Understanding Automotive Differentials: How Do They Work?

[HowlerMonkey]

Torque is the same while speed of each wheel is different.

 

[bugatti79]

If I understand the question correctly, you are OK with the idea that the left and right axle torque is the same. But I think you are confusing something I have said. I think the confusion is the statement that the sum of the left and right torque is that of the crown gear. So if the engine applies 100 N-m to the crown then the left + right shaft torque = 100 N-m. This doesn't mean that each wheel sees 100 N-m.

Think of a seesaw. Put a 30kg kid on each end. The pivot has to support 60kg. The crown gear has to apply enough torque for both wheels. If that torque split is even then the torque to each wheel is 1/2 that of the crown gear.

See attached my sketch. I have difficulty understanding how the torque is split 50:50 in the case of straight line driving in which the crownwheel/diff housing and diff gears are all rotating as one, ie just one rigid bar...

 

[mender]

Go the other way then; let's put the car in the air on a hoist and a guy at each wheel with a lever attached to the wheel. Pretend it's not a diff but is forced to act as a solid bar and the driveshaft is locked (automatic in park).

How much torque is there at the carrier if one guy is applying 40 N-m on one wheel? Now have the other guy exert 60 N-m in the same direction on the other wheel; how much torque is acting on the carrier now? Is it any different than if both guys were at the same wheel and still applying the same amount as before? The forces are balanced, so it's always 100 at the carrier because that's the total of the forces being applied.

If a mechanism allows the forces to be applied equally to two (or more) "resistances", the force applied at each is the original force divided by the number of resistances. That's why a four wheel drive doesn't accelerate any faster in four wheel drive than it does in two wheel drive; the power input is the same but the torque at each of the wheels in four wheel drive is half as much as at the rear wheels in two wheel drive. That's also what keeps the tires from spinning in four wheel drive when they would spin in two wheel drive.

 

[Averagesupernova]

bugatti79, I don't think you can assume that the axle shaft is solid from one side to the other in the case of straight line driving. The appearance is the same but it is not the case.
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I would say it is similar to hooking a chain from one corner of the front of the vehicle to the other corner of the front in a fashion that allows the chain to loop out ahead of the vehicle about 20 feet. Pull on it while allowing the loop to slide back and forth on whatever you are pulling it with and you can guarantee the force on the chain will be split in half. Anchor the loop chain to whatever is pulling it so it cannot slide and you no longer have that guarantee.

 

[slideways]

See attached my sketch. I have difficulty understanding how the torque is split 50:50 in the case of straight line driving in which the crownwheel/diff housing and diff gears are all rotating as one, ie just one rigid bar...

They aren't really rotating as one. Think of it as balanced. A seesaw with two 30kg kids is balanced but that doesn't mean the thing can't pivot. The only way it stays balanced is if the load on each side is equal.

Take a look at the video Tea Jay posted. Notice the part where he attached a single spoke to each wheel (3:45) and where he pushes each with a bar (4:20). What he is illustrating with the bar that can't pivot is a welded seesaw. He shows how holding one wheel will stop the bar. I think that should be easy to see. Because the bar is locked in the center, you can't spin it and both wheels turn at the same speed. This is the welded axle case.

At 4:51 he puts the bar on a pivot. Now it's like the seesaw. If the bar is going to not rotate the loads on either end must be equal. So if the load the bar applies to each side is equal then the torque to each wheel must be equal. Does that clear up the straight line case?

Now when we corner the bar (gear) does spin but it does so at a constant speed. Well let's think of angular acceleration. If the gear spins at a constant speed that means all the forces are balanced. The gear teeth on the left side have to push with equal but opposite force with respect to the right or the gear will not just spin but accelerate. So if it's not accelerating, then the forces must be equal thus the torques are equal. This second explanation is perhaps a bit to brief but I think if you understand the first one it's easier to see this one.

 

[bugatti79]

I think I am satisfied now where the 50:50 split is coming from for an open diff. I was making a big job out it more than I should have. See attached my interpretation with a FBD.

(I forgot to write that M_1=F_1y*r and M_2=F_2y*r but it doesn't change anything.

Everything Slideways says on LSD makes perfect sense. :-)

 

[slideways]

I think you are close to getting it but perhaps not quite.
In the picture the red arrow is the force the crown gear and diff housing applies to the spider gear. The two green arrows are the forces applied to the spider gear by the two axle drive gears (the bevel gears). The orange arrow represents rotation of the spider gear about it's own axis.

