[mallumowgli]

There has been lots of 'assumptions' vague reasoning and even sometimes wrong reasoning going on, so I re-read the whole thread again (thanks to moderators for spinning this on off!)

First things first -

(1) Engine does not produce torque - it produces force, downward force. And the torque is generated at the crank shaft, because the piston rod is connected not to the center of the crank shaft, but eccentrically. This torque rotates the crank shaft, and results in what we call 'rpm'. So rpm doesn't determine torque, but rather it is the other way round

(2) This torque and engine rpm combined results in power. Torque and power do not have a linear relationship, because there other physical factors involved like inertia, friction, heat losses, combustion, etc etc

Lucifer said at a particular rpm, torque is always the same. Theoretically yes, because a particular amount of torque has to produce the same net force - but we are not taking into account inertia of the vehicle and other parameters

When 1100D contradicted this statement - he is talking about downward force produced in the engine rather than torque. Heat energy produced in the cylinder getting converted to Kinetic energy

But in the above discussion we can take this assumption that at a particular rpm we have a fixed torque - because the engine torque and rpm is hardly a matter of contention here

It is the resultant torque and rpm - after the gears, for heaven's sake!!

Originally Posted by Samurai

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.

In the above statement - 'causes the object to rotate' is misleading. Torque will be there whether the object rotates or not

Originally Posted by 1100D

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.

Again wrong - with a differential splitting torque 50:50, assume 100Nm torque

- left (assume this to be slipping) gets 50Nm and spins the wheel at a particular rpm remember open differential - hence based on the torque, it rotates at a particular rpm (similar to what the torque available at the crank shaft makes it rotate)

- assume right wheel is stuck - hence the shaft just twists to the extent of the torque (50Nm). Shafts are designed to take this maximum torque plus a factor of safety.

Energy available at the left wheel rotates the wheel, whereas part of the energy at the right wheel is lost as heat, part is stored as a potential energy in the twisted shaft etc etc

- Now the consider the condition where the right wheel is not stuck, but has relatively more traction compared to left. This is the condition we have been discussing. Now the net torque available is just 50Nm, since the left wheel is using its 50Nm to just spin the wheels merrily. This 50Nm is not enough to move the vehicle forward - hence the vehicle stays put

-The third condition is when the differential is locked. Here the entire 100Nm (at a particular ratio - 50:50) is available to both the wheels and they cannot spin at their will. The wheel which is slipping cannot slip, hence some of the torque (twisting moment) is transmitted back to the other wheel and vehicle moves forward

Originally Posted by Samurai

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.

Four points in the above quote

1. Torque is the least force required - Wrong. Torque is the force available at the propeller shaft. Whether the wheel likes it or not - it has to be content with it

2. Yes, more traction means more torque is required. Question is whether it is available?

3. Wheel in the air means less torque is required. But more available torque means just more rpm at that wheel

4. Yes more torque means wheel starts slipping, but you can't do anything about it. Wheels will slip with the entire available torque

Originally Posted by 1100D

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).

You are right on one side - Wheel is slipping means no force applied on the ground. Means there is no friction, and hence wheel cannot apply any force to enable to move forward. But that does not mean that the shaft is not applying torque to the wheel

Originally Posted by ex670c

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

Correct, sir. But generally it is 50:50

Originally Posted by Samurai

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

The explanation given in that url is slightly wrong (not wrong - but not clear)

It says when the wheel is slipping on ice, the torque applied is zero : Wrong. It is actually, the torque required that is zero

Read the next line - it says Subaru uses a viscous coupling which heats up and transfers this unused torque to the front wheels.

So you need some extra arrangement which will transfer the unused torque to the other side

Originally Posted by Samurai

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.

The torque figure is for the engine at 2000rpm - whether under load or no load. What is transmitted is different. Under full load or up an incline, you use a gear sacrifice some of the power (reduce rpm) and increase the torque to take the incline (I think lucifer has explained, but there is a correction there too. I'll come to that)

Originally Posted by Samurai

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.

Yes, force. So engine needs to do more work to generate more force

Originally Posted by Samurai

In case of a fully locked differential, 100% torque goes to the wheel with traction:
Thus, Physics is satisfied. I rest my case.

Yes you are right, and for the reasons I've stated above

Originally Posted by Samurai

This statement defies mathematics too, let alone physics.

T = f(F) [i.e. torque is a function of force]

And then you say force varies, but torque remains constant.

Shall we wait for some real professionals answer this question?

It is generating just enough torque to free spin the wheels, no more. I was made to understand this stuff when I was learning to drive on icy/snowy roads of NJ, about 20 years back. Don't think the science has changed.

Can I request we wait for the professionals who actually understand the underlying theory to answer this question? Behram, Sutripta, Spike_Arrester, star_aqua are some of those who can speak with authority on this.

You seem to be stuck in the 'just enough' torque case. A wheel cannot decide the amount of torque it has at hand. It has to take whatever is given to it. More torque more spin. Simple

In muddy slippery you take your foot off the throttle, to gain enough traction so that the available torque can be put to good use

Originally Posted by lucifer1881

In other words, the torque to open a door is constant. The force varies depending on the distance from the hinges.

