I've started doing some working with diffs today and I could really do with getting my head around these mechanical wonders, they are one of the few parts of a vehicle I don't know how to correctly simulate.

So let's begin with the simplest of them all, the open diff.

Now an open diff cannot transfer torque, but simply giving each wheel on an axle 50% of the torque at the diff, while at first seems correct, isn't the whole story.

If this were true, then you would never get stuck when one wheel is on a low friction surface, or in the air. I remember when my Dad built his Westfield, he put the car on axle stands, started the engine, got me to hold one wheel still with just my index finger, then let up the clutch. The wheel I was holding did not even try to rotate, so clearly experience zero torque (or in reality, a very tiny torque), and the free wheel spun up to twice the rpm of the diff.

So what's happening? The explanation on wikipedia explains some things. Since the free wheel could not transmit any torque, the wheel I was holding got the same amount (nothing). So what happened to our torque? Surely it can't just disappear? For starters, the free wheel sped up, and to change the angular velocity of an object you have to apply a torque, so clearly some torque was still acting on this wheel? Then if this is the case, how can we say that torque was not transferred between wheels when clearly none was received by the other wheel?

I'm assuming the answer involves considering the ground reaction torque that a tyre on the ground would normally provide?

Incredibly old video

Get some Legos and build mechanical locking diffs yourself, you'll understand how it works in no time. No idea about viscos though, obviously Lego can't help with that.

€: Wow that guy would so piss Stewie off "hwheel"

unless i miscalcuated assuming the wheel is a pipe with zero wall thickness weighting in at around 5kg and 15" of diameter assuming your dad let go off the clutch within about 3 seconds you should have felt a torque of about 23Nm
ie hardly any torque to begin with

factor in the losses you have in car wheel bearings and almost none of that will have to have been exterted by your finger

Quote from morpha :

Incredibly old video

Nice. That doesn't get me any further to modelling it though.

Shotglass - care to elaborate on your figures?

Edit: I'm thinking, so if a tyre of 1m radius has 100N traction, and we are supplying 1000Nm at the diff, then 400Nm of the 500Nm it receives is "converted" into angular momentum, and the other 100Nm overcomes the tyres resistance, and it is this 100Nm, rather than the 500Nm, that must be shared with the other wheel (assuming of course the other wheel is not traction limited)? Then what has happened to the other 400Nm the other wheel should have had? As the other wheel is not accelerating? Surely I can't just disappear? Or does it somehow contribute to the first wheels angular acceleration?

first of all the wheel is assumed to be 15" in diameter
a high performance wheel will weight something in the range of 5kg and being a weistfield i presume the car had neither particularly large or particularly heavy wheels
to get a rough idea of the wheels moment of inertia i assumed it to be a pipe with 15" diameter 0 wall thickness and weighing 5kg which should roughly account for the tyres inertia at a larger diameter and the wheel hubs inertia at a smaller diameter

the inertia of such a pipe is mr^2 so that gives us
I = 5*(((15*2.54)/100)/2)^2 = 0.18 (whatever that unit is)

going with the westfield theme i went with the lx4s race s setup which has a first gear ratio of 3.4 and a final gear of 3.8 (roughly and i might have them mixed up but its close enough to a realistic value)
so with an idle rpm of 800 we get a wheel rpm of ~62
(btw this is where i made a grave error which results in even lower forces but well get to that in a bit)

so the wheels speed is
omega = 2*pi*62/60 rad/s
this is where i went wrong i got minutes and seconds muddled up and didnt divide by 60

assuming he let the clutch slip for 3 seconds and assuming the accerlation is constant we get
alpha = omega/3 = 2.2 rad/s^2

and finally the torque is
M = alpha * I = 0.4 Nm



the things you can learn from sewing eh?
have i meantioned the forum desperately needs a latex interpreter?

Hope this might help

Assuming you have all of the drive wheels off the ground and the engine is running at a constant RPM, the engine only transfers enough torque to overcome the resistances of the drivetrain.

Assuming you then press the throttle to the floor, the engine then makes max torque, overcoming the resistance (including inertia) in the drivetrain. However, this does not last very long because the engine very quickly reaches the rev limiter. When it hits that, you are again only transferring enough torque to counteract all the other resistances in the drivetrain, a fairly small number compared to the burst of torque a moment ago.

Things get a bit fuzzy for me when you start applying a resistance to just one wheel. Obviously, with an open diff and one wheel totally stopped, the other wheel spins twice as fast. However, I'm not sure how the wheel speeds are affected when you apply some resistance to one wheel but less than the input torque, such that neither wheel stops completely*.

*actually I'm not sure this is possible with an ideal open diff (zero internal resistance)

The engine isn't generating any torque (or very little - just enough to overcome friction), so your fingers can easily stop what reaches the wheel. The other wheel also isn't transmitting any torque. Thus it is balanced.

