A task on synchronization of two gears (2 x AC drives; 1 x encoder).

[Chelny]

hello everyone,
there is a task on synchronization of two gears. For details see sketch in the attachment.

Please any comments, points of view and so forth.

Of course, if those facilities aren't enough, the new one will be purchased. But let's start with a few.

Chelny

 

[Ingenieur]

drive 2 speed reference
= G1 teeth count / G2 teeth count x encoder signal
or
= G1 teeth count / G2 teeth count x drive 1 speed output signal

G1 teeth count / G2 teeth count = gear ratio

another method would be to use drive 1 torque signal x gear ratio as torque command input to drive 2

I assume the load is connected to gear 2

 

[Chelny]

drive 2 speed reference
= G1 teeth count / G2 teeth count x encoder signal
or
= G1 teeth count / G2 teeth count x drive 1 speed output signal

G1 teeth count / G2 teeth count = gear ratio

another method would be to use drive 1 torque signal x gear ratio as torque command input to drive 2

I assume the load is connected to gear 2

Did you mean teeth count by means of....?

 

[Ingenieur]

Did you mean teeth count by means of....?

the actual number/count of teeth per gear
if G1 = 60
and G2 = 46
ratio = 60/46 = 1.30
input rev/output rev 1:1.30
the gear ratio of the gear set

 

[Jraef]

What the heck would the point be in doing this? They are mechanically coupled, so one drive/motor is going to move the other.

If the point is to SHARE the load between the two motors, then it’s a simple torque follower scenario. One drive it the master, which is given the speed command, then an output torque signal of that drive is sent to the slave drive, and it matches the torque output of the master drive. There is no “synchronization” involved from a speed standpoint because that would be pointless.

 

[Ingenieur]

What the heck would the point be in doing this? They are mechanically coupled, so one drive/motor is going to move the other.

If the point is to SHARE the load between the two motors, then it’s a simple torque follower scenario. One drive it the master, which is given the speed command, then an output torque signal of that drive is sent to the slave drive, and it matches the torque output of the master drive. There is no “synchronization” involved from a speed standpoint because that would be pointless.

looks like a hw thought exercise


a similar scenario is very common on large mine belts
but always the same ratio

2 2000 HP motors each with its own gearbox, each box output shaft has a gear
those gears each drive a seperate bullwheel on a common shaft which
is the belt drive roller axle

iirc they use torque
possibly adding the 2 then sending a 50% command to each

 

[Chelny]

drive 2 speed reference
= G1 teeth count / G2 teeth count x encoder signal
or
= G1 teeth count / G2 teeth count x drive 1 speed output signal

G1 teeth count / G2 teeth count = gear ratio

another method would be to use drive 1 torque signal x gear ratio as torque command input to drive 2

I assume the load is connected to gear 2

By applying ref. speed you mentioned, to Drive2, this way, I will get equal angular speed of both gears. Am I correct?
But, I have some doubts. What will happen while approaching gear2 to gear1. I mean possible collision such as: 'one turning gear approaches to another one with same angular speed, and at certain moment a teeth of smaller gear hits a teeth of bigger gear'. There will be ideal synchronization :thumbsup:, but without teeth meshing. :? Not to mention the axis that moves smaller gear. Am I correct?

Chelny

 

[Chelny]

What the heck would the point be in doing this? They are mechanically coupled, so one drive/motor is going to move the other.

If the point is to SHARE the load between the two motors, then it’s a simple torque follower scenario. One drive it the master, which is given the speed command, then an output torque signal of that drive is sent to the slave drive, and it matches the torque output of the master drive. There is no “synchronization” involved from a speed standpoint because that would be pointless.

I'm interesting in both 'hecks':

• synchronization of gears that have conventional teeth meshing;
• synchronization of gears that haven't teeth meshing. So-called pseudomeshing, or when, if you wish, there's visible meshing but torque is not conveyed by gears.

Chelny

 

[GoldDigger]

What the heck would the point be in doing this? They are mechanically coupled, so one drive/motor is going to move the other.

If the point is to SHARE the load between the two motors, then it’s a simple torque follower scenario. One drive it the master, which is given the speed command, then an output torque signal of that drive is sent to the slave drive, and it matches the torque output of the master drive. There is no “synchronization” involved from a speed standpoint because that would be pointless.

