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ARB/Truetrac arrangement

ARB/Truetrac combo opinions

  • Truetrac/Truetrac

    Votes: 4 5.4%
  • Truetrac front/ARB rear

    Votes: 7 9.5%
  • ARB f/Truetrac r

    Votes: 14 18.9%
  • ARB/ARB

    Votes: 49 66.2%

  • Total voters
    74
  • Poll closed .
The way that torque works, there has to be a resistive force equal to the torque applied. If you put your vehicle on jack stands so both drive wheels are in the air, and mash the throttle, they will be only receving enough torque to overcome air resistance. Some engine torque also goes into overcoming driveline resistance, and driving accessories. The same happens if you have an open diff and put one wheel in the air.


That is why a dyno has to apply resistance to the wheels. It's not just freewheeling.


Here is an analogy... Think about tightening a bolt. When it's loose, the torque you can apply is almost 0. You physically can't apply more torque. Set your torque wrench and it just won't click. Only when it gets tighter can you apply more force.
 
The Truetrac doesn't just route torque. It needs resistance applied to one wheel in order to send 5x (or 3x or whatever) that torque to the other. If you put one wheel in the air it will just spin like an open diff.
 
http://www.houseofthud.com/differentials.htm

So what's wrong with the open diff? The problem is an open diff always tries to balance the torque. That's a hard statement to get a grasp on, but it means that if the spider gears are pushing on both drive gears and one of them offers lots of resistance (tire sitting on pavement) and the other side offers no resistance (up in the air, or sitting on a patch of ice), then it will find a happy balance where both sides are receiving almost no torque at all. All the rotational energy is guided to the side with the least resistance. In the end, that side spins very fast and the pressure on each drive gear is the same.. Almost no torque is needed to spin one wheel, and since the open diff always sends the same amount of torque to both output shafts, almost no torque is going to the other side as well. Anyone who's driven on snow or ice knows this trick.
Torsen
....
That means that a multiple of the torque that would have gone to the low traction side actually goes to the high traction side. So if 20 ft-lbs of traction is at the low traction side, something like 80 ft-lbs goes to the side that can actually use it. A ratio of 4 to 1 or 5 to 1 is common but changing the gear teeth angles changes the ratio.
....
Note however that without significant preload either torque biasing diff will not work well with a wheel completely off the ground. 0 ft-lbs time 4 is still 0. A simple braking trick helps though.

BTW the only reason any of this matters is it relates to how a Torsen functions
 
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I tried Richard, he's not getting it.

It requires torque to initiate and maintain rotation. One tire is spinning quickly, a manifestation of torque. a differential does not multiply torque, your equation makes no sense. By your reasoning, any wheel with traction sends 5x the amount of torque applied to the other wheel, which would in turn send 5x back the other way, which would then send 5x back the other way, and then 5x back the other way. The practical outcome of that lunacy is that any torque input into a Torsen diff quickly approaches infinity output, tires disintegrate and the rotational force quickly overpowers the rotational force of the globe gravity would then implode on itself becoming a black hole collapsing all space and time..

An open differential distributes power through the path of least resistance. This appears as an even distribution until torque>traction.

A locker distributes power evenly.

A clutch based limited slip differential also sends power through the path of least resistance, but forces some in the other direction.

A torsen based limited slip differential also sends power through the path of least resistance, but forces a percentage to the high traction side that is relative to the amount of resistance felt by the low traction side. More resistance felt = more power pushed to the high traction side. The equation can never equal more than 1 i.e. if half of the torque is going in each direction .5+.5=1 If one wheel is in the air, it is the "1" and as you apply the brake you subtract from that 1 and begin applying torque to the other side.
 
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If he actually said something that he thought for himself, a decent discussion might be possible. However, he just keeps regurgitating stuff he read off the internet and doesn't really understand what he read and now doesn't understand what he's saying. But, he's obviously convinced. He knows just enough to think he really knows.

