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Does anybody know how the power is split up between the front and rear?

as there is no resistance I would say there is little 'torque' created.
torque was defined as a measure of force.
very little force is generated freewheeling.
how much torque is generated with tranny in Nuetral? same idea.

I still go back to the 'torque' as being the wrong terminology to use.
 
I see what you are saying, although the motor is still producing torque when the tranny is in nuetral, it is just not transmitted down to the wheels (and in essence is wasted). When the tranny is in gear, that torque is transmitted to the wheels via the t-case and driveshafts.

Using the term torque as the ability to perform work.
Again I'm not saying who is wrong or right. I would just like a better understanding, and some of these examples (even MY earlier ones, don't seem to be adding up).
 
If not torque, then what?

To me it seems like the right term. If torque is a measure of rotational force - and force is the ability to do work - then even in the freewheeling scenarios, work is being done. We just have to remember that in terms of simply force and work - time doesn't matter. The same amount of work is done moving a tire one rotation on pavement regardless of whether it's 1 mph or 100. However POWER reflects how fast work is done.

With the theoretically frictionless ice, it still requires force to move the tires from a standstill, so it still requires a measure of torque, just much less than if it was on pavement.

It gets kind of confusing to me some times since torque and work have similar units, I always have to remind me that torque refers to the rotational force applied to a lever.

I love these discussions, this is where the best of CK5 comes out
 
people keep getting hung up on the "engine CAN deliver 300#feet of torque" part.
free wheeling or spinning madly the motor will not generate 300# torque.
it requires a load to work against to put up the numbers.
you are correct that it would require torque to spin the driveline.
the discussion isnt really about torque IMO.
it seems to be related in the beginning to powerflow in the typical transfercases in these old trucks.
in a 205 the front to rear sees 1:1 front:rear
as would a 203 in Loc
in normal mode the 203 would be fully variable with no fixed number.

though rereading the title and the post it almost seems like 2 questions.
the title suggests front to rear, the body question seems to ask the gearing advantage hi to low
so the title answer is 1:1 in Loc
and the body answer would be torque X low range gear ratio - frictional losses
 
I agree completely - available torque at a certain rpm and actual torque can be worlds apart. If it were otherwise, we wouldn't need chassis dynos, we could just hang the rear end in the air
 
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I still go back to the 'torque' as being the wrong terminology to use.

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Torque is easier for me to think about but is not the same as power. When someone states that there is a 40/60 front/rear ratio torque is what is ment (not power).

Power (energy) can never be created or destroied but can be changed from one form to another. That is "power in = power out"

Torque can be changed. Torque in does not have to equal torque out but is dependant on gear reduction.

HP = (RPM * Torque)/5252

If you want to talk about power
A 203 (not in loc)would have equal torque to each axle but not equal rpm (if one axle is spinning). The axle that is spinning would get more power sent to it (but the same ammount of torque) then the axle with more traction.

A 205 (or any other locking t-case) would have equal RPM at each axle. It seams that the torque to each axle is what is under question the most. Lets say you are driving up a rock slab (like in Moab...lots of traction). As everyone agruees if you lifted the front tires in the air you would have all of the torque going to the rear tires and none to the front (assuming no acceleration). But where would it transfer from say 50:50 to 0:100. If the front has 1000# on it and the rear has 5000# on it would there be equal torque to each axle? You would have to know the traction at each tire to figure that out (or at least the coeifficinet of friction). If you are driving up a dirt hill you will hit the point where you loose 50:50 alot sooner before the front end lifts into the air.

So in either case the front and rear axle do not have equal power...which to me makes it less usfull to talk about.
 
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When someone states that there is a 40/60 front/rear ratio

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...they are not discussing our stuff. so I file that under irrelavent until I can build an AWD ralley car.

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So in either case the front and rear axle do not have equal power

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they do have the potential/access to equal power.
missed the wording there before it hit the keyboard...
they dont see equal power but either end can have full power, or did I loss the whole train of thought again LOL
 
OK, so I am getting the terminology thing now. Torque isnt the term to use. At least in my mind, the question could be answered by saying that the torque outputs to each end would be equal with all things being equal at both ends. EG, both the front and rear tires are on exactly the same surface. They are exactly the same tires, on exactly the same wheels, with the same gears. The output of the t-case will be the same at the front as the rear. Right?

Now, the next question in my mind is this: Regardless of where the wheels are, the traction of whatever you are on at the time, the front driveshaft and the rear driveshaft have to spin at the same speed. In fact the front diff pinion and the rear diff pinion will turn at the same speed all the time. Is that right? I am talking about a 205 case. If this means that one wheel (open diff) has to spin out of control in mud while the other wheel is stopped on rock, that is what will happen. In a locker equipped diff, both tires must spin at the same speed or not at all. Am I totally off topic here now? I hope not.

Mike
 
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ok here is what is getting me hung up. "A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum." There is no mention of friction.

here is more "Notice that the torque units contain a distance and a force. To calculate the torque, you just multiply the force by the distance from the center."

I want to see a formula that includes the resistance being figured in.