]

You give a very good explanation of the problem of driving a load from two motors once the gears are in constant mesh.
Unfortunately the problem described in the PDF in post #1 is to take two turning shafts wit and synchronize them so that they will clash minimally when brought into mesh.

There will be no load on the smaller gear shaft initially, or if there is it is unrelated to the load on the larger gear shaft. That means that until the gears have completed meshing there will be no forced synchronization and no use whatever for torque feedback.

As only the large shaft has an encoder, I do not see any good way to preset the speed of the smaller shaft to minimize gear clashing other than to know and control the drive frequency to both shafts and make some, hopefully good, assumption about the slip from the corresponding synchronous rotation speed.

At least the encoder on the larger shaft, which is set to a constant speed, will eliminate having to guess about the slip percentage of the motor driving that shaft.
And if the smaller shaft is initially unloaded it might be possible to estimate the slip of its motor closely enough to be useful.

 

[Ingenieur]

in the OP's example 2 things need controlled
speed, since they differ due to gear ratio
and torque

speed: I would 'share' the encoder signal, just mult x gear ratio for drive 2

torque:
sum the T output from both drives
the motor on the driven load shaft would get: T sum/2 as a T command
the other would get T sum/2 x gear ratio

assume
a demand of 1000 Nm
a gear ratio of 2

first drive (on load shaft) would get a command of 1/2 x 1000 = 500
the other drive on the gearbox would get 500 x 2:1 =1000, net to shaft 1000/2 = 500
so drive 1 supplies 50%
drive 2 TRANSMITS 50% to the shaft

 

[Chelny]

Guys, what about understandable answer on my #7 post?

regards.

Chelny

 

[gar]

180212-1225 EST

Chelny:

I believe some responders have not understood what you want to do.

I believe what you want to do is to adjust rotation of a pair of gears, that are not yet engaged (teeth not touching), in such a manner that while rotating you can bring the gears together in mesh (no clanking, no teeth hitting on top of one another, and no tooth damage).

To do this means you need position information from both shafts.

1. Use a position encoder on each shaft, or
2. Use photocells or prox detectors on each gear to determine tooth relationship, or
3. Use a vision system looking at the tooth relationship.

.

 

[Jraef]

180212-1225 EST

Chelny:

I believe some responders have not understood what you want to do.

I believe what you want to do is to adjust rotation of a pair of gears, that are not yet engaged (teeth not touching), in such a manner that while rotating you can bring the gears together in mesh (no clanking, no teeth hitting on top of one another, and no tooth damage).

To do this means you need position information from both shafts.

1. Use a position encoder on each shaft, or
2. Use photocells or prox detectors on each gear to determine tooth relationship, or
3. Use a vision system looking at the tooth relationship.

.

Sorry, I didn't get that you were going to be meshing two gears together that are already turning.

So I agree with gar, although I think the effectiveness of #2 is likely untenable without some precision in the smaller drive, which then comes back to #1. It's one thing to count teeth with a sensor, it's another thing to know WHERE in the rotation those teeth are and for that, you will need an absolute encoder feedback on both shafts.

The vision system idea has merit though. I've never done it, but I've seen demos of it working.

 

[gar]

180212-1340 EST

Jraef:

I have built test equipment for checking the teeth in encoders used in automotive ABS systems. Using the sensor that goes with the encoder gear in the differemtial I was able to get good measurement information within a tooth to tooth cycle. These were generally dv/dt sensors (magnetic and coil) and thus have a phase shift. A Hall sensor would be better in this application. One can definitly get good tooth position information with a magnetic sensor where the speed is approximately constant.

.

 

[gar]

180215-0431 EST

Chelny:

It would be useful if you ask more questions, and provided more information about your gears.

From 30 year ago memory I believe the following information is about correct for magnet-coil pickup:

1. At 3 MPH we needed a minimum of about 1 V. Output voltage is proportional to speed. Thus, one uses zero crossing information. 3 MPH = 4.4 feet per second, because 30 MPH is 44 ft/sec.
2. Tire diameter about 30", and thus circumference = 94" = 7.9 feet.
3. Thus, axle RPM at 3 MPH = about 1/2 RPS.
4. Tone ring (gear) had about 100 teeth on a diameter of about 6". Thus, tooth pitch was about 0.25".
5. 100 teeth at 1/2 RPS is about 50 Hz.
6. Air gap tooth tip to sensor possibly 0.020".
7. Output somewhat like a sine wave.
8. Very reliable except for magnetic metal chips, but you don't want those in a differential anyway.

