Decoupling RKE from GPE, for fun and profit

[MrVibrating]

So presumably W3 in your profile refers to London W3, where I went to school (Gunnersbury Grammar). I did wonder.

Say what you like about the Rothschilds but Gunnersbury park is a personal oasis at any time of day or year. Totally overrun with parakeets these days, but whether the story's true or not they always remind me of Hendrix..

The grammar schools are bit before my time - i'm guessing it's now the International school down Gunnersbury Avenue past the cemetery?

One of the many cool things about Google Earth is that i can zoom in on my manor from outer freakin' space using just two keystrokes... "W3", for now, refers exclusively to a unique, if insignificant, location on the Earth's surface... just two characters... shorter than its place name or coordinates.

Or maybe that only works in the UK? Dunno.. I just like the brevity of it.

[MrVibrating]

Let's see if this will help Kaine. There are some similarities IMO.

In my thread I was investigating whether horizontal load masses could be accelerated by a falling driver mass which might lead to a complete (or near complete) transfer of momentum from driver to load. If so this would mean a gain in KE above GPE lost.

I was using the sim program and both geared pulley systems and then latterly the storksbill arrangements.

In the pulley systems the the driver' velocity increased downwards (albeit slowly increasing) and the loads velocity also increased by a fixed factor of the gearing ratio. The accelerations were 'fixed' and did not change because the gearing was fixed.

In the storksbill systems the gearing was variable. This allowed the driver to slow and reverse its acceleration while descending whilst the loads acceleration initially increased dramatically then began to reduce. Overall the load velocity continued to increase as the drivers velocity decreased eventually to zero. In sim world there was no excess KE apparently.

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

In Mr V's case he is using a double fusee (one up and one down) to create a variable gearing also, like a CVT. The hanging driver attached to the bottom fusee should be able to be slowed right down so it has very little KE, and the RKE of the 'disk' load (attached at the end of the top fusee) should be able to be sped up dramatically because of this constant gearing change brought about by the double fusee action.

Since angular momentum is a different beast entirely from linear momentum he is wanting to investigate whether a faux N3 break is created which could lead in principle to a gain in load energy (RKE) above driver GPE lost.

If I have that wrong in any way I'm sure Mr V will correct me with the detail.

Sorry for butting in Mr V.

No you're spot on. If we had an N3 break then we could accelerate a mass or system by an equal amount for each successive unit of input energy, whereas normally each additional unit of input energy would yield an ever-diminishing return of work.

To put it another way, under ordinary circumstances, achieving a constant rise in acceleration requires an increasing amount of input energy - increasing by the square of the rising velocity.

But Newton's 3rd law doesn't, itself, explicitly say anything about this - it just says that every action has an equal & opposite reaction (or whatever the exact wording).

As such, its role in enforcing the square rise in KE, or preventing a linear rise, is fairly circumstantial. It's causing us a practical inconvenience, sure, but there IS no law of nature explicitly proscribing linearly-rising input integrals. On the contrary, 1J can accelerate 1.414kg by 1 meter per second, with no regards to velocity.. so why not repeatedly? N3 might be blocking one route, but perhaps there's others..?

If so this might be a weak point in the CoE laws. Maybe Newton's 3rd is a bit over-extended here - it's certainly one of the things stopping us, but is it the only thing?

Hence anything that allowed us to keep yielding the same inertial displacement for the same input energy despite rising velocity would grant us all the energy benefits of a real N3 violation, even though the real thing is impossible. It would have no bearing on inertial propulsion, as a real break would, but we'd still have a free energy gradient.

And there wouldn't be a damn thing the CoE laws could do about it. Quite the opposite - they'd be our bitch, working for us. It's just a legal loophole.

So, it's not simply the fact that the rotor is geared up relative to the drop weight - more than this, we need a power train that destroys the rising displacement component of our input F*d work, keeping both F and d equal for each successive bout of acceleration, despite the rising velocity, and so reducing our input energy while the output KE as measured from outside the system follows the usual square of velocity.


Gotta say, it's ironic how i was so convinced Fletcher's concept was impossible because momentum and KE have different dimensions - they just seemed too mutually incompatible for a direct conversion.

And yet, if input energy can be maintained at m*V, never allowed to square up, as i'm proposing, then in terms of its dimensions it is functionally identical to converting momentum into KE. We'd literally be converting PE to momentum, to KE... with similar big gains.

