[DirtyDan]

Originally Posted by amit_purohit20

I
4) Once the differential is locked there is nothing called as distribution because the whole system moves as a single entity.So if one wheel is in the air and the other on hard rock. The wheel on the hard rock will get all the torque what the engine can generate because of the higher resistance it has to offer.

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?

[amit_purohit20]

Originally Posted by DirtyDan

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

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".

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!

[SS-Traveller]

Quick diagrams and question:

The diagram above shows a square bar being held in a vice at one end, a wrench turning the other end with a force / torque T(x), enough so that the bar becomes twisted into a spiral.

With the same bar held at both ends in 2 vices, and the wrench applied in the middle, do I need T(x) to twist the bar into a spiral, or do I need T(2x)? or any other quantity of T?

Now, if I release the vice at one end, obviously that half of the bar is not going to be twisted into a spiral. So to twist the other half into a spiral, do I again need T(x), or T(2x), or any other quantity of T?

[Samurai]

Originally Posted by amit_purohit20

Any object, water,fluid, even humans prefer the path of least resistance! But it is still wrong to use the word "least". Infact torque measured on a shaft is the max. torque it can transmit not the least. The least would be always zero! (Just referring to the torque definition and not our topic into consideration.)

I guess I'll stop saying least-force here onwards.

Originally Posted by DirtyDan

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".

Have you checked the very first post of this discussion? I said there was no distribution. But in reply (#3), you insisted on distribution.

[amit_purohit20]

Originally Posted by SS-Traveller

Now, if I release the vice at one end, obviously that half of the bar is not going to be twisted into a spiral. So to twist the other half into a spiral, do I again need T(x), or T(2x), or any other quantity of T?

Good and tricky question indeed! I refer to your cases as Case I and Case II and my cases as Case A and Case B.

CASE A

T= F X R so nowhere its mentioned where you apply it on the shaft.

Torque in this case here is dependent on the force you applied on the spanner and the length of the spanner. So torque will remain the same in both cases.


CASE B
Now there is a different catch you are talking about here - Creating the spiral effect or same angular deflections on both ends of the square shaft.

Torsional Deflection of Shaft



The angular deflection of a torsion shaft can be expressed as

θ = L T / (J G) (5)

where

θ = angular shaft deflection in radians

L = length of shaft (mm,in)

G = modulus of rigidity (Mpa, psi)

J= Second moment of area.

T= Torque

or T= (θ.J.G)/L

Courtesy:

http://www.engineeringtoolbox.com/to...fts-d_947.html

Say in Case I you have angular deflection of shaft of 1 Radian.

As in Case II you have put the spanner in the middle of the shaft. You also want to have the same torsional deflection of the shaft (θ) on both ends of the shaft ie. 1 Radian on left side and 1 Radian on right side.

Referring to the above formula in Case I you have full length L and in Case II you have Full length L/2

So in both cases the Torque required will be different.

In Case II the shaft is fixed on both sides and you are trying to twist it in the middle and you want same angular deflection ie 1 Radian on both ends of the shaft.As you require now same angular deflections twice at both ends ie 1 Radian at left end and 1 Radian at right end the torque required will be double.

The torque required to create a certain torsional deflection of the shaft is inversely proportional to the length of the shaft. Less the length of the shaft more the Torque required to twist it by a given angle.

If in Case II you only wanted 0.5 radian of deflection of the shaft at the two ends the torque required will be again the same as Case I

I hope I answer your query.

[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 and the rod is one of pliable material. With the spanner applying the force X (say at the centre) will it cause unequal twisting in the rod at either end ? I am inclined to think the twist at either end will be unequal.

[DKG]

Since an assertion has to be explainable in terms of the basic equation defining torque I am inclined to conclude that the resultant pivot point is not the same in case of a wheel on the ground and a wheel in the air. Since the wheel on the ground causes the pivot point to move further away from the point at which the force is applied more torque is applied to the wheel on the ground. So the answer to this dilemma is the resultant pivot point which is a function of the frictional contact between tyre and ground. It's not a fixed distance and the pivot point moves out. Greater the frictional bond between the tyre and the ground greater the pivot distance and greater the torque. In a fused rod ie a locked differential the force vector is the same across the rod but the resultant torque is different owing to varying lengths of the distance vector to pivot point which is a function of the contact patch of tyre and ground.

