Front Lockers On 3rd gen

The issue with your analysis is that no torque is being transmitted through the wheel with no traction to the ground so all of that torque is being transmitted through the wheel with traction to the ground. The only torque being transmitted to the wheel that is spinning is the necessary torque to spin that wheel, the rest is going to the ground through the hooked up wheel.

So you are right, its not 100% to the wheel with traction, but its not equal 50/50 when one wheel has traction and the other doesnt.

 

Thanks for clearing it up. I assumed that a locker, when locked, sent the tq 50/50 no matter what. Know I see it's just wheel speed but it can still transfer tq.

This video cleared it up in my head.

https://www.youtube.com/watch?v=_HOa0aRZYpw

 

As absurd as it sounds that's how it works. Torque at a wheel either must be reacted externally or else the wheel must accelerate. But with a locker a single wheel can't accelerate alone because speeds must match.

In your example if a locked vehicle has a wheel in the air and it's driving forward from the other tire that isn't slipping then that tractive wheel will get all the torque minus what's required to overcome friction and spin the wheel in the air. This would be about a 99:1 bias ratio situation.

To get 100% bias, wheels must be stationary while one wheel has zero traction. Think of loading up the drivetrain in a stalled situation with one wheel in the air.

In situations where both wheels have traction or are slipping then the bias is some ratio between 0% and 100% and again becomes a deformations problem not easily determined.

The point is a locker does not split torque 50/50. An open diff does.

 

I would like to install a front locker to my third gen, but i don't wheel hard enough or even often enough to justify it..

 

I get your point but tractive torque splits are really an academic issue. The point of a locking diff is to get both wheels on an axle to spin equally so you have the best chance of your vehicle's rotational force turning in to traction force. It's functionally irrelevant how that traction force is achieved. But your statement "It can bias torque up to 100% at one wheel" is factually incorrect since the diff or locking diff has no control of which wheel receives the rotational force (though an active LSD does). What you're saying above is traction dictates the torque to the ground which I agree with. But show me a vehicle with front and rear locking diffs where only 1 tire is spinning and I'll send you a dollar.

 

So I can assume you no longer stand behind this statement?

Yes, it's *possible* for a locker to bias 100%. It may seen academic but it's the fundamental reason a locker cannot be beat in offroad ability by any other device.

 

I'm not sure why this has to be so complicated and perhaps may be related to the general use or misuse of "torque"

The engine produces energy which is often measured in torque or Newtons, this energy is sent down the driveline through gears acting as torque multipliers to the differential and wheels. When locked that energy is sent to both wheels equally the left tire is receiving the same amount of energy as the right tire regardless if one has traction or not as the resistance or force on one tire affects the other. Through the use of Spider gears if the resistance of one tire is greater than the other energy is transferred to the path of least resistance.

 

I guess that's where I'm going. If you put your truck's rear axle on jack stands and lock the rear diff the rotational force would be sent to both wheels equally. Do those wheels have no calculated torque value until one or both make contact with a surface? Maybe the difference between drive torque and traction torque? Again, a kind of silly academic argument. If 100% traction torque was applied to one wheel and the other has 0% and not moving then neither are moving and you're stuck anyway.

 

He does a good job. He got a few minor things wrong, but overall he understands what's going on.

 

If you put the truck in 4wd and lock F/R diffs, than lift the truck up in such a way that only one wheel touches the ground and slightly press the throttle (not enough to slip and spin the tire touching the ground), 100% of the torque will go to the wheel touching to the ground even if no wheel is spinning. The moment wheel touching to the ground starts to slip and spin, the remain 3 wheel will spin at the same speed as that one. In other words, all 4 tires can only spin as fast as the slowest tire.

 

And here lies the problem, engine torque is applied to all wheels equally with rotational torque to wheel on the ground

 

I’ve ran a harrop in the front for a year and have had 0 issues. Lockers are awesome.

 

Did you install it yourself? how was the wiring? thanks,

 

No I ordered a built diff from ecgs w/5.29’s I did do the wiring myself it’s super easy switch and relay 2 wires to the diff.

