MorayMackenzie

Andy,

Here is a model that helps me visualise the problem. Please explain to me how it's wrong. Be gentle, as its all my own work.

Suppose we were to fit 3 wheels onto a shared axle, each wheel being free to rotate independently on the axle.

Now we drive peg through the outer edge of the centre wheel and the one on the left, thus linking them as a pair.

We then drive another peg through the wheel in the centre and the wheel on the right, only this peg is driven through midway between the outer rim and the axle.

Now we apply a driving torque to the centre wheel (aka the ring gear).

Then we measure the torques being excerted at the rims, or indeed the bearings, of both the undriven wheels. The figures will be the same on both wheels.

This, I believe, models the epicyclic diff operating with no axle speed differential.

Where am I wrong?

Hugs and kisses,

Moray

MorayMackenzie

A naff attempt at illustration of the "model".

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[Edited by MorayMackenzie - 3/14/2003 4:58:30 PM]

Andy.F

Moray

I'm going to 'cheat' and scribble a diagram on a bit paper then scan it in. This is due to my graphic design skill level falling way behind yours :P

Andy

Adam M

andy if you are thirsty, I suggest you get some water.

No offence but I think I was right about the bet, you simply will not be able to accept if you are wrong. This is my understanding, not pats, why would I believe him, he is an electronic engineer whereas I have a degree in physics and 2 A levels in maths including mechanics. No offence, but my own brain is enough to cope with this problem without leaning on his.

are you honestly trying to tell me that if the diameter of the planet carrier is small you get more torque than if it is large?

Oh dear.

I will agree that the measured torque at a greater distance from the centre is smaller, but then this is simply connected to an axle coaxial with that of the ring gear.

Morays example is a good one. Here is another.

Suppose you have a flywheel that you want to take an output drive torque from, you would connect an axle to its middle, such as a drive shaft. Are you saying that if you stick a big wheel on the end of the axle first, and then attach that wheel coaxially to the flywheel, then you get less torque trasmitted down the axle just because the wheel is further from the centre of rotation?

I dont know what more to say to convince you but one things for sure, you wont be getting any crates of beer form me, not on this one anyway.





[Edited by Adam M - 3/14/2003 5:28:09 PM]

Adam M

In essence, you think that

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is different from

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or have I got that wrong?



[Edited by Adam M - 3/14/2003 5:49:28 PM]

Andy.F

This is really amusing (Any Lager will do)

With those qualifications I guess I should just throw the towel in now then

As I watch in disbelief as you dig the hole you are in even deeper I feel I must just confirm that we are all talking about the same thing ?

The Type R subaru centre diff of single reduction epicyclic design, on 'open' or 'unlocked' setting. (we can discuss 'locked' later ;D) Drive to the annular ring from the gearbox. Outputs to the front diff from the sun gear and to the rear diff from the planet gear carrier.

We're not talking about 3 wheeled barrows with bolts or bevel geared Lego LSD's

Please confirm before I continue with my scribbled diagram

Andy

No, the opposite actually, torque split is related to the radius of the carrier pick up on the planet gears.


[Edited by Andy.F - 3/14/2003 6:13:33 PM]

Adam M

Andy,

yes that is what we are talking about, and I think I can understand where the problem lies.

Despite you lack of willingness to provide answers to the things above, there is some logic to what you are saying about being able to impart a torque without the gears turning.

You seem to think that this torque can come from the diameter of the planetary carrier, I disagree.

I do think it can come from the ratio of the teeth of the planet gear to the sun gear even when they are not turning.

The crucial difference is that you are arguing a point based on dynamic torque distribution based on a sudden increase in torque, I am arguing based on a static torque distribution based on constant speed down a straight grippy road )

If you impart a sudden torque change when the diff is unlocked then the ratio of the gearing (not the distance from the centre) will allow a greater torque to be transmitted to the rear wheels in this case, this explains why unlocked dccd equipped cars spin their rear wheels.

