Open Center Diff Query

[krawler]

Stolen from another forum, this question invoked some pretty good responses - but noone has really agreed on the answer. Here is the original question:

Ok...

In a car with an OPEN diffrential, what will happen when...

1) The car is in a stationary position (i.e. burnout)

2) One Wheel is NOT moving

3) One wheel is spinning on the spot without moving the car

Because effectively twice the power is being transmitted to that wheel, and the other wheel is locked still (hypothetically), does the moving wheel spin at the same speed as registered on the Speedometer, or does it move twice as fast (assuming a frictionless, perfect environment).

OR does the Diffrential absorb the 50% power loss to the other wheel?

Some of the following links where posted as reference for various arguments, which may be worth reading before happily posting away:

Basic definition:

http://www.rctek.com/general/differential_svm_basic_description.html

Suspected solution:

http://www.rctek.com/general/differentials_how_they_work.html

More suspected solution:

http://www.gti-vr6.net/library/transmission/Ians_diff_page/differentials.htm

More suspected solution:

http://auto.howstuffworks.com/differential3.htm

And more:

http://www.imajeep.com/faq/Diffs_Types.htm

And yet more:

http://www-unix.oit.umass.edu/~tcroy/artic...ifferential.htm

Basically the query ends up... If youre tailshaft is turning at 800rpm, and you turn a corner, we all know the inside wheel will slow down (say to 600rpm) and the outside wheel will speed up (say to 1000rpm)... This will still average out to the tailshaft speed of 800rpm (1000+600/2 = 800). Does this mean that in a burnout situation, you can have the inside wheel doing 0rpm and the outside wheel doing 1600rpm, yet the tailshaft still only doing 800rpm? Can the orbital gears act as a 2:1 gearing? And if so, will the speedo still say, say, 60kph (tailshaft speed of 800rpm), yet the wheel will be doing 120kph (1600rpm)?

Not really relevant to skylines as such, except the 31s and below I guess? Or anyone with a worn LSD But an interesting question and I'd like to see a few more thoughts on it

[NismoGirl]

... I guess noone?

[jas/slo32]

its the same speed the gearbox drives the speedo. in my vl turbo when i snapped an axle 100kph was 3000rpm exactly the same as before the axle snapped theres just less load spinning one wheel so normally you can make more speed than spinning both wheels more load.with a snapped axle my old car could spin one wheel at 100kph bcaus it had no load.

[SteveL]

Does this mean that in a burnout situation, you can have the inside wheel doing 0rpm and the outside wheel doing 1600rpm, yet the tailshaft still only doing 800rpm?

In a word.....NO

Tailshaft 'speed' is fixed by the gearbox ratio (ie of the gear your in). In turn, axle 'speed' is fixed by the diff ratio (ie CWP, or crown wheel and pinion). These things can't change no matter what the wheels are doing....they are all mechanically linked by fixed gears. The fact that one wheel stops and the other wheel keeps spinning is the result of the design of an open diff centre, but the spinning wheel will always spin at more or less the rpm required for the speed indicated on the speedo...

[krawler]

I disagree, based on this flash animation:

http://static.howstuffworks.com/flash/differential.swf

While turning, the inside wheel slows down but the outside wheel also speeds up. Assuming the tailshaft spins at a constant speed (which we know it does), this means the inside wheel is now turning slower than the tailshaft, and the outside wheel is turning faster than the tailshaft, right?

Keep tightening the turning circle.. The inside wheel will get slower and slower, while the outside wheel will get faster and faster. Take this to the extreme, and you have the "inside" wheel stopped and the outside wheel spinning at a rate faster than the tailshaft.

My only thought is that if you turn a corner, does the outside wheel speed up to a point faster than the tailshaft, or does it remain at the speed of the tailshaft while the inside wheel slows down to a point below? This would mean the outside wheel is taking the drive - obvious because you cant drive the inside wheel slower than the tailshaft is turning. But when you put the foot down on a corner in an open center car, its the inside wheel that starts spinning.. Does this mean when you accelerate the drive changes from outside wheel to inside? Does the outside wheel therefore slow down?

[SteveL]

No problem with disagreement .....

...but the fact remains that the diff ratio is fixed and determines driven wheel speed. The free-wheeling effect of the open diff centre merely compensates for differences in turning radius + has the added side effect of allowing one wheel to spin as the other stops completely.

[bjgtir]

For a normal type differential (ie with spider gears) the input speed is the average of the 2 output speeds.

Saying the same thing in a different way, the tailshaft rotates at the average of the two output (half/drive etc) shafts.

The spider gears allow for rotation difference between the housing and the spur (output) gears. It is not fixed as someone mentioned above.

Just reread the original post. Ythe question "Does the diff absorb the 50% power loss" is simply imposible. A domestic electric cooking oven would typically have be a maximum rating of 8kW. Given that we are talking about cars with at least 100kW to the wheels, a diff would be cooked if it had to absorb 50kW.

