Blood From Stone

[MrVibrating]

The prize, it is certain, is a gravitational interaction that gains momentum (rather than energy per se).

The cost of accumulating this momentum gain over successive cycles must be constant with respect to velocity - whereas normally, the energy cost and value of momentum squares with rising velocity, we want the value to square up, while the cost per cycle remains constant.

Below is a potential example of a first rung on that ladder - a gain in momentum from a closed-loop GPE interaction



Gravity off:




..no net change in momentum.



Gravity on:



..and we have a complete GPE interaction - the weight rises and falls equal distance - yet with a net gain in momentum.



Very simple. Utterly elementary.

Where did the momentum come from?

Where's the counter-momentum?


Just occurred to me in the last 30 mins - knocked this up quick-time while waiting for work to start, jobs coming in now so gotta go... back later.

[MrVibrating]

...so, is there some way of moving the weight back out, without giving back that momentum...

...or possibly even gaining more?

(NB: the empty arm on the right is just there to counter-balance the left one, showing that the momentum hasn't been gained from its GPE drop)

[ovaron]

No gain of momentum. The pivot point of the lever with the weight is descending. To close the circle, the gained momentum is needed again.

[cloud camper]

The motor does work lifting the weight which is then used to create momentum in the wheel. No gain that I can see.

But glad you're back!

[MrVibrating]

No gain of momentum. The pivot point of the lever with the weight is descending. To close the circle, the gained momentum is needed again.

There's a 3,000 kg-m^2-°/s gain in momentum when gravity's enabled.

Yes, the pivot point of the lever descends during the interaction. You're presumably aware that work is only done if the weight itself changes height.. not the point of application of a weight? Besides, we're looking at a momentum gain, not an energy gain - but even then, the second, empty lever perfectly balances the weighted one.

And besides which, the levers themselves are all but massless - the weight weighs 1 kg, the wheel is 100 kg, and the levers are 1 gram.

The levers and weight stop moving when it parks in the center.

In the no-gravity case, when the levers and weight stop at the center, all momentum and counter-momenta cancel out to zero. The system begins and ends with no net momentum.

But with gravity enabled, the same interaction yields this significant momentum rise.

Similarly, consider a conventional GPE interaction where a weight is lifted and dropped equal height - the momentum and counter-momentum applied to Earth normally all cancel out.

Here however, instead of pushing the Earth downwards in order to push the weight upwards, we're pushing the sides of a heavy wheel downwards instead - so the counter-momentum has been sunk into a heavy angular inertia, instead of being applied to Earth.

This is why the pivot point of the lever must descend. IE. it's descending because the wheel it's attached to is gaining counter-momentum.


For clarity, here it is again with tracking enabled, to highlight the weight trajectory:



...so the point is that we've lifted and dropped a weight equal height, so no net GPE work has been done, and the only net work done is the inertial interaction... That is, the momentum is not 'free' - we've paid input energy for it, via the motor powering the weight lever.

It's crucial to note that normally, neither an inertial interaction, nor a gravitational one, can yield a net change in momentum.

However performing both at once does accomplish this (a point i've demonstrated many times so far - this is just the most recent variation on the same thing..)


Given that an inertial interaction on its own cannot produce a net rise in momentum...

..and further given that a gravitational interaction on its own cannot produce a net rise in momentum...

..where has this momentum actually come from? Was it input by the motor, or by gravity? Neither, or both?

Simple interaction, fascinating results. But in order to refute the validity of the 3,000 kg-m^2-°/s being presented, you need to show where the other 3,000 kg-m^2-°/s of counter-momentum is hiding...

It looks to me like this can be repeated for a second cycle, will have a try later. If so, then the question becomes whether or not the cost of this momentum gain remains constant as RPM's mount up...

[MrVibrating]

The motor does work lifting the weight which is then used to create momentum in the wheel. No gain that I can see.

But glad you're back!

Cheers ears, but i'll never let this lie - the biggest secret is simply knowing it's possible, for which we have Leibniz et al to thank.


Leibniz, the very man who invented CoE - it's akin to having a videotaped interview of Albert Einstein lucidly recounting how he was abducted by aliens on multiple occasions..


Yes, the motor does work applying torque to the lever, and counter-torque back to the wheel.

The momenta induced by these torques are equal and opposite, and all cancel out to zero when the weight parks in the center of the wheel (as demonstrated in the no-gravity example).


