[amit_purohit20]

Originally Posted by 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?

Twisting of a shaft by a certain angle is fine.

But I don't understand how a single shaft can rotate at one end and not rotate at the other end?

By rotate do you mean the single shaft will achieve rpm at the free end?
Its next to impossible.

I am not getting your point...

[DirtyDan]

Originally Posted by amit_purohit20

Twisting of a shaft by a certain angle is fine.

But I don't understand how a single shaft can rotate at one end and not rotate at the other end?

By rotate do you mean the single shaft will achieve rpm at the free end?
Its next to impossible.

I am not getting your point...

If the shaft was made of rubber instead of hardened steel, it would be easier to visualize.....power out...gotta quit now.

[Samurai]

Exactly, the steel pipe will behave like rubber pipe under the circumstances.

The part between human and superman will rotate and the other side will twist and get mangled.

[amit_purohit20]

Originally Posted by Samurai

Exactly, the steel pipe will behave like rubber pipe under the circumstances.

The part between human and superman will rotate and the other side will twist and get mangled.

@Samurai- You cannot call this rotation it will only be a slight twist angle of certain degrees = Twist of the shaft on the other side.

We can use rubber to visualize but cannot imagine a rubber shaft rotating at certain rpm and also transmitting torque. The rubber shaft will only twist but not rotate.

Twisting is not rotation! Guys please at least get this straight.

Rotation means when the applied torque on the shaft exceeds the torque on the two ends of the shaft.

I really do not understand how this rotation funda is coming now?

The shaft will not rotate unless and until the applied torque crosses 1050 Nm. Below that the shaft can only twist but not rotate.

Isn't it absurd to say that half of the shaft rotates but the other half of the same shaft is stationary (or just twisted)?

[Sutripta]

Leaving aside a few minor quibbles, I agree with your expected observations to my questions

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?

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

the most relevant for this thread is the one for locked differential, one wheel on the ground, one wheel in the air

Originally Posted by amit_purohit20

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

A point I've been trying to say, without much success, on this thread. Now I'll leave the job of convincing people who think otherwise to you and others who agree with this!


************************************************** *****
Strictly for Amit
Lots to discuss about some other points you have raised, but in light of

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.

let me ask you a (mischievous ) question.

Consider that the mechanical arrangement is now a single shaft (axle) running from wheel to wheel. In the middle of the shaft the crown gear is welded on. No differential (locked or otherwise) anywhere!
Each end of the shaft (any position between crown gear and wheel) is instrumented as before to read torque.

Now let us repeat the experiment of one wheel on the ground, one in the air, and all other conditions same as before.

What torque value will we expect to see from our instruments? Why?

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

Leaving aside a few minor quibbles, I agree with your expected observations to my questions


the most relevant for this thread is the one for locked differential, one wheel on the ground, one wheel in the air

A point I've been trying to say, without much success, on this thread. Now I'll leave the job of convincing people who think otherwise to you and others who agree with this!


************************************************** *****
Strictly for Amit
Lots to discuss about some other points you have raised, but in light of

let me ask you a (mischievous ) question.

Consider that the mechanical arrangement is now a single shaft (axle) running from wheel to wheel. In the middle of the shaft the crown gear is welded on. No differential (locked or otherwise) anywhere!
Each end of the shaft (any position between crown gear and wheel) is instrumented as before to read torque.

Now let us repeat the experiment of one wheel on the ground, one in the air, and all other conditions same as before.

What torque value will we expect to see from our instruments? Why?

Regards
Sutripta

The instumentation (which uses shaft deflection for finding torque) on the side of the free wheel will read nearly zero torque and the other side instrumentation will read full torque as found on the Crown wheel gear.

By the way are you a mechanical engineer?

[DKG]

There was a time when i lived in a country where it snowed very heavily in winter and realised in no time that even a V8 engined car was useless as the torque simply didn't get applied. As torque is inversely proportional to speed revving simply didn't help, infact made matters worse.

To understand why in a locked diff a torque split occurs when one wheel is in the air and another on the ground you need to simply factor in the differential loads experienced at either end. Newton's third law of motion states that for every force exerted an equal and opposite force is exerted back.

The wheel on the ground has the load of its own weight plus its share of the car's weight to offer as resistance to the wheel to turn. Whereas the wheel in the air has only its weight to contend with. This variation in load at either end of the diff means whatever torque is applied through the pinion a greater chunk gets applied to the wheel with greater resistance load.

As offroad enthusiasts you may be interested to focus more on enhancing the coef of friction at the contact point of tyre and ground to ensure more torque gets applied at the wheel.

The least expensive way to replicate a locked diff effect in your 4x4 is to install a separate handbrake lever for each of the driving wheels. As you brake the slipping wheel more torque gets applied to the wheel which still has contact with the ground.

[ex670c]

Originally Posted by Sutripta

the most relevant for this thread is the one for locked differential, one wheel on the ground, one wheel in the air

A point I've been trying to say, without much success, on this thread. Now I'll leave the job of convincing people who think otherwise to you and others who agree with this!


************************************************** *****
Strictly for Amit
Lots to discuss about some other points you have raised, but in light of

let me ask you a (mischievous ) question.

