# Help designing bearing influenced by spring

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1. Oct 5, 2016

### Haim1404

Hello Guys,

I'm building a vertical wave machine.

I have a big bearing [2] (more like an eccentric) which is rotating by an electric motor with a constant speed of 0.359 rad/sec.
When 3/4 of the bearing touches bearing [1] it supposed to rotate it a 1/4 turn, which as a result the ladder with the rods should rotate 1/4 turn.
Once they are not touching each other (passing the green diameter), bearing [1] returns to its place a
due to the spring resistance force and creates wave movement in the statue.

I know the following parameters:
In order to to remove the rods from equilibrium I need torque of 30 Nm.
The motor output is 110 Nm and it rotate bearing [2] in constant speed of 0.359 rad/sec.
The spring have resistance force of 10N
Diameter of bearing [1] is 100 mm.

I need help in calculating the green line = the contact line= the big diameter of bearing [2].
If necessary I will be happy to provide additional data and info.

Thanks to all helpers.

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Last edited: Oct 5, 2016
2. Oct 5, 2016

### JBA

This is not a simple question because you must take into consideration how much time it will take to transfer the 1/4 turn base drive from the stack connection point to the top of the flexible assembly plus the time is required to allow the bottom of the stack to return to its next cycle start of rotation point, ie the stack's 1/4 turn windup +1/4 turn unwind total time. This will be controlled by the speed of rotation of the stack assembly during "bearing" contact and the spring factor and the inertia of the stack; and, could create some unusual and/or unanticipated stack oscillations and/or harmonics if this process is interrupted mid-cycle and can interact with the return spring as well. The natural frequency of the stack windup-unwind cycle and the natural frequency of the return spring must also be taken into consideration to insure that the drive will not match or closely match this frequency and potentially result in a catastrophic structural failure.

3. Oct 5, 2016

### Haim1404

Hi JBA,

1. Thanks for taking the time to look at the problem.
2. Any suggestions for an approximate solution? I will maximum do minor repairs in the field...

4. Oct 5, 2016

### JBA

Like I said, this is a very complex problem; and, while I am familiar with harmonics of conventional spring configurations a multimass/multi-cable arrangement such as yours is well beyond my expertise.

That said, many years ago, I was a member of a team that designed a spring sinusoidal drive oscillation machine that could demonstrate the effects of any coil extension spring rate/mass/damping combination from the low rate where the spring acted as a stiff bar up through the spring's natural frequency to the point that the input frequency was 180 degrees out of synchronization with the spring's natural frequency and all movement of the suspended mass ceased. As a result, I am still concerned about the matching of your periodic input impulses to your stack's natural oscillation frequency. Unless you exactly match your input frequency to the stack's natural frequency you will simply interrupt its natural oscillation by send an opposing rotation up from the bottom that will clash with the rotation returning from the top and that will either halt all oscillation or send reflected reversing oscillations up and down the stack from that point. What you will require is a continuous tightly controlled bottom rotational oscillation that very gently matches the stack's natural oscillation frequency while still controlling its amplitude to prevent a potential runaway oscillation, particularly in light of what appears to be a large scale structure.

What you are proposing is something very similar to the disc on a string loop toy where a disc with two holes with the loop strung through those hole and held at each end is spun in oscillation by holding each end of the loop and pulling and relaxing the tension on the ends of the loop. I suggest you try playing with one of these to get a sense of exactly how precise the input driving frequency must be for your device and what happens when it is not matched.

I realize this is a lot of input to digest but, while I cannot help you in solving your design problem, I want to make you aware of one the make or break elements in the design.

5. Oct 6, 2016

### Haim1404

I appreciate it, thanks.

6. Oct 6, 2016

### CWatters

I can't add much except that the geometry looks way off...

The contact point between cam and 100mm roller appears to be roughly on the line between centre of the cam and centre of roller. So the cam will force the axels apart rather than causing the roller/arm to rotate.

I'm no expert but perhaps one option might be to use circular gears with missing teeth instead of a cam arrangement?

7. Oct 6, 2016

### CWatters

You asked for the 100mm output gear to rotate 90 degrees (1/4 turn) when the cam gear rotates 3/4 of a turn. Using gears with missing teeth the "cam" gear actually has to be smaller than the 100mm gear. If the small gear has 1/4 of the teeth missing then the remaining 3/4 have to rotate the large gear 1/4 revolution. I make the diameters of the small gear about 33mm.

The main problem I see is the possibility of the gears jamming if they don't re-mesh correctly. Perhaps there is a tooth profile that can avoid this? Wear and lubrication is harder?

8. Oct 6, 2016

### Haim1404

Hi CWatters,
Thanks for your idea I will take into consideration.

9. Oct 6, 2016

### Baluncore

Is the bearing/cam/eccentric direction of rotation shown as backward? Does all the energy that goes into the wave step all come from the spring? Is the wave supposed to be a saw-tooth wave?
I would have expected a small diameter roller mounted on the arm that followed the eccentric cam, with the return spring holding the roller onto the eccentric cam.

10. Oct 7, 2016

### Haim1404

Hi, Baluncore

1. yes, the eccentric direction in the picture is wrong.
2. When bearing 1 rotates 1/4 turn so is the first rod. Once it is released, the spring brings it back, and the wave start. So yes it is possible to say that most of the energy that goes into the wave step come from the spring. (maybe even all, I dont sure).
3. no, the shape can be what ever, but of course some kind of circular pattern.
4. Diameter 1 must be bigger, becuese the motor output is 110 Nm, while in order to to remove the rods from equilibrium I need torque of 30 Nm.

