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Seamless Gearboxes


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#201 24gerrard

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Posted 28 February 2012 - 14:45

You just don't get it do you??

There is some 70 - 110 degrees for the lower gear to be disengaged after both gears are engaged before everything locks up.

There are people who read these and similar forums in a quest for knowledge, stop messing their minds up.


Keep studying the graph Cheapracer, try to find the 30 to 50 ms gearshift.
Stop looking at the one twentieth of a second bracket it shows.
The actual shift is not shown on the graph.
Almost all the changes in rpm and torque shown on the graph are not caused by any gearshift.

Edited by 24gerrard, 28 February 2012 - 14:55.


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#202 MatsNorway

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Posted 28 February 2012 - 16:09

Have you noticed 24 that no-one agrees with you,


Not correct. I believe there has to be a gap during upshift. its tiny but its there. In practical terms its probably not noticable..

Just because hes a bit crazy normally it doesn`t mean he is not correct this time.

And no one has commented about downshift! i would like to hear about that.


#203 ferruccio

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Posted 28 February 2012 - 16:29

The barriers to any radicaly new vehicle engineering concepts in F1 other than aero are obvious.

Really?? I did not know that :rolleyes: . Wow. But I do know this debate about F1 and radical engineering ideas have been discussed and raised at nauseum in this forum for years. Actually there is no debate at all since we know the sport of F1 well enough to know and accept it for what it is. Some of us may even be directly involved. I suspect you're probably in the wrong decade for following F1.

The point of the thread was very simple, to prove that the terms 'seamless' and 'zeroshift' do not apply to any form of stepped layshaft gearbox.

I'll let those with data do the talking. It would not hurt if you did the same. Many do come here to learn rather than hear someone hell bent on proclaiming they are more clever.

The fact that the layshaft stepped gearbox concept is the same today as it was in the 1890's is my real beef.
Tarting it up will not stop it from being long out of date.

The round wheel is far longer out of date yet it is still in use today.

This thread has run its course :down:


#204 Tony Matthews

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Posted 28 February 2012 - 16:35

Not correct. I believe there has to be a gap during upshift. its tiny but its there. In practical terms its probably not noticable..



And no one has commented about downshift! i would like to hear about that.

The ratchet is locked for downshifts so that the gear can drive the mainshaft in over-run.


Mats, if you don't want to believe an F1 gearbox specialist, or any other engineers on this thread bar one, that's up to you. Reading the post from rachael might help.

#205 cheapracer

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Posted 28 February 2012 - 16:52

Not correct. I believe there has to be a gap during upshift. its tiny but its there.


I'm fairly blank on gearboxes.



Well it would be rude for me to insinuate that you weren't telling the truth. :lol:

#206 NTSOS

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Posted 28 February 2012 - 18:07

Not correct. I believe there has to be a gap during upshift. its tiny but its there. In practical terms its probably not noticable..

Just because hes a bit crazy normally it doesn`t mean he is not correct this time.

And no one has commented about downshift! i would like to hear about that.


Mats, all you have to do is ask.......I would imagine that Zeroshift would gladly send you a detailed explanation of the graph that I posted to hopefully circumvent the mis-information!

Hint....that is a damped shift graph for a passenger car, but damped or undamped, there is no torque flow interuption or gap.

Also, please ask about downshifting......it is supposedly easier than upshifting with the Zeroshift mechanism!

John


#207 MatsNorway

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Posted 28 February 2012 - 18:47

Good news for you guys.. i have chanced my mind. I DO think its doable.

SO it works something like this then?

when the second arm hits the middle part the first arm gets released from transfering torque and can then be withdrawn within that small amount of space and time it has before the main part behind hits it. hence -> no gap -> no lift needed -> hugely impressive if so


john i have sendt a mail to Zeroshift company.

I also asked about what sampling rates we can do on our own equpiment that we use on trains and it was easily above 10 000 measurements pr sec.


Edited by MatsNorway, 28 February 2012 - 18:54.


#208 24gerrard

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Posted 28 February 2012 - 21:25

Mats, if you don't want to believe an F1 gearbox specialist, or any other engineers on this thread bar one, that's up to you. Reading the post from rachael might help.

The ratchet is locked for downshifts so that the gear can drive the mainshaft in over-run.


Only problem with that is that the mainshaft (output shaft) drives the gear in overrun.
Rachael has it the wrong way round.


#209 rachael

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Posted 28 February 2012 - 22:12

Only problem with that is that the mainshaft (output shaft) drives the gear in overrun.
Rachael has it the wrong way round.

This is correct (gerrard had to get lucky at some point) - in over-run the torque flow is from the mainshaft, through the ratchet into the mainshaft gear and then layshaft gear. Note the zeroshift system is not the same as that used in F1 boxes.

#210 NTSOS

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Posted 29 February 2012 - 00:05

Good news for you guys.. i have chanced my mind. I DO think its doable.