Consider that the mechanical arrangement is now a single shaft (axle) running from wheel to wheel. In the middle of the shaft the crown gear is welded on. No differential (locked or otherwise) anywhere!
Each end of the shaft (any position between crown gear and wheel) is instrumented as before to read torque.

Now let us repeat the experiment of one wheel on the ground, one in the air, and all other conditions same as before.

What torque value will we expect to see from our instruments? Why?

Originally Posted by amit_purohit20

The instumentation (which uses shaft deflection for finding torque) on the side of the free wheel will read nearly zero torque and the other side instrumentation will read full torque as found on the Crown wheel gear.

Hi Sutripta & Amit,

Taking in a Part Time 4WD.

1) In 2WD & RWD - Approx 100% of the torque is going to the Rear Axle.

2) Since the 2WD to 4WD Mechanism is basically a Shaft With Two Outputs, the Front Propeller Shaft rotates, with out positively engaging the Front Axle.

This scavenging effect has enough torque to Rotate a 7kg Propeller Shaft, and a 25Kg Differential Assembly at 500 Plus Rpm.

The above is also the Reason for Front Axle Disconnects and Free Wheeling Hubs.

3) However when the Front Axle is Engaged 50% of the Torque is sent to the Front Axle?

What dictates the above Split 50:50.
i) Type of T-Case.
ii) Differential Ratio


And if this split is 50:50, and the Front Wheels lose traction, Will the Rear Wheels lose traction?

If the Front Wheels lose traction, will most of the 90-100% Torque be sent to the Rear Wheels?

Regards,

Arka

[tsk1979]

Arka, diff cannot send torque anywhere. Torque is applied by the engine onto the diff. After that its like water. Water will flow where there is least resistance.
So with all 4 wheels having similar traction, you will have torque to all wheels.
As you turn, it will be like water flowing. you won't have any transmission windup or anything due to this nature of differentials.

If any wheels loses traction in open diff, only that will rotate.

When you lock diff, you have to understand there is no wheel.
The diff gets the torque. Its a locked thing. The torque that can be delivered in a locked diff is equivalent to the max possible to the wheel with most traction.


Think of it this way.
Imagine a wheel
You apply torque and turn the free spinning wheel. At constant rate of spin, only that much torque is needed as to overcome friction.

Now imaging this wheel has 4 different brakes.
If you apply one brake you need more torque to spin it. It does not matter that the other brakes are engaged.

As you keep applying more brakes you will need to supply more and more torque to keep spinning (use lower gear).

So if your 4WD with all diffs locked gets stuck with one wheel traction, its like this giant imaginary wheel in which 3 brakes are disengaged, and one brake engaged. The torque needed will be equivalent to what is needed to overcome traction of this one brake.

[ex670c]

Originally Posted by tsk1979

Think of it this way.
Imagine a wheel
You apply torque and turn the free spinning wheel. At constant rate of spin, only that much torque is needed as to overcome friction.

Now imaging this wheel has 4 different brakes.
If you apply one brake you need more torque to spin it. It does not matter that the other brakes are engaged.

As you keep applying more brakes you will need to supply more and more torque to keep spinning (use lower gear).

So if your 4WD with all diffs locked gets stuck with one wheel traction, its like this giant imaginary wheel in which 3 brakes are disengaged, and one brake engaged. The torque needed will be equivalent to what is needed to overcome traction of this one brake.

Hi Tanveer,

Nice explanation.

What is the Maximum Torque ONE WHEEL can deliver in a 4x4 with 3 Locked Differentials, Traction being Optimum.

a) approximately 50% of the Geared Torque (Gearbox/T-Case/Differential)

b) approximately 25% of the Geared Torque (Gearbox/T-Case/Differential)

And your reasons for the above choice.

Regards,

Arka

[tsk1979]

Arka, I could not understand your question. Wheel will not deliver torque. The engine will deliver torque to the diff.
Imagine the diff to be your giant wheel with 4 brakes.
Every time one of the brake is engaged, the engine will try to rotate the giant wheel against this braking force.

Imaging it this way.
Remember Joules experiment?

Now lets say you immerse your wheels in water jars and attach a brake to each wheel.
As you gun the engine, and apply just one brake, you will see water in that particular wheels jar heating up.
In a 0 loss system, as you burn 1000 calories in the engine, the 1000 calories will be used up in that jar only.
All 3 jars will be cold.

If you apply all 4 brakes equally, 250 calories will be transferred to each jar.

So with 3 locked diffs, if all 4 wheels have same traction, its 25%
If three wheels are free with no resistance, all the engine torque 100% will go in trying to spin the traction wheel.

[ex670c]

Originally Posted by tsk1979

So with 3 locked diffs, if all 4 wheels have same traction, its 25%
If three wheels are free with no resistance, all the engine torque 100% will go in trying to spin the traction wheel.

Hi Tanveer,

My questions was not all that complicated. You can interpret it as Torque at the Wheels, if it helps you understand.

If the Centre Diff is Locked, and Front Axle is locked and lost all traction, will 100% of the engine torque go to the Rear Axle?

