doubt on open differentials

[Thrilok]

Can anyone plase explain why it is said that open differentials supply equal torque in all situations??..I am asking this because as far as I understand in the situation explained below there is no equal supply of torque to the axles

If one of the tyre is on ice(no friction) and the other tyre is on a normal road(enough friction), the tyre on the ice starts slipping when power from engine is supplied, and there is zero power supplied to the tyre which is on the road..from this situation I understand that there will be torque transfer(power transfer through the spider gears) in between the axles when there is a difference in traction between the tyres, I would be thankful to anyone of you who can find fault in my understanding if i am wrong.

So I just want some one to help me in understanding the concept of equal torque bias in an open differential.

[Thrilok]

Can anyone plase explain why it is said that open differentials supply equal torque in all situations??..I am asking this because as far as I understand in the situation explained below there is no equal supply of torque to the axles

If one of the tyre is on ice(no friction) and the other tyre is on a normal road(enough friction), the tyre on the ice starts slipping when power from engine is supplied, and there is zero power supplied to the tyre which is on the road..from this situation I understand that there will be torque transfer(power transfer through the spider gears) in between the axles when there is a difference in traction between the tyres, I would be thankful to anyone of you who can find fault in my understanding if i am wrong.

So I just want some one to help me in understanding the concept of equal torque bias in an open differential.

[Luniz]

An open Diff sends equal torque to both wheels, but the thing is, that the total amount of torque transmitted is double the transmitted torque of the wheel with less friction. So in your example that would mean that the total transmitted torque is two times next to nothing. In this case you would need a limited slip differential, that can send some torque to the wheel with enough traction, even though the other wheel has very little traction. If the other wheel has no traction at al (lifted off the groud) even an LSD does not work. For this application (usually on offroad vehicles, where a wheel may be up in the air) a visco-lock differential is used.

[SNasello]

Say that the left and right tires are running on different surfaces with coefficient of 1.0 and 0.1, respectively. Assume that both tires have some constant (peak) sliding coefficient of friction (when the road surface has a mu=1.0). The driving torque that the right tire can handle is 0.1*mu_sliding*Fz*R_dyn and the left is 1.0*mu_sliding*Fz*R_dyn. If the engine supplies enough torque to the differential to exceed the torque capacity of the right tire, then it will spin.

The open differential cannot transfer torque to the opposite drive axle, it will have equal torque as the spinning axle, but because of the much higher friction coefficient, a much lower slip ratio. Since both wheels have the same amount of torque, but the right one is spinning faster, there will be more power transmitted to it than to the left tire. You are of course wasting the power because the tire is just spinning and not moving the car forwards. So you are right that there is power transfer, but not torque transfer.

If you take an open differential connected to a set of wheels, block one wheel from spinning, and spin the opposite wheel, (without any gear engaged on the engine) you will not feel any torque on the wheel that is blocked (except through friction in the gears).

FYI a spool is the opposite, in that it transfers an infinite amount of torque across the axle, but the wheel speed on both sides will be the same.

I will let you figure out how the various limited slip differentials operate.

[SNasello]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Luniz:
An open Diff sends equal torque to both wheels, but the thing is, that the total amount of torque transmitted is double the transmitted torque of the wheel with less friction. So in your example that would mean that the total transmitted torque is two times next to nothing. In this case you would need a limited slip differential, that can send some torque to the wheel with enough traction, even though the other wheel has very little traction. If the other wheel has no traction at al (lifted off the groud) even an LSD does not work. For this application (usually on offroad vehicles, where a wheel may be up in the air) a visco-lock differential is used. </div></BLOCKQUOTE>

A visco-lock is not the only type of differential that can do this, a salisbury differential will be able to transfer some torque as well due to the preload in the clutch plates and the torque input to the housing.

[Luniz]

I know... but the preload is not that much so the friction of the clutch plates is very little compared to the torque input on the housing. The torque input itself will also be not very high, because this input can only be as much as the total amount of torque transmitted to the wheels.

The preload helps a bit, but with a Salisbury diff you should avoid having a driven wheel in the air ;-)

[Thrilok]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by SNasello:
Say that the left and right tires are running on different surfaces with coefficient of 1.0 and 0.1, respectively. Assume that both tires have some constant (peak) sliding coefficient of friction (when the road surface has a mu=1.0). The driving torque that the right tire can handle is 0.1*mu_sliding*Fz*R_dyn and the left is 1.0*mu_sliding*Fz*R_dyn. If the engine supplies enough torque to the differential to exceed the torque capacity of the right tire, then it will spin.
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> </div></BLOCKQUOTE>
I understood the above explanation.
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> </div></BLOCKQUOTE>
The open differential cannot transfer torque to the opposite drive axle, it will have equal torque as the spinning axle, but because of the much higher friction coefficient, a much lower slip ratio. Since both wheels have the same amount of torque, but the right one is spinning faster, there will be more power transmitted to it than to the left tire. You are of course wasting the power because the tire is just spinning and not moving the car forwards. So you are right that there is power transfer, but not torque transfer.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> </div></BLOCKQUOTE> I understand the relation among power, torque and revolutions per minute is--[ power transmitted from 1st gear= torque applied on second gear * RPM of second gear]assuming there are two gears 1 and 2, 1 transmitting power to 2----and I am assuming that torque applied on second gear is the reason for the RPM of the second gear. So I understand it as-- when there is a power transfer to the wheel with lower friction its RPM is increased because of the torque applied on it, so when the wheel is rotating at an higher RPM there is an additional torque applied on it--as an analogy the velocity gained by a body(angular velocity or RPM in our case) moving in a straight line is due to the force(torque in our case) applied on it. please correct my way of understanding if i am wrong </div></BLOCKQUOTE>

