Zeroing in on Bessler's wheel

[agor95]

Could you please clarify a bit further? Perhaps you could provide a sketch to help illustrate your point.

Hi phi

If I may take that as a help possibility. Then I will keep you informed of progress of my illustration.

The objective is to have at least three compatible simulations using my bespoke creation an Algodoo and others proprietary software.
This will allow a virtual members review.

Of cause I need to confirm if the implementation is worth creating in other products.

You will find me posting in the Community section on this topic at the moment.

Regards

[agor95]

Hi phi

The objective is to have at least three compatible simulations using my bespoke creation an Algodoo and others proprietary software.

I imagine where a 2D simulations does not have a z index collision option.

Then three separate devices set side by side. With cog or chain like connections.
That would link all their rotations of each axle as if they were on the same axle.

I suppose the question is does Algodoo have a way too group objects; using a plain to use plane index?

Regards

[Robinhood46]

" imagine where a 2D simulations does not have a z index collision option."
It is possible to turn on and off collision layers, (or to swap from a to b and back to a for example).
Here is the command to copy and paste in the script menu of Algodoo, and a link to a video showing you how to do it.
(e)=>{
(e.this.pos(0) < 0.05) ? {
e.this.collideSet = 0
} : {
e.this.collideSet = 1
};
e.this.collideSet == 1 ? {
e.this.color = [0.83333331, 0.0, 0.0, 1.0]
} : {
e.this.color = [0.0, 0.66666669, 0.0, 1.0]
}
}
https://www.youtube.com/watch?v=M82sNK4Jpis&t=11s

[agor95]

" imagine where a 2D simulations does not have a z index collision option."
It is possible to turn on and off collision layers, (or to swap from a to b and back to a for example).

Thank you for your example. I did what the video.

I will hold this advice for when I think about using this software product.

I just do a quick study of the Rust Language. With relation to it's 3d & 2d graphics capabilities.

Rust looks promising in creating multi-platform executable code. Therefore it would be fast.

However the methods developed so far are more accessible in my opinion to viewers.

Regards

[WaltzCee]

I think agor is right. I should leave with no warning, no words of farewell----------------------------

Sam, agor's a parisite. Don't mind him.

Any progress?

[Fletcher]

Walt .. I think Sam probably saw something like this sim result in his real-world build. Added to this thread for educational purposes only.

** The wheel gets a push for one revolution (2.75 secs) then the extra rotational force ceases. Note the orange disk at 9 o'cl starts at its deepest position in the D trough when stationary. Note the tracks of all the disks - they all move left on the ascending side - this is due to Centrifugal (inertial) forces. Additionally the disks when nearing 12 o'cl hit the flat of the "D" and have to be raised upwards a small amount as said by T79 previously.

** All-in-all the greatest detriment to self-rotation is the system COM camped out to the left of the vertical from axle, which gives a NET negative torque - this gets worse as rpm increases - and leads to the wheel slowing down quickly and keeling as can be seen in the sim and the graph of rpm changes over time.



.......................

[WaltzCee]

I don't disagree, Fletcher.

I think that idea isn't new, yet it is cool. I termed it a liquid flywheel. A normal flywheel has no internal motion yet this one does

• Is there a gradient between some combination of those pieces of
  flywheel that might be (thru some transmission) added to the CoR
  to drive it?

If I were project manager, that's where it would be headed.

I'm beginning to repeat myself.

There are life things that need to be done but I have a solid design that needs to be built. I'm very optimistic, but what's new? ;)

Stay tuned!

[johannesbender]

Sadly yes , conservative forces are path independent , gravitational potential energy depends on the height and mass and gravity and no path taken to get there or from there changes that , kinetic energy does not depend on the path taken so no path taken from start to end position effects the final total kinetic energy.

So any path taken by a mass on the left side , versus any path taken by a mass on the right side , where both sides start and end at the same heights , have equal total kinetic energy and gravitational potential energy or potential energy states.