SO it works something like this then?

when the second arm hits the middle part the first arm gets released from transfering torque and can then be withdrawn within that small amount of space and time it has before the main part behind hits it. hence -> no gap -> no lift needed -> hugely impressive if so


john i have sendt a mail to Zeroshift company.

I also asked about what sampling rates we can do on our own equpiment that we use on trains and it was easily above 10 000 measurements pr sec.


What kind of trains?

Anyway, if this is correct........to disengage 1st gear you must first engage 2nd gear. The engaging of 2nd gear allows 1st gear to unload and disengage......no possible way for a torque hole with a properly functioning Zeroshift mechanism!

Although, I am not clear if a fluid damper between the gear dogs and the gear is used, would the undamped model's "fly-by-wire clutch control unit that regulates the speed difference across the clutch during the shift event, allowing the energy spike to be monitored and conditioned to suit the driving conditions" still be necessary!

John

Edited by NTSOS, 29 February 2012 - 01:32.


#211 gruntguru

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Posted 29 February 2012 - 00:25

I think everyone here would agree that the Zeroshift does not have both gears pushing the bike on the same time. When 2nd goes in, first gear slides out because of the ramping.

No there is a brief period where both gears are driving simultaneously. The torque transmitted by either gear is never static during this period, the lower gear has reducing torque and the higher gear has ingreasing torque.

If you understood Engineguy's analogy in post #178 and imagine the bumpers on the pushing trucks to be thick rubber, you will see what I mean/ Whether we are talking rubber or hardened steel, the result is the same - only the deflections and the duration of the handover period are different.

Alternatively look at Cheapy's post #180, diagram #2 with both pawls in conact simultaneously. Perhaps imagine them with rubber faces. I hope you get the idea.

#212 gruntguru

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Posted 29 February 2012 - 00:29

The only reason there is any gap in torque or speed on the graph (that I didn't post) is from metal elasticity, lubricant dispersing, dog's undercut engaging and/or engineered in momentary torque drop to protect the gear train from the "slam" when dark red pawl hits red dog.

I think the last on this list is the only explanation. If there was no slip allowed, the zeroshift would produce nothing in the way of a gap or dip - just a torque spike - immediately after the shift.

#213 TC3000

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Posted 29 February 2012 - 04:18

No there is a brief period where both gears are driving simultaneously. The torque transmitted by either gear is never static during this period, the lower gear has reducing torque and the higher gear has ingreasing torque.

If you understood Engineguy's analogy in post #178 and imagine the bumpers on the pushing trucks to be thick rubber, you will see what I mean/ Whether we are talking rubber or hardened steel, the result is the same - only the deflections and the duration of the handover period are different.

Alternatively look at Cheapy's post #180, diagram #2 with both pawls in conact simultaneously. Perhaps imagine them with rubber faces. I hope you get the idea.



Maybe it helps, to picture two small springs of equal stiffness in Cheapy´s illustrative sketch, to account for elasticity in the material and in the drive train in general.
In the first picture, the spring in front of the "blue square/dog", would be fully compressed, pushing the light blue trapez, transmitting the torque.

Now, when the "red square/dog" enters into the gap between the dogs, it´s spring is fully extended, but as it is faster (higher gear) it closes onto the brown trapez (dogring) in front of it.
In the moment, the spring of the "red square/dog" makes contact with the brown trapez (dogring) it´s spring will start to compress, and it will start to transmit torque.

For the same amount the spring infront of the "red square/dog" compresses, the spring in front of the "blue square/ dog" will relax.
As long as both springs are in contact, both will share the total torque tranfered.
There will be a point when both springs have the same length, this is the point where both dogs/gears a driving/transmitting equal torque.

After this point, the spring in front of the "red square/dog" will compress more (spring gets shorter) and the one in front of the "blue square/dog" relaxes futhers, get´s longer.
This process continues until, the spring in front of the "blue square/dog" is fully relaxeed/extented and the spring in front of the "red square/dog" is fully compressed.

If this point is reached all the torque to drive the car/bike, whatever is supplied by the "red square/dog" (higher gear) and the "blue square/dog" (lower gear) is fully unloaded.

Force at both ends of an spring can be defined as F = s x K ( displacement of the spring (s) multiplied by spring stiffness K) e.g.: 5000 N = 5 mm X 1000 N/mm
the example given is a case of two springs in parallel.
At the moment when both springs sharing the load it would be: 5000 N = (2.5 mm x 1000N/mm) + (2.5 mm x 1000N/mm)
Maybe this analogy makes it easier for some to understand the concept.

Edited by TC3000, 29 February 2012 - 04:27.


#214 cheapracer

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Posted 29 February 2012 - 05:50

SO it works something like this then?

when the second arm hits the middle part the first arm gets released from transfering torque and can then be withdrawn within that small amount of space and time it has before the main part behind hits it. hence -> no gap -> no lift needed -> hugely impressive if so


Yup, welcome to the dark side - we have cookies!


#215 24gerrard

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Posted 29 February 2012 - 09:07

Damn I must look at some 'springs' someone has worked out a way to transfer 100 percent of an F1 engine's torque with them.