Likewise, the momenta and counter-momenta intrinsic to a GPE interaction usually cancel out to zero - to push a weight upwards, we must propel the Earth downwards, and when the weight reaches its zenith and falls back down, the Earth falls back up in reciprocation; net change in momentum is nil.

So while i'm not necessarily disagreeing with your conclusion - the momentum is obviously produced by the confluence of the inertial interaction and gravitational interaction - we nonetheless have no net change in height (weight begins and ends at equal height), so we do not have the option of attributing the momentum gain to a drop in GPE - there's been no net output or input of GPE.

Yet if the momentum hasn't come from the GPE interaction, that only leaves the inertial interaction - the torque and counter-torques, and the momentum and counter-momentum they produced.

And again, as the no-gravity case shows (if a demo were even necessary), an inertial interaction shouldn't cause a net rise in momentum either..

Hence my assertion that the fact that both together do produce a momentum gain, is not trivial!

Acting in concert, the two interactions produce a result that neither on its own should ever be capable of!

It is disarmingly simple, and all the better if it seems perfectly straightforwards... but it also seems to be breaking a fundamental rule that neither interaction on its own is able to accomplish.


As ever, generating a net rise in momentum is only half the battle - the energy gain only comes from consolidating that momentum gain over successive cycles.


And then, only if its energy cost of operation remains constant - that is, if we can pay the same input energy to buy the same amount of momentum (wherever its coming from), accumulating it and building it up over repeated cycles, then mathematically:

• if we can consistently pay, say, ½ Joule per kg-m^2 of momentum, then 10 kg-m^2 costs 5 J

• yet according to KE=½mV^2, if our first kg-m^2 costs ½ J, then the second should cost 2 J, and the third, 4.5 J and so on; hence 10 kg-m^2 should ordinarily cost 50 J. But regardless of how much we actually paid for it, it's still worth 50 J - it'll perform 50 Joules of mechanical work, converting it to GPE or KE or sprung PE or whatever...

• Hence the mere act of accumulating momentum this way is an inherently over-unity process - our net efficiency keeps increasing, the more momentum we invest in.


I cannot stress enough that there is no other process in physics that does this!!!

What Leibniz et al were describing - Bessler himself - was excess mechanical energy. OU KE. Only an effective momentum asymmetry can accomplish this, and it is only possible in a wheel in which "everything must, of necessity, go around together"...

There is simply no way that Bessler could've deduced, from 1st principles known at that time, that "in a true PMM, everything must go around together" - the vis viva dispute hadn't yet been settled; Newton was saying mV, Leibniz mV^2, neither were exactly right or wrong - so he must be speaking from practical experience. He's experimentally nailed the decisive principle, that what we would today describe as an "OU motor" is essentially an ordinary motor in which the stator rotates along with the rotor.

In other words, the only type of mechanical OU possible is a motor in which torque is applied between a rotor and stator, but then the momentum imparted to the rotor is shared back with the stator, so that both are rotating together, and then this process is repeated, gaining more momentum each cycle. Ie. a motor hauling itself around by its own bootstraps, albeit with a lil' help from gravity..

[Fletcher]

Mr V .. just a suggestion or two (2 cents worth) to cross check what you are deducing with the sims. Haven't got time atm to do this myself.

1. turning the arrangement CW 90 degs so that the lever-weight pivot is at the top and the lw horizontal. Use the motor to swing the lw down and then upwards regaining GPE of the weight.

2. find a way to use two lws on the wheel base so that the lw action is fundamentally gravity driven (with perhaps a small help) rather than being energy consuming motor driven ?

Does it still show what you expect ?

[MrVibrating]

...coming back to the one-way wheels, which were under static torque when tied off stationary:

• With no stator to apply torque against, the only type of torque possible was "overbalancing" torque - ie. the wheel wanted to rotate because doing so lowered an internal weight.

• Yet OU is only possible from a motor in which the stator rotates around with the rotor. That is, it is dependent upon an asymmetric inertial interaction.

• Thus we've reached an impasse - an apparent paradox; the form of torque required for OU necessitates an inertial interaction, albeit an asymmetric one, somehow...

...yet the form of torque apparent from the 1-way wheel's static torque is exclusively consistent with a weight drop!


So there's only two possible solutions to this discrepancy:

• Either the inertial interaction happened during or after the GPE output - so the wheel rotates to lower a weight, and either at the same time, or afterwards, an inertial interaction is performed..

• Or else, the GPE interaction is the inertial interaction! Think about it; any GPE interaction is by default also an inertial interaction; pushing a weight upwards also pushes Earth's mass downwards, then they mutually fall back together, restoring their original positions.