Damn, this is the internet, why won't he believe us? :)



Pretty funny that power isn't generated by the motor, just by resistance. No resistance means no power, according to him. So, if the motor is turning 4000 rpm while a tire is in the air it only has enough power to spin the tire that fast. But then, if that spinning tire all of a sudden comes down to the ground and the axle breaks, now the motor all of a sudden has enough power to break an axle. (I understand that the spinning momentum has energy, but an axle has enough strength to stop a spinning tire, but maybe not enough strength to instantly absorb the torque put out by a motor at 4000 rpm.)

Been awhile since I've had an interaction with this much of an internet hard head. I should browse the mod tech forum more often. :)
 
...any torque input into a Torsen diff quickly approaches infinity output, tires disintegrate and the rotational force quickly overpowers the rotational force of the globe gravity would then implode on itself becoming a black hole collapsing all space and time..


Man...I really hate when that happens!
 
If he actually said something that he thought for himself, a decent discussion might be possible. However, he just keeps regurgitating stuff he read off the internet and doesn't really understand what he read and now doesn't understand what he's saying. But, he's obviously convinced. He knows just enough to think he really knows.

Damn, this is the internet, why won't he believe us? :)



Pretty funny that power isn't generated by the motor, just by resistance. No resistance means no power, according to him. So, if the motor is turning 4000 rpm while a tire is in the air it only has enough power to spin the tire that fast. But then, if that spinning tire all of a sudden comes down to the ground and the axle breaks, now the motor all of a sudden has enough power to break an axle. (I understand that the spinning momentum has energy, but an axle has enough strength to stop a spinning tire, but maybe not enough strength to instantly absorb the torque put out by a motor at 4000 rpm.)

Been awhile since I've had an interaction with this much of an internet hard head. I should browse the mod tech forum more often. :)

You don't understand the physics, and refuse to understand the physics of what I'm explaining to you. Instead you're attacking me and accusing me of "regurgitating" what I read on the internet.

The motor can only exert an amount of torque equal to the resistance. This is simple newtonian physics!! I don't know how much more I can explain this. I explained how a cutting brake works. I explained specifically how the lack of resistance on the lifted wheel leads to zero torque to the wheel with traction. I explained that a dyno needs to exert force on the wheels in order to cause the motor to create horsepower.

The lifted wheel you are talking about isn't exerting any torque on the driveline, except what it takes to overcome air resistance and driveline drag. The engine has POTENTIAL to make much more torque but ISN'T until that tire hits the ground and has resistance.

Think about my bolt torque analogy carefully. Tell me how much torque your arm muscles can exert on a free spinning nut. Go and try it yourself right now.

After you do that, find me a dynamometer that does not apply force to the wheels but is instead has free spinning rollers.
 
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I tried Richard, he's not getting it.

It requires torque to initiate and maintain rotation. One tire is spinning quickly, a manifestation of torque. a differential does not multiply torque, your equation makes no sense. By your reasoning, any wheel with traction sends 5x the amount of torque applied to the other wheel, which would in turn send 5x back the other way, which would then send 5x back the other way, and then 5x back the other way. The practical outcome of that lunacy is that any torque input into a Torsen diff quickly approaches infinity output, tires disintegrate and the rotational force quickly overpowers the rotational force of the globe gravity would then implode on itself becoming a black hole collapsing all space and time..

An open differential distributes power through the path of least resistance. This appears as an even distribution until torque>traction.

A locker distributes power evenly.

A clutch based limited slip differential also sends power through the path of least resistance, but forces some in the other direction.

A torsen based limited slip differential also sends power through the path of least resistance, but forces a percentage to the high traction side that is relative to the amount of resistance felt by the low traction side. More resistance felt = more power pushed to the high traction side. The equation can never equal more than 1 i.e. if half of the torque is going in each direction .5+.5=1 If one wheel is in the air, it is the "1" and as you apply the brake you subtract from that 1 and begin applying torque to the other side.