#216 24gerrard

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Posted 29 February 2012 - 10:05

This is correct (gerrard had to get lucky at some point) - in over-run the torque flow is from the mainshaft, through the ratchet into the mainshaft gear and then layshaft gear. Note the zeroshift system is not the same as that used in F1 boxes.


Sorry Rachael easy mistake to make. :kiss:

The slow low rpm shift shown on the graph is obviously not F1.
All the various types of mechanisms used for semi and auto shifts with layshaft gearboxes are similar though.
The effect produced at shift overlap is much the same for all of them, it is the time taken to control component rpm and torque that varies.

I dont know of a current system that uses control over the layshaft (with seperate gears free to rotate), at the same time as the mainshaft and gears though.
We played with that idea in the late 70's Rachael has it been tried again with the later selector mechanisms?
Should be a piece of cake with the electronic control available today.
Cuts the shift speed in half and reduces frictional losses and oil windage.
We also experimented with moving the gears along the shafts to assist engagement, using the side forces from the helical gears teeth.
That also speeded up the shifts.

#217 TC3000

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Posted 29 February 2012 - 11:43

Damn I must look at some 'springs' someone has worked out a way to transfer 100 percent of an F1 engine's torque with them.


well look at the two shafts as (torsion)springs, and you are on your way, most things will turn into a spring, or at least having a spring component to them in engineering terms if you look close enough/apply enough force.
May helps to understand where the "oscillations" in the (propshaft) torque graph come from.
springs/elasticity will help soften/lower the shock laod from impact, I don't think that this is so difficult to understand.

and while F1 gearboxes are interesting "piece of kit", they don't see all that much torque, compared to an LMP gearbox in an LeMans diesel for example.

BTW:

The effect produced at shift overlap is much the same for all of them


- intersting comment of yours, some could read this as two gears "engaged" at the same time. (note my use of the term "engaged" - not "fully engaged")

I see you understand the general principle:

Two gears are not FULLY engaged at the same time. The ratchets or dog rings may be, that is not FULL gear engagement.
In fact it is some 70 to 110 degrees away from FULL engagement.


but like to keep argueing for arguing sake, about the difference between "engaged" and "fully engaged" - fair enough, let's throw some more semantics into the mix.
At one point during the "handover" both gears are "equally engaged", when both springs have the same length, therefore both gears supplying 50% of the total torque.

Still, you will need to explain, where the "gap" in this process is, which you keep on banging/arguing about.

Edited by TC3000, 29 February 2012 - 13:19.


#218 cheapracer

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Posted 29 February 2012 - 12:29

Damn I must look at some 'springs' someone has worked out a way to transfer 100 percent of an F1 engine's torque with them.


I wish the ignored posts didn't show up in replies ....

There's barely a car, truck, motorcycle or any other vehicle from 2hp to 2000hp that doesn't have some form of "spring" designed into the drive, 99% of all cars on the road have springs in the clutch plate....

Posted Image

F1 rotating inertia is so low and ratios so close now that the shock would be very minimal and don't they have need for shock absorption although they certainly did up to the 70's F1 as with this '60's Lotus with it's large rubber donut drives...

Posted Image

and in the '70's the donuts ended up on the outsides for some reason (F3 shown)..

Posted Image

Edited by cheapracer, 29 February 2012 - 12:33.


#219 24gerrard

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Posted 29 February 2012 - 13:04

Hmmm.
So how much torque does it take to 'wind' these things up and how much torque goes into accelerating the vehicle at this 30 to 50 ms instant?


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#220 24gerrard

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Posted 29 February 2012 - 13:09

[quote]name='cheapracer' date='Feb 29 2012, 12:29' post='5556144']
I wish the ignored posts didn't show up in replies ....

There's barely a car, truck, motorcycle or any other vehicle from 2hp to 2000hp that doesn't have some form of "spring" designed into the drive, 99% of all cars on the road have springs in the clutch plate....

Posted Image[/quote]

I know it was me who suggested that Leyland fit a radialy sprung clutch disk to the Metro when they used the original A series power unit from the mini in it (the mini had a solid plate).
I played around with radialy sprung clutch disks and radial dampers of various types with our clutch flite development that resulted in the 25 speed bevel epicyclic unit we wanted to use in F1.


#221 24gerrard

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Posted 29 February 2012 - 13:28

[quote]name='TC3000' date='Feb 29 2012, 11:43' post='5556095']

but like to keep argueing for arguing sake, about the difference between "engaged" and "fully engaged" - fair enough, let's throw some more semantics into the mix.
At one point during the "handover" both gears are "equally engaged", when both springs have the same length, therefore both gears supplying 50% of the total torque.[/quote]

I dont think of it as argueing for arguing sake.
Engaged is engaged. If it is not it is either disengaged or partialy engaged.
Torque is indeed split between the gear disengageing and the gear engageing.
However just past the center of shift overlap, the gear disengageing no longer transfers torque (the bullets spring out) and for a very very short time the gear engageing is not driven by the laygear because the spring is still compressing.
There follows a torque spike at Full engagement.
If you magnify this point and measure at the output, you will see a reduction of torque transfer just before this spike.
A seam.