[krawler]

Just to clarify, the original post wasnt mine and I totally disagree with the theory the diff "absorbs" half the power

Also, you say the tailshaft is an average of the two outputs... Putting some figures down, say the tailshaft is rotating at 1000rpm. This means the wheels could be turning at 500rpm + 1500rpm (divide by 2 = 1000rpm)... Or to the further extreme, 0rpm + 2000rpm (divide by 2 = 1000rpm). Which is what my thinking is, no matter how badly I worded it along the way.

I still cant find a definitive answer from a "reliable" (as such) resource I can quote from to help prove my thinking Unless of course, you work for a diff lapper or engineer or similiar?

[PTR33]

For a normal type differential (ie with spider gears) the input speed is the average of the 2 output speeds.

   Saying the same thing in a different way, the tailshaft rotates at the average of the two output (half/drive etc) shafts.

The spider gears allow for rotation difference between the housing and the spur (output) gears. It is not fixed as someone mentioned above.

   

Give this man a prize.

He's the only one who knows what he's talking about in this thread so far.

Think about it this way.

1. Take your open diff car and put both rears on axle stands with the wheels clear of the ground.

2. With the engine off. Put the gearbox in gear (or the trans in Park).

3. Rotate one of the wheels by hand (say at 10 rpm) and watch the other one.

4. As if by magic, you will see the other wheel turn in the OPPOSITE direction (but the same speed) to the way you are turning the wheel on your side.

5. It's not magic.... it's the diff centre (bolted to the crownwheel) being held stationary by the pinion (bolted to the tailshaft) which is locked by the gearbox and stopped engine.

The right axle rotating clockwise will transfer its motion through the side gear in the diff via the spider gears and then to the opposite side gear which will rotate in the opposite direction.

THE AVERAGE OF 10rpm and -10rpm IS 0rpm. The speed that the tailshaft is doing in this example.

If you don't believe me, go out to the garage and try it.

If you are showing 50kmh on the speedo and you have an open diff and one wheel is stopped, the other one will be doing 100kmh.

This is the fact of the matter.

Believe it and be educated or don't and remain ignorant. (look ignorant up in a dictionary before you get insulted, it means "Lacking information or knowledge" not "Stupid")

If you think this is hard to visualise, just wait intil you have to understand a stacked planetary gearbox...

[PTR33]

P.S. I'm an Aircraft Engineer and have seen the insides of gearboxes that would give you nightmares (a diff is just a type of gearbox).

I have also rebuilt several car diffs over the last 25 years. (mostly the one from the POS Cortina I had back in the 80's)

[ocean]

bjgtir, PTR33

This is true for a diff ratio of 1:1.

It is more accurate to say that the average of the speeds of the output shafts is equal to to the input speed divided by the diff ratio.

-Patrick

[krawler]

Give this man a prize.

THE AVERAGE OF 10rpm and -10rpm IS 0rpm. The speed that the tailshaft is doing in this example.

If you are showing 50kmh on the speedo and you have an open diff and one wheel is stopped, the other one will be doing 100kmh.

So at the end of the day... I'm right?

[Brackez]

'But when you put the foot down on a corner in an open center car, its the inside wheel that starts spinning..'

I know the thread question has been answered, but have to throw my 2c in: the above is a result of weight transfer. Side G-loading through a corner inevitably places more weight on the outside wheel than the inside. Hence, the friction forces between the road and tyre are higher for the outside wheel. This means that with an excess of power to both wheels, the easier wheel to spin is the inside one. This is the point of a LSD really, to reduce the tendency of the inside/lighter wheel to spin when it breaks traction by limiting the transfer of power from one wheel to another.

[krawler]

That quote was more to point out why the thought I'd voiced out loud was wrong, in terms of the outside wheel taking the drive and the inside wheel simply slowing down. That wasnt what my thinking was, just thinking some other theories out loud

Definately agree with it being weight transfer that makes it spin, as in an open center diff "the amount of torque is limited to the greatest amount that will not cause a wheel to slip". The weight taken off that side reduces traction and causes the wheel to spin. Once it starts spinning, any additional power will simply spin it faster as opposed to putting power down to the other wheel (as in an LSD).

[Brackez]

Ah, re-read your first post, gotcha . Bloody clever things LSDs, seen all the pics attempting to explain how they work, think I'd actually have to pull one to bits and stuff about with it to fully understand them tho.

[PTR33]

bjgtir, PTR33

This is true for a diff ratio of 1:1.

It is more accurate to say that the average of the speeds of the output shafts is equal to to the input speed divided by the diff ratio.

-Patrick

Yep, well spotted, neglected that bit of the equation. D'oh.

[V8skylineMAN]

as for the inside wheel spinning, thats is only a result of the weight of the car shifting off that wheel, hence less traction