The irony here is that you simply don't understand the reality of what I'm trying to explain, yet you two are taking this high and mighty "he's an idiot, he this, he that" attitude. I've tried to explain without being condescending, and in return I'm getting insulted.

Being a gearhead doesn't necessarily mean you understand the physics of automotive systems, and this is the perfect illustration. Go and read up on differentials, please. Everything you read about Torsens will talk about torque bias, and every professional explanation of open differential action will explain equal torque split. There is one thing I forgot to tell you--- the Torsen's torque multiplication is toward the slower spinning wheel. That is explained in the article I posted. So there is no implosion of the space time continuum.

If you don't believe me, ask an automotive engineer.
 
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BTW here is an explanation by the same Jerry Bransford guy, specifically about torque distribution. If you read it, I think you will understand, then the Torsen torque bias will make sense.

http://www.jeepforum.com/forum/f27/4x4-torque-answers-130169/
Since questions about the frequently misunderstood subject of 4x4 and how torque is delivered to the wheels is a common one, I thought I'd make this a "Sticky". :)

All four wheels are driven and pulling equally if all four tires have equal traction. If all four wheels have equally good traction, then all four will pull equally well. So even if your Jeep does not have a locker or limited slip differential, each of the four wheels will receive approximately 25% each of the torque from the engine, when traction is equal under each of the four tires.

A standard open differential always (always!) splits the torque 50:50 between both sides. The problem is that when one tire starts spinning due to poor traction, that reduces the amount of torque seen by either side by an exactly equal amount. So if one side starts slipping, the other side's power (torque) is reduced by a corresponding amount which usually means insufficient power to keep you moving. The engine, via the axle, can deliver no more torque than the wheel with the least amount of traction can accept before slipping.

So the short answer is 'yes', all four tires pull in 4wd. But the moment one side starts slipping, neither side gets enough power to keep you moving even if the non-slipping side still has traction.

Why does the power (torque) get reduced to both sides and not just the side that is slipping? Again, because the differential will always (!) split whatever torque it sees 50:50 to both sides (few people know that). Why does the amount of torque get reduced at all? Because the engine only develops torque when it is working into a resistance. Run an engine with a torque meter connected to it and watch what happens to the amount of produced torque as resistance to the engine is varied. When the engine is working into zero resistance and just running free, it produces nearly zero torque. Apply a braking action to the engine and the amount of torque will increase in direct proportion to the amount of resistance it is seeing. The more braking force applied to the output shaft, the more torque the engine produces.

So because the engine only produces torque when seeing resistance, a spinning tire reduces the amount of power (torque) the engine produces because the engine is only seeing the nearly zero amount of resistance produced by the spinning tire. This works the same way with 2wd and 4wd, it's just that with 4wd, you have more of a chance that at least one of the two axles will have enough traction to allow the engine to produce enough torque to keep you moving. Which is why you get stuck in the first place... when one side is spinning, there is insufficent power being delivered to the other side to get or keep you moving.

And all of this is why a limited slip differential (LSD) can help since it helps to "couple" (via a clutch or a gear-based device) the resistance the side with good traction is seeing to the other side with poor traction, increasing the amount of resistance seen by the engine... thereby increasing the amount of torque delivered to both sides. Which is why the simple technique of stepping on the brakes a little when a tire is spinning can often get you unstuck... because it too increases the resistance the engine is working into which increases the amount of torque that is delivered to the wheels. Or if a rear tire is spinning constantly, pulling the parking brake up a couple clicks can often help by acting as a poor-man's limited slip differential. Pulling the parking brake up a few clicks when you already have a limited slilp differential will help improve its operation. A key drawback to a LSD whether it be clutch or gear based is that when one tire is spinning up in the air, it doesn't usually help create enough torque for the other tire still on the ground to keep you moving. Without you helping it by stepping on the brakes or pulling the parking brake handle up if it's a rear tire that is spinning, a LSD just doesn't do much for you. On flat terrain they can help a lot, but on uneven terrain when both tires can't always be on the ground, a LSD is not very helpful at all. That's a situation where a locker reigns supreme over a LSD.