The torque required varies with the distance from the hinges at which you are applying the force. If the torque required is more, then your applied torque should also be more. Hence you require more force

The torque to open a door is constant only at a particular distance

For a wheel spinning at X rpm, the torque required is constant whether the wheel is free-spinning or on tarmac or stuck in the mud. The force required to generate that torque changes.

The difference is torque required and torque generated. But we cannot say whether the torque required to move the vehicle is constant. Varies with conditions. The torque available at the wheels at a particular engine rpm at a particular gear remains roughly constant

Similarly, if an engine is rated for 240Nm at 2000 rpm it will generate that torque irrespective of whether the transmission is engaged or in neutral. The engine, in case of transmission being in neutral, would be running on almost no load since the force required to generate that torque is very low. If the transmission is engaged and the car is moving forward then the engine will be under load conditions depending on weight of the car, traction of the wheels, etc. This is one of the reasons why stop-start traffic gives you lower mileage because the engine has to work on higher loads to get a car moving from standstill than to keep a moving car in motion. This is because coefficient of rolling friction is less than coefficient of static friction.

Here again, saying 'force required to generate torque' is wrong. Torque will be generated based on the force, not the other way around. We give less throttle when less torque is required. But the torque generated at the crank doesn't depend majorly on the load - though inertia has a major role to play in practical conditions

Originally Posted by 1100D

The torque generated by the engine at any RPM is also a function of the quantity of fuel being fed. That quantity of the fuel is based on calculations (from dataset) as well as the throttle position.

Torque generated by an idling engine is much less than the torque generated by the same engine at the same rpm, engaged in gear and rolling, so is the fuel feed.

The much lower torque generated by the engine, when the driven wheels are spinning (or in air) goes to counter the rolling friction of the powertrain components (GB, Propshaft bearings, diff, axles etc), this torque is anyway there when the wheels have traction, and the tractive effort is added to this (rotational frictional component).

As an experiment, please raise both driven wheels of any vehicle on jacks and try to rotate them using engine power and see for yourself, how much little throttle inputs are required to accelerate the spinning wheels. Compare with the acceleration felt, when the vehicle is on road,
the answer will be very clear.

True to an extent - more fuel is fed to produce more torque in case of increase in load. But the load does not directly decide the torque - rather it is the engine and the energy produced. Once the energy is produced, resultant torque is always achieved

Originally Posted by Samurai

Let me summarize now. You agree that force depends on the load on the engine, and that load depends on traction. I will shake your hand on that.

Force depends on Load - We agree
Load depends on traction - We agree

Now look back at your torque formula: T = F x L x Sin(A) [where L and A are constant for a vehicle]

That means Torque depends on Force, directly. If you agree with this, we are then in full agreement. That means torque depends on force, which depends on load, which depends on traction.

Though simplistic - this is a reasonable conclusion, but is not directly related to torque distribution

Originally Posted by Sutripta

Do let us have the details of the dynamometer setup, and the MLD. The dyno owner might need some convincing. See that the setup is safe, esp if on an inertial roller.

Does the dynamometer gives the torque output based on rpm? how does the calculation work?

Originally Posted by star_aqua

Which mean the entire torque or the twisting force from the propeller shaft multiplied by diff ratio would be shifted to one axle which offer full traction.
The one which lost the traction would not share the torque from the other axle, hence would not bear any stress. Am I right?

Wrong in the first statement. Torque will not be shifted unless there is some external arrangement is made to do so

The stress on the slipping wheel is less only because there is no friction

Without load, the torque cannot be measured

No idea about the dynamometers, but torque in a free rotating shaft can be measured

Which also mean the torque will not be produced without having an opposite force or the load. Torque definition remains same as we all studied in physics. When it comes to producing the torque, there should be always an opposing force. For an example of tightening a nut or bolt, initially need less torque and as it gets tighter, would need higher torque. What does a torque wrench do?

Physics has never stated that torque is generated only when there is an opposing force. It is just a measure of a twisting force. But Newton's law says that every action has an equal and opposite reaction - think only in terms of momentum here!!

Originally Posted by lucifer1881

Example: You are driving at 100kmph in fifth gear at 2500rpm. You approach a steep incline. Without down-shifting, you continue up the incline and floor the accelerator. You will notice that if the incline is steep enough then the engine rpm reduces and so does your speed. This is because the load on the engine increases. The engine first needs to overcome the additional load due to gravity and then impart the rest of its power to the driven wheels. Consequently, the torque being generated at the flywheel and, by extension, the driven wheels is lower. However, if you down-shift and manage to attain 2500rpm you will achieve 100kmph again. The torque is the same. The engine is simply having to work harder to generate that torque.

An addition : You reduce gear, torque available increases. But more load to the engine subsequently

To repeat, torque is not force. Torque is a function of force. Torque on the wheels is determined by how much usable power the engine is able to transmit to the wheels. This usable power is determined by the load on the engine.
Phew!

Correct absolutely - but again does nothing the the torque distribution argument

Originally Posted by Sutripta

If the genset engine is connected to the alternator by means of a shaft which is instrumented to measure the torque it is transmitting, what torque values will I see. If I've understood your argument it means that as long as the engine rpm is constant, so is the torque. So the torque reading I get from my instruments should essentially remain the same irrespective of the load on the alternator as long as the engine rpm remains the same?

Regards
Sutripta

Originally Posted by Samurai

He means any automobile engine. Since Thar CRDe engine appears very close to that number, let's consider that.