If you have one wheel on the grass (let's say virtually zero grip, and hence virtually zero torque) then the other wheel can only apply the same torque. The wheel on grass therefore spins up (this requires very little torque compared to what the engine could provide against a brake), and so you get no traction. It is only the existing velocity of the car that enables you to get back on the road and accelerate again.

Quote from Shotglass :

whatever that unit is

kgm²

Anyway, I'm assuming your calculations are for the resultant torque generated by conservation of angular momentum? Which would apply against the vehicle body, and not the other wheel? Or does just the smaller change in angular velocity (in situations where both wheels are accelerating, but at different rates) get applied against the body, and the difference between the two to the opposing wheel?

I'll give my idea a test tomorrow and see how it drives, I think what everyone else has contributed has confrimed my suspicions. Annoying that I can explain a diff in words no problem, but not mathematically with numbers.

my dad explained it to me

make way for the 14 year old to teach you something

open, you hit the nail on the head

locked, the exact amount of torque and power is applied to each wheel at all times no matter what (unless something breaks)

clutch. you can set up amount of locking in this when there is power, and without, so when you have power, the clutch puts however much pressure you set to make the wheel keep that much locking, and with power off, the engine tries to still keep the wheels staying at however much you set the wheels to stay at. if you apply more force then the clutch is using to keep it semi-locked, it will overpower the clutch

viscous, this is not very complicated, the more speed at the lsd, the more it keeps it locked, so at low speeds it is nearly open, and at breakneck speeds it is nearly locked. you can also set how much force is maintaining this....should i make a guide?

Quote from logitekg25 :

locked, the exact amount of torque and power is applied to each wheel at all times no matter what (unless something breaks)

Try again.

:dunce:

Jeezz… for anyone posting how stuff work or simplistic general idea videos. Of course Bob knows in general how open diff works.
Title states clearly that he is seeking a solution from the mathematic point of view.

Quote from logitekg25 :

:dunce:

In locked diff both wheels rotate at the same speed but not necessarily both wheels transfer the same amount of torque to the ground. If it’s a traction limited case then the wheel that has more traction delivers grater torque to the ground. If both wheels have sufficient traction, then the torque delivery of each wheel depends on the relative distance each wheel rolling to in the same amount of time. Because both wheels have the same angular velocity, if they are not following the same tracks (eg in a turn) they deliver different amount of torque to the ground at the same time.
Anyway this is an example of useless information in this thread. I might be able to understand how a diff works but that doesn’t mean I can explain that in mathematic formulas.


I just want to add a +1 to that.

Quote from Forbin :

[snip] the engine only transfers enough torque to overcome the resistances of the drivetrain.

Quote from tristancliffe :

The engine isn't generating any torque (or very little - just enough to overcome friction),

As an answer to “where the other XX amount of torque goes” It isn’t going any ware. Any engine provides at a given rotation speed as much torque as it’s asked to give. Till a point when the engine reaches it’s maximum torque delivery limit. (Common sense. I just write it down so the middle sentence doesn’t look illogical.)

So stop worrying about it and start working on the important stuff. (how power-angular velocity is transmitted between the two wheels.)

Now adding some general thoughts, I hope will help as a methodology in trying to solve this problem.
Generally we know that given specific circumstances a certain amount of torque is needed to overcome the traction of the less loaded wheel. While this happens we know that, speaking about an ideal diff with no internal friction, at both wheels is applied the same torque, limited from the wheel that is using the most power. (the one spinning in a greater speed).
If you know how much torque is needed to overcome the friction of the more likely to break traction wheel, you can calculate the power it uses each moment to increase it’s angular velocity, overcoming any friction and angular inertia. Then you know the torque applied to the other wheel and can calculate the power that is delivered to it. (if any in case it doesn’t rotate at all).
So the required power for this to happen is the sum of the power delivered to both wheels.
Then you can calculate the specific amount of torque that is transferred from the engine to the diff.
Using power because that way you don’t need to know the relation between the “planet” gears and the gears attached to the axels in order to calculate the specific amount of torque delivered from the pinion gear to the crown wheel, when a wheel is spinning faster than the other and final gear ratio has nothing to do with it anymore…
(a bit exaggerating by saying "nothing" but it is almost nothing if you don't know the relation between all gears in the diff when wheels are rotating in different speed.)