#222 TC3000

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Posted 29 February 2012 - 13:37

Hmmm.
So how much torque does it take to 'wind' these things up and how much torque goes into accelerating the vehicle at this 30 to 50 ms instant?


only a problem, in a conventional gearbox.
in this case it does not matter, how long it takes to "wind up the spring" as the lower gear, with it's fully compressed spring still does the driving
at this point.

As the spring in the higher gear "winds up"/compresses it takes off load from the spring in the lower gear, which is still contributing at the time
unless it's spring is fully unloaded/uncompressed, at which point all the driving is done by the higher gear.

mm spring compression for two 1000 N/mm springs and total force:

lower gear					  higher gear		   lower gear						higher gear			  total
   5 mm							  0 mm			   5000N								  0N				 5000N
   4 mm							  1 mm			   4000N							   1000N				 5000N
   3 mm							  2 mm			   3000N							   2000N				 5000N
   2 mm							  3 mm			   2000N							   3000N				 5000N
   2 mm							  3 mm			   2000N							   3000N				 5000N
   0 mm							  5 mm				  0N							   5000N				 5000N


#223 TC3000

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Posted 29 February 2012 - 13:46

However just past the center of shift overlap, the gear disengageing no longer transfers torque (the bullets spring out) and for a very very short time the gear engageing is not driven by the laygear because the spring is still compressing.


A partly compressed spring is still transmitting force, so where is the point when the force goes to "zero"?
This would only occur if there is a "gap" time when the first spring is fully extented, but the second spring, does not start to compress.
The situation, you have with a "normal" gearbox, but with not in the case of "zeroshift" or an F1 gearbox with two shift barrels, which can
much two dog rings into the "engage" position at the same time.
That does not mean that both gears are "fully engaged" at the same time, but two "partly engaged" gears can still transmitt force, just
as two "partly compressed springs in parallel" do.

#224 24gerrard

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Posted 29 February 2012 - 14:32

A partly compressed spring is still transmitting force, so where is the point when the force goes to "zero"?
This would only occur if there is a "gap" time when the first spring is fully extented, but the second spring, does not start to compress.
The situation, you have with a "normal" gearbox, but with not in the case of "zeroshift" or an F1 gearbox with two shift barrels, which can
much two dog rings into the "engage" position at the same time.
That does not mean that both gears are "fully engaged" at the same time, but two "partly engaged" gears can still transmitt force, just
as two "partly compressed springs in parallel" do.


I totaly agree.
So how much torque is driving the vehicle and how much is winding up the spring and the output components?

#225 24gerrard

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Posted 29 February 2012 - 14:40

only a problem, in a conventional gearbox.
in this case it does not matter, how long it takes to "wind up the spring" as the lower gear, with it's fully compressed spring still does the driving
at this point.

As the spring in the higher gear "winds up"/compresses it takes off load from the spring in the lower gear, which is still contributing at the time
unless it's spring is fully unloaded/uncompressed, at which point all the driving is done by the higher gear.

mm spring compression for two 1000 N/mm springs and total force:

lower gear					  higher gear		   lower gear						higher gear			  total
   5 mm							  0 mm			   5000N								  0N				 5000N
   4 mm							  1 mm			   4000N							   1000N				 5000N
   3 mm							  2 mm			   3000N							   2000N				 5000N
   2 mm							  3 mm			   2000N							   3000N				 5000N
   2 mm							  3 mm			   2000N							   3000N				 5000N
   0 mm							  5 mm				  0N							   5000N				 5000N


3.69 ft Ib then.
I dont think the car will accelerate much on that.
When the high gear is fully engaged however, there is a spike, showing torque transfer.

Edited by 24gerrard, 29 February 2012 - 14:41.


#226 NTSOS

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Posted 29 February 2012 - 20:12

Posted Image

This graph shows a damped shift for a passenger car. There is a fluid damper between the gear dogs and the gear.

Prop shaft torque is the blue curve.

At its lowest point it drops to 225Nm. It never drops to zero. The actual shift (first contact of 2nd gear) takes place at 6.53 seconds.

This is easy to see because that is when the damper starts to move. The damper is the red curve.

The damper slows down the time it takes for the handover from 1st to 2nd. For a short period they are both driving the car but at the end of the damper stroke it is 100% 2nd gear.

There is no torque interruption not even for one tenth of a millisecond. None at all.

Immediately after the shift event, (first contact for 2nd gear) the prop shaft torque starts dropping down to the 2nd gear level.

This is a damped process so it takes from 6.53 seconds to 6.57 seconds (40 milliseconds) to drop to the 2nd gear level. This time to drop is best seen on the red curve.

If you look at an undamped shift curve you will see the initial torque spikes up after the shift as the inertia hits. That is the reason for the damper on passenger cars.