Thar CRDe engine specification says 247 Nm @1800-2000 RPM. Does it produce 247Nm at 1800-2000 rpm, irrespective of load?

As I said earlier - rpm is the result of torque. But for easy understanding, and easy measurement manufacturers give the value based on rpm. Torque of 247Nm is the maximum torque the engine can produce and at that value the rpm will be around this value, always!! But when the vehicle is in motion, the inertia of the vehicle comes into play and the rpm becomes a factor of this too. When you suddenly reduce the gear at higher speed, then the rpm shoots up because in this case the wheels begin to drive the gears instead of the other way round, but engine wouldn't produce that much energy or torque - or you can say the engine struggles and there are lots of power losses and damages!!

[ex670c]

Originally Posted by star_aqua

In an extreme off road situation, assume rocks or sand, one of the axles(left or right) would be more likely to break in 4low when lockers are applied. Why? What could be the reason. And the axle which breaks would be the one driving the wheel which offer more traction.

Originally Posted by Sutripta

In an open diff, the torque on each halfshaft is equal, and thus limited to the lower value of the traction provided by the two wheels. This acts as a safety valve.

With a locked diff, one can land up with one wheel (and thus the corresponding halfshaft) doing all the work. So all the stress is on that.

Originally Posted by star_aqua

Which mean the entire torque or the twisting force from the propeller shaft multiplied by diff ratio would be shifted to one axle which offer full traction.
The one which lost the traction would not share the torque from the other axle, hence would not bear any stress. Am I right?

Hi Guys,

In 4x4 RWD, in 2WD, 100% of the torque is sent to the Rear (OPEN) Differential, Which multiplies and distributes, it 50:50 to each Wheel (depending on traction).

What happens when 4WD is engaged?
50% Torque is sent to the Front Differential and 50% is Sent to the Rear Differential.

The Open Differentials, multiply and split it another 50:50 LHS:RHS, depending on Traction, so it is 25% per wheel.

What Happens when 4WD is Engaged and Rear Differential is Locked?
1) Both the Rear Wheels will spin at the same speed.
2) Torque will be distributed equally to both wheels 50:50 or 25:25 (overall perspective of 4 Wheels)

In a locked differential both the wheels will spin at the same speed irrespective of surface conditions, then why will there be a torque transfer, to the wheel with maximum traction?

When a vehicle with a locked differential, is on one wheel, the chances of breaking the axle maybe due of the vehicles weight resisting the torque on the Axle Shaft.

Where as in an open differential vehicle the wheel with least traction, will spin depriving the wheel with traction of the torque needed to move.

THINK of the DIFF-LOCK as a 4x4 High Between the Left and Right Wheels

Why is is recommended to gently deliver power on icy roads?
IIRC, excessive torque will cause the wheel to spin, due to the low co-efficient of friction on icy surface.

Why is it recommended to gently deliver power on sandy terrain?
IIRC, the excessive torque will, cause the wheel to dig into the soft terrain, but the sandy surface provides a high co-efficient of friction.

If 4WD is engaged, and the front wheels are lifted off the ground will the rear wheels ALONE move the vehicle under its own power?

Regards,

Arka

[Jeroen]

Sorry, but the notion that an engine produces the same torque at a given rpm is simply not true. The figures you find in those tables/specification are the maximum ones, under normal operating/driving conditions..

Torque is the result of a unit of force (newton) per unit of distance or arm (meter). The arm is determined by your engine crankshaft and is fixed. The force is essentially determined by the amount of fuel injected into the cilinder and the subsequent "explosion", which creates pressure on top of the. Less fuel is less pressure is less torque.

If the gear is in neutral and you rev to whatever the RPM for max torque is, the engine wont be developing that . It will only develop whatever torgues are necessasy to overcome the internal friction in the engine. Same with the example of wheels slipping. The engine needs to be loaded up to be able to produce that particular max torque. Consider this. You drive down a shallow hill, engine is at the max torque RPM. Now you get to the bottom, turn and drive up that hill, again at the max torque RPM. So up and down a hill at the same RPM/speed. I think most of us would agree you would be using less fuel going downhill than uphill. And less fuel equates to less pressure in the cilinder, ie less torque.

Lets look at the hill scenario a bit more in a situation we are all familiar with. You drive on a straight and level road at a steady speed/rpm. Obviously, under those circumstance the engine produces a steady output/torque. Now you get to the bottom of the hill. If you keep your right foot in the exact same position the car will start decelerating. So you put your foot down to maintain the same speed. That means more fuel gets injected, more combustion and thus higher torque at the exact same speed/RPM.

Earlier in this thread reference was made to a generator set. Obviously this produces a wide range of torque all at the exact same RPM. It was dismissed as being different from car engines, for this particular topic, but this not correct. For both the genset engine and the car engine its the exact same thing. The torque depends on the load. No load no torque. For a genset load is how much electrical power gets drawn. On a car the load is depending on several factors, air resistance, rolling resistance, acceleration to name a few. But to revisit the slipping wheels, if they slip there are no torgues. Anybody that owns a torque wrench will be able to confirm this.

Enjoy,
Jeroen

[SS-Traveller]

Originally Posted by lucifer1881

Example: You are driving at 100kmph in fifth gear at 2500rpm. You approach a steep incline.
...if you down-shift and manage to attain 2500rpm you will achieve 100kmph again.

