Open Car Differential Burnout Debate

[peachmango]

I would really appreciate if you could help me with a question I have. I was having a debate with someone I work with about whether or not a car with an open diff can spin both wheels at the same time (aka do a burnout). Let me know if my logic is incorrect. If a RWD car is on ice (low friction) and it tries to accelerate, both rear wheels will slip and rotate at the same speed because the torque at the wheels is enough to overcome the static coefficient of friction. Using the same idea if a very powerful RWD car that is on asphalt (consider even friction across entire surface, therefore equal traction at both tires) applies its front brakes and accelerates with full force both rear wheels will spin in place at the same speed because the torque of the car is so high it can overcome the higher friction of the surface the car is on. Obviously this is an ideal situation where the frictional forces are equal at both rear wheel to allow for equal speeds, but it is possible correct? Thank you so much for your help.

 

[Mech_Engineer]

Theoretically its possible, as long as both tires have the same traction available. More typical is one tire loses traction before the other, and spins in a "one tire fire."

If the car has traction control or other electronic shenanigans, all bets are off. See here: https://www.carthrottle.com/post/wh...ut-in-an-open-diff-car-with-torque-vectoring/

 

[Averagesupernova]

Also, if there is side slide it is much more likely that both wheels will spin. Think taking a corner fast and hitting the gas. Sprint cars as well as many other race cars drift through the corners in this manner and of course they have a solid rear axle so it is more controlled. Alternately, it is easier to slide sideways if the wheels are spinning (in a burnout). Lose traction in one direction and you are likely to lose it in a different direction as well.

 

[OldYat47]

Although it is theoretically possible (all factors identical, including torque-correcting weight placement) I've never seen it done in decades of fooling around with cars, no matter how much torque was available.

 

[peachmango]

Thanks for your replies! So if this can only theoretically happen, what happens on a even layer of ice in real world conditions where friction is low and relatively even?

 

[Ranger Mike]

the right rear tire rotates forward and the left rear tire rotates backward. You can burn rubber on the right rear tire only. Too many country roads show this one tire stripe on saturday morning. In the old days the slush , in slush box automatic transmission cars, you could put an open differential car in reverse and back it up about 10 MPH then slam it into gear and " write Js" on the pavement. This action would make a J pattern on the pavement.

 

[Mech_Engineer]

Thanks for your replies! So if this can only theoretically happen, what happens on a even layer of ice in real world conditions where friction is low and relatively even?

In the "real world," one tire loses traction and spins, the other stays in contact. I've seen this time and time again while driving in snowy/icy conditions.

 

[jack action]

It is theoretically possible to have both tires spinning with an open diff, although practically impossible to achieve. And it is not an increase of torque that will do it.

For this to happen, both tires must unhook at exactly the same time (can't put enough emphasis on exactly).

With an open diff, the torque is the same on both wheels, no matter what. When torque increases, it goes against the tire friction force. When the static friction force barrier is achieved, the tire starts spinning and the coefficient of friction goes down to the kinetic value. This means that the reaction torque goes down in proportion with this new - lower - friction force. The torque being equal on both wheels, it also reduces on the other wheel that did not begun to spin. So now, this wheel only produces a torque equivalent to the kinetic friction force, so it will never be able to reach the static friction barrier it needs to reach for the wheel to start spinning.

See, the torque at the wheels is dictated by the friction force at the tire-road contact patch, not by the torque output of the powertrain. Any extra torque sent will be converted into acceleration of the wheels (or wheel).

So, unless both tires unhook at the same time, you would need to either reduce the normal weight $N$ on the stationary wheel or its radius $r$ to increase the friction force to achieve the static coefficient of friction $\mu_s$ needed to resist the wheel torque $T$ ($\mu_s < \frac{T}{Nr}$). Unfortunately, when suspension design allows lateral weight transfer, it is usually the reverse that happen weight-wise and the radius change is never enough to compensate. Worst, when the normal weight drops, the tire's coefficient of friction usually increases.

There are two forms of friction, kinetic and static. If you try to slide two objects past each other, a small amount of force will result in no motion. The force of friction is greater than the applied force. This is static friction. If you apply a little more force, the object "breaks free" and slides, although you still need to apply force to keep the object sliding. This is kinetic friction. You do not need to apply quite as much force to keep the object sliding as you needed to originally break free of static friction.

Some common values of coefficients of kinetic and static friction:

    Surfaces         µ (static)  µ (kinetic)
Tire on concrete       1.00         0.80
Tire on wet road       0.60         0.40
Tire on snow           0.30         0.20

These values are approximate.

source: http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/Ben_Townsend/StaticandKineticFriction.htm

 

[Randy Beikmann]

From personal experience, I can tell you that a car with an open diff can spin both tires at once, on dry pavement. It's just not very predictable, and most of the time one will spin. Even an open diff has some friction in it, so maybe the resulting partial lockup is part of the cause.

 

[Averagesupernova]

From personal experience, I can tell you that a car with an open diff can spin both tires at once, on dry pavement. It's just not very predictable, and most of the time one will spin. Even an open diff has some friction in it, so maybe the resulting partial lockup is part of the cause.

I have found the same. In my wilder days I left a lot of rubber on the pavement and dug a few trenches on dirt/gravel roads.

 

[256bits]

Doing donuts, both tires have to spin!

 

[Rx7man]

On solid axle cars it's very unlikely to spin both tires at once, reason for this is the applied torque from the driveshaft effectively tries to turn the entire rear axle about the axis of the driveshaft.. of course it can't, but it will apply more downward force to one tire while lifting the other... the tire that's getting 'lifted' will of course have less friction between it and the road, and will be the first to start spinning.
In most cases it's the right side tire, because most driveshafts (and engines) are clockwise rotation when viewed from the front.. so when viewing from the front, it 'plants' the car's left tire, and unweights the right, which is why it spins first

 

[Randy Beikmann]

On solid axle cars it's very unlikely to spin both tires at once, reason for this is the applied torque from the driveshaft effectively tries to turn the entire rear axle about the axis of the driveshaft.. of course it can't, but it will apply more downward force to one tire while lifting the other... the tire that's getting 'lifted' will of course have less friction between it and the road, and will be the first to start spinning.
In most cases it's the right side tire, because most driveshafts (and engines) are clockwise rotation when viewed from the front.. so when viewing from the front, it 'plants' the car's left tire, and unweights the right, which is why it spins first

The way to counter the right rear tire lifting is to do donuts turning to the left, evening out the load!