Decoupling Per-Cycle Momemtum Yields From RPM

[MrVibrating]

If only piecemeal momenta can cut the PE cost of a net momentum / KE value, yet they cannot be sequenced in time..

..then they must be coincidental.

Q: what advantage might four simultaneous accelerations have over one larger one of equal net momentum rise?

It has to be that all four benefit from the 'standing-start' PE / momentum efficiency, right? IE. so still circumventing the progressively-squaring cost of continuously-rising velocity, only now without necessitating four dissipative collisions to divide up the accelerations; all four benefiting from the same base-rate efficiency at the same time / given system RPM..

Could be half an idea here... just need to mull it over a while..

[MrVibrating]

i've played extensively with these kinds of interaction

but would be happy to give feedback on any specifics.

As i have already said, i do have difficulty following your posts. I left school with a trade (woodmachinist). My level in science leaves a lot to be desired.
When you played with this kind of interaction, bearing in mind we are only talking about "the buyer" in Bessler's words. Were you able to establish through tries or mathematically, which configuration was the most cost effective?
From what i understand, which could be complete nonsense, is that evenly distributed "buyers" are less cost effective than unevenly distributed buyers.

Evenly distributed buyers have a COG (of all the buyers) allways offset from the center axis one side. Basically hovering where the moving weight is. The one doing the buying, whatever it is it buys.

Unevenly distributed buyers have a COG, of all the buyers, that hovers to the left and to the right of the central axis. Yet the movement of the buyer is identical.

Does this not mean that we get the same energy return at a lower energy cost? Or am i missing something?
You don't need to dumb it down for my cat to understand it, but dumbing it down just a little would be helpfull.
Thanks.

Mate i have zero quals, just trying to grab the bull by the horns; the only thing to be buying is momentum - B's wheels accelerated - and the only place selling it to statorless wheels is gravity & time. KE is bound to the ground / harnessing FoR so a rotating system of masses can only have 'the right' KE, so only a PE discount on the cost of accumulating momentum can break PE:KE symmetry.

Energy squares with velocity so ten purchases of one unit of momentum at half a Joule each spends a total of 5 J for a ten-unit velocity rise (assuming net mass / inertia was constant, 'momentum' = 'velocity'), hence 50 J of KE, and 10x OU.

So we need to keep the input / relative value of V² to its min. (ie. pseudo-static) value when buying..

..yet while sitting on the nest-egg of its squaring cumulative value in the ground / KE frame.

Upon manifesting gain we obvs. then need to reinvest it internally..

If your idea is essentially to cause a clockwise vs CCW momentum-from-gravity asymmetry that's precisely the right objective, and the only sane course to investigate.

It's the input / output efficiency that obviously matters tho - how low you can buy, relative to the fixed market value.

Without telemetry, a rig either demonstrates a clear momentum or energy asymmetry, or it doesn't, i'm afraid.. without replicating in WM there's little more i can say (not least with one arm ATM)..

[Robinhood46]

Time is the only thing we can play with. it is the only explaination to PM being possible.
The only way to play with time, is to play with different FOR.
The way i see it, momentum is what we have to spend.
The momentum of one buyer on the one system, (all the buyers) with the small weights. To create the excess weight on the other system ( the wheel/frame) with the heavy weights.

The buyer spends some momentum and buys enough lateral displacement of the total COG to cover the costs and cause acceleration.
The uneven distribution of the buyers is what brings the cost down.
The falling back of the buyer is what seperates the two systems which allow the advantage of the time difference between the two systems to come into play.
Are we saying the same thing in different ways?

[MrVibrating]

..poss. Euraka moment (back's still wet); could it be this simple..?

• spool a hanging weight off a wheel via a rope, thus applying constant OB torque & dp/t

• nest a series of equal MoI's within it, interconnecting via motors

• the weight doesn't drop, instead suspended by the net counter-angular momentum * time of the nested angular accelerations

• so if each MoI = 1 kg-m² and each motor applies a 1 rad/s accel using 1 N-m over however many seconds as required, is that still ½ J of torque * angle each? IE. so for ten-deep, 5 J net PE, for 50 J of KE..? Versus a weight that never actually descends, just counter-balanced by the net dp/t of the nested rotors versus its own dp/t as a function of its static weight..?

Crazy-simple, no? Will test it shortly.. ☺


ETA: if this works then we could just go four-deep and solve the riddle without dissipative losses or stacking force * displacement.. IOW near-instant OU.. 100% conversion of work, zero GPE or inelastic collision losses..!

..gotta be too good to be true, no..?

[MrVibrating]

Time is the only thing we can play with. it is the only explaination to PM being possible.
The only way to play with time, is to play with different FOR.
The way i see it, momentum is what we have to spend.
The momentum of one buyer on the one system, (all the buyers) with the small weights. To create the excess weight on the other system ( the wheel/frame) with the heavy weights.

The buyer spends some momentum and buys enough lateral displacement of the total COG to cover the costs and cause acceleration.
The uneven distribution of the buyers is what brings the cost down.
The falling back of the buyer is what seperates the two systems which allow the advantage of the time difference between the two systems to come into play.
Are we saying the same thing in different ways?

Gravity's an ambient, constant, background time rate of change of momentum, or 'delta-p (momentum) over time' - dp/t.

An OU wheel must somehow buy momentum from that source, while spoofing a lower-than-actual (and thus, that much cheaper) velocity.