A car doing 100 km/h in 5th gear at 2500 rpm will again achieve 100 km/h at 2500 rpm after downshifting, i.e. in 4th or 3rd gear?

You can't be serious!

[Jeroen]

Originally Posted by SS-Traveller

A car doing 100 km/h in 5th gear at 2500 rpm will again achieve 100 km/h at 2500 rpm after downshifting, i.e. in 4th or 3rd gear?

You can't be serious!

It must be one of those new constant rpm and speed gear boxes that gives you the same speed at the same RPM and speed in any gear under all circumstances!
Jeroen

[mallumowgli]

Originally Posted by Jeroen

Sorry, but the notion that an engine produces the same torque at a given rpm is simply not true. The figures you find in those tables/specification are the maximum ones, under normal operating/driving conditions..

But to revisit the slipping wheels, if they slip there are no torgues. Anybody that owns a torque wrench will be able to confirm this.

Enjoy,
Jeroen

Yes you are right. Apologies for mixing it up earlier - Peak torque shall be at load - torque tapers away as the engine overcomes all resistance, including the load and starts to run smooth

But again would differ on torque availability -

Power = rpm X torque (approx)

So when there is no torque means no power? When the wheel slips the available torque gets converted to power and there is no torque experienced/measured. Means input torque from the propeller shaft is still there. If not the wheel will not spin.

[Sutripta]

Originally Posted by 1100D

Wrong, absolutely, wrong.

Originally Posted by Jeroen

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

Just us stating this, no matter how strongly we word it, is not going to convince the naysayers. We have to show them that it leads to all sorts of weird extrapolations, logical inconsistencies etc. Wish we could hold a demo in front of any large diesel with individual bottle type injectors, fuel rack couplings exposed and visible, connected to a Woodward PSG/ PGA etc.

Let this (engine rpm and torque) be resolved (to other peoples satisfaction) before we proceed to more problematic topics. Like torque distribution and differentials.

@Arka: Don't agree with you on a number of basic points. But can we leave that for later. As I said before, one issue at a time.

Regards
Sutripta

[Samurai]

Originally Posted by mallumowgli

There has been lots of 'assumptions' vague reasoning and even sometimes wrong reasoning going on

Yes, I very much agree on that point.

Originally Posted by mallumowgli

so I re-read the whole thread again (thanks to moderators for spinning this on off!)

You are welcome.

Going forward I am only quoting the parts we disagree on:

Originally Posted by mallumowgli

(1) Engine does not produce torque - it produces force, downward force. And the torque is generated at the crank shaft, because the piston rod is connected not to the center of the crank shaft, but eccentrically.

Last time I checked crankshaft was part of the engine. Will you tell a sprinter that it is his legs doing the running and not him?

Originally Posted by mallumowgli

Lucifer said at a particular rpm, torque is always the same. Theoretically yes, because a particular amount of torque has to produce the same net force - but we are not taking into account inertia of the vehicle and other parameters

You as well as lucifer are talking about torque based on resultant force. I don't believe engine torque rating is based on resultant force. This is a major bone of contention of this thread.

Originally Posted by mallumowgli

But in the above discussion we can take this assumption that at a particular rpm we have a fixed torque - because the engine torque and rpm is hardly a matter of contention here

It very much is... We cannot assume torque is constant at a given rpm. You are again assuming resultant force. How can an engine be rated based on resultant force when you don't know the opposing force. Say your engine can generate X amount of force. If you know that the opposing force is Y, you can say resultant force Z = X - Y. But you are saying engine is rated for Z. How do you get Z when you don't have Y?

Originally Posted by mallumowgli

It is the resultant torque and rpm - after the gears, for heaven's sake!!

Yes, I know you mean resultant torque. But how can you know resultant torque when the load (opposing force) is unknown? If you say resultant torque of Thar engine is 247Nm, then what is the absolute torque?

Originally Posted by mallumowgli

In the above statement - 'causes the object to rotate' is misleading.

Please send your complaint to Physics department of University of Guelph (Canada). I took that definition from their web: http://www.physics.uoguelph.ca/tutor...que.intro.html

Originally Posted by mallumowgli

Torque will be there whether the object rotates or not

Yes, but can you measure it? I can imagine measuring torque only when it moves something. I guess that is why they say ‘causes the object to rotate’.

Originally Posted by mallumowgli

Again wrong - with a differential splitting torque 50:50, assume 100Nm torque

1100D was refering to MLD, not open differential in that statement, Remember where this discussion started.

Originally Posted by mallumowgli

1. Torque is the least force required - Wrong. Torque is the force available at the propeller shaft. Whether the wheel likes it or not - it has to be content with it

Let’s take a human example again since the mechanical ones are really confusing you. Take a old fashioned manual borewell pump.



You have to apply force at the end of the handle to force the water up. If you press it lightly, the lever doesn’t even move. We can’t measure this torque since no work was done. Only when you apply enough force to push the lever down, it makes the pump to draw water. This is what I mean by least force, the minimum force that was required to get the work done. As long as you supply that minimum force, the pump will work and you get to drink water.

Originally Posted by mallumowgli

The torque figure is for the engine at 2000rpm - whether under load or no load. What is transmitted is different. Under full load or up an incline, you use a gear sacrifice some of the power (reduce rpm) and increase the torque to take the incline (I think lucifer has explained, but there is a correction there too. I'll come to that)

On this we will continue to completely disagree.

