mechanical paradox? an illusion?

[jonnynet]

Here is an idea I thought many hours about and because I can't find the reason where the problem is here I ask you. IMO it seems to be a mechanical paradox!?

At first let me give a description of how the parts should operate. As you can see this is a 'wheel in the wheel'-design. Georg Künstler built a similar one - his idea inspired me. But here, the smaller wheel rolling inside the larger wheel drum is special. It has 8 roller-pairs installed on the rim whereby the blue rollers are heavier than the green ones. These pairs are pivoted and connected among each other (not shown in the drawing).

The question is now, what would happen when the wheel will leave to its own? I'm sure the principle of this design is so simpel that it mustn't work. At the moment I'm blind - I simply can't see how the masses can reach a centered position. Please help!

[KAS]

Jonnynet,

It first appears that you may have something here.
Then I noticed that you say the blue wheels on the bogies are heavier than the red ones.
In that scenario, the blue wheels will fall and the red ones will raise, aligning the bogies vertically and cancelling any benefit.
Unless there is something I have missed.


Kas

[jonnynet]

In the way you describe, the blue rollers can't fall. They all can only move synchronous because they're interconnected (I wrote that the connection is not shown in my drawing!). But look at the lowest rollers on the bottom - they can't fall (or rise) so the others can't do it also.

[Vic Hays]

The interconnectedness itself may cause a countertorque.

[LustInBlack]

Good Idea, but, IMO, the counter-torque will destroy the effect wanted.

I believe the inner rotor will shift forward, until the blue roller hits the outer wheel, where the inner rotor will settle down ..

From there, the outer wheel will turn a bit..

The inner rotor will now rest on two blue rollers, all rollers now pointing at about 45 degrees (approximated grossly) ..

The outer wheel will have some inertia, it might oscillate back and forth..

The inner rotor will roll on it's blue rollers, and I think it will stay in place on it's blue rollers.


Giving a push on the wheel might make it turn a bit.. But I believe the effect will be similar as a bottle of coke filled with half water that you push horizontally on a table..

It will give pulses and go forward and eventually stop ....

But the effect is interesting, maybe I am wrong!

[rlortie]

IMO it is not going anywhere!

In truckers language, the tandem weights although of different mass are connected to a "walking Beam" that is pivoting on a third or dead axle.

This pivot point is symmetrical in the radius of the inner ring. It will be acted upon as one simple weight hanging on the pivot point. Tying them together will not change the weight on each pivot point.

Ralph

[KAS]

Yeh, I think Ralph is right. It is the connectors that we can't see that would cause the problem.

I imagine that the blue roller on the bottom is the one that is taking the strain and that it is connected to the other blue rollers holding them in their horizontal position.
If that is the case then there is a hidden third wheel made up of connecting beams that balances the whole structure.

I don't believe the idea is dead though. There may be another way to retain it's asymmetry. Its a good concept IMO.

Kas

[Fletcher]

Jonnynet .. I had to look at this one for a while & I'm still not sure if I understand it yet ? I'm guessing that both the set of blue roller weights & the green set of roller weights are independently connected by, lets say, ropes, with no slack, so that they form their own counterpoised circular systems. The two (blue & green) are only connected via the shared pivot point of the bogies.

Because there is no slack between each roller weight, if one is horizontal they are all held horizontal etc ?!

The blue set (which is heavier) would want to hang vertically (if there was no outer rim to rest on & the inner wheel had its own axle) & at the same time raise up the counterpoised green set. So far it appears just like an ordinary pendulum arrangement in that regard.

But it cannot hang down because it is a floating system & all is supported by the outer rim it rests on.

I think that the uneven weight distribution would cause the whole assembly to rotate CW a little bit until all the weight was evenly distributed left & right under the axle.

Perhaps with some spring tensioning system this hamster may yet find legs to walk around the cage ?

[jonnynet]

I believe the inner rotor will shift forward, until the blue roller hits the outer wheel, where the inner rotor will settle down ..

From there, the outer wheel will turn a bit..

The inner rotor will now rest on two blue rollers, all rollers now pointing at about 45 degrees (approximated grossly) ..

Your description seems realistic and here is the point where I expect a reset of the roller-weights so that the outer wheel doesn't rotate back (CCW). I'm not sure if this would happen at that time.

@Fletcher
You understand well how the wheel's operation is desired. You can imagine a solid third ring for the connection of the blue rollers.

I think that the uneven weight distribution would cause the whole assembly to rotate CW a little bit until all the weight was evenly distributed left & right under the axle.

I hope this case wouldn't occur. It's not easy to estimate the real operation and it remains the question how about the self-reset. Will it happen early enough? I don't know.

[Jon J Hutton]

Although this comment/idea/method is very simply it has helped me to rule out hundreds of wheels that do not work that I have come up with.

Before I had wm2d I had to use an old $14.00 architect program to do this with, but it saved me much shop time.

Draw a line through the axel from top to bottom. measure how far away the weights are from the line on the left side then on the left. The highest number wins the rotation. If the weights are different multiply the distance by the ration of difference. Rotate the drawing by no more than 3 degrees and recalculate.

This will only work to test an Over balanced wheel and does not take into account momentum of shifting weights.

From many, and I many hours of this tedious work I finally realized that I had to place complete attention on getting at least 5 times horizontal displacement/movement to 1 part vertical displacement / movement

Like I said it is simple. I also saw this on a 1700's drawing.

JJH

[KAS]

I am changing my opinion slightly to "there's possibilities" with this design.

The more I look at it, the more I can visualize a paradox occurring.

Imagine as jonnynet says that there is a third ring attached to the blue wheels. What gravity would want to do is to align the blue roller at the bottom vertically with the axle. But this can't happen! If it does, the third imaginary ring (if it is weighted) will want to fall to the right until the next roller touches the outer ring ??? This is then true OOB.

I am puzzled and can easily see why jonnynet used the word "Paradox" in the title.

I don't think we should dismiss this design out of hand, but I do think we should try and study the effect of the inner system more closely.


Kas

[jonnynet]

To make the thing easier, I decided to rotate the drawing by 7 degrees CW so that we've a situation where the total mass of the inner wheel should be centered nearly. Because the green rollers have their own weight, even if it's little, the visualized third ring might be still more on the right a bit.

Now then the wheel should be in balance and there are two possibilities: Either the wheel does nothing more than oscillating for a little while and stop then (situation like in drawing 2) or the inner wheel begins to fall so that all roller-pairs aligns horizontally again and the third (weighted!) right ring is shifted to the right like in the first drawing. Decide what will happen.

[Fletcher]

A couple of points to ponder.

The weight of the internal system must be applied thru the roller weights. These are circular running against the circular inner rim which carries it. Although weight acts vertically the transmission of the weight is at an angle i.e. perpendicular to the rim. If you were to draw a line from the axle of the blue roller weight to the contact point with the rim this would be how the weights acts downwards. In other words there is not the full compliment of vertical weight applied only to consider.

Secondly, the inner structure [where it rests on the rim] is supported by the bogies central pivot. This is to the left of the vertical central imaginary line. This means the weight distribution or more correctly imo the force distribution is balanced in or near this position, IINM.