Since the spider gear can turn we can see that the two green forces must be equal or the spider will start turning. If it turns the two wheels must spin at different speeds. If we are going straight and not spinning the tires we know that the two outputs are spinning at the same speed thus the green arrows are the same and thus the force applied to the spider gear = 2x the force applied to one axle. Add some radius values and now we have torques.

So what about the case when we are turning? Well the 50:50 still holds true EXCEPT for the case when we are changing our yaw rate.

Looking at the spider gear we can see that it's rotation is going to be controlled by something like τ=J dω/dt. (sum of torques = inertia* angular acceleration). The left and right green arrows of course create the torques on the spider gear. If we are making a steady right turn the left wheel will spin faster than the right but the speeds will remain constant. So let's say we are making a steady right hand turn thus the left axle is spinning at 30 RPM while the right spins at 27. You can see the spider gear will spin in this case. Making some assumptions about the size and directions of the gears I'm going to say the spider spins at 3 RPM clockwise as seen in my picture. So in that case what is the sum of the torques? Look at τ=J dω/dt. J, well it's something bigger than zero. dω/dt, well it's 0 because the rotational speed of the spider gear is constant. That means τ must sum to 0 thus the left and right green arrows must be equal.

Where this actually breaks down is when you change your rate of turn. At that point because you have a change in ω you must have a net τ. However I think if we had real numbers we would find the numbers are small enough to not mater.

Hope that further clarifies.

 

[bugatti79]

I think you are close to getting it but perhaps not quite.
In the picture the red arrow is the force the crown gear and diff housing applies to the spider gear. The two green arrows are the forces applied to the spider gear by the two axle drive gears (the bevel gears). The orange arrow represents rotation of the spider gear about it's own axis.

Since the spider gear can turn we can see that the two green forces must be equal or the spider will start turning. If it turns the two wheels must spin at different speeds. If we are going straight and not spinning the tires we know that the two outputs are spinning at the same speed thus the green arrows are the same and thus the force applied to the spider gear = 2x the force applied to one axle. Add some radius values and now we have torques.

So what about the case when we are turning? Well the 50:50 still holds true EXCEPT for the case when we are changing our yaw rate.

Looking at the spider gear we can see that it's rotation is going to be controlled by something like τ=J dω/dt. (sum of torques = inertia* angular acceleration). The left and right green arrows of course create the torques on the spider gear. If we are making a steady right turn the left wheel will spin faster than the right but the speeds will remain constant. So let's say we are making a steady right hand turn thus the left axle is spinning at 30 RPM while the right spins at 27. You can see the spider gear will spin in this case. Making some assumptions about the size and directions of the gears I'm going to say the spider spins at 3 RPM clockwise as seen in my picture. So in that case what is the sum of the torques? Look at τ=J dω/dt. J, well it's something bigger than zero. dω/dt, well it's 0 because the rotational speed of the spider gear is constant. That means τ must sum to 0 thus the left and right green arrows must be equal.

Where this actually breaks down is when you change your rate of turn. At that point because you have a change in ω you must have a net τ. However I think if we had real numbers we would find the numbers are small enough to not mater.

Hope that further clarifies.

Very good, I like that. Although I am curious where you think I don't get it? From my diagram I have Tx coming from the crown wheel which translates into the red arrow force you have, ie from the housing onto the spider..

I believe things make perfect sense to me now.

 

[slideways]

Very good, I like that. Although I am curious where you think I don't get it? From my diagram I have Tx coming from the crown wheel which translates into the red arrow force you have, ie from the housing onto the spider..

I believe things make perfect sense to me now.

I couldn't follow your written description (it wasn't that easy to read the photo) and I wasn't sure if we were saying the same thing. Either way, if what I just posted is in line with what you were thinking all is clear.

 

[A.T.]

I found this very cool demo explaining differential gears. I post it here for future reference:

https://www.youtube.com/watch?v=K4JhruinbWc

Click here to jump directly to the explanation.

 

[Tea Jay]

I found this very cool demo explaining differential gears. I post it here for future reference:

https://www.youtube.com/watch?v=K4JhruinbWc

Click here to jump directly to the explanation.

Did you find it on page 1 of this thread, where I posted it?

:D



It is a great video though.

:D