On the first possibility - that the inertial interaction is a separate discrete interaction; it cannot occur during the GPE interaction, since it would also be applying torque whilst the wheel was stationary, which in turn would've lifted the internal stator against which that torque was being applied.

Yet if it occurred after the weight reached BDC instead, we have to keep the inertial and gravitational interactions synchronised as the speed builds up, without a proper (ie. actually stationary) stator to lean on..

Furthermore, what of the 'other' weight, that must be re-lifted somehow? It would mean this process can only begin once the first weight's dropped, and we've subsequently completed our asymmetric inertial interaction, then netted the momentum gain and converted it into a GPE rise...

..it's not simply that it gets so mechanically complicated as to be all but unassailable; as RPM's build up, there simply isn't the internal time or space to keep these actions synched without a proper stator, and Bessler not only denied anything hanging from his axles, but accurately explained that this would preclude OU in the first place..


Hence Occam's pointing to the latter option - that the GPE interaction is the asymmetric inertial interaction.

Here, we're instantly on much firmer ground:

• This 'OB torque' from gravity has no instantaneous counter-torque - we only get the counter-torque when we come to re-lift the weight, but it presents none on the way down.

• Hence it would seem most elegant if the form of output torque was actually exclusively OB torque - since it remains 'reactionless torque' until we need to relift the weight.

• Hence the implicit proposition that there must be some way of re-lifting a weight without applying the usual counter-torque:

- either the counter torque can be cancelled, or reversed / inverted.

• So we'd be looking for some way of re-lifting a weight, that nonetheless torques the wheel in the same direction as the OB torque when the weight's descending..


...and this is what hit me before work this morning. Twist'n'lift, applying positive torque for both the lift and drop.

As you can plainly see, it's not only possible, but so simple as to appear utterly trivial. Weight goes up, and back down, both strokes generating torque of equal sign and magnitude, for a clear'n'easy 3,000 kg-m^2-°/s rise in momentum.

It really is the same shite i've been doing for the last year or so - coupling a GPE interaction with an inertial one... neither on its own can do anything special, but paired up in the right way, these mechanically simple motions produce challenging results.. namely, unilateral momentum, in an ostensibly-closed system of interacting masses..

And that right there is our meal ticket. The only one that'll ever be on the table; the "true PMM"..

[MrVibrating]

Mr V .. just a suggestion or two (2 cents worth) to cross check what you are deducing with the sims. Haven't got time atm to do this myself.

Always most valued..

1. turning the arrangement CW 90 degs so that the lever-weight pivot is at the top and the lw horizontal. Use the motor to swing the lw down and then upwards regaining GPE of the weight.

You've just precisely described the 'reset stroke' i've been envisaging all day - with the weight parked in the center, let the system coast around 180° so that the weighted lever is hanging, then rotate it out and back in for a second complete GPE interaction and accompanying asymmetric inertial interaction.

This means we get two 'power strokes' per full wheel cycle - just as B. intimated of his own designs ("stampers are lifted and dropped twice per cycle").

2. find a way to use two lws on the wheel base so that the lw action is fundamentally gravity driven (with perhaps a small help) rather than being energy consuming motor driven ?

This is the confusing part - surely if the weight's travelling equal height up as down, there can be no net output of work from gravity! We can neither look or hope for one, it's simply not sur la table...

We have to have an inertial interaction, superimposed upon the GPE interaction. We have to pay for the inertial interaction (some other store of PE besides the OB weight is required). The best we can aim for is simply an inertial interaction, in which the balance of momentum and counter-momentum is skewed, somehow, by gravity, resulting in a non-cancelling sum and thus net rise.

Furthermore this cost of operation - how much energy we have to spend to gain how much momentum per cycle - has to remain constant, not squaring up with rising velocity.

Only then do we break CoE.

Does it still show what you expect ?

I'm a complete idiot who'll think and say all kinds of mad shite, but i think the second stroke you suggest is going to add even more momentum, from wherever it's coming from..

..if so then best case scenario is we'll end up with ~6,000 kg-m^2-°/s of momentum (not as much as it sounds in these degree-based units, i'll switch to radians for the energy accounting stage).

So the objective then will be to measure the energy cost of each stroke as a function of torque * angle applied by the motor/s, and compare this to the actual RKE we've generated.


It may turn out to be just a conventional motor, albeit needlessly Heath Robinsoned, in which RKE is only ever equal to input energy. That was the result i got from the prior scheme in which a mass was levitated by thrusting another mass or angular inertia downwards - it generates unilateral momentum, but its cost of production never diverges from its actual value.

