Team-BHP - Torque generation and distribution
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Quote:

Originally Posted by DirtyDan (Post 3324749)
Upon actuation the axles are locked as one, hence "Locker". Whatever torque there is from the driveshaft is spread to both wheels equally as there is, in effect, one solid axle.

Well, I know what you are saying. But let's look at the definition of torque:

Torque is a measure of how much a force acting on an object causes that object to rotate.

If the axle is acting as one piece, then the torque is basically used to rotate that axle. There is no distribution.

Quote:

Originally Posted by Samurai (Post 3324066)
Found these statements little odd.

He is describing open differential here. If MLD worked like that, if one wheel loses traction, both wheels get almost no torque and the vehicle gets stuck.

So it doesn't fully become free even at highspeed?

There was a discussion on the working of the MLD in this thread, had some explanatory video's too.

http://www.team-bhp.com/forum/4x4-ve...ml#post2747100

It becomes fully free. Performs as an open differential, no issues.sounds resistances on tarmac on sharp turns.

Quote:

Originally Posted by dhanushs (Post 3324250)
In "Distribute torque equally" , the torque should be, @the propeller shaft. Not the other wheel.

Quote:

Originally Posted by dhanushs (Post 3324614)
With this Eaton MLD, the torque available at the propeller shaft, is distributed evenly to both tyres, irrespective of their traction.

It does not distribute torque equally, say one wheel is slipping, there will be zero torque at that wheel, whereas the wheel with grip will be the one that will be transmitted the power.

Quote:

Originally Posted by Samurai (Post 3324636)
This is exactly what an open differential does.:D

I think Eaton MLD turns both wheels at the same speed irrespective of their traction. But torque would be different, if the traction at each wheel is different.

Exactly. Its interesting, if both wheels are slipping the torque at the propeller shaft will probably be spent accelerating the locked/unlocked rear axle! Anyway Off-topic!

Quote:

Originally Posted by Samurai (Post 3324760)
Well, I know what you are saying. But let's look at the definition of torque:

Torque is a measure of how much a force acting on an object causes that object to rotate.

If the axle is acting as one piece, then the torque is basically used to rotate that axle. There is no distribution.

Or, if you like, the distribution is equal. Same same. I choose to put it this way because the "object", the axle, can be split into 2 pieces upon unlocking and you can get "distribution" of torque. Then the complexity of "distribution" is compounded by the introduction of sundry clutches, sensors, and all that other stuff.

I don't want to get into a 3 way semantic quibble with me, you and Dhanush.:) Serves no purpose. Suffice it to say, my rear differential in my Invader is making some ominous noises. This may be the time for me to sally forth and locker-ize.

Quote:

Originally Posted by 1100D (Post 3324801)

It does not distribute torque equally, say one wheel is slipping, there will be zero torque at that wheel, whereas the wheel with grip will be the one that will be transmitted the power.

No, sorry to disagree. The torque is a constant and is still very much there at the slipping wheel. Were it not so, the wheel/axle would not be spinning wildly. What is missing is traction, not torque.

Lord Siva, please send recovery instructions to the half dead brains cells that used to house my high-priced college physics courses. Shouldn't be hard, there is only a handful of them.

Quote:

Originally Posted by DirtyDan (Post 3324831)
No, sorry to disagree. The torque is a constant and is still very much there at the slipping wheel. Were it not so, the wheel/axle would not be spinning wildly. What is missing is traction, not torque.

I too have to disagree then.:) Torque is the least force required to turn the wheel. If traction is high, more torque is required to turn the wheel. If there is no traction (wheel in air), very little torque is required to turn the wheel. One can never deliver more torque than required, that is once the wheel starts slipping.

Quote:

Originally Posted by 1100D (Post 3324801)
It does not distribute torque equally, say one wheel is slipping, there will be zero torque at that wheel, whereas the wheel with grip will be the one that will be transmitted the power.

If ^^ is right: Then assume one of the rear axles is not connected to one rear wheel and is spinning freely, if so, with the Eaton locker locked, and with only one axle in function, the vehicle should accelerate just like it would do with two wheels powered?