So while i began by asking why KE squares, there's probably more than a sprinkling of cross-pollination here - and the same thing goes for JC's recent thoughts re. his own work... i gave up looking for an N3 violation pertaining to odd numbers of mechanisms (which i'm still intrigued by nonetheless), but couldn't shake the consistencies between Bessler's wheel and the effects of an N3 violation.

Just thought a little hat tip was in order.. (as in spreading the blame if this is yet another brain fart).

[WaltzCee]

The only question that matters here is:

Why does KE square up, and does it have to (as in really, really have to)?

It's as fundamental as doubling the radius and getting 4 times the area.

[ME]

It's another fundamental: a basic integral.

a=acceleration[m/s^2], h=height[m], m=mass[k], v=velocity[m/s], t=time[s]
Drop a mass, under influence of acceleration
As it makes a 'louder bang' when dropped on some wood, let's call this Energy: Ep=m*a*h
We know v=a*t and the integral: h=(1/2)*a*t^2 ... + vt+h(0), note: v^2=v*v, v=Sqrt(v*v)
With h=(1/2)*a*t^2--> t=Sqrt(2*h/a), note v(0)=0, h is a delta, so h(0)=0
With v=a*t and t=Sqrt(2*h/a) --> v=Sqrt(2*h/a)*a = Sqrt(2*h*a)
So, v= Sqrt(2*h*a) --> v^2=2*a*h --> (1/2)*v^2=a*h
Make it more familiar in the form of energy, and reintroduce mass
With (1/2)*v^2=a*h and m --> (1/2)*m*v^2=m*a*h
or Ek=Ep

[Fletcher]

These are some of the implicit circular associations and constraints (so far) ...

mgh = 1/2mv^2

gh = 1/2v^2

g = v^2/2h

h = v^2/2g

v = sqrt(2gh)


f = ma

WEEP => energy = f x d :. energy = mad = mgh = 1/2mv^2

ETA: ME beat me to it.

[pequaide]

I think you see your improper use of J instead of F as a small or insignificant matter; but it is not it is a huge matter.

Mr V; quote: To put it another way, under ordinary circumstances, achieving a constant rise in acceleration requires an increasing amount of input energy - increasing by the square of the rising velocity.

No such concept would come from Newton, Newtonian physics has nothing to do with J it is all Force. Gravity is a F.

F = ma; a is v/t so we have F = mv/t; or Ft = mv.

The only form of motion conservation you will find it your books is mv. The Law of Conservation of Momentum states that the mv in equals the mv out (in a closed system). You will never see KE in equals KE out; because it is not true.

Since Ft = mv; you could also state that Ft in equals Ft out for The Law of Conservation of Momentum.

This means that you can accelerate (with a F) a large mass and then transfer the motion to a small mass and you will have to pay the same Ft to get the motion back out. Getting it back out would mean a very long high energy rise.

Ten newtons applied for ten second on a 10 kilogram mass will give you 100 units of momentum and a velocity of 10 m/sec for the mass. Transfer that mv to a one kilogram mass and it will have to have a velocity of 100 m/sec; 1 * 100 = 10 * 10; this is the same mv and it will take the same Ft to get the 1 kilogram mass to stop. But the 10 kilograms moving 10 m/sec is 500 Joules and the 1 kilogram moving 100 m/sec is 5000 Joules. The one kilogram will rise 509.68 m; and the 10 kilograms will only rise 5.0968 m.

Your sims might give you KE in equals KE out but you will never find it in a science book, or in a mass transferring experiment. Newtonian physics always works and it needs no frames, heat, sims, or mirrors.

[WaltzCee]

These are some of the implicit circular associations and constraints (so far) ...

mgh = 1/2mv^2

gh = 1/2v^2

g = v^2/2h

h = v^2/2g

v = sqrt(2gh)


f = ma

WEEP => energy = f x d :. energy = mad = mgh = 1/2mv^2

ETA: ME beat me to it.

When I have more time I'll clean your thoughts up. :)

[ME]

(Sorry Fletcher :-) perhaps your write-up is more clear)

Because mathematically Ep=Ek (excluding the practical conversion losses) and basically a wheel goes as it goes... and then stops, we need to do something else.

Is it possible to have some stored Ep at 6-'o (perhaps in some fusee) and convert it to some Ek at 12-'o; and when things go down to 6-'o again it restores Ep?