Originally Posted by mallumowgli

You seem to be stuck in the 'just enough' torque case.

Already explained this one. You can apply more force to get things faster.

[DKG]

I use a torque wrench regularly when working on cars and to me it's quite obvious that unless there is a load the fastener is exerting the wrench won't apply the full torque it is set to. What this means to me is that torque is a property not independent of the load it is applied to. Therefore in a locked differential where one wheel is in the air and the other on ground a certain distribution must take place where just about enough torque is applied to the weight of the wheel in air and it's resultant load and more torque is applied to the wheel on the ground which is exerting the load of it's weight plus the frictional force it encounters with the ground.

In summary I am inclined to agree that more torque is applied to the wheel on ground vis a vis in the air, just that I wouldn't say it's hundred percent as the weight of the wheel in air still exerts some load and utilises some torque.

[Sutripta]

In the spirit of one concept at a time, lets concentrate on 'in an engine, rpm determines torque'. While we wait for Lucifers viewpoint on say what happens to the now missing power in the genset case, my comments on some other posts.

Originally Posted by mallumowgli

First things first -

(1) Engine does not produce torque - it produces force, downward force. And the torque is generated at the crank shaft, because the piston rod is connected not to the center of the crank shaft, but eccentrically. This torque rotates the crank shaft, and results in what we call 'rpm'.

Treat the engine as a blackbox with a shaft coming out of it. Internal details not really important for this discussion. You can also have a Wankel!

So rpm doesn't determine torque, but rather it is the other way round

Neither.

(2) This torque and engine rpm combined results in power. Torque and power do not have a linear relationship, because there other physical factors involved like inertia, friction, heat losses, combustion, etc etc

Torque, rpm and power are pretty well tied together. If we know any two, the other is known. Exactly.
If there is scope for confusion, define the place where this measurement is taking place (like shaft HP) and the conditions under which it is generated and measured (SAE, DIN, Gross, nett etc)

Lucifer said at a particular rpm, torque is always the same. Theoretically yes,

No, but we'll wait for Lucifer to explain his reasoning.

But in the above discussion we can take this assumption that at a particular rpm we have a fixed torque -

Assumption is wrong.

because the engine torque and rpm is hardly a matter of contention here

It is very important in the other part of the discussions:- Torque to wheels. But we'll cross (or burn) that bridge when we come to it.

Regards
Sutripta

[amit_purohit20]

Originally Posted by ex670c

Hi Guys,

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

Originally Posted by Samurai

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

Originally Posted by DirtyDan

When auto manufacturers write the specs for their vehicles, they list torque as a function of the ability of the motor to turn a shaft. They do not write torque specs as the ability of a wheel to move...if it has x,y,z modes of traction and power. However, this does not mean that the concept of torque as defined such is useless or wrong.

Originally Posted by Samurai

Can the engine generate maximum rated torque at no load? Can you send loads of torque to a shaft with no load?

I agree with Samurai. Only one thing I do not agree is the definition of the torque being the least amount of Force X Distance.

Samurai can be a good teacher indeed, the way he explains with simple analogies!

Lucifier has got it wrong by equating rpms with fixed torque. Also adding too many variables in his explanation is more confusing and doesnot help bring clarity to the situation.

Well Facts first:
1) Engine produces torque depending on the demand (resistance) from the wheels. If there is no resistance from the wheels the torque generated would be very minimal.
2) Torque is not dependent on rpms.
3) An open differential distributes torque equally to the two wheels when both feels face the same resistance from the ground. ( In real world this never happens even if going over a plain road, there is always a small difference in resistance from the both wheels of the axle.)
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.
5) The differential distributes the torque based on the amount of resistance the wheels have to offer. (Because torque is generated because of resistance and resistance only!) So the wheel having higher amount of resistance will get more torque transferred to it upto the max limit of what the engine can generate under maximum load (Load again here is created because of resistance at the wheels.)

Let me start with a different but simple analogy to understand what torque is-
Consider a water pump (any type be it a centrifugal/reciprocating/gear pump). The outlet of a pump is a horizontal pipe at the same level of the pump on which a tap (valve) is mounted.

A pump never creates pressure. Period! It only displaces fluid. The pressure is created only when it finds any resistance to the flow!

Pressure = Force / Area.

If the tap is open and the pump is switched on it only displaces the fluid and the fluid comes out of the pipe through the tap. Leaving aside the losses there is no pressure created. Now if we start closing the tap there is a resistance created. The pump has to work against this resistance because it needs to displace the fluid volume against this added resistance. This is where the pressure is created. So without resistance no pressure is created.

(Note: Dont go in smaller details in the above example like fluid resistance and other resistances. Use the KISS (Keep it Simple) principle!)
So if you have a pressure guage mounted on the pipe before the
tap, when the tap is fully open the guage will show minimal pressure (due to some unavoidable losses). As you start closing the tap, you would start finding the pressure guage showing higher and higher readings. The pressure in the system would go on increasing to the point where the load on the pump increases to such a point that the pump stops rotating. This is the max pressure the pump can generate.