What does a locker do? It mechanically locks the left and right wheels together so when one turns, they must both turn at the same speed. This arrangement prevents one side from spinning uselessly while the other side does nothing. Automatic lockers are always locked but they unlock automatically to allow the outside wheel in a turn to 'ratchet' faster as the outside tire must do through the turn. Once the turn is completed, the locker re-locks both sides together. A manual locker is one that doesn't lock the left and right sides together until it is actuated either via a push-button or lever. An ARB Air Locker is air-pressure actuated, others like the Detroit Electrac are electrically actuated. The Ox Locker is actuated via a lever and cable. Manual lockers have an advantage for on-road driving in that when unlocked, they act like an "open" axle... i.e. one without a locker or limited slip differential... which means it drives like an unlocked vehicle until the locker is activated via the push-button or lever.

Part-Time and Full-Time 4x4 systems...

A part-time 4x4 system called Commandtrac is in all Wranglers together with low-end Cherokees and Liberties. A part-time 4x4 system locks the front and rear driveshafts together inside the transfer case so they drive the front and rear axles together in lock step. Because they are locked together, the front and rear tires must rotate at the exact same rpms. However, the front tires must rotate faster than the rear tires during any turn so a part-time system fights that... which makes a part-time system inappropriate on a paved road because the high level of traction on a paved road prevents the tires from slipping which would otherwise allow the front and rear tires to grudgingly rotate at different rpms. Offroad this is not a problem since the poor traction of an offroad trail allows the tires to slip as needed. But when they try to slip/rotate at different rpms on a high-traction surface, the entire drivetrain is stressed which is bad for it. This problem is called "wind-up".

In reality however, the front and rear axles really don't even turn exactly the same RPMs when you're in 4wd so you still get "wind-up" if you drove in 4wd on the street even if you drove in a perfectly straight line. Why? Because 1) you can't drive in a perfectly straight line and 2) the front and rear axle ratios are usually .01 different from each other. Like a 3.73 and 3.74, 4.10/4.11, etc.. Why the .01 ratio difference between the front and rear axles? Because the front and rear axles usually have different ring gear diameters which makes it nearly impossible for the gear manufacturers to economically make the front and rear axle ratios exactly the same. And no, they are not made .01 different on purpose to make the front or rear pull more when in 4wd, that is an old wive's tale.

Finally, a full-time 4x4 system like Selectrac is available on Grand Cherokees, Cherokees and Libertys couples the front and rear axles together, but they are not mechanically locked together like they are with a part-time 4wd system. The front-to-rear axle coupling can be done via either a differential like the Selectrac system uses (just just like what is in the center of an "open" axle) or a fluid (viscous) coupler. The benefit to a full-time 4wd system is that because the front and rear axles are not mechanically locked together, the front and rear tires/axles can rotate at different rpms from each other. This allows a vehicle with a full-time 4wd system to drive in 4wd "full time" on a paved road without problem since there is no 'wind-up' problem to harm the drivetrain. You cannot get a full-time 4x4 system in a Wrangler from the factory.

Hope this helps! :)
 
I'm bagging on you because all you can do is repeat what you've heard. If you understood it we'd be able to find common understanding and move on. It's basically semantics, since the result of both explanations is the same. If I was talking to this Jerry guy, since I disagree with part of his explanation, he and I could come to terms with the semantics and reach common ground. All you seem to be able to do is quote someone else saying what you're saying. And don't get on your high horse, you have clearly stated that we don't know what we're talking about and if we listened to you we would "get" it. So, yes, I'm bagging on you, while you are talking down to us.......I don't like to be talked down to.

