Perpetual Motion is Impossible

[ME]

The momentum of one weights moves into the other weight. A very simple situation. There is no extra velocity. There is instead extra kinetic energy available.

So basic elastic collision is ruled out?

[eccentrically1]

I'm posting garbage, right.
The formulas for momentum and kinetic energy only have two parts, mass and velocity. To magically transform the net kinetic energy of the two weights to a greater quantity requires one of those parts to increase.
If there isno magical transformation of mass, and no extra velocity, from whence does the extra kinetic energy originate?

Got the white space fixed. Much better that way.

[jim_mich]

Twelve days back, on Dec 14, in post #139011 I posted the following...

As an example. Assume you have two weights moving a same speed. Transfer momentum from one weight to the other weight. This can be done spontaneously in a rotating environment of a wheel with little or no effect on the rotation. Do the math. Assume both weights were moving at say 20 feet/second and both weights have a same mass of two units. The weight's KE is 1/2 × 2_mass × V^2. Since 1/2 × 2 equal 1, we can make the formula very simple as KE = V^2. Thus KE of each weight equals 20^2 = 400 KE units. Total KE combined of two weights is 800 KE units.

Now one weight spontaneously gives up half its velocity to the other weight.
Weight #1 V = 10 ft/s. Its KE = 10^2 = 100 KE units.
Weight #2 v = 30 ft/s. Its KE = 30^2 = 900 KE units.
Total KE combined of two weights is 1000 KE units.

Thus we gained 200 KE units simply by allowing motion to transfer from a weight that is slowing down to a weight that is speeding up. We added no energy. Momentum was conserved. The weights changed their velocities spontaneously without consuming any energy.

WAS THIS EXAMPLE TOO HARD TO FOLLOW ???
The formula for inertial momentum is IM = m × v.
The formula for kinetic energy is KE = 0.5 × m × v^2.

The result is that as the IM increases in a linear fashion, then the KE increased exponentially.

If momentum doubles, then kinetic energy increases four times. And if momentum triples, then kinetic energy increases nine times.

The two do not directly track one to the other.

[ME]

WAS THIS EXAMPLE TOO HARD TO FOLLOW ???
The formula for inertial momentum is IM = m × v.
The formula for kinetic energy is KE = 0.5 × m × v^2.

The result is that as the IM increases in a linear fashion, then the KE increased exponentially.

Probably I overlook something..
L=I*w; I=m*r^2; w=v/r; L=m*v*r
(radius dependent)

It really is not that difficult of a concept, except it takes knowledge of HOW to PROPERLY move momentum.

I don't Know How you move your momentum Properly.
(perhaps others do? - At this point I need a picture for that.)

This can be done spontaneously in a rotating environment of a wheel with little or no effect on the rotation

The two do not directly track one to the other.

So transfer happens when the initiator-mass shifts radius.
When such weight basically freewheels and centrifuges straight ahead, it will loose speed relative to the radius it is at and on an increasing angle...
I'm not sure all the formulas line up as nicely as you say;
-but perhaps eventually they just do, and I'm clearly missing something-

[Fletcher]

I wouldn't worry about it ME .. you're probably not going to crack the formula's with only a few days left (and neither am I btw).

Just sit back and enjoy the occasion. History could be in the making and you might get to witness it and tell your grandchildren.

[Furcurequs]

http://star.psy.ohio-state.edu/coglab/Miracle.html

[ME]

I wouldn't worry about it ME .. you're probably not going to crack the formula's with only a few days left (and neither am I btw).

Just trying to understand that "easy" base premise which Jim is trying/willing to explain.
I don't expect to get a pre-glimpse of his mechanism; I hope for Jim it works, and for us to be revealed.
And when we assume it does work and we eventually know the mechanism, that mystifying easy-base-premise is gone - once you see it, it's hard to be unseen.

[MrVibrating]

Any interaction creates thermal energy. You can't build an ideal machine, 100% efficient, much less a machine that is more than 100% efficient, powering other machines.

Any interaction is asymmetrical in the "loss" column. So no, it's not possible.

You're conflating two incompatible measures of efficiency - our principle concern is input vs output efficiency, not the coefficient of performance of either stroke - confuse 'em and we may lose the baby with the bathwater.

Input vs . Output is what efficiency refers to and that is what I meant. I don't think what I said referred to the coefficient of performance of either stroke.
I'm not confused.

You said that "any interaction is asymmetrical in the loss column" - this is a confused statement, belying muddled thinking, so i'm trying to disentangle these precepts.

Entropic (ie. dissipative) losses are wholly irrelevant to interaction efficiency.

Non-dissipative losses would be an essential consideration, but these only arise in time-varying systems.

To calculate interaction efficiency we first plot the input work integral. Ideally this should be considered as notionally lossless. However if losses do need to be included then we plot them as a separate integral, or indeed several more if necessary, and these all stack up on the left side of the equation. These products are then summed together to fully describe our input energy.

