On 10/4/2018 at 5:40 AM, Edy said:

I've implemented a setup like that. This one uses that exact tree of 7 differentials chained:

However, in reality these vehicles use a so called H-drive drivetrain. There's a single center differential, then the drive wheels at each side are linked together.

So lucky to be noticed by a famous Unity developer

Definitely gonna drive that around the Nurburgring and set a record

In the vehicle physics asset you're developing, do you also model the torsen differential? Most people seems to steer away from it and go  for clutch type instead. 

By the way, what is your opinion on the importance of modelling different types of differential? Even in a notable sim like Assetto Corsa, there are only a few most importance options like coast, power and preload to mimic the effects of different types of differentials.

And one thing that I've heard around the drifting corner of sim community is that a "sensitive" diff, which is a diff that locks up even with low torque difference between 2 wheels and low power input from the engine, would help break out the rear easier, because when the diff locks, the inner wheel would lose traction and cause oversteer. However, in my understanding, differential lock up is used to prevent oversteer. Locking up means the inner wheel would receive less torque and rotate at the same speed as the outside wheel. An open diff would make oversteer easier since the torque is always 50-50, so when the inner wheel lose traction, it would receive all the torque and keep on spinning without traction. With the racing and drifting experience you have, would you mind clear this confusion up for me? (I saw your drifting video on youtube, and it was awesome)

10 hours ago, Edy said:

Here I'm talking on "standard" differentials with symmetric geometry providing the same torque to both ends. I won't cover asymmetric or epicyclic differentials, which provide different torques to each end.

I guessing an example of asymmetric differential would be the center diff/torque splitter in AWD that split torque front-rear based on requirements (Audi's Quattro 60-40 front-rear)?

10 hours ago, Edy said:

Viscous differential: a constant amount of locking effect, no matter the state of the wheels or the transmission.

I thought Viscous has looking effect based on wheel speed difference? Or maybe I'm getting the wrong idea about locking effect. In my understanding looking effect means the amount of resistance torque from the differential (friction from clutch patch that applied to the spinning wheel with less traction, opposite to the rolling direction of that wheel. Maybe what you're saying about viscous is that the clutch patch get pushed by the liquid but never locked up, and the resistance torque is just constantly applied to the spinning wheel every time a wheel loses traction and spin, without an activation (throttle, engine brake)

10 hours ago, Edy said:

When a differential is fully locked the car tends to go straight. The rear axle becomes a rigid rod that fights against the car steering. However, the locked differential makes easy to break the adherence of both wheels simultaneously, which is useful in some driving styles.

I think I'm slowly getting the (maybe wrong) picture. A locked diff would make the car understeer when cornering when coasting, but once you get on the throttle hard,  both wheels would loose traction because of the locked diff, and also because the outer wheel gets all the torque,. This allow for easier to initiate and maintain a drift. So coasting: understeer, accelerating: oversteer

10 hours ago, Edy said:

An open differential facilitates the steering, but the car might lose traction in some situations due to one wheel slipping when applying throttle or engine brake.

An open diff would make the car corner easily when coasting due to the free rolling motion of each wheel, inner wheel can spin slower to rotate the car around, but when the engine braking is too strong, you can get oversteer. However, when you get on throttle, the inner wheel would get all the power (torque remains 50-50 between each wheel), while the outer wheel still rotates slowly, resulting in understeer. So coasting: oversteer, accelerating: understeer.

11 hours ago, Edy said:

It turns out that most LEGO Technic cars come with differential.

Do all Technic cars have differentials? What kind of diff do they have? I'm thinking of getting one now.

Thank you very much for your help with a huge amount of information

On 14/4/2018 at 7:11 AM, Edy said:

I'm not sure that an open differential would such defined effect on handling. An open diff may reduce the performance due to some wheel getting most of the power. 

I came up with the understanding based on wheel load alone, so it was not at all correct. For open diff, my thinking went like this:

- Car steers into turn, outer wheel gets more load, inner less load. Torque is split 50-50, so the outer wheel accelerates slower, inner accelerates faster, lead to inner wheel losing traction and spinning, while the outer wheel sitting duck. Result: understeer and/or losing power when cornering. But when you go for too much throttle, the outer wheel can also loose traction and the result is oversteer, and this oversteer is hard to control due to unpredictable traction lost between wheels. While in a locked diff it would be more predictable. 