[Sutripta]

@amit
Arka (and lots of others) think that when a diff is locked, the two halfshafts essentially become one. As such the same torque will prevail all over, and so torque at the two wheels will be the same.
You agree

Originally Posted by amit_purohit20

For a locked differential its again the case of a single shaft and there is no such thing such as half shafts in locked mode!

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). Once again

Agree with you Sir!

Please do make up your mind.

Regards
Sutripta

[SS-Traveller]

Originally Posted by amit_purohit20

Creating the spiral effect or same angular deflections on both ends of the square shaft.
I hope I answer your query.

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

What would be the T then?

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 and the rod is one of pliable material. With the spanner applying the force X (say at the centre) will it cause unequal twisting in the rod at either end ? I am inclined to think the twist at either end will be unequal.

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.

[lucifer1881]

Originally Posted by lucifer1881

...However, if you down-shift and manage to attain 2500rpm you will achieve 100kmph again...

What made me say that, I will never know. Please excuse. I must have had a terrible moment.

Originally Posted by Samurai

...Can you please provide some technical web resources on the same to convince people who disagree with you on this?

Originally Posted by 1100D

Wrong, absolutely, wrong. At any given RPM the torque produced by the engine is a function of many parameters like throttle position...

Originally Posted by Sutripta

Just us stating this, no matter how strongly we word it, is not going to convince the naysayers...

Originally Posted by Jeroen

Sorry, but the notion that an engine produces the same torque at a given rpm is simply not true...

Image Source: http://forums.edmunds.com/discussion...rid-diesels/p3

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.

I got into this discussion to only clear up the concept of torque. Torque is a cruel mistress and is known for its wily ways.

[Samurai]

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.

That graph is showing max torque that could be achieved at different rpms by the engine. Max torque specs for this engine would be 280Nm @ 2000-2200rpm.

Originally Posted by lucifer1881

I got into this discussion to only clear up the concept of torque. Torque is a cruel mistress and is known for its wily ways.

I'd say torque is misunderstood rather than cruel.

[mallumowgli]

Originally Posted by Samurai

Exactly, it is a twisting force Then why are people saying torque is not a force at all and laws of conservation of force doesn’t apply to it?

Samurai - i never said law of conservation of force (rather energy) is not valid. I was implying not to give the same understanding for torque that we give for force/energy -since torque is a result of force

And in the light of the lucid explanation given by amit_purohit20, I stand enlightened.

But still not convinced that in an open differential, if one wheel slips the entire torque or even a portion will be transferred to the wheel with traction

[DirtyDan]

Originally Posted by Sutripta

@amit
Arka (and lots of others) think that when a diff is locked, the two halfshafts essentially become one. As such the same torque will prevail all over, and so torque at the two wheels will be the same.
You agree

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).

Regards
Sutripta

Originally Posted by Samurai

Have you checked the very first post of this discussion? I said there was no distribution. But in reply (#3), you insisted on distribution.

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

[Samurai]

Originally Posted by mallumowgli

Samurai - i never said law of conservation of force (rather energy) is not valid.

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

Originally Posted by mallumowgli

Remember Torque is not energy so the laws that energy cannot be destroyed does not apply here.

Originally Posted by mallumowgli

But still not convinced that in an open differential, if one wheel slips the entire torque or even a portion will be transferred to the wheel with traction

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.

[Jeroen]

Originally Posted by lucifer1881

I cannot explain any further than this. So I will rest my case.
.

Please do!
I'm only a humble engineer, but some parts of this thread should be a case study on human behavior. What will it take to convince somebody they have got it wrong! Im certainly done trying. As they say, ignorance is bliss!

Jeroen