Perhaps the steady state thing should be made clearer, as once you are talking about dynamic torque torque distribution becomes a moot point. Am sure you are well aware that it is possible for these diffs to exhibit between 99:1 and 1:99 depending on the state of slip of the front and rear wheels.

Dont know how to put this, I think you are wrong on one point (regarding the diameter of the planetary gear carrier), but will accept that the other bit was based on an undefined battlefield.

Will say I think I was wrong in that stationary gears can transmit a torque to the shaft (even if they dont do any work in the process) but you were wrong in thinking this torque came from their distance from the centre of the sun/ring, and not from the teeth ratio.

How about a pint at elvington (if my pile of bits is ready!)
Not an admission of defeat, just figured out that we are arguing about different things.

Andy.F

So far so good We're half way there

The radius of the planetary carrier is the radius at which the force on the planet gears is applied to the output shaft.

The same force at a different radius will result in a different torque output.

The radius of the planetary carrier centreline is obviously dictated by the size of the gears so the two are linked.

Do you agree with the part where I claimed that if you have the same gear size on the sun as the planet that yo do not get a 50:50 split ?

Andy

So am I

As I said earlier in the thread you can even go beyond that...into the -ve's
ie 120 : -20


[Edited by Andy.F - 3/14/2003 6:52:29 PM]

Adam M

no I do not.

I still see that from the point of view of mine and morays diagrams above, in that it doesnt matter where you remove the torque from.

The torque is still the force times the distance from the centre of the circle, the distance starts at the drive point of the ring gear, and ends in the middle of the axle. It doesnt matter how it gets there.

I can accept that the distribution of this torque between sun and planet carrier is affected by the ratio of teeth, but they are the only thing I can see as yielding a mechanical advantage, the radius of the centre line of the planetary carrier in my mind, would only have an effect because it is determined by the ratio of the teeth of the gears it is carrying to the teeth of the sun gear.

Try me with your force diagram. You never know, there is always a chance I am wrong.

Adam M

get what you are saying, but I dont believe you can separate the radius of the carrier from the teeth ratio. The radius is merely consequential. ie radius of run plus radius of planet.

negative torque is fine, but the reason I limited my numbers is that even when fully open, the lockup clutch does not allow the diff to be totally free.


Thats it, I have had enough. I am buying a modena centre atb diff, screw this dccd ****!



[Edited by Adam M - 3/14/2003 7:25:56 PM]

Andy.F

Consequential I agree but fundamental to the calculation as it is the 'lever' length for the rear torque moment. Think of the planet gear as its own lever, doubling the force on the carrier.

You still think equal size sun and planet gears will give 50:50 torque......hmmmm...sketch needed


I'll buy your centre diff regardless of how it splits the torque

[Edited by Andy.F - 3/14/2003 7:18:42 PM]

Adam M

and I'll buy your a six pack, and a crate if you can convince me!

Andy.F

A pair of spur gears is represented in the diagram by their pitch circles, which are tangent at the pitch point P. The meshing gear teeth extend beyond the pitch circle by the addendum, and the spaces between them have a depth beneath the pitch circle by the dedendum. If the radii of the pitch circles are a and b, the distance between the gear shafts is a + b. In the action of the gears, the pitch circles roll on one another without slipping

If a gear of pitch radius a has N teeth, then the distance between corresponding points on successive teeth will be 2ða/N, a quantity called the circular pitch. If two gears are to mate, the circular pitches must be the same. The pitch is usually stated as the ration 2a/N, called the diametral pitch. If you count the number of teeth on a gear, then the pitch diameter is the number of teeth times the diametral pitch. If you know the pitch diameters of two gears, then you can specify the distance between the shafts.