I guess NO. The performance is halved. ie, the turning capability coming from the prop shafts is divided equally and given to two axles.
Quote:

Originally Posted by Samurai (Post 3324760)
Torque is a measure of how much a force acting on an object causes that object to rotate.

Quote:

Originally Posted by Samurai (Post 3324851)
Torque is the least force required to turn the wheel. If traction is high, more torque is required to turn the wheel. If there is no traction (wheel in air), very little torque is required to turn the wheel. One can never deliver more torque than required, that is once the wheel starts slipping.

Well, think of torque as turning capability. In this case, the turning capability the prop shafts is multiplied by the final drive gears and divided by two (with a locked diff). One half to one wheel, and other half to other wheel.
Quote:

This is exactly what an open differential does.:D
The purpose of a differential (open) is just to let the two wheels spin independently.

I guess what you have in mind is based upon the final work done?

Very simple. What is torque, its turning force, measured by force exerted at a point multiplied by the distance to the point from where the turning is generated. If a wheel is spinning freely, its not exerting any force at the point of contact (periphery), hence the product of that "into" the "rolling radius" of the wheel is thus "Zero" (almost zero practically, as there is some work done in turning the rotational components and the tyre contact point slipping against the loose soil).

Anyway, I have also tested the m-Locker with one rear wheel lifted on a rolling hydraulic jack.

Quote:

Originally Posted by dhanushs (Post 3324878)
If ^^ is right: Then assume one of the rear axles is not connected to one rear wheel and is spinning freely, if so, with the Eaton locker locked, and with only one axle in function, the vehicle should accelerate just like it would do with two wheels powered?

I guess NO. The performance is halved. ie, the turning capability coming from the prop shafts is divided equally and given to two axles.

And what happens to the force that is applied to the free axle? Remember, energy cannot be destroyed. See, this goes against Physics. Consider this experiment. Hold a pen loosely in your left hand palm. Now rotate the pen using your right hand at one rotation per second (60rpm). Obviously you have to apply very little torque. Call this torque X. Now try applying 10 times more torque, say 10X while keeping to same rpm. Can you? Once the pen is rotating freely, you can't apply any more force, except to make it go faster.

In other words, the free spinning wheel needs very little torque compared to the wheel with traction. So MLD can't send the same torque to spinning wheel as well as wheel with traction, when both are turning at same speed. You are really sending most torque to the axle with traction.

Quote:

Originally Posted by dhanushs (Post 3324878)
Well, think of torque as turning capability. In this case, the turning capability the prop shafts is multiplied by the final drive gears and divided by two (with a locked diff). One half to one wheel, and other half to other wheel.

This is true only in case of open differential. In case of LSD, more torque is delivered to the wheel with traction. In case of MLD, all torque must go to the wheel with traction.

Ok, another experiment:

Hold a 6ft long metal pipe at the center, horizontally to the ground. Have a strong guy hold one end tightly. And there is nobody at the other end. Now try rotating the pipe at the center. Now, whom do you think you are fighting against? Just the guy at one end? Are you also fighting the non-existing guy at the other end? if the torque is distributed equally, only half of your energy is spent on the strong guy at one end. The other half is spent on the non-existing guy at the other end. Is this possible? I think not.

Quote:

Originally Posted by dhanushs (Post 3324878)
The purpose of a differential (open) is just to let the two wheels spin independently.

Not exactly. The purpose of the open differential is let the two wheels turns at different speeds. It is not independent. The average speed of the two wheels have to be equal to prop shaft speed X axle ratio.

Example:

Prop shaft speed = 500rpm
axle ratio = 5.38
Average speed of two wheels = 500/5.38 = 92rpm

Quote:

Originally Posted by dhanushs (Post 3324878)
I guess what you have in mind is based upon the final work done?

I am just verifying the physics. Whatever mechanical components are used inside the Eaton MLD, it has to satisfy laws of physics.

Help us, Behram! I think we are stuck in the mud!