If you take an open differential connected to a set of wheels, block one wheel from spinning, and spin the opposite wheel, (without any gear engaged on the engine) you will not feel any torque on the wheel that is blocked (except through friction in the gears).
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> </div></BLOCKQUOTE>
In the above situation if we dont block the wheel and rotate one wheel then the other wheel rotates in the opposite direction, so from this i understand that the other wheel is rotating because of some torque applied on it, so can you please correct me if i am wrong in my understanding.

[JasperC]

Whether or not the other wheel will spin in the opposite direction will depend on which components in the drivetrain have more friction. If you take the chain off the car and then spin one wheel, you will also spin the entire differential, and the other wheel will spin in the same direction (the differential gears will not rotate w.r.t. each other).

However, with the chain installed, there is more friction in the chain than inside an open differential, so that's why the diff housing (and chain) will remain still and the other wheel will rotate in opposite direction.

Keep in mind that in these cases you are dealing with VERY small torques, so it doesn't necessarily say much about what happens when you push the pedal to the metal.


In the case where one wheel has less grip, there is indeed some torque tranfer to that wheel, but only that torque which is needed to accelerate the wheel (i.e. angular acceleration times inertia of the wheel). The rest of the torque supplied by the engine is used to accelerate the engine itself. So when you have one spinning wheel and you keep applying the throttle, the wheel will spin faster and faster and the engine's RPM will increase until you hit the rev limiter.

So, yes - some torque is transferred to the spinning wheel. And yes, the same torque will be transferred to the other wheel as well. But again, this torque is so small that it is negligible.

Kind regards,

Jasper Coosemans
Chief Drivetrain 2009-2010
DUT Racing Team
Delft University of Technology

[dazz]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Thrilok:
Can anyone plase explain why it is said that open differentials supply equal torque in all situations??..I am asking this because as far as I understand in the situation explained below there is no equal supply of torque to the axles

So I just want some one to help me in understanding the concept of equal torque bias in an open differential. </div></BLOCKQUOTE>

Draw a diagram of the diff gear set. You'll note that the two axles are geared together at a ratio of 1:1 (with a reversal).

Simple, there are two equations/ relations:
1 - the two axles are geared together at a (reversing) ratio of 1:1 therefore whatever torque is resisted by the left hand tyre must also be resisted by the right hand tyre.

2 - torque input to the diff carrier must EQUAL torque from the left hand axle PLUS torque form the right hand axle (neglecting gear, bearing and seal friction and rotational inertia)

3 - since from (1) LH torque = RH torque and max resisted axle torque is when one of the two start to slip/spin, we have:
diff carrier torque = 2x the torque reacted by whichever tyre starts to slip first

So, if you have one tyre on ice that requires 'X' amount of torque to spin then the other tyre is transmitting 'X' amount of torque to the road. Might not be enough to move the car forward, but it IS there! And the input to the diff carrier is 2 times 'X'


Long winded practical explanation in grey:
<span class="ev_code_GREY">- So, if you lock the carrier housing, and have a mate hold one axle while you try to turn the other - you're mate will have to be equally as strong as you are to stop you from turning the axle, ie the torque is transfered at 1:1 (neglecting bearing, seal and gear friction losses). Axle1 = Axle2
- Now lock one axle and rotate the diff carrier one turn, what happens? The free axle will rotate two full rotations - a ratio of 2:1. So the torque at the turning axle is 1/2 of the input torque to the diff carrier. Axle1 = 1/2 Diff Carrier
- Next get another buddy to try and turn the diff carrier while the other two try to hold the axles. The two guys on the axles should have it easy because the torque that the bloke is applying to the diff carrier is being split evenly between the two guys on the axles. Axle1 + Axle2 = Diff Carrier
- Finally onto your scenario. Have one guy try to turn one axle (this would be the wheel on ice) while you and the other bloke try to hold the other axle (wheel with traction) and the diff carrier (engine power) stationary. Whoever is trying to hold the diff carrier stationary will lose. Why? Because the guy trying to hold the carrier stationary must be strong enough to resist the torque input from the guy turning the axle PLUS the torque input from the guy holding the axle stationary. still Axle1 + Axle2 = Diff Carrier but in real world is Diff Carrier = 2x(lowest of Axle1 or Axle2) (if you are using a car diff for this, you will have to take into account that there is about a 3:1 ratio into the diff carrier) so that would then become Diff Input = 2x(lowest value of friction_force*tyre_radius of Axle1 or Axle2)/3 where friction force is determined by the vertical load on the tyre that varies due to undulating roads and weight transfer effects multiplied by the co-efficient of friction that varies due to road conditions, tyre temperature and percentage of slip</span>

[DougMilliken]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Thrilok:
Can anyone plase explain why it is said that open differentials supply equal torque in all situations?? </div></BLOCKQUOTE>
See Chapter 20 in "Race Car Vehicle Dynamics", SAE book # R-146.
We include a lever analogy on p.736 (not mentioned in this thread), this may help some people to visualize the various types of diff.