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Old 26th December 2013, 13:44   #31
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Default Re: Eaton MLD (M-Locker) now available off the shelf

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Originally Posted by lucifer1881 View Post
L & A do not change for a vehicle. They are constant.
So... if L and A are constant in a formula that says T = F x L x Sin (A), then T is directly proportional to F. How can T remain constant when F changes? Do explain the mathematics here.

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I used the formula to explain the concept of torque. The first thing to understand is that torque is not force. It is the moment of force.
I know torque is not same as force. But torque is often described as twisting force or rotational force. I meant in that way. If it bothers you that much, I will say rotational/twisting force instead of saying just force.

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If a wheel is stuck in the mud, then the engine will be working on higher load to impart any rotation to the wheel.
Stuck how? with traction (no spinning), or without traction (freespinning)? If the vehicle is stuck with traction, then I agree with you.

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Originally Posted by lucifer1881 View Post
I understand I made one gaffe in my earlier post:

I should have stated that the load on the engine to generate the force required to achieve the torque to spin the wheel changes.

This load depends on how much traction the wheels have.
Finally we agree on something.

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Originally Posted by lucifer1881 View Post
The torque required to spin the wheels does not depend on traction.
You lose me again.

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Originally Posted by lucifer1881 View Post
Another example. In my car I achieve 80kmph in fifth gear at 2000rpm. Whether I drive uphill or downhill, 2000rpm in fifth gear results in 80kmph.
True. I agree on this.

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Originally Posted by lucifer1881 View Post
Torque required is the same in both cases.
We disagree again.

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However, the engine load is higher uphill than downhill.
And we agree again.

--------------------------------------------------

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.

Last edited by Samurai : 26th December 2013 at 13:48.
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Old 26th December 2013, 14:03   #32
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Default Re: Eaton MLD (M-Locker) now available off the shelf

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...
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.
If L & A are constant, then Torque generated is directly proportional to force applied. We are in agreement here.

What we disagree on is simply this:

For a wheel to spin at 10 rpm, it needs the same torque whether it is in air, vacuum or dipped in custard.

Let us analyse this statement.

T = F x L x Sine (A)

We keep L and A constant to simplify the mathematics of the explanation. This implies T is proportional to F.

In vacuum, less force is applied to spin the wheel than in custard. We will agree on this too. This is where things get a little geeky. Bear with me .

Let us assume we are in space far, far away where there is no significant impact of any gravitational force or frictional force or any other force. In such case, all force applied to the wheel will be utlised to generate the torque to spin the wheel.

Now let us assume, the wheel is submerged in custard. In this case, for the wheel to spin, the forces applied must be first greater than the forces that act against the wheel.

In custard, a wheel experiences the frictional force that wants to prevent it from spinning. Let us call this force f1.

We apply a force f2. The resultant force is rF = f2-f1.

It is this resultant force that is important. This rF is to be used in the formula for Torque, because effectively it is this force that is going to act on the wheel.

Torque is proportional to Resultant Force.

So while it may appear that we are applying more force to spin a wheel in custard than in vacuum, in effect the resultant force needed to spin the wheel at a given rpm is the same in both cases. In Vacuum, rF is very nearly same as the applied force. In real world conditions, rF is less than the applied force. Hence it appears to us that we are applying more force where as in actual fact the rF needed to spin the wheel at a given rpm is constant.

I hope I have been lucid enough. This is not a concept that can be easily explained on a discussion forum. College professors spend hours explaining this to their students in classrooms.
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Old 26th December 2013, 14:52   #33
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Default Re: Eaton MLD (M-Locker) now available off the shelf

hi lucifer1881,
when you say we apply force f2 did you meant engine apply? i think we should consider f2 while calculating torque because that is the total force which has to be applied to rotate the wheel/s and that is coming from the engine.
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Old 26th December 2013, 14:59   #34
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Default Re: Eaton MLD (M-Locker) now available off the shelf

Does the force (F) in the following formula refer to absolute force (F2) or resultant force (rF)? I think it is the former (F2).

T = F x L x Sine (A)

This is why I prefer to wait for a professional to answer this question. You didn't have to waste time & effort explaining resultant force. I understand that concept. But I don't believe the above formula refers to resultant force.