With locker the diff is out of the equation and rear axles become one piece. So no torque "distribution" only torque "transfer" from drive shaft to "one" axle takes place.

@DirtyDan
MLD will help you of of the mud. LOL

Quote:

Originally Posted by sudev (Post 3325064)
With locker the diff is out of the equation and rear axles become one piece. So no torque "distribution" only torque "transfer" from drive shaft to "one" axle takes place.

@DirtyDan
MLD will help you of of the mud. LOL

Yeah, I understand that. You are merely repeating what Samurai said. I don't see any logical difference between saying that a locked axle distributes torque equally to both wheels or saying that power is "transferred". I confess, I do not know what the engineers and physicists lingo is these days. From what little I can glean from google, even the definition of "torque" is not set in concrete between physics and engineering applications and even varies between countries. So I am not going to toss and turn (and torque out) over this.

Hi Guys,

IIRC.

In an Open Differential Tranfers torque 25%: 25%(Geared Torque at Transfercase) in deal conditions, with optimum co-efficient of friction.

If the co-efficient of Friction Reduces, the traction will also reduce, as traction is a result of (Vehicle Weight X Contact Patch of Tyre X Co-Efficient of Friction)

In a Locked Differential the Torque is split 25%:25% irrespective of the co-efficient of friction, even if one wheel is in the air or in deep mud.

Regards,

Arka

Quote:

Originally Posted by ex670c (Post 3327068)
In a Locked Differential the Torque is split 25%:25% irrespective of the co-efficient of friction, even if one wheel is in the air or in deep mud.

I can't agree with this statement. If the diff is locked and one wheel is in air, all the traction necessary for torque generation is coming from the other wheel. So we have to admit that the entire torque from the propeller shaft is going to the wheel with traction.

Check the last paragraph under diff locks in here: http://www.4wdonline.com/A/Diff.locks.html

I don't know mechanical stuff like you, I am entirely depending on Physics here.

Quote:

Originally Posted by Samurai (Post 3327266)
I can't agree with this statement. If the diff is locked and one wheel is in air, all the traction necessary for torque generation is coming from the other wheel. So we have to admit that the entire torque from the propeller shaft is going to the wheel with traction.

Check the last paragraph under diff locks in here: http://www.4wdonline.com/A/Diff.locks.html

I don't know mechanical stuff like you, I am entirely depending on Physics here.

You are coupling "traction" and "torque" in a way that Dhanush has pointed out to you, is not the way a lot of other people think about it. I don't think a pure physicist would do that, usually,....but I think this discussion is more of a "semantic quibble" than an empirical or logical disagreement on points of fact. The torque ultimately comes from the motor turning the driveshaft. Traction may be there or may not be there. But the torque is still there regardless of traction because the cause of torque in this application is the motor. That is what Dhanush, Arka and I mean by torque.....I think.....you are all giving me a headache, a curse on you all. We could be in the old dilemma, "If a tree falls in a forest and there is nobody there to hear it, is there still noise." kind of argument...and its variant, "If a man utters something in a forest and his wife is NOT there to hear it, is he still dead wrong."

It sort of depends on how you define "noise"....it sort of depends on how you define "torque".

Dan, this is not a quibble. This thread is under technical section, so we can get technical here.

Torque is a physics concept, it can't be re-interpreted in many ways like art. There is no way to have different opinions about it.

Torque is the least force required to turn something on an axis.

Engine has the potential to generate X amount of torque. But only when required. Consider an engine that can generate 240Nm at 2000rpm. If the gear is in neutral, then the engine will only generate enough torque to spin the flywheel at 2000rpm. So you will never see it generate 240Nm of torque at neutral. It won't generate 240Nm while driving downhill or straight road either. That may only happen when the vehicle is climbing uphill with 10 people on-board.

Consider a human example. You may have the strength to lift a 50kg bag of sand. But if you are asked to lift a 100gram bag of sand, you can't apply the same force as required by the 50Kg bag.

What we really need here is a physics expert to put an end to the speculation about torque. Where is our resident Physics professor Sutripta?


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