Fly by wire clutch control is still used with the damped version to slightly reduce clutch pressure before the shift. This prevents the damper from needing to be large enough to cater for engine inertia. The clutch control is also used for launch.


#227 24gerrard

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Posted 29 February 2012 - 21:34

You could always start by asking why the green trace, output shaft rpm, remains constant from one end of the graph to the other.
The only 'drop' is at the shift overlap.
For the shift shown I could use a broom handle in a bowl of custard to do a better job.

#228 gruntguru

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Posted 29 February 2012 - 22:32

I totaly agree.
So how much torque is driving the vehicle and how much is winding up the spring and the output components?

Since the components being wound up are in series with the drive to the vehicle the answers to your two questions are 100% and 100%. The only way torque can be lost along the path is in acceleration of significant rotating masses.

#229 gruntguru

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Posted 29 February 2012 - 22:37

You could always start by asking why the green trace, output shaft rpm, remains constant from one end of the graph to the other.
The only 'drop' is at the shift overlap.
For the shift shown I could use a broom handle in a bowl of custard to do a better job.

The green trace is not constant. It is rising with a slope of about 1/3 that of the input shaft trace (since the gear ratio is about 3:1) and about 1/2 after the shift.

#230 24gerrard

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Posted 29 February 2012 - 23:05

The green trace is not constant. It is rising with a slope of about 1/3 that of the input shaft trace (since the gear ratio is about 3:1) and about 1/2 after the shift.


I beg to differ, it is constant.
Near as makes no difference anyway.
Means the vehicle is barely accelerating.
The engine rpm is decaying with clutch slip which either means electronic control to reduce engine rpm or a trailing throttle.
There is some 2000 rpm difference between the engine rpm and the input shaft rpm, that is NOT a slight smoothing it is a disengaged clutch.
The engine rpm will come together with input shaft rpm off the graph when the clutch re-engages.
Until then engine rpm cannot be controlled by the gearbox.
Rotational inertia must make up most of the torque input for this shift.
I suppose the rear brakes could be binding or there might be some lead shot in the boot/trunk.

Edited by 24gerrard, 29 February 2012 - 23:34.


#231 NTSOS

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Posted 29 February 2012 - 23:16

You could always start by asking why the green trace, output shaft rpm, remains constant from one end of the graph to the other.
The only 'drop' is at the shift overlap.
For the shift shown I could use a broom handle in a bowl of custard to do a better job.


Well, of course you could, but I have noticed that you "seamlessly" transitioned away from torque flow interruption to torque flow reduction...some progress there! Sorry, not going to argue with you and I'm not going to do your homework for you!

-The actual shift is not shown on the graph.
Almost all the changes in rpm and torque shown on the graph are not caused by any gearshift.

-Hahaha and there I was thinking you clever guys could read a graph.

-Take a good hard look.
The gear shift on the graph shown takes one twentieth of a second.
Which accounts for you guys being unable to see the drop to zero torque on the output.
BECAUSE THE ACTUAL SHIFT TAKES 50 MS.


-This would mean you are saying the lower gear is fully engaged at the same time as the next higher gear.
This is impossible.

-You will have to explain to me one day how you trick physics and engage two gears at the same time.

-There is NOT an uninterupted flow of torque through the shifts with one of these gearboxes.

-There IS a gap in torque 'flow' in so called 'seamless' layshaft gearboxes.

-I do not care if the gap is half that in time as for the 'big bang', it is still a gap in torque transfer.

-Think again. The shift is very fast but the lack of a gap is an illusion .

-Shift completed and NO GAP in torque delivery.
Incorrect

-The bullets engage with the cut out teeth in the gears.
During the shift the ends of the bullets 'slide' along the gaps between these cut outs.
There has to be a gap in torque to allow the lower gear driving bullets to disengage.
This has GOT to happen before the higher gear is FULLY engaged, otherwise you would be trying to put max torque through two gear sets rotating at different speeds.
During this friction contact, torque is only used for gear shift actuation not to drive the vehicle.
It is a very very smal time period but it has got to exist.

John





#232 gruntguru

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Posted 29 February 2012 - 23:44

I beg to differ, it is constant.
Near as makes no difference anyway.
Means the vehicle is barely accelerating.
The engine rpm is decaying with clutch slip which either means electronic control to reduce engine rpm or a trailing throttle.
There is some 2000 rpm difference between the engine rpm and the input shaft rpm, that is NOT a slight smoothing it is a disengaged clutch.
The engine rpm will come together with input shaft rpm off the graph when the clutch re-engages.
Until then engine rpm cannot be controlled by the gearbox.
Rotational inertia must make up most of the torque input for this shift.
I suppose the rear brakes could be binding or there might be some lead shot in the boot/trunk.

As best as I can measure off the image, the otput shaft speed is 1100 rpm at the beginning of the trace and 1146 rpm at the start of the shift.
Over the same interval the engine speed goes from 3284 to 3418 and the input shaft from 3284 to 3359. This yeilds a gear ratio of 2.95 at the beginning and 2.93 or 2.98 at the start of the shift depending on whether you believe the engine speed sensor or the input shaft sensor.