The velocity ratio r of a pair of gears is the ratio of the angular velocity of the driven gear to the angular velocity of the driving gear. By the condition of rolling of pitch circles, r = -a/b = -N1/N2, since pitch radii are proportional to the number of teeth. The angular velocity n of the gears may be given in radians/sec, revolutions per minute (rpm), or any similar units. If we take one direction of rotation as positive, then the other direction is negative. This is the reason for the (-) sign in the above expression. If one of the gears is internal (having teeth on its inner rim), then the velocity ratio is positive, since the gears will rotate in the same direction.

If the arm is fixed, so that it cannot rotate, we have a simple train of three gears. Then, n2/n1 = -N1/N2, n3/n2 = +N2/N3, and n3/n1 = -N1/N3. This is very simple, and should not be confusing. If the arm is allowed to move, figuring out the velocity ratios taxes the human intellect. Attempting this will show the truth of the statement; if you can manage it, you deserve praise and fame. It is by no means impossible, just invoved.

The trick is that any motion of the gear train can carried out by first holding the arm fixed and rotating the gears relative to one another, and then locking the train and rotating it about the fixed axis. The net motion is the sum of the two separate motions. To carry out this program, construct a table in which the angular velocities of the gears and arm are listed for each, for each of the two cases. The locked train gives 1, 1, 1, 1 for arm, gear 1, gear 2 and gear 3. Arm fixed gives 0, 1, -N1/N2, -N1/N3. Suppose we want the velocity ration between the arm and gear 1, when gear 3 is fixed. Multiply the first row by a constant so that when it is added to the second row, the velocity of gear 3 will be zero. This constant is N1/N3. Now, doing one displacement and then the other corresponds to adding the two rows. We find N1/N3, 1 + N1/N3, N1/N3 - N1/N2, 0.


Just to clarify things

Andy


[Edited by Andy.F - 3/14/2003 8:53:22 PM]

Adam M

that didnt help much to be honest. Not too hard to follow, just doesnt bring me to a conclusion that helps.

I still have an issue which I think it would be helpful if you explained to me directly.

The lockup clutch. The biggest hurdle that stood in my way of understanding this properly was the ability to transmit torque from ring to planets/sun without them all turning relative to one another.

I am fine with that now, but something doesnt add up still.

The lock up prevents the drive speeds of the two axles from differing, so prevents the torque over coming grip of the rear tires and thus prevent sit from spinning. but since you have made it clear that relative rotation is not required for a torque split, how does locking of the clutch change the torque distribution to 50:50 when it is locked?

As far as i can see from what has been said here, it doesnt, it just stops things getting out of hand as spinning wheels are no good to man nor beast (useful to speedway riders though ).

Does this mean that changing the torque distribution was never an intention of the dccd?

MelTypeR

iTrader: (4)

Great debate but i'm still .


Mel.

Andy.F

Adam

When the diff clutch 'locks' it takes over the control of the torque transfer from the gears. The gears have no role to play as the torque is all transmitted via the locked clutch which has no mechanical advantage. You could actually remove the gears from the cluster and the car would still transfer torque.

When the diff lock is in the midway position then you are adjusting between 66:34 split and locked, so on a grippy surface this will tend towards 50:50.

The key part here is that during diff lock running the torque split is determined by the available traction whereas when it is 'open' there will always be a 66:34 split regardless of conditions (disregarding drivetrain inertia and landing forces after a yump for this simplification)

Andy

Glenn Coombs

Great discussion so far. Not sure I'm following everything though. Can somebody just say if the following is true:

An active center diff equipped scooby is more likely to spin the rear wheels whereas a "normal" scooby is more likely to spin the front wheels.

That is how I thought the active center diff worked but after reading this thread so far I'm not sure if that is true after all. Seems like it should be as otherwise I can't see the point of having an active center diff...

Adam M

oh no, andy, its all gone away from me again!!!

you said,

"The key part here is that during diff lock running the torque split is determined by the available traction whereas when it is 'open' there will always be a 66:34 split regardless of conditions"

this brings me to what I was saying before about steady state torque distribution. Surely if an axle is splipping then there is no reaction force (cf. torque) from the road, which means that 100% of the available torque imparted by the gearbox to the ring gear will go through the wheels which have drive?