.

 

[RumRunner]

Guys, what about understandable answer on my #7 post?

regards.

Chelny

If this is still a work in progress (ie.) being designed to suit your specific needs. . . this synchronization dilemma can be addressed far more than what is presented in the drawing. Torque, speed and positioning of the gears and even holding the gears in place when machine is in idle.

Accurate positioning and easy control by using pulse signals from a controller along with servo to provide feedback signal from the output shaft of the smaller gear.

Check out stepper motors applications with some high torque in a compact body for ease of integration into your machine.

Both servo and stepper motor can be mounted at the input and output shaft.

I've use smaller versions of these in small-scale robotics.

 

[RumRunner]

As an aside, I assume the bigger gear is the input?
Since both shafts are marked AC drive, and probably having its own motor. . . an auxiliary planetary gear (not shown in drawing) is where the work is being performed. The two gears are there to perform one common task?

Some drawings, when it comes to controllable speed motors--the VFD controller is sometimes labeled VFD Drive referring to the controller itself not the motor.

Is this common in Russia?

 

[RumRunner]

If this is still a work in progress (ie.) being designed to suit your specific needs. . . this synchronization dilemma can be addressed far more than what is presented in the drawing. Torque, speed and positioning of the gears and even holding the gears in place when machine is in idle.

Accurate positioning and easy control by using pulse signals from a controller along with servo to provide feedback signal from the output shaft of the smaller gear.

Check out stepper motors applications with some high torque in a compact body for ease of integration into your machine.

Both servo and stepper motor can be mounted at the input and output shaft.

I've used smaller versions of these in small-scale robotics.

I

 

[Ingenieur]

it can be done
engagement speed is a factor
assume land and tooth the same on both
both gears turn same speed

the land has an angle that sweeps across g2
as the land ang on gear 1 approaches perpendicular to gear 2
start moving gear 2 torwards gear 1
time gear 1 land to be perpen to gear 2 when they meet
gear 1 ang should sweep at least 2 gear 2 teeth
land vs gear position are arbitrary

I need to calc
engagement speed as a function g1 swept ang on g2
if it misses the first g2 tooth it will snag the next

need to do math
too burned out now lol

 

[Ingenieur]

think about this (assuming he is linearly and radially engaging the gears)

g1 dia = 15, roots = 60, teeth = 40
g2 = 10, 40, 27
typically the root is 1.5 x tooth area per rev
let's say 60% root and 40% tooth
and a gear ratio of 1.5:1
assume tooth to tooth vel for g1 is 100
and root a bit less (smaller dia) 98
both gears ~ the same since
vg1 = D1 x Pi x v1 rev/sec
vg2 = 2/3 x D x Pi x 3/2 x v1
same

set the encoder up so it always starts to engage g2 when a root is pointing/perpendicular towards/at g2
so consider it stationary

your odds are
g1 is always 1 or 100% root since it is positioned by the encoder
g2 root 60% and tooth 40%
that you hit a root 60%
that you hit a tooth 40%
if you hit a tooth good
if you hit a root we need to adjust, but how do we know? we have no feedback

you need to make sure that by the time the next root on g1 is 'in position' we have a tooth
in line on g2

I assume g2 is driving the load
and g1 is xmting power to it
so let us run g2 a bit slower if required
g1 root to root 98/60 = 1.63, same as g2
g2 tooth to tooth 100/60 = 1.67, same as g2

if it is synch'ed will continue to strike at the same point, root-root
g1's next root will be in position again 1.63 later
but if g2 is run a bit slower it will have a tooth in position
root to land delta 1.67-1.63 = 0.04 (using vel, easily converted to time, deg, etc)

so if g2 is run 0.04/1.67 x 100% = 2.4% slower than g1
if it doesn't engage the first tooth it will get the second
speed of linear engagement = tooth depth/0.04 in this case