Imagine the path on the left is closer and the path on the right is further , or the path on the left is straight up and the path on the right is further , for any weight moving on the path of the right side ,that weight must apply the same total energy to complete the path on the left side , anything less than that would mean the weight cannot complete the path to the same positions again.

Throw in losses because real mechanical machines have efficiency less than 1 or unity , it becomes obvious the path cannot close from start to end positions again.

So to have it complete the path continuously where the path would start and end at the same positions , would require , more kinetic energy converted from gravitational potential energy on the right side ,than the total required on the left , one side would have to gain more than what is required to expend on the other .

[Robinhood46]

Sadly yes , conservative forces are path independent , gravitational potential energy depends on the height and mass and gravity and no path taken to get there or from there changes that , kinetic energy does not depend on the path taken so no path taken from start to end position effects the final total kinetic energy.

So any path taken by a mass on the left side , versus any path taken by a mass on the right side , where both sides start and end at the same heights , have equal total kinetic energy and gravitational potential energy or potential energy states.

This is all well and good, but how do you apply this logic to wheels which don't have the weights fixed to specific sections of the wheel?
The energy being transferred from the weights, (because of the effect gravity is having on them (accelerating positively and accelerating negatively)) to the wheel is not symmetrical, which is the case with the weights fixed to specific sections of the wheel.
Trying to get weights or trying to get a wheel to miraculously create energy from nowhere is impossible. We need to try getting the interaction between the weights and a wheel give us a tiny fraction of the forces in play, it cannot happen if the moving weights are fixed to the wheel.

[johannesbender]

free body diagrams demonstrate the law holds up , where a single particle/point mass unrestricted or not attached to anything else , would have the same result because conservative forces are path independent , is this what you mean RH?

[Robinhood46]

My thoughts are that the mass will have the same interaction with gravity in the two cases. The wheel will also have the same interaction.
To apply my thoughts to your last diagram, you would need to move point B.
A falling weight from A to 6 o'clock will raise to B, irrespective of whether it is connected to a wheel or not. (A and B being the same height).
By connecting the weight to a wheel, the weight will accelerate less, because it is causing the wheel to accelerate, but it will still raise to B. The process will simply take more time. Gravity will be putting energy into the wheel and taking energy out of the wheel at a slower rate because of the additional mass of the wheel. A light wheel will make little difference and a very heavy wheel will make a much greater difference, to the time needed for the weight to go from A to B.
The COM of the weights and the wheel, combined, also behave in the same manner, accelerate positively from A to 6 o'clock, and accelerate negatively from 6 o'clock to B.
One cycle of the wheel must be 360°. If one cycle of the weights is also 360°, there is nothing to be gained. Therefore the only place a gain can be found is by changing the number of degrees in the cycle of the weights.
Therefore to come back to your diagram, the wheel goes from A to B but the weights only go from A to B -5cm ( for example).
B -5cm then becomes the new A for the wheel, and the weights, and the cycle restarts.

[JUBAT]

Great animation Fletcher. I'm no expert, but would there be any advantage gained by fine-tuning this arrangement with an off-set center of rotation? My idea is that potentially it would create a super-lopsided mechanism that really swings down, but then sort of coasts back up while in balance or less off-balance than the initial power swing.

Over the years I've started drifting towards off-set CoR as a possible means to solving Bessler's wheel. I have an image in my head of how an off-set CoR coupled with the mechanism inside could make the wheel super-lopsided, yet on the upswing be balanced to try and harness the momentum gained on the trip down.

Of all the attempts made at solving the wheel that have failed, I'm seeing a common theme of everyone wanting to use a centered axle (even weight distribution) and although the axle was apparently centered on Bessler's wheel, one can create a pseudo off-set CoR by how the weights/mechanisms are distributed in the wheel.

It's almost like having a pendulum within a pendulum set in opposition to each other. If one were to lock all the weights/mechanisms in a wheel of this sort and then spin the wheel, then the weights would by default be taking a path closer to the axle and then farther away. Start letting the weights move and then you have the potential for a double-offset going down and a balanced state going up.