Similarly an engine will produce more and more torque as the resistance to the rotation of the wheels goes on increasing. At one point of time the engine will nearly stop rotating (or die). This is the peak or max. torque it can produce. As stated in earlier examples this is 247 N-m. Off-course needless to say that this max. torque is generated at only one rpm of the engine that is called as the Max. Torque rpm.

Consider another simple analogy to understand torque.

Consider a bolt with a nut on it which is welded onto it. Now if you use a spanner of length 1 metre. And the max force which I can apply at the end of the spanner is say 10 Newtons then the max torque which can be recorded using a guage is -

Torque = Force X Radius (Distance of the Force applied from the centre of rotation)
T= 10 N X 1 m = 10 Nm.

Point to be noted here is that the nut was welded and so I could apply the max. force I am capable of ie. 10 N. Welded Nut means infinite resistance ie nearly fixed to the bolt itself. So the Max. Torque -10 Nm here is limited by the amount of Max. Force I can apply at the spanner. (Spanner length remaining unchanged).
But the same torque of 10 Nm can also be generated by applying only 5N Force with a 2metre length spanner!

T= 5N X 2m = 10 Nm.

If the Nut was not welded and there was less resistance to rotation of the nut I could get less torque generated!

Lucifier's example of fixed torque required to open a door is constant is right!

So if you try to open a door near the hinges you require more force (as shown in 1st example -10N) and as you move away from the hinges you require less force (as shown in the 2nd example -5N with increased radius of 2m). I really dont know why lucifier used the Sin A in the equation. It unnecessarily complicates the things. In the definition of Torque its always the normal force applied to the spanner end and not an inclined force which we talk about so Sin A is always 1. Let us drop this Sin here itself.

In our further course of discussion we will not bring the Radius into picture because it is constant for a particular make of vehicle. So Torque generated is directly proportional to Force applied.

In an engine during combustion the charge temperature rises causing the gas to expand. This expansion of gases is restricted by the resistance of the piston to move downwards (Bottom Dead Centre). This resistance causes the pressure to rise in the combustion chamber on the top side of the piston. Pressure = Force x Area (Of Piston)

We get the Force on the piston in the downward direction from the above equation. This force is transmitted through the connecting rod to the Crank Pin.

So the force on the Crank pin (Not equal to the force on the piston due to geometry changes) multiplied by the crank radius (offset) gives the torque.

Torque= Force x Radius. (Nowhere in this equation rpms are mentioned!)

Work Done = Force x Displacement. (If there is no displacement work done is zero.)

Power means how fast the work can be done. Here time factor comes in.

Power= Work Done/ Time.

Also Power (Engine) = 2 * Pi * N * T /60

where T is Torque developed by engine and N means rpms of engine.

2*Pi/60 being a constant... Power = Constant * N * T

So Power is directly proportional to either Torque or Engine rpm.

A engine can have higher power by above two methods ie. higher engine rpms like our passenger cars or higher torque like our tractor engines.

So power remaining constant we can either get high torque or high rpms.

So rpms come into picture in the Power equation and Not Torque equation.

Torque is nothing but a twisting force. The max twisting force which can be applied is called Max. Torque. Its the maximum capability of an engine to generate torque which can be reached only when there is maximum resistance. Torque cannot be generated without resistance similar to pressure.

By resistance here I mean resistance to the rotation of the wheels.

Dont confuse Torque and Traction.

Traction has nothing to do with the torque, but torque transmitted has a lot to do with the traction! Because traction creates the resistance to the rotation of the wheels.

T (Torque on Axle) = Force (applied on the road by the wheel) x Radius of the wheel.

Force applied by the wheel onto the road (Fw).
Force applied by the road onto the wheel (Fr) - Based on the Newtons third law.

For the wheel not to slip at least Fr= Fw.
If Fw> Fr The wheel slips

As Lucifier said Fw- Fr = Resultant Force. There is nothing called as resultant force during threshold conditions of slip/traction because Fw will not be considerably more than Fr.

So max. torque that can be generated (Tg) depends on the engine but the max. torque that can be transmitted (Tw) via. the wheels is limited by the traction available. Tw is also called as the Wheel Slip Torque.

Max. Torque that can be transmitted through the axle to the wheels = Tw= Fr x Radius of wheel.

Torque generated can never be more than the max. torque transmitted (Forgetting other losses). As Tg becomes greater than Tw wheel slips starts and it limits the Tg.
So as soon as the Torque generated by the engine exceeds the max. level of torque which can be transmitted to the wheels (traction) it stops generating any extra torque.

So if traction is limited your vehicle engine might never reach its true potential of generating max. engine torque! It can be only reached in a Dynometer. What we do in a dynometer is nothing but we increase resistance on the wheels (or reistance on the flywheel of the engine) till the time engine dies. The Max Torque generated before the engine dies is called the Max. Torque generated.

Coming back to Traction.

Consider a Block of Weight W (W) on ground. It is being pushed by applied force Fa, then by Basic Frictional Formula ...

Fa= Mu (Coefficient of friction) x W.
Mu is always less than one for road tyre at max it is considered 0.85
So Force required to move the block (Fa) is always less than the weight of the block.

This Force Fa here is the same force as Fr in our above equation of calculating wheel slip torque.

So Fa ie. Fr X Radius of wheel = Tw (Wheel slip torque at axle.)