The winning formula is very specific:

• Generate 'reactionless' momementum (effectively reactionless from within the closed system of interacting masses; obviously, a working system will be applying counter-momentum to Earth via gravity: that is, the system is not actually thermodynamically closed)

• Accumulate this momentum, building it up over successive cycles.

With those conditions met:

• Its energy cost of production must be speed-invariant - so, not squaring up with rising velocity. In practice, the cost may vary... provided it doesn't go so far as to square with increasing RPM, we still have a gain margin in the difference.

The final step - cashing in - is simply the vanilla KE=½mV^2. Like i say, if we can generate 10 kg-m^2 of momentum at 10 * ½ Joule per kg-m^2 (the mathematical minimum price) then we have 50 J of KE for an outlay of 5 J, hence 10x OU.

But even if it's 1 J / kg-m^2, that's 10 J in for 50 out. If it's 2 J per kg-m^2, that's 20 J in for 50 out, etc. etc. In fact, because KE squares with velocity, then whatever the unit energy cost of momentum, there will be some threshold velocity below which we're under-unity, and above which we're over. The only requirement for success is that the cost does not square with rising system velocity.



So it's the same old same old, with a slightly different mechanical approach. It's the same OU energy gradient i'm trying to access, a divergence between the cost of generation of momentum, and its conventional ½mV^2 value, hinging upon an effective violation of Newton's 3rd and thus 1st laws - to wit, successively gaining momentum in an otherwise closed-system of interacting masses, which is an inherently over-unity process.

As ever, even if it works, i've no idea how to implement it in a real system. My first and only priority for now is to prove such a mechanism is possible in principle, if not yet in practice. Either torque * angle on the motors is going to equal net RKE, or it isn't..

[cloud camper]

Before you start typing pages of text why don't you just calculate the GPE gained by the weight and then determine the RKE of the wheel.
Dollars to donuts they will be equal!

I'm too lazy to do this but you can save everyone reading walls of text by doing this simple calculation.

[MrVibrating]

...there is no net change in GPE - it rises and falls equal distance, therefore equal work has been input to and output by gravity.

The main measure of work is going to be the torque times angle of the motor, relative to the RKE of the base wheel. I've no reason to presume they're unequal at this stage - probably a perfect match.

The exploit i'm attempting is a momentum asymmetry, not an energy asymmetry, in the first instance.


In short, we're all familiar with the fact that gravitational interactions are energy-conservative - that G*m*h = ½mV^2, hence 'symmetry' between GPE and KE... however there's also the issue of momentum symmetry - the fact that it's impossible to alter the net momentum of a system of interacting masses via the internal expenditure of work, only an externally-applied force can do this, which in turn necessitates applying that force against some other mass, which now becomes part of that 'net system' for which the net momentum remains constant.. all gravitational interactions are also inertial interactions, and it is the balance of momentum from this inertial interaction that i'm trying to manipulate.

Like i say, the issue is not the unit energy cost of momentum per se, so much as its speed-invariance - in an OU motor, the cost of generating more momentum is unaffected by the rising system velocity. Whereas, in a conventional motor, in which torque must be applied against a stator, the unit energy cost of momentum increases per ½mV^2.

It'll probably turn out that the weight here is just acting as an ordinary stator, that the input energy will have to square up with rising RPM and hence precluding anything interesting happening. Dunno. Let's see if i can repeat the cycle, adding more momentum, then i'll tot it all up and see what's what..

[cloud camper]

You are just fooling yourself, not anyone else.

You become mesmerised by your walls of text and ignore the obvious.

Nothing happening here!

[sleepy]

Gotta agree with CC. The amount of energy it will cost to reset the weight, no matter where in the cycle you reset it, will always equal,if not exceed,the positive gains. I have always appreciated your forward kind of thinking, but IMHO, this is a non-runner.

[MrVibrating]

Like I say, all that matters is whether or not the torque * angle of the motor squares with velocity.

I suspect it does, since the motor will have to spin faster with rising RPM - hence it's basically just a novel type of stator..

It would seem fundamentally no different to a crank & conrod torquing against a conventional stator..

Still, started so I'll finish. Will do the torque plots when I get home later..

[cloud camper]

The torque angle of the motor has nothing to do with it.

The only thing that matters is how high the weight was lifted (GPE) and how fast the wheel turns (RKE)

Everything else is smoke and mirrors.

But thanks for trying!