Understanding Automotive Differentials: How Do They Work?

[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