Aside from that paragraph above, I am still a little uneasy with what you have said about the diff lock.

I was the one all along who was under the impression that if there is no relative motion of the output shafts then there could be no torque. I was of the impression that all components of the epcyclic diff were spinning at the same rate then you could replace them by a rigid gear. This I now know is not the case, but you said above, that when you lock the diff you are keying the otput shafts together so that in that situation you can substitute the epicyclic for a rigid gear.

Again if you consider a straight flat grippy road, two identical cars, in one car, a locked dccd, in the other an open dccd. Both accelerating hard, but neither breaking traction so that at any point, the front and rear axles are always turning at the same speed.

from what you have said, the open diff has a 64:36 torque split, but the locked one can be considered as a rigid gear with a 50:50 torque split. The job of the clutch is to prevent relative motion by mechanically locking front and rearn output shafts together, but if there is no slippage then its job is done for it by the road, so we have a redundant lock up clutch. (or do we? - I think this may be the crucial point).

In reasoning this out in my head, I think it is all becoming clearer. The sun gear and carrier may well be imparting different amounts of torque to the front and rear shafts, but because they can no longer be considered independent due to the viscous lock up clutch the torques they contribute can merely be added together, the fact that different proprtions are being trasnmitted via different gears is totally irrelevant if they are transmitting to what is effectively a rigid axle.

I really hope this is right, I would much prefer it if you could put some over complicated maths that made no physical sense in front of me and prove a final point, I know its odd, but if the numbers could work it would make more sense to me, and might actually click. As it stands its hard to visualise.

In one case the road/tire combination is effectively making it a rigid axle, in the other a lockup clutch does the job, but in the first case, the axles can distribute torque unevenly, and in the second they are fixed to produce the same torque.

is this because the effective axle coupling provided by the road is occuring after the axles get their share of the torque?

Andy.F

You got it Adam

Coupling the axles via the road only stops relative rotation and has no effect on torque transmitted. Coupling the DCCD via a clutch eliminates the torque multiplication from the epicyclic gears (they are locked together)
The torque transmitted then becomes proportional to the reaction (traction) at either axle.

Andy

Adam M

yay.



thanks andy.

its like a splinter in my mind has finally been tweezered out!

that definitely deserves the beer/lager.

name your brand, and I will see you at elvington (unless the engine rebuild god smites me again (for the thrid time).

Andy.F

No worries Adam I work for a Brewery so no need really ........hic

Andy

MorayMackenzie

This is an interesting page on subarus own site... you get to see the actual internals of one of these diffs. like the way teh planet gears are made up of two gears on a shaft.

http://www.subaru-global.com/about/parts/06.html

If it isnt immediately apparent, choose Centre differential gear on the menu to the left. I liked the quote about it being a difficult unit to understand, even for mechanical engineering grads. Fell less stupid for taking time to understand it now.

Moray

[Edited by MorayMackenzie - 3/24/2003 2:07:33 PM]

Dave R

John Felstead's written many a post about this. Do a search on centre diff/DCCD etc...

Basically, with the the diff unlocked (lower green setting) the centre diff sends the majority of the engine torque to the rear wheels. As you wind the lock on, it steadily reduces the torque biasing action of the centre diff, so that on full lock, the diff sends equal amounts of torque to the front and rear axles, BUT this solid centre diff will now not act like a diff. This means it will now not allow the rear wheel to rotate slower during corners, so producing understeer/driveline wind-up.

On dry tarmac, generally leave it totally unlocked, but if the surface becomes wet/slippery/icy, wind the lock on 1,2 or max 3 clicks. I would only use more in snowy conditions or on gravel roads, or as I did last month, use it to reverse out of a ditch I had to use as an escape road when the bird in front stopped and I hit ice.

Hope this makes sense and John concurs.

Cheers
DR

[Edited by Dave R - 3/12/2003 9:46:31 AM]