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Gravity, not friction. The one pound weight is always pulling on the weightless bar due to gravity, thats your resistance or opposing force.
 
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If you had a 203 (not in loc) with lockers front and rear you would have the same torque going to the front and rear axles (almost none). The fronts would spin like mad and the rears would slowly move.

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"almost none"....I know mj agreed, but where does the torque go assuming the motor is twisting. Most of the torque is being applied to the axle that is slipping. The torque is "wasted" and the reason for going to fluid couplers in the newer designs. Unless torque vanishes into thin air, where does it go as energy is neither created nor destroyed with your explanation?


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Like MJ said your getting hung up on numbers. Your motor may put initially out 300 lb/ft of torque to over come the rotational inertia of the drivetrain, but once its up to speed a couple of things can happen, you keep your foot in it, motor output torque stays up, causing the rpms go through the roof and you blow your motor -or- you back way off the gas, motor output torque drops to "almost none" and so does the torque on both the driveshafts.

So torque is not wasted or destroyed, once the wheels are spinning only enough is generated to over come any friction (traction), that is assuming you let up on the gas to save the motor. Your motor cannot generate 300 lb/ft of torque unless it is pushing against something that can resist 300 lb/ft of torque.

They have already nailed it, for our tcases locked in 4 wheel it just doesn't matter: there is no torque split or biasing or whatever you want to call it, the driveshafts turn at the same speed. How much torque is on either driveshaft changes from moment to moment and depends on tire traction and tire rotational speed (like turning a corner when the fronts want to go a little faster or mismatched tire sizes).

Now if you really want a brain ache, lets get an engineer to explain the how those rally cars on Speed Channel work!
 
Good point. For those still confused about where the torque "goes".

Another way to think about it is this. You have a torque wrench and you're turning a bolt. Just because you know that you're able to exert 150 foot pounds of torque, doesn't mean that you do when the nut is spinning freely - it's just taking you a lot less effort (i.e. energy). Only once the bolt is tight are you able to exert the full 150 ft lbs. A spinning wheel is like the bolt before it's tight. Likewise, undoing a bolt is like the wheels coming off the ground. Once the bolt loosens you need a lot less ofrce (torque) to spin it - or if you apply the same amount of force you spin it much faster (like an air ratchet or the like).

Power is just an expression of how fast work can get done - the amount of work being done divided by the time it takes to do so - so with the viscous couplers, when one drive axle breaks free more power is sent to the axle with traction since it takes more force to turn those tires. Since it takes more force, more work is being done at that end over the same period of time and hence more power.

For the creation vs. destryed idea - remember that the energy we're talking about is coming from burning gasoline, so as was mentioned, if you kept applying the same amount of gas (energy) in neutral or on ice, you'd spin things faster since the same amount of energy is translated into rotational motion - so, you let off the gas.

I'm beginning to digress, so I'll quit with one last thought. Torque does not 'require' friction for a given value, it's like straight force. If you try to twist an anchored wooden pole with a lever, even after you've twisted it some and stopped, you still need to apply a force to keep your lever from springing back at you. No movemennt is involved, so there's no friction, bbut you're still applying torque.
 
Similarly to as above, it takes 50lbs of force to lift a 50lb weight. You can not lift a 50lb weight with 100lbs of force. This is where your resistance comes in. An applied force can not be greater then the resistance of what it is applied to. You can apply as much power as you want though to lift that same weight. Power = Force X Velocity. The more power or energy you use to lift a weight the faster it moves because the Force is constant and the force comes from it's resistance. It's the same way with a differential. The torque is equal to both sides of a differential but the power is what is different and that is because of the speeds are different. When a differential is locked the speeds are the same but the torques ain't because the resistances are different. Either way power is different on both sides of a differential. Locking it just changes the variable from speed to torque.

Reasons for the front driveshaft being smaller then the rear: Used a lot less then the rear therefore less fatigue stress, Generally there isn't as much force on the front as there is on the rear, and the rear shaft is a lot longer and needs to be bigger because of that.
 
I see that now. Thanks /forums/images/graemlins/thumb.gif
Also a fun/discussional/informative link explaining HP/Torque: http://member.rivernet.com.au/btaylor/BMWText/technical/EnginePower.html

Torque is nothing without work being done (actually it is work done). A free spinning motor is producing a lot less torque at 4000 rpm, than the same one propelling a vehicle at 4000 rpm, although since it is the same motor, in both instances there is the same potential to do work. The instance of a free spinning motor, most of the energy produced is potential energy (slight kinetic), and in the instance of propelling a vehicle, the majority of potential energy is converted to kinetic energy.

As mj stated (and I am beginning to understand in a different light), the driveshafts of a fulltime case have the potential of equal power at ALL times (50/50 or 100/100 depending on how you look at it), but it is the forces at the end of the shaft that determine how much of the available torque is "used" at each end. And assuming no slip, it was my fault for saying the rest is wasted, it is just potential energy that is not used.

I guess at first I didn't see the forest for the trees!

Thanks guys!
 
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