[Tejas@perioimpl]

Alright,

Let's understand torque in a layman term.

What is torque? It is a force that is generated against a resistance.

Let me give you an example:

1. I tell you to lift a 10kg dumbbell. You will lift it easily.

2. Now I give you a 30kg dumbbell. You will strain and lift it.

3. Now I again give you the 10kg dumbbell and ask you to lift it with the same effort as the 30kg one. Even though you are capable of generating "torque" to lift a 30kg dumbbell, you will not be able to generate that same strain and effort because the load on your hand is different.

Let's take a car related example.

You are in a 2wd car and are driving on a trail.

The torque needed so slowly propel this car for example is 100N and you are nicely crawling till you come to an are where the right wheel looses traction and you don't have enough momentum to go through this obstacle.

So what happens? The engine only produces torque on demand and the differential always acts to equally split the torque but also take the path of least resistance.

Which is the path of least resistance? The wheel in the air.

What kind of torque does it need when you see it rotating at full speed? Only enough torque to overcome the weight of the rim and tyre and components and frictional forces.

Let's assume that is 10N. The differential will equally split that 10 to the left wheel with traction as well. But that 10N is not enough the move the car forward. So the engine is just generating only 20N which is not enough and you are effectively stuck.

[Samurai]

Originally Posted by Blooming Flower

Let's consider a situation where two wheels are getting same amount of torque in an open differential config.

That's like saying let's consider earth is a sphere. Yes, I can consider that.

Originally Posted by Blooming Flower

one wheel is spinning and the other isn't. If both of them are having same torque where does the torque at the stationary wheel vanish into? Doesn't it defy the basic 'conservation of energy' law?

It converts into heat. Let's smash a big sledge hammer into a steel anvil. The anvil doesn't move, are you saying no work was done? Just touch the surface of the hammer and the anvil. It would be warm. It is not obvious in most situations, that's all. Try pushing a concrete wall, it doesn't move. Is that violation of conservation of energy law?

Originally Posted by Blooming Flower

I would like to get my fundamentals rectified if I am thinking wrong.

Kindly read this post, and let me know your objections to it.

[RadixLecti]

Originally Posted by Tejas@perioimpl

1. I tell you to lift a 10kg dumbbell. You will lift it easily.

2. Now I give you a 30kg dumbbell. You will strain and lift it.

3. Now I again give you the 10kg dumbbell and ask you to lift it with the same effort as the 30kg one. Even though you are capable of generating "torque" to lift a 30kg dumbbell, you will not be able to generate that same strain and effort because the load on your hand is different.

I understand the point you are making, but to be pedantic, you could easily use the same force to lift the 10kg dumbbell as the 30kg one. You would just end up accelerating the 10kg one 3 times as fast. An analogous situation exists with torque and angular acceleration.

Originally Posted by Blooming Flower

Let's consider a situation where two wheels are getting same amount of torque in an open differential config.; one wheel is spinning and the other isn't. If both of them are having same torque where does the torque at the stationary wheel vanish into? Doesn't it defy the basic 'conservation of energy' law?

I would like to get my fundamentals rectified if I am thinking wrong.

It doesn't vanish. It twists the drive shaft by a small amount. Just like when you are trying to undo a screw with a screwdriver, but you haven't yet applied enough torque to break the resistance of the screw. The screwdriver twists (sometimes quite visibly), and you can feel that you are expending energy, but nothing moves. Your twist also generates a little heat by straining the material of the screwdriver.

OT: even a freely spinning shaft twists, to a greater or lesser degree. On the large engines I work on, we measure torque (and hence power) by measuring the twist in the driveshaft.

[Tejas@perioimpl]

Originally Posted by RadixLecti

I understand the point you are making, but to be pedantic, you could easily use the same force to lift the 10kg dumbbell as the 30kg one. You would just end up accelerating the 10kg one 3 times as fast. An analogous situation exists with torque and angular acceleration.