If 100% Torque is delivered to the Rear Axle, then this is as good a Normal 2WD Mode.

If the Centre Diff is Locked, the vehicle is in Low Ratio and Front Axle is locked and lost all traction, will 2.46 X 100% of the engine torque go to the Rear Axle?

If 2.46 X 100% engine torque is sent to the Rear Axle, then this works like 2WD-Lo ratio, and the Rear Axles will break.

What happens when we engage the Front Axle or Lock the Centre Differential?

Regards,

Arka

[tsk1979]

Ok, now I see.
Yes, if your front wheels go free, and offer no resistance, engine power will work to move the rear wheels.
Its like on your giant wheel 2 brakes are not engaged, but 2 are engaged.
So all torque will work against overcoming the 2 brakes. Powerful engine, weak parts will result in breakage.

But remember what breaks is the weakest link in the chain.
If all 4 wheels have traction, and lets say rear prop shaft is flimsy, it will break, front will not.

To better understand, imagine you have 4WD with center diff locked.
Imagine rear prop shaft connecting at rear diff breaks.
You can still keep driving by engaging 4WD lever. No torque to rear diff as prop shaft is now free. All torque to front wheels.

It will be like normal front wheel drive.

[Samurai]

Originally Posted by ex670c

My questions was not all that complicated. You can interpret it as Torque at the Wheels, if it helps you understand.

The question is not entirely simple. That is because you have not explained the meaning of 100% torque. Do you mean 100% of what the engine is capable, or 100% of torque that is required by the wheels?

Traction or load is the cause, torque is the effect. Engine will not deliver its max torque unless it is required.

Originally Posted by ex670c

If the Centre Diff is Locked, and Front Axle is locked and lost all traction, will 100% of the engine torque go to the Rear Axle?

Let's say the rear axle requires 100Nm of torque to turn the wheels. And the front axle needs just 2Nm to spin freely thanks to zero traction. Engine is capable of 250Nm. I hope this describes your situation. Then the engine will generate 102Nm of amount of torque, 100Nm will go to the rear and 2Nm will go to the front. So it is not 100% of engine ability, but nearly 100% of generated torque that goes to the rear. So it is like 2WD.

Originally Posted by ex670c

If the Centre Diff is Locked, the vehicle is in Low Ratio and Front Axle is locked and lost all traction, will 2.46 X 100% of the engine torque go to the Rear Axle?

Even though you switched to low ratio, the wheels still need the same torque to overcome the collective traction, that is 102Nm. Now the engine needs to generate 102/2.46 = 41Nm of torque. Therefore, engine will generate 41Nm, the TC will multiple it and send 2Nm to front and 100Nm to the rear.

Originally Posted by ex670c

If 2.46 X 100% engine torque is sent to the Rear Axle, then this works like 2WD-Lo ratio, and the Rear Axles will break.

Why will the axles break considering you are delivering the same torque, the amount required to turn the wheels. There is no cause.

But you are on the right track, it can break if the torque required is much much higher. So let us provide the cause, that is load.

In the above example, let's say the Jeep is stuck in a very steep angle and traction is excellent. That means rear wheels will not slip, but needs 500Nm to rotate. Instead of 100Nm, the rear axle requires 500Nm to rotate. So the opposing force is 500Nm + 2Nm = 502Nm. Now the engine will deliver 204Nm (502/2.46), the TC will multiple to get 502Nm. The first 2Nm will be sent forward, the remaining 500Nm will be sent to the rear. If the rear axles can't handle 500Nm of twisting force, they will indeed break. This won't happen in the High ratio, because the engine can only send 248Nm (250-2) to the rear, and that is not sufficient to break the axle or turn the wheels.

[Samurai]

Originally Posted by amit_purohit20

I really do not understand how this rotation funda is coming now?

The shaft will not rotate unless and until the applied torque crosses 1050 Nm. Below that the shaft can only twist but not rotate.

Isn't it absurd to say that half of the shaft rotates but the other half of the same shaft is stationary (or just twisted)?

Let's consider a modified pipe.



This is a soild pipe, with a much strengthened center. Center is where the Superman is holding it. The thinner parts of the pipe will start twisting at 500Nm of torque, but the center won't twist until 2000Nm of torque.

If the human grip can be overcome with 50Nm and Superkid's grip requires 1000Nm to overcome, what happens when superman starts turning the center? I would say the Human side will start turning since the effective torque on that side is 50Nm (too less to twist the pipe). The superkid side of the pipe will be twisting because it can't handle above 500Nm of twisting force. This twist won't leak to the other side because the much bulkier center can easily withstand 500Nm of torque. Visualise, don't apply formulas.

If you still can't believe one side can rotate and other can twist... let's change the condition without changing the applied torque.

Let superman hold the center stationery. Let the human try to turn the pipe applying upto 50Nm. Let Superkid also try to turn the pipe, but he can go up to 1000Nm. As you can see effective torque everywhere is same as before. Now, the human side remains stationery, and the superkid side will twist once he exceeds 500Nm.

In the both the above cases, the human side will follow the superman (rotate or stationery), the superkid side will twist.