As soon as the Torque demand from the wheel increases the engine rpms start dropping. We try to nullify this drop by pressing the throttle. With more fuel to feed, the gas pressures rise again in the engine and the engine starts generating more and more torque. It does so till Torque generated = Torque demanded from the wheels. Once this is reached the rpms stabilise and stop increasing. If you further press the throttle the Power generated increases thus increasing the engine rpms also.

But please note in the above statement that we have here not crossed the threshold boundaries of traction.

Once you cross the threshold boundaries of traction the engine can no more generate more torque (because resistance is not increasing) So as the torque not increasing and the power being increased (due to increased throttle) the rpms start increasing!

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.


If you have a LSD with one wheel in air the max. torque transferred to the wheel on hard ground will depend on the frictional torque created by the Clutch plates inside the differential.

I hope I clear the confusion.

I remember I was once asked in my Viva-
There is a hill and you have two vehicles with you one is a scooter and the other a motorbike. Both vehicles have Torque at axle constant.
Which one you would prefer to climb the hill?

T=F X R ... Torque at axle remaining constant value of F will be high for Scooter as Radius of wheels is less and for Bike Value of F will be low as wheel radius is bigger.

I want more F to climb a wheel but I did still choose the Motorbike because normally its not the torque generated which is the limiting factor its the torque which can be transmitted which is the limiting factor!

And torque transmitted depends upon Traction which in turn again depends on the area of contact patch - Bigger for a motorcycle wheel.
(Although Friction formula never says friction/traction is dependent on area in contact. This is a debatable topic and I would not like to debate here as it will be offtopic!)

Similar is the story of a Tractors rear tyre being bigger- Offtopic!

[ex670c]

Originally Posted by amit_purohit20

3) An open differential distributes torque equally to the two wheels when both feels face the same resistance from the ground. ( In real world this never happens even if going over a plain road, there is always a small difference in resistance from the both wheels of the axle.)
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.
5) The differential distributes the torque based on the amount of resistance the wheels have to offer. (Because torque is generated because of resistance and resistance only!) So the wheel having higher amount of resistance will get more torque transferred to it upto the max limit of what the engine can generate under maximum load (Load again here is created because of resistance at the wheels.)


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.


If you have a LSD with one wheel in air the max. torque transferred to the wheel on hard ground will depend on the frictional torque created by the Clutch plates inside the differential.

Hi Rohit,

If a rod/shaft is being rotated by X amount of torque, what will be the torque on either ends of the rod?

X or X/2

If it is X at each end, that mean the Shaft is providing a Two times mechanical advantage.

This is the working of a simple T-Case 2WD to 4WD, also the case for a Differential Lock.

Regards,

Arka

[Samurai]

Originally Posted by DKG

In summary I am inclined to agree that more torque is applied to the wheel on ground vis a vis in the air, just that I wouldn't say it's hundred percent as the weight of the wheel in air still exerts some load and utilises some torque.

You are right, it is never 100%. But in science discussions, it is common to ignore the insignificant losses due to self weight or friction, to get the main point across.

Originally Posted by amit_purohit20

I agree with Samurai. Only one thing I do not agree is the definition of the torque being the least amount of Force X Distance.

Samurai can be a good teacher indeed, the way he explains with simple analogies!

Thanks, I happen to be a teacher as well. It is one part of my work, to teach technology to junior staff. I always try to give simple real life examples to explain complex ideas, and have found it quite effective. I enjoy teaching and have been enjoying it since 1985.

Nice to see you are an automobile designer.

Regarding the “least force” comment, I know it is not the correct way to define torque. But I have been using that for a long time. In the last 2-3 days I have been wondering how I came up with it. I think now I understand why. I started using it to describe why less torque is delivered when wheels are spinning in slush.

Question: Why even the most powerful vehicle can get stuck in slush?
Answer: Engine will deliver just enough torque to rotate the wheels. It doesn’t care whether the vehicle is moving or stuck. While trying to describe this to physics challenged people, I started saying torque [delivered by the engine] is the least force required to turn the wheels.

This helped me explain why Thar CRDe could not generate more torque than my lowly CJ340 in slush. I could roughly say things like if 50Nm can overcome the traction offered the slushy terrain, then both CJ340 and Thar CDRe engine will develop just 50Nm and get the wheels spinning, assuming everything else is cet par. In other words, torque delivered by the engine is the least force required to turn the wheels. After a while I dropped the qualification of “delivered by the engine”. I realise it an imperfect definition, but it was helpful as a teaching tool.

Originally Posted by amit_purohit20

5) The differential distributes the torque based on the amount of resistance the wheels have to offer. (Because torque is generated because of resistance and resistance only!) So the wheel having higher amount of resistance will get more torque transferred to it upto the max limit of what the engine can generate under maximum load (Load again here is created because of resistance at the wheels.)

I am a bit confused here. In case of an open differential, isn’t the net resistance equal to twice the resistance offered by the wheel with least resistance? If left wheel offers resisting force of 10N and right wheel offers resisting force of 40N, what is the net resistance offered to the propeller shaft? I thought it would be two times the lesser of the two, that is 20N. Pardon me if I used wrong units to represent resisting force, I am after all a software guy.

Originally Posted by mallumowgli

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

Originally Posted by amit_purohit20

Torque is nothing but a twisting force. The max twisting force which can be applied is called Max. Torque. Its the maximum capability of an engine to generate torque which can be reached only when there is maximum resistance. Torque cannot be generated without resistance similar to pressure.