* can I solve this using energy equations? if so how?

11. Oct 7, 2016

### CWatters

Can I check something....

Is the idea that it acts like a cam and follower or a friction drive that slips when the diameter of the cam reduces?

12. Oct 7, 2016

### Haim1404

kinda more like friction drive that slips, since the axis of bearing 1 is fixed, unlike normal cam that usually have spring.

13. Oct 7, 2016

### CWatters

Ok so it's not to dissimilar to the gear with missing teeth approach.

3/4 of a revolution of the cam = 1/4 revolution of the output.

So the ratio of the diameters when driving must be ...

3/4 : 1/4
or
1:3

So if the output is 100mm diameter the cam must be 33mm diameter over the 3/4 of its circumference when it's driving.

For the 1/4 of its circumference when it's not driving the diameter just needs to be sufficiently less that slipping occurs. Probably not critical. Depends how hard the rubber is.

14. Oct 7, 2016

### CWatters

The ratio 1:3 also increases the motor torque so if I've understood correctly the output torque available would be 300Nm which should be plenty (you only need about 40Nm?).

The output rotates at 1/3 the motor rpm when driven.

15. Oct 7, 2016

### CWatters

I think you may need to make the output wheel bigger than 100mm to allow a larger cam wheel. 33mm diameter is quite small and I think it will be hard to set it up so that it only drives for 3/4 of a revolution accurately.

16. Oct 7, 2016

### Haim1404

Thanks.

17. Oct 7, 2016

### Baluncore

I think the mechanism is required to gradually rotate the two pins until it is suddenly released to generate the step wave. That cycle should repeat about every 30 seconds.

If friction is used to drive between the two discs then there must be some give in the structure that will maintain contact pressure. Over time there will be significant frictional wear at the point where the release occurs.

I expect the suggestion of two gearwheels, with the driver missing a quadrant of teeth, would be a better solution than friction.

Another alternative would be a stepped dog clutch that is pushed apart to separate when rotation reaches a set limit.

It may also be necessary to also consider how the two pins are controlled after release. They might be immediately re-captured and held firmly while the wave propagates, or the reflected wave might take control before the eccentric re-engages.

18. Oct 8, 2016

### Haim1404

Thanks Baluncore,
I will think about it to.

I fought that I will need bearing / gear 2 to be bigger then 1 so that the output torque will be smaller and more close to what I need (30 Nm),
If the toruqe will be much bigger (300Nm as CWatters mentioned) will it not take out of control the generated wave?

19. Oct 8, 2016

### Baluncore

The size ratio of the friction wheels is irrelevant if the friction drive does not work reliably.
First specify the twist angle against time of the wave you want to generate. Then find a reliable mechanism to make it happen.

20. Oct 8, 2016

### CWatters

Thing is you asked for a 3/4 turn of the cam to produce a 1/4 turn of the output. To do that the cam must be smaller than the output.

If it takes 40Nm to rotate the output then that's what the motor will deliver even if the motor can generate more due to the gearing. If something stalls the output the motor will increase its output upto it limit (I calculated about 300Nm). This is all fine.

As I described above the output will revolve for 90 degrees at about 1/3 of the motor speed. Then it will slip back/unwind as fast as the spring can manage sending a pulse/wave up the tower.

That wave may go up and down the tower several times or die out quickly if there is sufficient damping.

Eventually the drive re-engages twisting it again ready for the next pulse/wave.

Last edited: Oct 8, 2016
21. Oct 8, 2016

### Baluncore

I think the terminology needs to be revised. A disc here is being driven by a profiled cam. The side forces between the shafts and housing is carried by internal lubricated bearings.

If friction drive is used, what force is needed between the cam and the disc to prevent slipping?
How is that force applied accurately?
What implications do the cam and disc material friction coefficients have on the cost and size of the bearings needed?

If a gear tooth drive is used instead of friction, how strong must a single tooth be? The cost and size of the bearings will be a function of maximum torque and gear contact angle, until the moment the last tooth begins to skip when some flexibility will be needed somewhere in the mechanism.

The phase velocity of a wave machine is slow so it might be better to drive the wave with a reversible geared servo motor.

22. Oct 9, 2016

### CWatters

Good points. 30-40Nm at 0.1m is 300-400N which i think is a lot to transmit by friction through a device this size.

23. Oct 11, 2016

### Haim1404

I must understand something that CWatters said:

"If it takes 40Nm to rotate the output then that's what the motor will deliver even if the motor can generate more due to the gearing"

So lets say I need only 1 Nm , and the motor after the gearing is giving me 100Nm,
the motor can "feel" that I need only 1 Nm and that what it will give me after the gearing?

I'm asking this because I fought that the ratio of the torque will force the ratio of bearing and the cam...
In this case what affecting the ratio? only the fact that I wont a 3/4 turn of the cam to produce a 1/4 turn of the output - 1:3?

24. Oct 11, 2016

### JBA

Since the stack will have its own oscillating frequency (similar to a pendulum) as the 1/4 turn lifts the weights; then, as the spring pulls it back it is going to start resisting the spring at some point of rotation as it tries to start it natural return rotation and that is the point at which the engaging force on the drive system will be at its minimum.

25. Oct 11, 2016

### CWatters

Correct.

If there was no load (no torque required to turn the output) then the motor would only have to deliver enough torque to overcome losses in the gears. In short the motor might be capable of delivering 100N but at any one time it will only deliver what the load and any gearbox losses need.

This is one reason why electric motors can be dangerous.... If you caught your finger in the mechanism the motor will automatically increase it's torque up to its maximum capability (usually this is called the stall torque in data sheets).