 

No, that's not the way it works. I'm not sure what way of explaining it will get the point across. I'll try to come up with another way.

 

It is whether you like it or not, power is applied through the driveshaft and the differential, it doesn't matter whether wheel is on the ground or not. period. All the energy to turn your tire is coming from the driveshaft.

If I remove my right tire and sit the axle on a stand with the left wheel still touching the ground and spin my drive shaft the torque from the drive shaft goes to both axles equally, the resistance is shared across the entire axle. I can also put a wrench on my right wheel hub and turn it and the torque needed to rotate my right wheel hub is equal to the torque required to rotate the left tire which is touching the ground.

If it takes 150 ft-lbs to rotate my left tire it doesn't matter what the resistance is, 150 ft-lbs is coming out of the driveshaft to turn that tire because you have a single power source, torque isnt split between the tires because you would need two independent power sources for each.

The reason everyone is missing your point is because you don't really understand it.

"If you can't explain it simply, then you don't understand it well enough. " -Albert Einstein

 

It's hard to break someone's intuition but i'll try one more time using your example.

The example is locked axle on stand, left wheel on ground, right tire removed. We're going to make the example completely static to take speed out of the equation to simplify the concept, which I know is confusing you guys. We can measure the torque applied at each wheel/hub by strain gauging the axle shafts. We do this all the time to measure torque. Any torque that reaches the wheel must be carried through the axle shaft. So we strain gauge both axle shafts. Now, for this example, if you go crank on the driveshaft by hand with a huge pipe wrench you can log the strain (twist) in the axle to measure torque. The left axle shaft is trying to twist between the side gear in the carrier and the reaction of the left tire on the ground. It will read some level of torque. The right axle shaft, however, is just chilling with nothing reacting anything at the hub. There is zero twist created in the axle shaft and therefore no torque at the hub to report.

This simplifies our equation. You find that the only two torques are the input and what's reacted at the left wheel. Once you correct through the reduction of the ring and pinion you find that the all the torque at the carrier is applied 100% to the left wheel at the ground.

That's it, this is how it works, and it's basic mechanics and statics.

 

doubled

 

So while locked there is effectively no right or left axle, its just a single axle at that point, it would be no different that if the carrier were welded or spooled the torque is applied equally. Like a similar example above, there are solid axles ATVs with an attached sprocket and chain.

I think here you are considering rotational torque of the tire and its resistance which is why we see axle shafts break with lockers, an axle will break due to resistance of applied torque

As I mentioned earlier there are two considerations. I see what you are saying but it's only half the whole story.

 

In your example there is torque in the freely spinning wheels, but it's not a lot it's just what's required to accelerate the inertia of the wheel and overcome friction past the differential. Torque applied to the wheels must go 1 of 3 places. 1) Overcoming friction in the system such as through brake drag, bearing drag, etc. 2) Reacted externally at the tire's contact patch. 3) Accelerating the wheel and increasing its angular velocity.

Thought experiment. Let's say, for example, torque is sent 50/50 to opposite wheels and one wheel has the grip to handle 100 ft*lbf of torque and the other is in the air. Let's say it would take 200 ft*lbf of torque to get moving so we're not going anywhere yet, just sending torque down the driveline. Let's also say we apply 100 ft*lbf of torque at the differential carrier. So we'd have 50/50 split with 50 ft*lbf at the tire with grip and 50 ft*lbf at the tire in the air. The tire with grip reacts the 50 ft*lbf at the ground, no problem, as there's 100 ft*lbf of grip available. Now, we also have 50 ft*lbf of torque at the tire in the air. Where does that torque go? Well, friction may take 1 ft*lbf, there's no reaction at the contact patch, so we're left with 49 ft*lbf which MUST increase the angular velocity of the wheel. But wait, this won't work because it's locked to the opposite wheel so it cannot physically accelerate. Therefore, there cannot be equal torque at the wheel in the air as it's physically impossible by the design of the locked diff.

Make sense?