.

The moment you consider these three shafts as separate I agree to your Torque Distribution theory!

Originally Posted by Sutripta

Assume the standard theoretical conditions of perfect mechanisms, no friction/ drag/ power loss etc.
Assume 1:1 diff ratio. (Not really necessary, introduced for simplicity).
Vehicle with automatic GB/ torque converter.
Chuck the non driving wheels so that the vehicle cannot move. (Actually nothing changes if the vehicle is moving, but is far simpler to visualise if the vehicle is stationary. Which necessitates introduction of the torque converter. Better than a slipping clutch!)

I agree stationary things are more easy to imagine and thats how we were taught in engineering to analyse things.

Originally Posted by Sutripta

Open differential.
Put the vehicle in D and gently rev the engine.
What expected (steady state) observation be for each of the wheel-halfshaft combination?
Now jack up one of the driving wheels so that it is in the air.
How will the observation change?

1st Case- Both half shafts will carry equal torque. (Considering resistance on both wheels of the axle is the same.)
2nd Case- The half shafts will carry only that much torque required to spin the wheel which is near zero.

Originally Posted by Sutripta

Now repeat the same procedure with the diff locked.
What expected observations now?

By observation I mean torque and rpm. Qualitative, not quantitative

Regards
Sutripta

1st Case- Both half shafts carry the same torque.
2nd Case- Both half shafts will carry different torques. The one with the wheel spinning will be nearly zero and the one with the wheel on ground will carry the max. torque which can be transmitted based on traction.(Also based on throtte inputs, It can also be less based on throttle inputs. Let us not confuse here on the throttle part.)

Please note that for second case above I am considering two half shafts as different shafts so talking about distribution.

If I consider the two half shafts and the complete differential case and gears as a single system as they are locked and dont have relative motion then I would say the all parts of this system possess a single torque value.

@SS-Traveller One more thing to be noted here is that in a power transmission system, shafts are loaded at the ends only. Very rarely we will find such examples where a single shaft is resisted at both the ends and the driving torque is applied at the centre.

Originally Posted by mallumowgli

Isn't it "will transfer torque equivalent to that (or twice that) of the wheels having least resistance"?

Say In and Unlocked Front Differential the least resisting torque is 10 Nm, then the Front Propeller shaft should carry 20 Nm.

Now say In and Unlocked Rear Differential (of the above vehicle) the least resisting torque is 20 Nm so the Rear Propeller shaft should carry 40 Nm.

But it will not carry 40 Nm because the Unlocked Centre Differential will transmit the least of the Front and Rear Propeller Shafts which is again 20 Nm.

Ah! You guys made me think a lot, very rare of me using my brains nowadays.

Thanks to all of you to get my fundamentals straight , I hope now no one again twists it .

[DirtyDan]

Originally Posted by DirtyDan

Torque to one wheel of a locked axle and not the other?

If the diff is locked and the whole system moves "as a single entity" you are saying that this "single entitiy" both has torque and does not have torque at the same time? No contradiction there I suppose....

If you are saying that this axle is one and I cannot talk about "distribution"
of torque between two wheels because of that then I must ask you to stop talking about torque being different between the two wheels for the same reason, we are talking about a "single entity". If you persist, I am going to have to turn you over to the contradiction police.....and I warn you, you are subject to arrest and a FULL body search. ...yeah, yeah, I know some of you are looking forward to that.

There sure is a lot of "stuff" flying around in this thread. I wonder what's at stake?

Originally Posted by amit_purohit20

I was expecting someone to catch me on this and nothing better than to be caught by the DirtyDan himself . (Lot of wise souls fooling around here ).

Blame it on my english skills or lack of concentration. What I wanted to say that a locked differential makes the two half shafts into one which means the whole entity behaves like its a single shaft.

So you are right, the torque on a single shaft is same everywhere be it on towards the wheel which is on the hard ground or in the air.

Nice catch Dan!

Amit, my argument above (in the form of "reductio ad absurdum") does not prove that torque is the same everywhere on a locked axle. It only makes a case for the concept of "distribution" of torque between two wheels of a locked axle. This concept is logically necessary for us to consider torque being different between two wheels of a locked axle.

[Samurai]

Originally Posted by amit_purohit20

But still I disagree with his usage of words like torque distribution on a single shaft. It cannot have different torques at different places on the same shaft.