On 14/4/2018 at 7:11 AM, Edy said:

I'd describe the effects of a locked diff in more wide terms:

So the point of LSD is to increase stability by introducing a bit of understeer when the diff is semi-locked (and also helps you getting out of sink holes)? The understeer is due to both wheels spinning at the same angular speed, but the heavy loaded wheel (outside wheel) can receive all the power it needs to propel the car instead of being stolen by the lighter loaded inner wheel. But when you're not careful and when full throttle happy (or the engine decides that the car is moving too fast and brake it heavily), the heavy loaded wheel can also loses traction, resulting in both wheels losing traction

I also heard the term Torque Bias Ratio being tossed around when talking about LSD, like 2:1 or 4:1. The definition is "The amount of torque that can be sent across from one side of the diff to the other." So for example 2:1 means 2 times the torque of one side (wheel) can be tossed to the other side (wheel).

So does that mean for example, car is turning, outer wheel have heavier load/higher wheel reaction torque, so the torque from the engine is spllited unevenly and the outer wheel receives maximum of  2 times the torque as the other wheel?

Or is it that the wheel reaction torque ratio between two wheels can only be maximum to 2:1, and the clutch torque from the diff is applied to the lighter load wheel, opposite to the rotation direction to increase its reaction wheel torque, keeping wheel torque to around 2:1, thus limiting its angular speed?

Here is where I got it from: https://oppositelock.kinja.com/your-differential-and-how-traction-works-1661277563

On 14/4/2018 at 7:11 AM, Edy said:

The Porsche model includes not only differential, also a dual-clutch gearbox with paddle shifters (!!)

Oh boy how far LEGO has evolved. Back in my childhood days the most complicated LEGO I've done was the Creator Jet (LEGO was kind of an Unicorn my hometown, they didn't have a lot around). That 911 is dope though, even the engine is wheel modeled.

On 24/4/2018 at 11:15 PM, Edy said:

I don't think the wheel load alone is a good criteria. It's much more important the grip understood as the ability of the wheel to receive torque without slipping. We may simplify the grip as wheel load multiplied by the coefficient of friction of the tire.

Thanks so much for your help. I took a step back after your last reply and realized my mistake right here. Watching Youtube videos on Assetto Corsa and Project Cars differential setup helped a lot in understanding things visually and in terms of real world application.

Imagining an extreme situation helped me understand better, for example when one wheel is on tarmac and the other is on ice. The one on ice, let's say, has no grip at all, so it keeps on spinning. The other one on tarmac, receiving the same torque percentage, but that torque is not enough to overcome friction reaction torque (or the wheel is being held firmly) and roll the wheel, so the car is sitting duck. This is due to the spider gear setup that allows 2 wheels to spin independently. When one wheel is spinning while the other has traction, the spider gear would spin on its own axis, allowing the spinning wheel to keep on spinning and taking power from the traction wheel. That's why in road cars (even some super cars like the McLaren P1) with open diff, they would have a "poor-man" LSD where the brake of the spinning wheel is applied to transfer power to the dead wheel. The braking reduce the rotation of the spider gear and allow less independent rotation between two wheels

In LSD this is overcame by using clutch packs that connects the housing case to the axle. The housing in turns connected to the ring gear. In the clutch packs are steel plates connected to the case and friction disks connected to side gear (on the axle), so when they are pressed, the axle would rotate with the housing case and the ring gear. Because the spider gears are not connected directly to the housing, but rather through a pair of pressure rings )the rings are connected to the housing). When there is an acceleration from the engine to the ring gear, the case and the pressure rings, the spider gears are "lagged" behind and slid along the pressure rings, pushing the rings away, in turns pushing the clutch packs, creating locking force between axles. When there is no acceleration, the spider gear would sit quietly inside the pressure rings, rotate around their own axis and allow independent rotation between 2 axles

Found this on youtube, very helpful btw:

On 24/4/2018 at 11:15 PM, Edy said:

However, going for too much throttle does not cause the outer wheel to slip in an open differential (again, leaving inertial effects aside). All power is routed to the slipping wheel, leaving the adherent wheel producing the same -reduced- force as the slipping wheel. 

The reason I had that idea is thinking of the engine torque being splitted 50-50 between 2 wheels in a constant function like this: leftTorque = rightTorque = engineTorque * 0.5f. So if the engine torque coming from the engine to the pinion gear increases, the amount of torque coming to both wheel would also increases, thus giving the dead wheel with traction more torque to work with and overcome traction reaction torque and rotate. But I realized it's much more complicated than that.

On 24/4/2018 at 11:15 PM, Edy said:

This means that if one wheel is in the air the LSD differential behaves exactly as an open differential: the adherent wheel receives no torque (only the torque caused by the inertial effects, to be exact).