Last edited by Samurai : 26th December 2013 at 15:01.
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Old 26th December 2013, 15:35   #35
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Default Re: Eaton MLD (M-Locker) now available off the shelf

Let me pitch with a small clarification, if it helps

There are three things at play here - Torque, power and force. Engine produces power, and pushes the piston downwards. This is force and this force generates torque, rotates the crankshaft and produces power. Power is roughly the product of Torque and rpm. The torque (and hence power) is transmitted after amplification/reduction at the gear-box.

This amplified/reduced torque available at the end of the propeller shaft may be called available torque. Can we call it T(a)?

Depending on the design of the differential and shafts the torque is split between the wheels and this is always constant in the normal differential.

In a 2WD rear drive vehicle, with conventional differential -the split is always fixed. Mostly 50:50. So T(left) + T(right) will always be T(a). And Ta is fixed (at the end of the propeller shaft, remember) for a particular engine rpm.

So what happens when one wheel loses traction - or is raised?

Or better, consider when you have jacked up both the driven wheels? what happens to the torque? It is simply driving both the wheels generating power. (product of rpm & torque)

Now apply friction and bring one wheel (right one) to a stop. Whats happening? The left wheel continues to rotate with the available 50% T(a). Note the rpm of wheel doesn't increase since it is limited by the rpm of the propeller shaft. The right wheel has 50%T(a), but is not generating any power since there is no rpm. (The torque produces a bending moment on the right half shaft, though)

Remember Torque is not energy so the laws that energy cannot be destroyed does not apply here. The energy provided to one wheel is just lost as heat due to friction and other losses

Now there is a need to create a differential torque in the event of a slip, to enable one wheel to generate more torque when the other is simply spinning freely. This is what is done in things like limited slip differentials, viscous coupling (if I am right) etc. Of course this theory of limited slip differentials and the like are quite known to off-roaders

@Samurai - Reg your question about applying load to the center of a long shaft, holding one end fixed and the other free : The torque at both ends are fixed (depends on the characteristic of the shaft). What you are producing is bending moment towards one side and no significant bending moment at the free end. Torque available at both ends remain same (need not be 50-50, but maybe 50-50)

I am just writing from a memory of a long forgotten auto-engg elective!! Newer/fresher graduates can explain the theory better

PS : The torque mentioned in the first paragraph is the rated torque of the vehicle whereas the power mentioned there need not be so - different manufacturers take different approaches as far as power (bhp/ps) is concerned

Ah, and one more thing - no need to get confused with forces f1/f2/fn etc. It is the resultant torque and power which matters

Last edited by mallumowgli : 26th December 2013 at 15:39.
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Old 26th December 2013, 15:35   #36
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Default Re: Eaton MLD (M-Locker) now available off the shelf

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Originally Posted by i.1979 View Post
hi lucifer1881,
when you say we apply force f2 did you meant engine apply? i think we should consider f2 while calculating torque because that is the total force which has to be applied to rotate the wheel/s and that is coming from the engine.
Engine applies f2. This f2 determines the load on the engine. f2-f1 is what the wheel is experiencing. Hence when stuck in mud, engine needs to apply greater f2 to counter the greater f1 experienced.

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Originally Posted by Samurai View Post
Does the force (F) in the following formula refer to absolute force (F2) or resultant force (rF)? I think it is the former (F2).

T = F x L x Sine (A)

This is why I prefer to wait for a professional to answer this question. You didn't have to waste time & effort explaining resultant force. I understand that concept. But I don't believe the above formula refers to resultant force.
Let us consider Work.

Work = Force x Displacement

The two of us push a cart in opposite directions with forces f1 and f2, and the cart displaces by d.

Work = (f2-f1) x d [could be f1-f2 also, depending on which direction you choose to represent positive displacement]

I apply 100N force while the wheel is spinning freely in the air. This results in the wheel rotating at 200rpm, for example. But if I apply the same force when the wheel is under water, it rotates at, say, 100rpm.

Where did the additional rpm go? Some of the force applied to the wheel went in countering the greater resistance to movement offered by water. Hence it is the resultant force that is responsible for any spin the wheel undergoes.

The F in the formula for torque represents applied force only in vacuum. That is how most physics formulae are. The speed of light, for instance, is approx 3 lac km per second in vacuum. The formulae are for ideal or lab conditions.