Jerry, in the first two paragraphs, refers to traction and power to the ground, which is exactly the point I was making. I agree with him there. I don't agree with him about the 50/50 split of an open diff. I understand why he's explaning it that way, I just don't agree that it is the best, or most accurate way of explaining what's happening, and I would tell him that. His point about an engine only producing enough power to overcome the resistance it is working against is generally true, but there are too many variables in the on and off traction conditions we wheel in to make that universally true, or to unconditionally apply it to this disussion. The assumption is that the throttle position changes instantly as resistance changes, which in a perfect traction condition it would, like on the street. But, too many times offroad the traction comes and goes much faster than a driver changes throttle position, and yes, the engine will account for some of that change in resistance (or load) by changing rpm, but it's not a perfect world. If you've ever seen a rock buggy on a full throttle assault of an obstacle bouncing with all four tires off the ground you'd get my point. According to yours and Jerry's explanation, since all four tires are now off the ground, even though the throttle is to the floor, the engine all of a suden isn't putting out any power. Tell that to the axles as the tires come back into contact with the ground, possibly over and over again. But, like I said, much of it is semantics, and I have no doubt that Jerry and I would quickly come to an understanding.

BTW, he never says anything about a Torsen diff multiplying power. A diff can't produce or multiply power, it can only channel the power that it receives. A limited slip diff uses clutches or gears to resist the two axles turning at different speeds. This resistance is what channels part of the power to both axles regardless of the traction, or resistance. In a clutch LSD, anytime one axle spins faster or slower than the other the clutches slip (including going around a corner). In low traction situations the friction of the clutches routes the amount of power to the low traction wheel that is required to overcome the friction of the clutches. In a gear LSD like the Truetrac there are small pinion gears that sit in the carrier case, and when power is applied in a low traction situation pressure is applied to those gears pushing them against the case and resisting the two axles turning at different speeds. Since this is sensitive to the power applied, more pressure can be exerted than what there would likely be with clutches so it can provide more power to the wheel with less traction, or more "limited" slip. But, it can never multiply the power going into the diff from the driveline, (which you are saying, but I don't think you really mean) only increase it's resistance to "slip" as power is applied to each axle as traction/resistance changes.

Really, it all doesn't matter, because the end result of what we're all saying is the same thing, it's just which is a clearer explanation of what is going on.
 
The issues I see with a Truetrac are that it's expensive, still a full carrier diff, and they really aren't rated for use in the rear. We've seen them break in rear axle applications.

Where you have seen the rear true trac fail, what was the tire size, and did the breakage occur when modulating the e-brake? They are made for the c8.25 and IIRC the dana 35, so I would have to suspect the fact they aren't rated for rear applications would be conditional.
 
BTW, I appreciate the effort you guys (Goat, wink, and goodburbon) put into expressing yourselves clearly with all appropriate attention to sentence structure, grammar, diction, and logical flow.

Makes for easy reading!
 
I'm bagging on you because all you can do is repeat what you've heard. If you understood it we'd be able to find common understanding and move on. It's basically semantics, since the result of both explanations is the same. If I was talking to this Jerry guy, since I disagree with part of his explanation, he and I could come to terms with the semantics and reach common ground. All you seem to be able to do is quote someone else saying what you're saying. And don't get on your high horse, you have clearly stated that we don't know what we're talking about and if we listened to you we would "get" it. So, yes, I'm bagging on you, while you are talking down to us.......I don't like to be talked down to.

You talked down to me first buddy.

I'm not sure why you're disagreeing with the facts of differential operation. You're basing your opinion on what happens with an open differential, which we all understand, but you are trying to deny the physics behind it.
I'm NOT repeating what I've read-- I actually understand what is going on. I didn't always understand. I had to have it explained to me too.



