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Cars differentials, open, locked and preload


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#1 vadevaman   Members   -  Reputation: 157

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Posted 25 April 2014 - 01:43 AM

Hello,

I've ran into some problems when trying to do locked differentials. I quite dont understand how to use preload torque on velocity difference based differentials. So what I know, is that:

  • a open differential has equal torques and unequal velocityes (lighter wheel spins more).
  • a locked differential has equal velocityes and unequal torques (lighter wheel has less torque).

So what I dont know is: a preload torque, does it mean that the differential stays locked until some point and after that point the differential becomes unlocked (so its action is quite reversal to an lsd differential right?) now after the differential becomes unlocked the lsd part comes to play and try to match up the speeds?

...this all seems a bit jibberish to me and un logical. Since a lsd differential has angled wedges in them wich allow the differential to lock up. The simplest way I could do a close to a locked differential is velocity difference method. Like this:

velocityDifference = (leftSpeed-rightSpeed)
leftTorque = driveTorque - (velocityDifference * 2000)
rightTorque = driveTorque + (velocityDifference * 2000)

2000 is a magic constant. Currently this constant gives enough torque to lock up the diff almost 100% (some small velocity difference still exists) So my question is,

Is my constant "2000" the preload torque? If so where comes the lsd part in??? Another question would be, where to use road reaction torques? From racer's doc's(http://www.racer.nl/tutorial/differentials.htm) I read that I should use the feedbackTorque from both sides to define if the differential is locking or not, but the preload part doesn't make sense to me.


Edited by vadevaman, 25 April 2014 - 01:48 AM.


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#2 vadevaman   Members   -  Reputation: 157

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Posted 27 April 2014 - 11:59 PM

Anyone?



#3 LorenzoGatti   Crossbones+   -  Reputation: 2511

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Posted 29 April 2014 - 03:48 AM

You are completely failing to explain the actual problems you have, apart from being perplexed by a new and unfamiliar subject or mixing up different models of different designs of differential. Are the cars in what I suppose to be a a racing game doing anything inappropriate?  

 

There is a wide range of useful locking and torque distribution policies you can adopt in your software-simulated differential, without even worrying about actual mechanical implementations.

As long as the behaviour of your differential is physically sound (e.g. you don't drive wheels with friction in the clutch, with preload springs, or with torques out of thin air) and it conforms to some basic principles (unlocked for small wheel velocity differences, locked when one wheel appears to be slipping) you can do anything you want with parameters such as a "preload torque": unlike actual physical quantities such as wheel angular velocities they are only an abstract description of what the differential does.


Produci, consuma, crepa

#4 vadevaman   Members   -  Reputation: 157

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Posted 29 April 2014 - 10:00 AM

My problems are clearly stated in my post.

And I'm not going to point them out once again when you clearly dont know anything about this topic, when you are saying things like "you can do anything you want with parameters such as a "preload torque""
coding wise this sentence is correct. But I'm simulating cars and originate from real life data. So i cant simply "do anything" when I need to follow speciffic logics.



#5 CombatWombat   Members   -  Reputation: 425

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Posted 14 May 2014 - 01:29 PM

First you need a more general description for the velocities.

 

A differential is a bevel gear implementation of a planetary gear train. 

For an automotive type differential where the bevel gear ratio is 1:1, the velocities must satisfy:

(VL - VH)/(VR - VH) = -1

where VL = left drive shaft velocity

VR = right drive shaft velocity

VH = housing & ring gear velocity

 

You can see the derivation of this here:

http://books.google.com/books?id=QNj87rJZIpUC&pg=PA373&lpg=PA373&dq=KINEMATIC+EPICYCLIC+GEARING+DIAGRAMS&source=bl&ots=w9OLebHeMw&sig=1Ut2U15xCnXa53MPhdU8bEHbI_A&hl=en&sa=X&ei=RYZzU7HyHc6zsAT9-4H4DA&ved=0CDsQ6AEwAzgK#v=onepage&q=KINEMATIC%20EPICYCLIC%20GEARING%20DIAGRAMS&f=false

(Or refer to any kinematics of machinery text)

 

Add up net moments on the components:

The ringgear/housing has a moment on it from the pinion gear.  (This comes from your engine/transmission model).

A reaction moment comes from the pin which mounts the center bevel gear.

The center bevel gear in turn applies moments to each output shaft.

The output shafts have reaction moments from the tire & brake models.

The output shafts have reaction moments between them to enforce the velocity rule from earlier.

 

The various locking & lsd differentials add more forces/moments to these components.  Exactly which components are effected depends on the specific type.

Simple example would be viscous style which has a locking effect as a function of velocity difference between the shafts.

Add another moment to each axle.  Essentially (leftAxle Velocity - rightAxleVelocity) * someConstant.

 

To directly answer your question on "preload".

Certain types of lsd utilize friction clutches to couple between the two output shafts.  Usually these are angled to be sensitive to the torque input to the differential.  "Preload" means that even with zero torque transmitted, there is still some coupling torque between the output shafts. 

 

A good search term would be "torque in epicyclic gear trains".

 

 

 

 

 






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