You would lift it faster but you wouldn't strain as much even if you tried to. #beingpedantic

[Blooming Flower]

Originally Posted by Samurai

der that.
Kindly read this post, and let me know your objections to it.

Yes. No objection!

Just need to get some more light on the following quotes.

Originally Posted by dhanushs

With this Eaton MLD, the torque available at the propeller shaft, is distributed evenly to both tyres, irrespective of their traction.

Originally Posted by Blooming Flower

If MLD kicks in, it essentially prevents the intrinsic action of the differential. In this case, the 100% torque from the propeller shaft is distributed evenly at both the drive shafts connected to the left and right wheels, so that they turn at the same speed irrespective of the resistance arising from the traction at each wheel.

Do these two speak anything differently?

Originally Posted by Samurai

MLD when engaged, locks the differential fully, which can result in nearly 100% torque transfer if one wheel is totally in the air. If one wheel has 3 times more traction than the other, that wheel will get 75% of the torque, while the other gets 25%.

[Samurai]

Originally Posted by Blooming Flower

Just need to get some more light on the following quotes.

Do these two speak anything differently?

Come on, you are quoting dhanush from 2013, when we were all still trying to understand MLD?

Even I said some wrong things in the early parts of this thread in 2013. Once corrected, we learned and we moved on.

[RadixLecti]

Originally Posted by Tejas@perioimpl

You would lift it faster but you wouldn't strain as much even if you tried to. #beingpedantic

Sorry, but you could. F = m x a
Force required to accelerate a 10kg mass at 3 m/s^2 = 30N
Force required to accelerate a 30kg mass at 1 m/s ^2= 30N
You are exerting a force of 30N in both cases.
Assuming you move the weights by the same distance you are also doing the same amount of work, W = F x distance

This relationship works with torque as well, T=Iɑ, and W=Tϴ

[Tejas@perioimpl]

Originally Posted by RadixLecti

Sorry, but you could. F = m x a
Force required to accelerate a 10kg mass at 3 m/s^2 = 30N
Force required to accelerate a 30kg mass at 1 m/s ^2= 30N
You are exerting a force of 30N in both cases.
Assuming you lift the weights to the same height you are also doing the same amount of work, W = F x distance

This relationship works with torque as well, T=Iɑ, and W=Tϴ

My last OT post on this topic: let's hook someone to a heart rate monitor and see the strain that I mentioned while lifting the 2 weight options. And then ask him to strain using the light weight and see what the monitor says. Real world vs theoretical.

From the first post I'm talking about strain and exerting forces.

[dhanushs]

Originally Posted by dhanushs

With this Eaton MLD, the torque available at the propeller shaft, is distributed evenly to both tyres, irrespective of their traction.

Originally Posted by Blooming Flower

..
Do these two speak anything differently?

I was wrong there Blooming Flower. Check the torque meter on your Thar while driving and you will understand. You will find similar engine behavior for different torque demands.

[Jeroen]

Originally Posted by RadixLecti

Sorry, but you could. F = m x a
Force required to accelerate a 10kg mass at 3 m/s^2 = 30N
Force required to accelerate a 30kg mass at 1 m/s ^2= 30N
You are exerting a force of 30N in both cases.

Not on earth would you move, or rather lift, a mass of 10 kg, resp 30kg, with a force of 30N!.

In layman term, 30N is 3kg! 3 kg won’t lift 10kg, unless you are in a weightless environment, or at least an environment with significant lower gravity pull than 9,8m/s2.

Earth acceleration constant is give or take 9,8 m/s2. It is, give or take the same everywhere on earth.

So you need to factor this in, assuming you are an earthling of course!

We can do it a bit more scientific and use your formulas.

A 10 kg mass exerts a force of 10 x 9,8 = 98 N on whatever it is resting on.
A 30 kg mass exerts a force of 30 x 9,8 = 294 N on whatever it is resting on.