Jerry, in the first two paragraphs, refers to traction and power to the ground, which is exactly the point I was making. I agree with him there. I don't agree with him about the 50/50 split of an open diff. I understand why he's explaning it that way, I just don't agree that it is the best, or most accurate way of explaining what's happening, and I would tell him that. His point about an engine only producing enough power to overcome the resistance it is working against is generally true, but there are too many variables in the on and off traction conditions we wheel in to make that universally true, or to unconditionally apply it to this disussion. The assumption is that the throttle position changes instantly as resistance changes, which in a perfect traction condition it would, like on the street. But, too many times offroad the traction comes and goes much faster than a driver changes throttle position, and yes, the engine will account for some of that change in resistance (or load) by changing rpm, but it's not a perfect world. If you've ever seen a rock buggy on a full throttle assault of an obstacle bouncing with all four tires off the ground you'd get my point. According to yours and Jerry's explanation, since all four tires are now off the ground, even though the throttle is to the floor, the engine all of a suden isn't putting out any power. Tell that to the axles as the tires come back into contact with the ground, possibly over and over again. But, like I said, much of it is semantics, and I have no doubt that Jerry and I would quickly come to an understanding.
You're mixing up torque travelling through the drivetrain to the ground with the momentum of the drivetrain. If you put your vehicle up on a lift so that none of the wheels have traction, and floor it, the engine won't make maximum horsepower or torque. It will rev up to max rpm, but it will be putting out very little torque. The most torque will be what it makes to accelerate the driveline, and that energy is stored as momentum. If you drop the vehicle off the lift somehow, that momentum will convert to forward motion. And as resistance is applied to the wheels, the engine will create more torque. You'll have to increase throttle or else it will bog down.

You ignored my analogy of the free spinning nut, so here's another one. Imagine a 10lb dumbell. If you are holding one up, how much force can you exert? 10lb. Now imagine a 30lb dumbell. How much force can you exert to hold it up? 30lb. Now imagine no dumbell. How much force? Only the amount it takes to hold up your arm. And it doesn't matter how strong you are.

Here's another analogy. Turn a bike upside down and spin the pedal with your hand. Have a friend apply the brake. What happens to the force you felt at your hand? Does it stay the same or does it increase with the resistance?

The point, which I've been saying, is that the force you can apply is always exactly the same as the resistance force. Therefore it's impossible for a lifted wheel to get more than a very small amount of torque.


BTW, he never says anything about a Torsen diff multiplying power. A diff can't produce or multiply power, it can only channel the power that it receives. A limited slip diff uses clutches or gears to resist the two axles turning at different speeds. This resistance is what channels part of the power to both axles regardless of the traction, or resistance. In a clutch LSD, anytime one axle spins faster or slower than the other the clutches slip (including going around a corner). In low traction situations the friction of the clutches routes the amount of power to the low traction wheel that is required to overcome the friction of the clutches. In a gear LSD like the Truetrac there are small pinion gears that sit in the carrier case, and when power is applied in a low traction situation pressure is applied to those gears pushing them against the case and resisting the two axles turning at different speeds. Since this is sensitive to the power applied, more pressure can be exerted than what there would likely be with clutches so it can provide more power to the wheel with less traction, or more "limited" slip. But, it can never multiply the power going into the diff from the driveline, (which you are saying, but I don't think you really mean) only increase it's resistance to "slip" as power is applied to each axle as traction/resistance changes.

Really, it all doesn't matter, because the end result of what we're all saying is the same thing, it's just which is a clearer explanation of what is going on.
You misunderstood what I said about torque multiplication. The torque that the torsen multiplies is the resistance to the wheel with less traction/ more speed. That's why I'm trying to explain an open diff's equal torque split. An open diff is 1:1 torque bias. A Torsen is 1:3, or 1:4, or 1:5

Remember what I said about the cutting brake? If you apply 100lbft torque to a lifted wheel using a cutting brake, 100lbft goes to the wheel with traction. Do you dispute that? I wouldn't think so. (maybe that makes the equal torque split more clear)

Now with a torsen, if you apply 100lbft with the cutting brake, 500lbft goes to the wheel with traction.

Since most vehicles don't have cutting brakes, let's look at what happens if you use the handbrake or regular brakes.
You apply 100lbft to both wheels, left and right of the Torsen diff. The wheel with traction gets the same 500lbft, but 100 of that has to overcome the force of the brake. So it gets 400lbft (max) to move the vehicle. No matter how much gas you give the engine, that 400lbft will be the maximum that wheel can get.
 