Newton's first law states that a body at rest will remain at rest and a body in motion will remain in motion unless acted upon by an external force. To an observer, this is not so because we all know that a ball rolling on the ground will eventually come to a stop. The unseen force here is frictional force.

Similarly, in real world conditions, in formula for torque one must account for the unseen forces of resistance. Hence it is the resultant force that is used for calculating torque.
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Old 26th December 2013, 15:57   #37
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Quote:
Originally Posted by lucifer1881 View Post
...
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.....
Thanks for the excellent (and, needless to say, correct) explanation, lucifer1881

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...I hope I have been lucid enough. This is not a concept that can be easily explained on a discussion forum. College professors spend hours explaining this to their students in classrooms.
I agree wholeheartedly to this point .
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Old 26th December 2013, 17:23   #38
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Default Re: Eaton MLD (M-Locker) now available off the shelf

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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.
Is the confusion because, one person is saying "it will" generate whereas other person is saying "it can"?

This discussion started from torque transfer and now it is heading to torque generation
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Old 26th December 2013, 17:39   #39
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Default Re: Eaton MLD (M-Locker) now available off the shelf

I was anyhow going to dyno my Jeep from Coimbatore, after a ~week, when the work gets over.

Now, I guess I'll have to dyno it the normal way and also with diffs locked, one wheel in the air and one on the dyno.

This confusion will be sorted out with solid proof then.
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Old 26th December 2013, 18:35   #40
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Now, I guess I'll have to dyno it the normal way and also with diffs locked, one wheel in the air and one on the dyno.

This confusion will be sorted out with solid proof then.
Answer is right here i guess. What kind of dyno will you be using?

And dyno's are connected to all the driven wheels isn't it? Why is it so? Why dyno is not done only from one wheel of the driven axle?
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Old 26th December 2013, 20:39   #41
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Originally Posted by dhanushs View Post
I was anyhow going to dyno my Jeep from Coimbatore, after a ~week, when the work gets over.

Now, I guess I'll have to dyno it the normal way and also with diffs locked, one wheel in the air and one on the dyno.

This confusion will be sorted out with solid proof then.
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.

Guys, this thread has so many point of view, (and obviously I disagree with 50% of those!) that I don't think any one or two posts is going to change ones mind. How about going back to basics we can relate to.

Most of us have an intuitive understanding of (linearly applied) forces and their interaction with objects (masses). Because we deal with it everyday in our real world. Whether it is pushing a table out of the way, throwing an object, or opening/ shutting a badly fitting drawer or door. We can easily extend and apply this intuitive understanding to torque and rotating objects if we make the following correspondences (simplistic, but I think a very good starting point):-

Force <-> Torque
Mass <-> Moment of inertia
Velocity <-> RPM, angular velocity.

And apply Newton's laws. Esp. the first two!

Then we can go onto what happens when a wheel breaks traction. These two together can handle the case when a diff is locked.

Differentials:- that's a topic I'll leave to others (and for another day. For the moment, lets stick with locked differentials)!

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Originally Posted by lucifer1881 View Post
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.
What is the torque at the flywheel in case of a genset engine which operates at (I assume) constant speed but varying load? Say we connect the engine to the alternator with a shaft which we instrument. (Transmission dynamometer principle.)

Regards
Sutripta

Last edited by Samurai : 26th December 2013 at 23:07. Reason: back-to-back post
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Old 26th December 2013, 22:36   #42
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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.

Last edited by star_aqua : 26th December 2013 at 22:39.
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Old 26th December 2013, 23:32   #43
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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.
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.

Regards
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Old 27th December 2013, 09:04   #44
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Default Re: Eaton MLD (M-Locker) now available off the shelf

How do you measure torque? Does measurement of torque necessarily involve "load"?

Measuring something and defining it are often two different things, but I would like to know just to satisfy my curiosity. Experts? Quasi-experts? Bruce Lee?

Last edited by DirtyDan : 27th December 2013 at 09:12.
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Old 27th December 2013, 10:00   #45
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Originally Posted by Sutripta View Post
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.

Regards
Sutripta
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?

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Originally Posted by DirtyDan View Post
How do you measure torque? Does measurement of torque necessarily involve "load"?

Measuring something and defining it are often two different things, but I would like to know just to satisfy my curiosity. Experts? Quasi-experts? Bruce Lee?
Yes, this is what made everyone confused here. The definition and how you measure it. Without load, the torque cannot 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?
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