Your pathetic 30N ain’t going to accelerate these masses here on earth!

Jeroen

[RadixLecti]

Originally Posted by Jeroen

Not on earth would you move, or rather lift, a mass of 10 kg, resp 30kg, with a force of 30N!.

In layman term, 30N is 3kg! 3 kg won’t lift 10kg, unless you are in a weightless environment, or at least an environment with significant lower gravity pull than 9,8m/s2.

Earth acceleration constant is give or take 9,8 m/s2. It is, give or take the same everywhere on earth.

So you need to factor this in, assuming you are an earthling of course!

We can do it a bit more scientific and use your formulas.

A 10 kg mass exerts a force of 10 x 9,8 = 98 N on whatever it is resting on.
A 30 kg mass exerts a force of 30 x 9,8 = 294 N on whatever it is resting on.

Your pathetic 30N ain’t going to accelerate these masses here on earth!

Jeroen

Thanks, I realise that. I also realise that from one post to the other I moved from discussing 'lifting' the weights to simply 'moving' them. My mistake. My original numbers would only work for horizontal movement on earth.

Though the member I was discussing this with does not wish to continue as it is OT, I do stand by the gist of my point - one can use the same force to accelerate a smaller mass faster or a larger mass more slowly (relative to each other). This holds for acceleration upward as well.
500N would accelerate both these masses upwards - on earth - at different rates.

[Blooming Flower]

Originally Posted by Samurai

Isn't BLD acting on an open differential? In an open differential, there is zero torque transfer. There is only 50:50 torque split. It is like fiddle brake, by applying it, you increase the load on the engine, forcing it to generate more torque, which will be split 50:50 to each wheel. There is no transfer of any kind.

Originally Posted by dhanushs

I was wrong there Blooming Flower. Check the torque meter on your Thar while driving and you will understand. You will find similar engine behavior for different torque demands.

Does Mahindra's technical literature call for any correction?

[Samurai]

Yes, it does. Hopefully someone will inform them.

[alpha1]

Originally Posted by Samurai

First thing you need to understand is Torque is a function of load. If there is no load, there is no torque. That means torque generation does change with load.

Let me examine an even more fundamental question ... the torque that we speak of is the twisting moment at the crank shaft.
What causes this? Is it the load/resistance or is it the fuel being burned in the cylinder?

(This is with respect to carbureted engines that I understand) When I keep accelerator pressed/throttle opened at fixed position, it is the engine vacuum which sucks the fuel+air mix into the cylinder. If the throttle is fixed, then the fuel+air mix is also fixed.

Since, the fuel has fixed calorific value, and the quantity of fuel is also constant (at fixed throttle position) - it should generate constant explosive force on the piston during expansion stroke ... creating the constant force moment (torque) at the crank shaft.

[Samurai]

Originally Posted by alpha1

Let me examine an even more fundamental question ... the torque that we speak of is the twisting moment at the crank shaft.

It is the torque as mentioned in the first post. Please ignore the last line of that post, I was wrong there. I learned otherwise in the next few days.

Originally Posted by alpha1

What causes this? Is it the load/resistance or is it the fuel being burned in the cylinder?

It is both.

When you are idling, the A-pedal is not pressed, very little fuel is spent, just to keep the engine running. When you press the A-pedal at neutral, you will spend little more fuel to keep the engine running at higher rpm. The torque generated is just to keep the engine running, in both cases.

Once the gearbox is engaged, and car starts moving, fuel is being spent to do the extra work of moving the car, and even accelerating the car.

Less load means less torque and less fuel requirement. More load means more torque and more fuel requirement.

Each engine also has an upper limit.

Originally Posted by alpha1

Since, the fuel has fixed calorific value, and the quantity of fuel is also constant (at fixed throttle position) - it should generate constant explosive force on the piston during expansion stroke ... creating the constant force moment (torque) at the crank shaft.

Is the resistance to expansion stroke (aka load) constant?