I am trying hard to make any sense.

wow, i diddnt see the (mathematically) in there :dunce:

sorry for the useless info

Quote from Bob Smith :

Anyway, I'm assuming your calculations are for the resultant torque generated by conservation of angular momentum? Which would apply against the vehicle body, and not the other wheel?

uhm no... or yes actually
what im calculating is the torque nescessary to accelerate a freely spinning wheel in idle
as per conservation of momentum the same torque acts on the cars body (more than that actually)

and as per the continuity condictions inside the diff its the torque you felt during your little experiment... well half actually since i forgot to take into account that the spinning wheel was going twice its idle speed

Quote from logitekg25 :

my dad explained it to me

make way for the 14 year old to teach you something

open, you hit the nail on the head

locked, the exact amount of torque and power is applied to each wheel at all times no matter what (unless something breaks)

clutch. you can set up amount of locking in this when there is power, and without, so when you have power, the clutch puts however much pressure you set to make the wheel keep that much locking, and with power off, the engine tries to still keep the wheels staying at however much you set the wheels to stay at. if you apply more force then the clutch is using to keep it semi-locked, it will overpower the clutch

viscous, this is not very complicated, the more speed at the lsd, the more it keeps it locked, so at low speeds it is nearly open, and at breakneck speeds it is nearly locked. you can also set how much force is maintaining this....should i make a guide?

No. It would be wrong and simplistic.

I've got two solutions that both appear to work but are not mathematically equal. So one of them must be wrong.

Method A:

Method B:

Where DiffTorque is from the engine/gearbox, the InputTorques are from the tyres pushing back on the road, and the OutputTorques are then applied against the wheels (and the difference between input and output torques causes angular acceleration)

Maybe some else can see which one (or both?!) is wrong and why.

I frankly have no idea what physical principle the first one tries to use for the calculation. Also I wonder about your second aproach because there is some sort of torque split in the equations.
However I can't make a judgement. I just wanted to point out that I wouldn't start with an equation of torque because you're mixing torques with very different directions. So don't be caught out when saying torque A + B must equal torque C, since they don't necessarily point in the same direction (vectorially speaking).

To me the mathematical discussion becomes easier when talking in terms of power and conservation of energy. The open diff:
1. Conserves energy and doesn't save it. Thus input power must always be output power.
2. Exerts equal torque on both wheels.

E.g., if you were to stop one wheel from moving and then added input power to the diff all of it would have to be transferred to the free wheel. The torque exerted on the free wheel (and thus also on the fixed wheel) would change over time as the wheel spins up, even though input torque remains constant.

Vain

Quote from Vain :

To me the mathematical discussion becomes easier when talking in terms of power and conservation of energy.

Totaly agree

Quote from Vain :

I frankly have no idea what physical principle the first one tries to use for the calculation.

It decides what the maximum torque is that can be applied to any wheel, sends this to both, then sends the extra torque to the wheel of least resistance, where we hava already sent exactly enough torque to match the resistance, so the extra torque is what results in angular acceleration.

Quote from Vain :

Also I wonder about your second aproach because there is some sort of torque split in the equations.

Less so than the first method, actually.

Both methods seem to give an open diff behaviour, i.e. the inside wheel spins up, but the second method is actually better at keeping the tyre torques equal, and was the one given to me to use, whereas the first method I invented.

Unfortunately, I understand the first method, and how I could then adapt it to make an LSD, whereas I'm clueless about what to do with the second method. I have some code for a TBD but I haven't got my head around it yet.

im not sure what the first one is trying to achieve but from what i can see both are wrong since they result in different torques on the lest and right drive shaft which just isnt possible with an (ideal) open diff

not sure if this info will help you Bob, but in 15 years of oval racing, we were restricted to mandatory open rear diffs in one of the divisions i built cars for, so we had to figure out how to apply upwards of 300 horsepower to the rearend, all the while maintaining minimal wheelspin and maximum forward momentum. not an easy thing to do, especially when required to use a bias-ply tire, which when heated grows like a pimple on prom night.

you need to take into consideration the amount of downward force applied to each tire contact patch(static weight), the difference in circumference between the rear tires (stagger) if any, the approx. radius of the turn, and the crossweight percentage change relative to the amount of static caster setup in front wheels.

an example, we were required to weigh no less than 3400 lbs., with a max. 54% left side, 49% max rear weight.
ideally, we would use the crossweight to load the left rear/right front tires, atempting to get as close to 1000 lbs of static weight on the left-rear. by doing this, and also by adjusting the pre-load using shims, we were able to manipulate an open rearend to act like a locked rearend, driving both wheels with what seemed nearly the same force. now had we not attempted to force so much static weight on the left rear, wheelspin would result, regardless of the amount of pre-load in the diff.

no this is not a mathematical answer, but maybe the example will help you reach your destination......

Quote from z-ro 8 :

especially when required to use a bias-ply tire, which when heated grows like a pimple on prom night.

sorry for the useless bump, but that's a hilarious analogy.

Quote from bunder9999 :

sorry for the useless bump, but that's a hilarious analogy.

you should be racing not trolling