This is the only part where I still didn't get about LSD, but I think I do now. Say when one wheel is in the air, when there is acceleration coming in, the ring gear instead of "lagging" and pushing the pressure ring, would spin on their own axis and allow the wheel in the air to keep spinning and taking power from the dead wheel. The reason for the ring gears to not "lagging" is because of the very high traction reaction force from the dead wheel holding the wheel (and that axle) back, and the other wheel have no resistance at all, so when the spider gears rotate with the pressure rings, the gears don't receive any resistance from the spinning wheel to make them "lag" behind and push the pressure ring, thus making the spider gears spinning on their own axis, allowing for independent rotation of 2 wheels. This is actually better to understand using a modified LEGO diff with spider gears inside pressure rings, and hold one wheel while spinning the pressure ring. One would see that even though there is acceleration coming into the pressure ring, the resistance from the spider gears is not enough to push the pressuring away, and instead the power makes the spider gears rotate on their own axis

Is the above explanation correct?

On 24/4/2018 at 11:15 PM, Edy said:

However, locked and viscous differentials route the torque in a way so the torque produced by the engine is always transformed, entirely or as a fraction respectively, into tire forces among both wheels.

I realized this now. Locked diff is pretty obvious, engine torque is always going to both wheels since they move as a single unit. Viscous is basically a progressive locking diff based on velocity difference, so the locking force will progressively increase with speed difference until it is high enough to actually lockup the 2 axles.

On 24/4/2018 at 11:15 PM, Edy said:

A bias ratio of 2:1 means that while a wheel is slipping, the adherent wheel will receive twice as torque as the slipping wheel, thus allowing it to produce twice as force. This reduces the performance loss that would be produced in an open differential, as described above. A ratio of 4:1 means up to four times the force of the slipping wheel. 

So an example of implementation: First I would calculate the difference in tire forces between wheels, and figure out clutch locking force by multiplying that difference with clutch locking friction strength. Then apply this amount of locking force (turn it to torque first) to the wheel with the less tire force opposite to the rolling direction of the wheel, which in other words, reducing the torque from the spinning wheel. This locking force/torque also apply to the adherent wheel same as the rolling direction of that wheel, or in other words, taking torque to the adherent wheel. For bias ratio, if the ratio of tire force between 2 wheels is less than or equal to the bias ratio, then use that difference in tire force to calculate locking force, otherwise, uh, where do I go from here?

Again thank you so so much. Having a conversation like this really clears things out for me. I got a hang of the mechanics behind the diffs now, what's left is actually come up with an implementation that suits the current drivetrain model I'm doing.

Oh by the way, I ditched Unity WheelCollider and wrote my own tire model, because having no way of extracting tire force is really frustrating. And really I don't know what the people doing PhysX is thinking, locking away a very important aspect like tire force.

On 5/12/2018 at 7:38 PM, Edy said:

It's not about the acceleration. The system could be running at constant speed and the clutch packs being pushed providing locking force. The effect is caused by the reaction torque of the output with less resistance. This also explains why when the less resistance is 0 (wheel in the air) then there's no locking force at all.

Looks mostly correct to me. Change "acceleration" with "resistant torque" or "reaction torque" (you name it).

I rather not discuss about implementation details, sorry. Each implementation depends on how the specific drivetrain model is designed: considered variables, torque flows, inertias, analytic vs. numeric resolution, etc. 

You tell me!

Hello.

I'm sorry for digging this topic up. But I'm still having trouble understanding the math model (or rather, the programmer way of modeling) the differentials. I figured the mechanical stuffs out, but still having some questions:

1 - For the case of open diff, in an extreme case where one wheel have 0 road reaction force/torque to push the car forward, is it neccessary that ALL the POWER will go to the slipping wheel? Because my understanding is that torque coming in from the drivetrain will always be split 50-50 between wheels so:

- Let's say the torque coming in is 1000Nm, 500 to left wheel and 500 to right wheel. Left wheel is on air so no road force. Right wheel can put down 500Nm it receives to the road. Ignoring wheel radius and exact car mass for the moment, let's say the neccessary amount of torque to push the car forward is 500Nm, exactly equal to how much the right wheel with traction can put down. So now should the car still move? Because although the left wheel is slipping and not producing any force to push the car forward, the right wheel is still having enough traction to push the car. Which means some power is still going the the wheel with traction, not all to the slipping wheel, correct?

2 - For clutch type LSD, is there a way for calculating the amount of locking torque that would be sent to each side? According to the wikipedia page on LSD, the amount of torque going to each side would be: TorqueIn * 0.5 + lockingTorque (side with traction) and TorqueIn * 0.5f - lockingTorque (side slipping). Is the magnitude of this locking torque the same for both sides?

Also, can the clutch type LSD be completely locked up like a spool? Or there would always be some slipping? Does the diff work by behaving like a spool when the torque bias between sides is less than or equal to the diff's bias ratio (or the torque difference is within the maximum locking torque of the diff)? Or does it constantly applying a locking torque that is equal to the maximum locking torque?

Thank you very much.