Then we do exactly the same thing for the output energy, on the right side of the equation.

Then we simply add the net energy input to the net output energy - if they have a non-zero sum (positive or negative) then we have an asymmetric interaction.

So the "symmetry" in question compares input to output energies of a complete interaction.... not just the work conversion efficiency of one half of it.

Just because either side of the interaction has losses included does not mean that a single side of the exchange can be "asymmetric".

Your comment implied you're thinking that "symmetry" refers to the conversion of work vs entropic losses. By that metric, perfect symmetry would imply a 50% efficient conversion of input energy to input work.

But that's not what the word symmetry refers to. Even if the conversion of input PE to input work was only 1% efficient, we still put that 1% on the left column, with the 99% loss underneath it, also on the left, and then we do the same for the output column on the right, and compare the two totals on either side.

Yet all of this is more information than we need; everything we need to know here can be distilled into a concise catch-all requirement:

- we're looking for a system in which net input (whatever its forms or terms) is unequal to net output (whatever its forms or terms); this is a classical asymmetry.

And that is impossible. Bessler was subject to the same laws as we are. His wheels were not outputting more than they were inputting.

LOL but that's what a classical asymmetry means! It's what over-unity means. It's what an open thermodynamic system means.

Without an I/O asymmetry we have a closed system; there is no excess work potential and at best, Bessler's wheels would've coasted for a while.

I agree with what i believe is your underlying sentiment - that energy is ultimately conserved, and cannot be created or destroyed. We have no end of evidence supporting this, and in terms of epistemology any proof to the contrary is intrinsically impossible - we can only conclude than an asymmetric interaction is indicating an open thermodynamic system, and that the excess or deficit of energy is accounted for by whatever was responsible for manifesting the force field or its variation in time.

So to pull all of this together into a simple example, suppose we have some candidate mechanism we wish to test. The input energy could be anything, but will inevitably conform to the work-energy equivalence principle - so it'll be describable in terms of a graph that plots force magnitude on one axis, and displacement on the other axis.

From this we get some kind of line - could be a curve if the force is changing non-linearly with displacement, or a straight diagonal line if the change in force is constant relative to displacement. If the force change is nil or negligible then we get a flat line. But whatever the shape, the area under the curve is called the "line integral" - it represents the net energy of the system at the end of the input stroke, not including losses (which at this stage would usually be a needless complication).

Then we have an output stroke, plotted in exactly the same way. The two strokes could be in series or parallel, and the latter may even precede the former. All that matters is that we can clearly delineate them.

Then we compare the two integrals - the net area under the two curves. If they're equal then we have unity, and some kind of lever mechanism, however extravagant. The system is thermodynamically closed.

But if they don't cancel - if their sum is non-zero, then we have an asymmetry - an open system. Its net energy is not conserved. Energy is being sourced from or sunk to somewhere, depending on whether we have a loss or gain.

Bessler's system produced an excess of work. Work = force times displacement. It was a rotary system so all such displacements must be closed loops (the weights can't progressively work their way out of the wheel). So if the "d" part of the F*d work integral is constant, then the only available variable is F.

So we have an input F*d, and another for output. The two d's must be equal, therefore if we have a loss or gain then F must be varying either during or between input and output strokes. If we gained energy then net input force was lower than net output force over the same distance. If we lost energy then output force was lower than input force over that same displacement.

Classical conservation of energy does not and cannot be applied to time-varying systems, since their net energies are a function of internal displacements relative to external (ie. passive) variations in force magnitude. Such systems are thermodynamically open. You may get a unity result, or you may not, depending on the particulars of the inbound vs outbound displacements relative to the prevailing force magnitudes, as described by their respective integrals.

So yes, Bessler was subject to the same laws of physics, but i think he'd agree with my main point here, too - which is that the physics is telling us as much what to investigate, as much as what not to bother with. When you get right down to it, the first half of MT is thoroughly illustrating the many ways in which a closed loop through a static field yields zero energy. In AP he makes this same "friendly word of caution".

The only consistent resolution, for an asymmetry requiring equal inbound vs outbound displacements, is a passively time-varying net force. Either the input workload, or the output workload, or both, are a function of dynamic, not just static, forces.

The only other alternative is that it is, in fact, the "d" part of the work integral that is time-varying, rather than the "F" part. But if input and output forces are equal, then how can displacement vary, without the location of the masses progressively drifting out of the system over successive cycles?

And this is one reason why the possibility of an effective N3 violation is so intriguing - it would "create" energy by keeping the relative input displacements to a low constant value, due to the reaction mass remaining in the accelerating reference frame; whereas normally the distance over which a given force must be applied to achieve a given acceleration increases by half the square of the accumulating velocity. In effect we'd be circumventing the usual requirement to raise the "d" part of our input work F*d integral in step with rising speed, while the net system energy as measured from the external static frame remains a function of constant rotational inertia times rising angular velocity (rising displacement / time). Hence the internal input energy per cycle would remain equal to a constant MV, while external output energy remains a function of the accumulating net velocity.