Strange that a transmission expert with your years of experience could eyeball that chart and reach such an incorrect conclusion.

Edited by gruntguru, 29 February 2012 - 23:47.


#233 gruntguru

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Posted 01 March 2012 - 00:05

I beg to differ, it is constant.
Near as makes no difference anyway.
Means the vehicle is barely accelerating.

This may be unfamiliar but lets take an engineering approach. During the time prior to the shift the output shaft has accelerated from 1110 rpm to 1146 rpm ie 3.2%. Lets assume the 2.95 gear ratio is first gear and 3,250 engine rpm is 30 km/hr. So the vehicle has accelerated from 29.07 kph to 30 kph in 0.23 sec. This is equivalent to 0-60 in 14.8 sec ie a moderate acceleration rate (not "barely accelerating" as you stated) which would also be consistent with 3,000 rpm shift points in the transmission.

Edited by gruntguru, 20 March 2012 - 23:33.


#234 24gerrard

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Posted 01 March 2012 - 09:36

Propshaft rpm only shows an increase on this graph after the shift overlap as the clutch begins to re-engage.
Torque measured at the propshaft also measures torque from output inertia, this masks what is going on at the shift overlap.
The graph (if it were measuring torque at the shift overlap, which it is not) is also at to large a scale to show the torque drop to zero at the overlap, after the low gear bullets disengage and the springs wind up just before the full engagement spike of the high gear bullets. (or ratchet teeth)

Edited by 24gerrard, 01 March 2012 - 09:38.


#235 24gerrard

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Posted 01 March 2012 - 09:52

This may be unfamiliar but lets take an engineering approach. In the time prior to the shift the output shaft has accelerated from 1110 rpm to 1146 rpm ie 3.2%. Lets assume the 2.95 gear ratio is first gear and 3,250 engine rpm is 30 km/hr. So the vehicle has accelerated from 29.07 kph to 30 kph in 0.23 sec. This is equivalent to 0-60 in 14.8 sec ie a moderate acceleration rate (not "barely accelerating") which would be consistent with 3,000 rpm shift points in the transmission.


As near as dammit all of this propshaft rpm increase happens after the shift as the clutch re-engages.
It is also interesting to see that the propshaft rpm decreases at the shift overlap.
How can torque be transfered through the shift components to the output when the propshaft is trying to reverse its rotation?

Edited by 24gerrard, 01 March 2012 - 09:58.


#236 gruntguru

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Posted 01 March 2012 - 10:13

As near as dammit all of this propshaft rpm increase happens after the shift as the clutch re-engages.

Can't you read? Didn't you even bother to verify my measurements? The entire rpm increase I measured was PRIOR to the shift.

It is also interesting to see that the propshaft rpm decreases at the shift overlap.

Not the propshaft rpm, the "output" rpm. Connected to the propshaft sure, but the propshaft is a major source of elasticity in the system. The system is experiencing a dramatic drop in torque during the time interval you are looking at. The reduction in torque is acompanied by an "unwinding" of the drivetrain and therefore a temporary reduction in speed at the output shaft. The roadspeed will certainly be increasing at the same time and all parts of the drivetrain in between the output shaft and the road will be accelerating (+ve and -ve) at values in between.

How can torque be transfered through the shift components to the output when the propshaft is trying to reverse its rotation?

Ever seen a truck going up a steep hill - engine providing massive torque and all components of the drivetrain decreasing in speed.


You need to brush up on the physics of transmissions.

Edited by gruntguru, 01 March 2012 - 11:09.


#237 GreenMachine

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Posted 01 March 2012 - 10:37

You need to brush up on the physics of transmissions.


Ouch!  ;)

#238 cheapracer

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Posted 01 March 2012 - 11:06

You need to brush up on the physics of transmissions.


Anarchist.


#239 24gerrard

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Posted 01 March 2012 - 11:34

Posted Image

propshaft rpm 1100rpm start of graph, 1125rpm (start of shift bracket, not overlap) 1125rpm at end of shift bracket, 1200rpm end of graph.
propshaft rpm 1000rpm at shift overlap.

Edited by 24gerrard, 01 March 2012 - 11:36.


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#240 24gerrard

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Posted 01 March 2012 - 11:43

Ever seen a truck going up a steep hill - engine providing massive torque and all components of the drivetrain decreasing in speed.

You need to brush up on the physics of transmissions.


So tell me why the torque is also dropping at this point?
You are probably right I should brush up on transmission physics.
I conceed it is the output rpm and not propshaft rpm.
It would be nice to see the torque figures at the output gear shift ring and bullets (ratchet) rather than at the propshaft.
It would also be nice to see a trace for engine torque and what is causing the drop between the start of the graph and the shift bracket.
It doesnt look like a power on upshift to me.
You could work it out with the gear ratios grunt (not very high is it) but that doesnt give enough data for the shift.

Edited by 24gerrard, 01 March 2012 - 12:28.