Lets again take the example of 50 Nm + 49 Nm = 99 Nm.
We cannot say here that the torque on one end of shaft will be 50 Nm and the other end will be 49 Nm. The max torque when measured on the shaft will be 99 Nm all throughout the shaft.

If you have a device which can measure torque placed at any random position on this rotating shaft you will get 99 Nm. Normally such devices are strain rosettes used for data measurement.

Now you have forced me to supersize my human example. According to a recent Superman movie, Superman has a kid. Now I need them both.

Let's take the superhuman example.

Let Superman hold the same 6ft long metal pipe at the center, horizontally to the ground. Have a regular human hold one end tightly. And Superman's kid will be holding the other end. Let's say we need 50Nm to overcome the regular human and 1000Nm to overcome the superkid.

Now Superman will start rotating the pipe, remember that Superman is lot stronger than his kid. Superman needs to apply 1050Nm to rotate the pipe, according to our previous discussion.

The end held by the regular human will rotate... but the other end will start twisting! That is because the pipe can't handle any where near 1000Nm of twisting force. Let's say it starts twisting when 500Nm is applied to it. So when Superman applies 550Nm of torque, the human side of the pipe starts rotating (50Nm), the superkid side of the pipe will twist (500Nm) into a spiral. This means torque is indeed distributing even in the single shaft....

[amit_purohit20]

Originally Posted by Samurai

Let's say it starts twisting when 500Nm is applied to it. So when Superman applies 550Nm of torque, the human side of the pipe starts rotating (50Nm), the superkid side of the pipe will twist (500Nm) into a spiral. This means torque is indeed distributing even in the single shaft....

Good example.

First the human end of the pipe will not rotate as you said.

I think you meant that the human end of the pipe will turn (and not rotate) by an angle= angle of deflection created by the twist on the other side (SuperKids side) of Supermans Hand. I agree.

The twist angle between Supermans Hand and Super kids hand will be different than between Supermans Hand and Human hand. I agree here.

So if you are saying you put torque measuring sensors (which calculate torque based on the twist/angular deflection of shaft) one between Human hand and Supermans hand and the other between Supermans hand and Superkids hand, you will surely get two torque values. I agree.
Here the torque sensor is measuring torque based on the angular twist caused by Supermans Hand.
But I would not call this torque distribution, because...

Lets say now you measure the torque on the shaft using a dynamometer (by increasing resistance on the shaft).

If you place the dynamometer in two different cases on the either side of Supermans hand you will will get the same reading of 1050 Nm or infact more till the time Superman can not generate more torque say 2000 Nm.( Offcourse condition remaining the shaft/pipe doesnot break). This is the Max. Torque the pipe can carry (because of Supermans limitations)
The Minimum Torque required to rotate the shaft is 1050 Nm.

Normally for engineering calcualtions we use the above method of designing shafts- the Max. Torque one so I always say that a shaft carries a single torque.

In your case its the different resistances at the different ends of shaft which is causing different twists in the same shaft.

The shaft by its own property is not distributing the torque differently to different ends. Its the different resistances at the two ends which is causing this to happen.

For eg. the Superman now applies torque at one end of the shaft and the other is fixed in a wall (Infinite resistance). So the torque throughout the shaft is equal to the Max. Torque what the Superman can apply that is 2000 Nm. Thus its not the property of the shaft to distribute the torque but it depends on the loading conditions.

In the earlier case the superman was applying torque in the middle of the shaft and in the above case at one end of the shaft. This causes the difference in torque findings.


If you call this distribution... hmm, I dont know what to say. I need to think more!

Originally Posted by DirtyDan

Amit, my argument above (in the form of "reductio ad absurdum") does not prove that torque is the same everywhere on a locked axle. It only makes a case for the concept of "distribution" of torque between two wheels of a locked axle. This concept is logically necessary for us to consider torque being different between two wheels of a locked axle.

Yes I got your point, we have discussed this in the above posts. Its all about the different resistances at different ends of the shaft causing the shaft to behave so.

[Samurai]

If the net torque between human and superman is only 50Nm, why won't it rotate? Let's say Superman's grip is so hard that it doesn't let any twist to transfer to the other side.

Further, what if the human lets go... Won't that side rotate?