In other words, the reason the masses wouldn't be progressively working their way out of the system despite the asymmetric distrubution of input to output displacements is because the net change in position is angular, not linear.

But whether we try to modulate F, d or both, what we're looking for is an interaction where the F*d of the input work is unequal to the F*d of the output work. Entropic losses would not need including in the evaluation of a principle for asymmetry - you may choose to measure them in a real build, but if a principle asymmetry isn't there in the theory then losses are irrelevant, while an actual asymmetry persists regardless of losses - again, if input workload is 90% efficient, wasting 10% as heat, and the output workload is only 40% efficient, losing 60% of its energy to heat, this has no bearing on the symmetry of the interaction - its net efficiency. If net input is unequal to net output then whatever its distribution of macroscopic or microscopic displacement (heat), the sum of input pennies and pounds is unequal to the output sum.

So we have two options - two potential lines of attack; find a passively time-varying force against which to manage a closed-loop trajectory, or else manipulate Newton's 3rd law to eliminate the accumulating displacement that normally arises between mutually-accelerating masses.

And although i initially considered the latter option the outlier, having pretty much exhausted the former (like everyone else here) i'm now seriously reconsidering it. So is JC. And i've seen Steorn do it successfully with EM systems (via Lenz's law). Plus there was a recent post here about someone claiming to have solved the Bessler case, also implicating N3 symmetry as the decisive hurdle.

But whatever the blend, OU = input F*d < output F*d. An odd pair of integrals. Each in their own right is fully consistent with, and indeed dependent upon, conservation of energy applying instantaneously; both integrals have exactly the right amount of energy for their respective, but different, conditions. Their non-zero sum is, to quote Sean McCarthy, a "logic trap" for the conservation laws - basically gaming the system, turning the terms of its own rules against it by exploiting a legal loophole.

There can be no paradoxes... the ultimate source or sink in an asymmetric interaction is defined by the same terms - an excess of force and/or displacement has been paid for by whatever powers a force field or its variation, whether inertial or gravitational.

As a closing thought, lurking around in the muddy depths of this puddle are profound questions on the nature and role of time in mediating the conservation of energy - we often consider the implications for the nature of the vacuum in classical asymmetries, but i suspect (as others) that validation of such systems will inevitably lead to further revelations for our conceptions of time.. could it have a latent energy - a momentum, driving things forward, symmetry or no? Or perhaps it's just a dumb croupier, letting the chips fall where they may.

"Diverging timelines" is also something Sean McCarthy's mentioned recently.. but the role of time has always been central to their working theory. However there may be interesting implications for manipulating relative rates of time if it is indeed powering the forces: an asymmetric system would be temporally accelerating; the direction depending on the sign of the asymmetry (loss or gain).

Big "IF", but a fascinating compliment to the usual considerations...

[jim_mich]

Blink... Blink... Blink...
My eyes hurt. You make things way too complex.

The solution is minimal input to get things rotating. Then no more input. Only output. Perpetually. Until motion is stalled by excessive load or the component parts fail.

[Trevor Lyn Whatford]

Hi Jim_Mich,

The solution is, first get something that works! Do not expect people to eat crow, when all you are feeding them is BS.

[ME]

I agree with Jim's recent summary of MrV's dissertation: OU = input F*d < output F*d

Can anybody tell why a gradient gravitational field would be beneficial in achieving pepertual motion?

[jim_mich]

Can anybody tell why a gradient gravitational field would be beneficial in achieving perpetual motion?

Yes, why would a gravitational field be beneficial in achieving perpetual motion?

I can see no benefit. Any gravitational input must be re-used to re-lift the weights back upward, thus no rotational gain whatsoever.

[ME]

I was just wondering the same thing.
I though I read such statement somewhere, don't know where / could find a relevant search... (lazy me thinks it's easier to just ask)

[Trevor Lyn Whatford]

Hi ME,

Your Quote,

Can anybody tell why a gradient gravitational field would be beneficial in achieving pepertual motion?

No, Not a gradient gravitational field, it would take too much energy to make one, if we had the technological, so without that, the only other way is distance, so what we are looking for hear it would be no good.
Unless you have found a extremely low energy input gravity shield that is.

Edit, whereas using a constant gravitational field, there would be a usable force input through out the device, that can be manipulated by a number of mechanical systems.

[eccentrically1]

Mrv, jim is right , your posts are unnecessarily wordy and over-analytical for Bessler's wheels.
When I said any interaction is asymmetric in the "loss" column, what I meant was any interaction will produce friction. This energy is dissipative and guarantees any interaction will have less output than input, i.e. < 100% efficiency, continuing the argument I was making. Sorry I confused things there.
I think, but I'm not sure because I 'm not going to wade through your entire post, my eyes are old too, that we are saying the same thing in different ways.