#241 gruntguru

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Posted 01 March 2012 - 22:28

propshaft rpm 1100rpm start of graph, 1125rpm (start of shift bracket, not overlap) 1125rpm at end of shift bracket, 1200rpm end of graph.
propshaft rpm 1000rpm at shift overlap.

OK, some more calculations here so read slowly. We will use your figures to avoid arguments ;) . Output shaft decelerates during shift from 1125 to 1000 rpm and back. So average speed of the output shaft during this period is about 1063 rpm. Assume the other end of the driveline (the road) continues at constant speed equivalent to 1125 rpm at the output shaft. The "unwinding" of the driveline will be delta rpm x time = (1125 - 1063) x 0.044sec x (1min/60sec) = 0.045 revolutions = 16.4 degrees. Sounds about right to me. Go chain a car to a wall. Apply 250Nm to the gearbox output to tension the chain, increase it to 450 Nm and measure the rotation during the increase.

EDIT. The oscillations in oputput shaft speed after the shift are confirmation of the elasticity in this system and the disturbance generated by the shift.

Edited by gruntguru, 01 March 2012 - 22:31.


#242 gruntguru

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Posted 01 March 2012 - 22:45

So tell me why the torque is also dropping at this point?

Deliberate clutch slip and damper windup. You can calculate how much by integrating the difference between engine speed and input shaft speed - similar to the calculation in my previous post. From the graph looks like about 70 degrees of it is in the damper.

It would also be nice to see a trace for engine torque and what is causing the drop between the start of the graph and the shift bracket.

Engine torque dropping. Hard to say if this is the part throttle torque characteristic of the engine, or the transmission control unit is reducing the engine output for the shift.

It doesnt look like a power on upshift to me.

It probably is. Engine speed is climbing throughout and a lot of energy is going elsewhere (clutch slip and damper windup)

#243 24gerrard

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Posted 02 March 2012 - 09:47

Thanks grunt I agree with your last two posts.
From the output damper to the propshaft there is sufficient damping to reduce outputshaft speed by approx 250 rpm when the torque comes of the shift components, correct?
You must add to this the propshaft wind up (depending on the position of the measuring device) and the axle and tyre damping.
This is a large amount of damping.

Because the torque is read at the propshaft (probably near the gearbox, if the damper theta is to be believed, the damping conveniently masks what the torque is doing at the shift components.
It is the torque trace for the high gear ring and bullet assembly I would like to see.
Imagine it exaggerated.
A huge soft baloon joining the gearbox output shaft to the drive axle pinion.
If you measure torque at the output the damping will smooth the trace.
Couple this to the obvious input rpm reduction before the shift and the engine parting company with the gearbox at the beginning of the shift brackets and you have covered up what is happening at the actual shift overlap.

It is all done to smooth out the shift and to allow the shift components to work without to much wear or damage but it also MASKS the operation of the shift system.
What I am saying is that if torque at the high gear shift ring/bullet assembly (ratchet etc) were to be shown in a big enough scale.
As the low gear bullets(ratchet) disengages as the input torque goes off it, there is a very very short period of time when the input torque is being used solely to compress the spring system on the high gear ring/bullet assembly, against the damping.
For this brief instant no torque is transfered from input to output.

Edited by 24gerrard, 02 March 2012 - 10:08.


#244 gruntguru

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Posted 02 March 2012 - 10:50

If you disregard all the damping, clutch slip, driveline elasticity etc, the zeroshift transmission will upshift with no torque "gap" or dip" - only a brief spike above the thrust curves produced by engine torque alone in the same two gears.

With no damping, clutch slip or driveline elasticity the torque spike would be infinitely high and brief.

With zero inertia on the input shaft (engine etc) the torque spike would not occur and the output would follow the theoretical thrust curves exactly.

I suppose I should have just said "no - your last paragraph is wrong".

Edited by gruntguru, 02 March 2012 - 10:51.


#245 24gerrard

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Posted 02 March 2012 - 21:44

This 'is' a graph for a 'zeroshift' gearshift.
The shift overlap occurs because the input torque is removed and the low gear bullets disengage.
A two thousand rpm difference between the engine rpm and the gearbox input rpm is not a slip, it is a disengagement.
There is a gap in the transfer of torque before the higher gear drives the engaged sprung loaded dogs.
This is cleverly masked by the content and the way in which the graph is drawn.

It is no good trying to ignore the clutch 'disengagement' and the obvious torque control of the engine during the shift.
This gearbox obviously needs this control to operate.

An F1 gearbox will have the advantage of very low weight rotating components and a light vehicle.
This will speed the shift overlap and reduce the need for output damping, clutch engage control and engine torque but these things are still going to be needed.
Trying the shift without proper control and you will grenade the box, I have known this since the 1970's.

To see this you need a torque graph measured at the shift components, not at the damped propshaft.

Edited by 24gerrard, 02 March 2012 - 21:52.


#246 gruntguru

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Posted 03 March 2012 - 02:47

This 'is' a graph for a 'zeroshift' gearshift.

Correct.