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You ignored my analogy of the free spinning nut, so here's another one. Imagine a 10lb dumbell. If you are holding one up, how much force can you exert? 10lb. Now imagine a 30lb dumbell. How much force can you exert to hold it up? 30lb. Now imagine no dumbell. How much force? Only the amount it takes to hold up your arm. And it doesn't matter how strong you are.

If you apply more force upward than gravity is applying downward, the dumbell moves upward. Apply 50# to a 10# bell and it's called "throwing it". Bad analogy.



I think I understand what you're trying to say, but let's put theoretical numbers to it to test it.

engine output is relative, right?

Brake is applied 10lbft, torsen forces 5x10 or 50lbft to opposite wheel. 10 is used to overcome braking force leaving 40 to initiate and maintain forward momentum.

So you've got 10lbft on one wheel and 50lbft on the other, engine output is 60lbft.

Seems simple enough right? Lets try some more numbers.

Engine output still relative, right?

let's say I put a 40 lbft load with the brakes. torsen then forces 5x40 or 200ftlb to the other wheel 40 is used to overcome the brakes and you're left with 160 to move forward right?
but what if my engine output was only 100 ftlbs max. Evenly distributed the 100 is easily enough to overcome the 40# on each wheel since 40+40 is only 80. How can you explain that your theory just created a 200ftlb force from a <100ftlb input?
 
I really want to move on here, this is getting boring. I'm trying to throw out an olive branch in saying that it's mostly a matter of semantics, and how each of us is explaining it, and I think my explanation is better and more accurate than yours. However, you keep insisting that we don't agree with you because we don't understand the subject matter. And the comment about you needed it expained to you too is insulting.

You are getting better about explaining the Torsen style diff. And yes, we are talking about the power transmitted through the drivetrain......that's what eventually gets to the wheels. A limited slip has the ability to transfer a certain amount of the available power to the wheel with the least resistance/traction, and it does this by essentially binding itself up. A clutch LSD is generally accepted that it can transfer 30% of available power to the wheel with most resistance (when the clutches are in good shape). A gear LSD like the Truetrac can transfer a higher percentage, like 40-50% of available power........like you are saying in a 1:3, 1:4, or 1:5 ratio. But, it isn't really a ratio like most would understand, because the transfer is created by resistance inside the diff (resistance to the axles turning at different speeds) so it is more easily explained as a percentage of the power that is applied through the diff. This is why I so strongly disagreed with you saying the diff multiplies torque up to 5 times. I understand what you were trying to say, but it just doesn't explain what's going on very well, in fact is misleading.

By applying a cutting brake to one wheel, you are changing the resistance of that wheel and so changing the power distribution, exactly the same as if that wheel actually had traction. You are sort of faking out the diff and forcing it to send more power to the opposite, higher traction wheel. You are not sending 5 times the power to that wheel, you are simply changing the percentage of available power that is going to each wheel, increasing the percentage to the wheel with traction. I understand that you are also trying to say that as the traction/resistance increases that the motor is also now putting out more power and so now the power is increased to the wheels, and I agree that this is a factor, but I disagree that this applies automatically, and I disagree that it helps to explain how the differential works. The diff doesn't control how much power it's getting, the driver does with how much throttle is being applied and how many rpm's the engine is turning.

Using your weight lifting analogy to make my point, a buggy at full throttle bouncing with all the tires off the ground is like holding that 30 lb weight, but with your muscles flexing/tightening so more weight can be added without exerting more strength, energy, or whatever you want to call it. And you can most certainly tighten your muscles to exert more force than required, like I do when I tighten my stomach muscles and let my grandkids wail on my stomach. If I waited for the resistance to then apply more force it's too late, the force must be there before the resistance if I don't want to be hurt. Just like I'm saying happens with engine power in various changing traction situations in the real world. Engine power is not just rated in torque, it is also rated in horsepower, which is a combination of torque and speed, and an engine running at 4000 rpm with minimal resistance still has speed and is doing work and that power is being applied through that diff. I believe that an engine can have a reserve of power that is more than the work that it is doing at that time. I also understand your point that an engine will put out the power required to meet the resistance that is applied to it, but I think that is a basic principle in an ideal setting and not true unconditionaly.