One, obvious (ahem), way to try this is to simply nest the acceleration within another..

..a less obvs. solution, perhaps; kiiking - the ice-skater effect causes speed changes that affect up and down-swing exposure times to gravity's constant dp/t, thus causing net momentum changes over a full swing or rotation.

Classic OB likewise subsidises the time spent rotating upwards, shedding momentum back to gravity, with radial lifts instead, the weight thus spending more time per cycle over-balancing than balanced back in the opposite direction, and so net momentum rising.

Not an oracle mate, soz, limited time and a bust arm here, please go easy on me..

Besides i wanna try this latest idea..

ETA: to grasp the speed / energy / momentum relationship study the KE and momentum formulas, they're only using the same two vars. - it really ain't rocket surgery..

[MrVibrating]

NB. @ all:

• resolution of the vis viva dispute is absolutely central & pivotal to Bessler's IP

Anyone not intimately au fait with its solution simply doesn't, seriously, have their head in the game, or much prospect of ever understanding his principles..

[MrVibrating]

Tried it, five-deep:



No cigar; it only works as envisaged if the motors are controlled for 'acceleration' (1 rad/s²), and as anyone more sensible might've guessed, simply compounds the net work being done by each underlying motor, for perfect unity.

If they're controlled for 1 N-m of torque instead, only the top rotor actually accelerates / rotates at all - all further torques and counter-torques beneath perfectly cancelling.

For good measure i tried doubling, then halving, the drop weight, to cause it to fall or rise instead, again finding perfect unity in every case.

So this approach - while initially seeming to satisfy the riddle's conditions - can't be right..

On reflection, the one aspect that perhaps didn't perfectly fit was that there's nothing intrinsically 'quadratic' about it; if it did work, you could go four-deep or ten.. tho just two would suffice. Perhaps the same is true for whatever Bessler was describing and 'four' was just a nice round number - perhaps for rhyming reasons, even - but given the weight of importance he ascribes to the riddle, i'm given to think it's probably not arbitrary.. perhaps even relating to a minimal implementation as per 'one crosspiece'..

Will just have to think of summink else... summink inherently involving a 'factor of four', for some non-trivial reason..

[Leafy]

There is something mysteriously wonderful about the ice skater effect. Suppose I have a rotating disk of large inertia as the figure shown. Point A is at the outer rim and point B is half the radius inward. The velocity of A is twice the velocity of B. However, according to A and B frame nothing is moving.

Now let’s move a mass from A to B. The energy when the mass arrives at B equals the input energy + the speed difference since the large inertia disk does not speed up much. Suppose we extract this energy, then the mass now move with point B. Now we let the mass slide back to A. As the mass slide back, the disk impart energy to the mass so it would be at the same speed.

So you see, if we repeat the process, we keep gaining energy. The energy is taken from the disk. What if the disk is Earth. That means it is possible to extract energy from the Earth.

[ME]

Now let’s move a mass from A to B.
...
Suppose we extract this energy

Then you will not get from A to B.
It's like saying, we kick a ball upwards but don't kick it to save the effort.
Or, when it does arrive at B it no longer has that amount of kinetic energy as you put in at A.

What if the disk is Earth. That means it is possible to extract energy from the Earth.

The Earth spins so we all fling a bit outwards. Gravity is stronger, so as a net effect we actually won't.

We could use gravity to create a potential difference; a battery.
And the lifting of huge amounts of mass impacts the rotation of the Earth.

Three Gorges Dam (Yangtze River, Hubei, China) (lifting 175 m) :: Deviation in the Length of Day: 0.06 µS/yr
https://www.besslerwheel.com/forum/view ... 110#161110

[Leafy]

Well, if from A a 1kg mass has speed of 1m/s. If it moves to B it’ll have 2m/s. The energy input is 1.5J. The disk speed at B is .5m/s. We can extract 1.875J.

[ME]

Hard to verify those numbers without specifics.
But, as the main culprit, how do you bring it up to rim-speed again when it arrives back at A?

[Leafy]

The disk will bring it up to rim speed.

Overall, can we not extract energy from the disk being on the disk?

[agor95]

Hi Leafy

When I first read your post I skipped it; as members do when the concept is not in-line with their current ideas.

However walking in another members shoes is a good way to see if you need some new shoes.

The conjecture you are putting forward is plain too see.

There needs to be work done to move the mass-A to position of mass-B.

This will cause mass-A's velocity too come down and mass-B velocity up.
So when they are at the same location they will be doing the same speed.

Note. It would be good to either put the masses on the front and back of the disk.
So they can be in the same location.
Also look at the results of having them at 180 degrees radial from each other.

I appreciate you are putting forward the disc has a high inertial mass.
So mass-B velocity changes slightly while mass-A velocity changes the greatest.

Mass-A when released from location B will travel tangentially back to it's radial position
without doing any work by an external method.

From this you see work being done to move mass-A in and no work to move it out then more work to move it in again.

The disc rotation rate goes up then down again. However the large inertial disc mass make this very small.

This is my understanding of a non-gravity outcome.

All the Best

[Leafy]

Thanks

This is a simpler way to see. As the mass move back and forth there exist Coriolis force. We can use Coriolis force to do work. The disk will slow down as the result.

[agor95]

Hi Leafy

This is an interesting point, Coriolis force, as the mass-A spirals in and out as it moves.

I did read some analysis on the subject of an outward spiral track that was rotating.

I look forward to your research in this area.

All the Best