A two thousand rpm difference between the engine rpm and the gearbox input rpm is not a slip, it is a disengagement.

The maximum difference I see is about 1400 rpm. Throughout the period of slip (almost 0.5 sec visible on the graph alone) there is consistent torque at the output shaft - THERE IS NO DISENGAGEMENT.

There is a gap in the transfer of torque before the higher gear drives the engaged sprung loaded dogs.

There is no gap visible on the graph and anyone who reads and understands the description of operation can recognise that there will be no gap - ZERO GAP - NONE - NOT A MICROSECOND. The rest of us can read it and understand it - whay can't a "transmission expert" understand it?

By the way, the higher gear starts driving at about 6.54 seconds on the graph.
- At no point prior to this does the propshaft torque drop below the 450+ Nm available in first gear at this throttle setting.
- At no point after this does the propshaft torque drop below the 250'ish Nm available in second gear at this throttle setting.

This is cleverly masked by the content and the way in which the graph is drawn.

The graph looks like an unadulterated presentation of the data to me - sampled at 800 Hz or more. Any gap in torque longer than 1 ms would appear on the trace.

It is no good trying to ignore the clutch 'disengagement' and the obvious torque control of the engine during the shift.
This gearbox obviously needs this control to operate.

Nonsense. The control is only required to smooth the inertial torque spike - inevitable with a gapless ratio change. Whether a full-power shift without torque reduction would grenade the box depends entirely on the sizing of the bullets. I have seen motorcycle dog-boxes modified (with bevelled leading-edges and increased spacing) for full power clutchless shifts in drag racing bikes and I am 100% confident in saying that the Zeroshift system would be stronger for a given size dog. Of course the dog box still has a very small gap in torque delivery while the Zeroshift does not.

Edited by gruntguru, 20 March 2012 - 23:37.


#247 24gerrard

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Posted 03 March 2012 - 10:12

Go find a car with a manual gearbox and a rev counter, drive it along the road, press the clutch down until the rev counter reads 1400rpm above the rpm when it is fully engaged, your figures.
Find out how much you have to lift the throttle to prevent the engine running away.
Find out that the vehicle is no longer accelerating.

Even a 1400 rpm difference between the engine rpm and the input shaft rpm is a disengagement NOT a slip.
How do you transfer torque through the shift mechanism when there is no input shaft torque??????

The graph should show:-

Engine torque
Input shaft torque
Torque at the input and output of the shift mechanism
Torque figures at the propshaft are a result of many factors,

However I do conceed that the propshaft figures show a smoothed out shift which could be described as zerosurge.
No way is 'zeroshift' acceptable as a technical description, perhaps an ill conceived marketing term or a company name.

Edited by 24gerrard, 03 March 2012 - 10:21.


#248 24gerrard

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Posted 03 March 2012 - 11:17

[quote]name='gruntguru' date='Mar 3 2012, 02:47' post='5562405']
Nonsense. The control is only required to smooth the inertial torque spike - inevitable with a gapless ratio change. Whether a full-power shift without torque reduction would grenade the box depends entirely on the sizing of the bullets. I have seen motorcycle dog-boxes modified (with bevelled leading-edges and increased spacing) for full power clutchless shifts in drag racing bikes and I am 100% confident in saying that the Zeroshift system would be stronger for a given size dog. Of course the dog box still has a very small gap in torque delivery while the Zeroshift does not.[/quote]

All you need to exceed the shift speed of a zeroshift mechanism is to operate a dog ringed box with a two barrel two selector at the same time on alternate baulking assemblies mechanism.
You would end up with a much faster shift overlap.
Fit a bigger damper on the output and the shift result would be nearly as smooth.
Wear and damage would be the problem, not shift efficiency.
There would be a very small gap even with FULL on power upshifts just as there is in All shift mechanisms on layshaft gearboxes.

Reading torque at the propshaft would again allow the damper and other torque factors at the rear of the complete powertrain to mask the shift overlap gap in torque delivery.

Edited by 24gerrard, 03 March 2012 - 11:38.


#249 24gerrard

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Posted 03 March 2012 - 11:23

The graph looks like an unadulterated presentation of the data to me - sampled at 800 Hz or more. An gap in torque longer than 1 ms would appear on the trace.


Yes yes yes it is and yes yes yes a 1ms would definitely show on this graph, yes yes yes.

The gap is lost before it can be read at the propshaft!

#250 24gerrard

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Posted 03 March 2012 - 11:28

By the way, the higher gear starts driving at about 6.54 seconds on the graph.
- At no point prior to this does the propshaft torque drop below the 450+ Nm available in first gear at this throttle setting.
- At no point after this does the propshaft torque drop below the 250'ish Nm available in second gear at this throttle setting.


The clutch is still disengaged at this point the torque read at the propshaft cannot be from the engine.
True; torque is gradualy applied from the engine from the high gear engaged point (at the point where the engine rpm starts to drop) on until full clutch engagement occurs off the graph.

Edited by 24gerrard, 03 March 2012 - 11:40.