I hope there is some redeming value to a few who might still be reading all of this. I don't see this discussion as a matter of who's right, I see it as a matter of a better explanation. Technically, most of what has been said in this discussion is correct from a particular perspective.

Happy wheeling, :)
 
I think I understand what you're trying to say, but let's put theoretical numbers to it to test it.

engine output is relative, right?

Brake is applied 10lbft, torsen forces 5x10 or 50lbft to opposite wheel. 10 is used to overcome braking force leaving 40 to initiate and maintain forward momentum.

So you've got 10lbft on one wheel and 50lbft on the other, engine output is 60lbft.

Seems simple enough right? Lets try some more numbers.

Engine output still relative, right?

let's say I put a 40 lbft load with the brakes. torsen then forces 5x40 or 200ftlb to the other wheel 40 is used to overcome the brakes and you're left with 160 to move forward right?
but what if my engine output was only 100 ftlbs max. Evenly distributed the 100 is easily enough to overcome the 40# on each wheel since 40+40 is only 80. How can you explain that your theory just created a 200ftlb force from a <100ftlb input?

You're right, I didn't explain what I meant by the torque multiplication until the last post.

The multiplication only relates to the faster moving side and the slower moving side of the diff. As I understand, the ratio is constant. So the torque that the transmission actually sends into the diff gets split by that ratio. If you apply 40lbft, which would cause 200lbft, and the transmission couldn't give you 240lbft, I would think the engine would stall.

You just wouldn't be able to apply enough brake to get that 200ftlb to the wheel with traction. If the transmission can deliver 100lbft max for a given amount of throttle, the most you could brake would be 16lbft before stalling. So about 80lbft would get to the wheel with traction, and 64lbft would move the vehicle forward.


BTW with the dumbell analogy, throwing it would be equivalent to revving up and storing momentum in the drivetrain. A better analogy would be that you could apply a lot of force and spin it like a top... but once it's spinning it doesn't take much force to keep turning it.
 
I don't need to quote my source, because equal torque split is how a differential works, and multiplying torque is how a Torsen LSD works. That's why made it a point to say that the TrueTrac is a Torsen not a clutch LSD.
Here is a source though http://www.jeepforum.com/forum/f27/

You're thinking about the open diff wrong. Each side gets equal torque. If one wheel is in the air, it gets almost 0 ft lbs-- let's say 0.1 lb ft to overcome air resistance. That means 0.1 lb ft can go to the other side. No matter how much you rev your engine, the torque won't be greater than that. If you apply a cutting brake to the wheel in the air, the other wheel will get an amount of torque equal to the force you apply with the cutting brake.

A torsen is different in that it can send several times the torque to the wheel with traction. If one wheel is totally in the air, the other one won't get torque either (because 0 * 5 = 0), unless you apply brake.
The torque applied to the differential (an open one) will always take the path of least resistance first. That means the wheel with no traction gets the torque. It's true you can redirect some torque to the wheel with traction but it will never exceed the torque of the wheel with no traction. If you apply the brakes, it's the same as if the wheels started to get traction, the torque will still go to the wheel with the least resistance to turning.
 
Goatman, my point with saying that I had to have it explained is to say that I didn't get the unintuitive equal torque split at first either.

I didn't insult anybody until I was insulted. You accused me of regurgitating what I read instead of coming up with it on my own. Of course I didn't come up with it, because this is stuff that has been known since the differential was invented. But I learned it and understood it.

I'll reply to the rest of your post later but yeah it does seem like a semantics argument now because we agree on how these things work.
 
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