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?

Originally Posted by ex670c

Hi Rohit,

If a rod/shaft is being rotated by X amount of torque, what will be the torque on either ends of the rod?

X or X/2

Arka, it entirely depends on the resistance offered by either ends of the rod. Unless you supply that info, it is an incomplete question. BTW, he is Amit, not Rohit.

[Sutripta]

Originally Posted by ex670c

Hi Rohit,

If a rod/shaft is being rotated by X amount of torque, what will be the torque on either ends of the rod?

X or X/2

Assuming 1:1 diff ratio, and no losses etc. the sum of the torques of the two half shafts = torque at propeller shaft.
In case of an open diff, the torque at the two halfshafts will be equal. So X/2 + X/2 = X
In case of a locked diff, the torque at the two halfshafts can be anything, but the sum part still holds.

Originally Posted by Samurai

I am a bit confused here. In case of an open differential, isn’t the net resistance equal to twice the resistance offered by the wheel with least resistance? If left wheel offers resisting force of 10N and right wheel offers resisting force of 40N, what is the net resistance offered to the propeller shaft? I thought it would be two times the lesser of the two, that is 20N.

You are right. (Assume 1:1 diff ratio. Not important for this discussion, but to keep the nitpickers at bay).
Unit is Nm.
Your example:
Open diff.
40 Nm is the maximum of what that wheel is capable of. 10 Nm is what is being transmitted and resisted right now. Add these (10 + 10) and you get 20.

Regards
Sutripta

[amit_purohit20]

Originally Posted by ex670c

Hi Rohit,

If a rod/shaft is being rotated by X amount of torque, what will be the torque on either ends of the rod?

X or X/2

If it is X at each end, that mean the Shaft is providing a Two times mechanical advantage.

This is the working of a simple T-Case 2WD to 4WD, also the case for a Differential Lock.

Regards,

Arka

If shaft carries a torque X it carries the same torque everywhere let it be left or right wheel end its the same. Its like providing resistance at any point on the shaft the centre, left, right anywhere. The max. resistance the shaft can overcome is X. So there is no case of mechanical advantage.

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!

Originally Posted by Samurai

Nice to see you are an automobile designer.

Thanks . Wish all Indians had teachers like you. Other wise teachers become very theoretical and monotony sets in their way of delivering knowledge. If a teacher cannot generate interest in the students he is not a good teacher! I was lucky to have some good teachers and good seniors who could generate interest in some topics. For those subjects which I had not so good teachers I am still weak! Eg. Thermal Engineering.

Originally Posted by Samurai

Regarding the “least force” comment, I know it is not the correct way to define torque...

Question: Why even the most powerful vehicle can get stuck in slush?
Answer: Engine will deliver just enough torque to rotate the wheels. It doesn’t care whether the vehicle is moving or stuck. While trying to describe this to physics challenged people, I started saying torque [delivered by the engine] is the least force required to turn the wheels.

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.)

It is the fault of the differential that it tries to give power or transmit torque to the wheel having the least resistance.

Originally Posted by Samurai

I am a bit confused here. In case of an open differential, isn’t the net resistance equal to twice the resistance offered by the wheel with least resistance? If left wheel offers resisting force of 10N and right wheel offers resisting force of 40N, what is the net resistance offered to the propeller shaft? I thought it would be two times the lesser of the two, that is 20N.

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?

Laws of physics do apply to it! And yes 20 Nm is the right figure which the propeller will have to work against.

The least amount of resistance found on consideration of both wheels of an axle is the resistance which is available at both wheels which can be put to use.

Originally Posted by Sutripta

Assuming 1:1 diff ratio, and no losses etc. the sum of the torques of the two half shafts = torque at propeller shaft.
In case of an open diff, the torque at the two halfshafts will be equal. So X/2 + X/2 = X
In case of a locked diff, the torque at the two halfshafts can be anything, but the sum part still holds.


You are right. (Assume 1:1 diff ratio. Not important for this discussion, but to keep the nitpickers at bay).
Unit is Nm.
Your example:
Open diff.
40 Nm is the maximum of what that wheel is capable of. 10 Nm is what is being transmitted and resisted right now. Add these (10 + 10) and you get 20.

Regards
Sutripta

Agree with you Sir!

Practically even on a plain ground and going straight there is not equal resistance from both wheels of the axle. The one which has least resistance is the wheel which is getting power and equal amount then being sent to the other wheel.

The planet pinions of the differential are always in motion. There is hardly any time when they are not rotating, except when the differential is locked.

Thats why I did always prefer a Locked differential than a LSD because the torque which can be transmitted is often limited to the amount of resistance provided by the friction plates inside. Although LSD is easier to use and it engages and disengages automatically without any intervention from the operator.
Also if I remember in old LSD's the friction plates use to create problem during turns as it used to sense that as a outer wheel spin! But I think this problem has been taken care in present generation LSD's as they come into picture only when there is sufficient amount of wheel spin. Its a tricky job to design an LSD. I had worked on a LSD but unfortunately couldnot get it tested. So my fundamentals are still not so strong in LSD.

Also one note of appreciation to all you guys who are not mechanical engineers or automotive designers that you have your fundamentals so strong indeed! Really feels happy to be a part of this forum.
I doubt even a high percentage of automotive designers in India understand these fundamentals correctly!