Perhaps that has been tried before and it doesn't work.

[JUBAT]

Another thought if we don't want to go down the rabbit hole of off-center CoR - is there anything to be gained by sticking your wheel on top of a 4-pendulum MT21 that fills the gap in between the semi-circle compartments? This too is another rabbit-hole I've been playing with recently - stacking different mechanisms out of sync to each other. No success yet, but it sure looks cool in my pile of kinetic art.

I've been thinking of going through all of the important MTs of Besslers - printing them all out in fact - and marking exactly where their stall points are. I'm old-school so using a clear film/acetate would be helpful. Then start overlaying different mechanisms with each other to compare where their stall (keel) points are. At that point it just becomes a method of layering more than one MT to make sure the wheel never can reach its keel point.

Worth a shot I guess.

[johannesbender]

My thoughts are that the mass will have the same interaction with gravity in the two cases. The wheel will also have the same interaction.
To apply my thoughts to your last diagram, you would need to move point B.
A falling weight from A to 6 o'clock will raise to B, irrespective of whether it is connected to a wheel or not. (A and B being the same height).
By connecting the weight to a wheel, the weight will accelerate less, because it is causing the wheel to accelerate, but it will still raise to B. The process will simply take more time. Gravity will be putting energy into the wheel and taking energy out of the wheel at a slower rate because of the additional mass of the wheel. A light wheel will make little difference and a very heavy wheel will make a much greater difference, to the time needed for the weight to go from A to B.
The COM of the weights and the wheel, combined, also behave in the same manner, accelerate positively from A to 6 o'clock, and accelerate negatively from 6 o'clock to B.
One cycle of the wheel must be 360°. If one cycle of the weights is also 360°, there is nothing to be gained. Therefore the only place a gain can be found is by changing the number of degrees in the cycle of the weights.
Therefore to come back to your diagram, the wheel goes from A to B but the weights only go from A to B -5cm ( for example).
B -5cm then becomes the new A for the wheel, and the weights, and the cycle restarts.

Perhaps an image would help me understand RH , i presume you mean instead of going all the way to the top , the weight should go to the top - 5cm for example ?

This gives a clearer picture of what i meant , there is no path from A to B on the right side where the path has more energy , than what is required in energy to move from B to A on the left side , where A and B start and end on the same positions (each the same height from A to B and B back to A) , in the most efficient resistance and friction free scenario , now if we add our real world losses , the scenario becomes even more tainted.

However , there are other variables too , there is total mass from A to B and total mass from B to A that can be designed to differ which we could call overweight , but in my opinion overbalance (force by distance on one side vs the other side) is no go.

[Robinhood46]

I'm thinking more along the lines of A being at 1 o'clock and B being at 11 O'clock.
A to BDC (Bottom Dead Centre, or 6 o'clock) is = to BDC to B. The effect gravity has on a weight, and the effect that gravity's effect on the weight has on the wheel, is equal. One side being positive acceleration and the other side being negative acceleration, they cancel each other out. As you said yourself, the path the weights take also cancel each other out, because the height is always the same. All the time the weights rotate around the same frame of reference, at the same rotational speed, everything cancels everything out, irrespective of the path taken by the weights. That is in a hypothetical environment without any other forces in play, which doesn't exist, and things only get worse in the real world.
Therefore, a weight causing a positive acceleration of the wheel from A to BDC, puts enough energy into the wheel, for it, the wheel AND the weight, to raise the other side so the weight finds itself at B. There is no miracle that can move the weight from B (11 o'clock) to A (1 o'clock), but we don't need to move the weight. It is the COM of the whole thing that needs to "go over the top", not the weights themselves. The weights will go over the top, but not at the end of the cycle that they come down on, but at the beginning of the next cycle which is just starting.
An OOB wheel will turn, it will raise the COM to the same height as where it falls from. There is no need for the weights to go over the top, as long as the COM does.
An OOB wheel that cycles every 359° will probably not be able to overcome all the forces in play that need to be overcome. As the cycling arc decreases the power of the wheel increases.