Ep[Wheel, 12'-o] + Ek[Mech, 12'-o] <==> Ek[Wheel, 6-'o] + Ep[Mech, 6-'o]

Or is this in the end just a more complicated way to lose energy?

Marchello E.

[Fletcher]

I am a little reluctant to write this as this is Mr V's thread and I don't want to dilute his focus and energies by waffling on in different directions.


Pequaide raises an additional observation about f = ma.

He is correct AFAIK that Newton wasn't concerned about energies (KE & GPE etc) but was concerned about forces and by deduction from rearranging the f = ma formula finding linear momentums.

To quote pequaide ...

f = ma and a = v/t therefore f = mv/t => ft = mv

So from the force formula momentum can be derived as either mv or ft.

He gives the example of 10N's of force being applied to a 10kg mass for 10 secs. So we can work out easily the momentum by either mv or ft at 100 units. That means that if time is 10 secs force must be 10N's. And if mass is 10kgs velocity must be 10m/s.

And he is right that if all momentum was transferred to a 1kg mass (or to stop it when it is doing 100m/s) would require the same momentum of 100 units. And if time is 10 secs then force must be 10N's.

So we know that linear momentum is conserved in that example, provided we have a mechanical way to fully transfer that momentum between two unequal masses. N.B. so far personally I haven't found a way to mechanically do that because each mechanical system is anchored in some way to the earth surface and 'absorbs' some of the momentum we wish to transfer making it less than 100% efficient transfer.

I think pequaide is of the opinion that the yo-yo despin device that nasa has used to stop satellites spinning on their axis achieves this. I guess it has its own reference frame being out in space and not anchored to earth.

Anyways, back to the full mv transfer example. Whilst mv = ft and is the same for the 10kg mass and the 1kg mass the displacements are different by a factor of 10. The 10kg mass travels 50meters and the 1kg mass travels 500meters. Force x distance = Joules and is energy equivalent under the Work Energy Equivalence Principle (WEEP), therefore although both masses would have the same momentum they have vastly different KE's as pequiade points out.

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

Back to Mr V's hypothesis.

Here he is also looking to throw out WEEP. By saying that in some circumstances the equivalence does not hold true. In linear relations it does because there is always a displacement factor proportional to speed etc. But we note that when gearing is used in pulley experiments that the inertia of the masses must be 'adjusted' by a factor dependent on relationship between masses and the gearing squared. This really means that gearing increases the distance or displacements. And since the mass must travel further faster then its inertia (or MOI) must also increase. We know this is true because otherwise we wouldn't have different 'Power' for different gearing options. N.B. Power (energy/time) is the rate of doing Work (f x d). e.g. we all know that taking off in 4th gear has a higher top speed but we will get to the finish line slower than the guy in 2nd gear from a standing start when the finish line is 100 meters away. We do use the same energy though! That means for the same force we cannot get across the finish line any faster regardless of what gear we are in (to use the analogy), but we can get there quicker.

This increase in inertia causes CoE (KE's against GPE) to be apparent and so far inescapable.

Mr V is postulating that the MOI of a flywheel is unchanged at any rpm. Therefore it may be possible with the use of tandem fusees or some such to take a linear fall of GPE and thru constant variable gearing of fusees cause the flywheel to accelerate (gain RKE) up to factors of the GPE lost by the driver. That's because the MOI of a flywheel does not change.

It is not known by me whether the variable gearing of the fusees will run into the same problems as the inertia increase of linear systems with gearing ? Mr V articulates his hypothesis well and I would like to see the results of his experiments.

I support anyone who has a new idea and is willing to share and lay it out for others to follow. Especially when they are willing and able to spend resources building real world models and collect the data (Inputs and Outputs etc) and share it with us. More power to their elbows I say and maybe we'll learn something.

[Grimer]

So presumably W3 in your profile refers to London W3, where I went to school (Gunnersbury Grammar). I did wonder.

Say what you like about the Rothschilds but Gunnersbury park is a personal oasis at any time of day or year. Totally overrun with parakeets these days, but whether the story's true or not they always remind me of Hendrix..

The grammar schools are bit before my time - i'm guessing it's now the International school down Gunnersbury Avenue past the cemetery?

...

Yep. That's the one. You can tell it was once a catholic school by the two Celtic crosses on the top of the buildings. I often look out for Canon McCliment's office, the end room of the jutting out single story bit (now hidden by a hedge in street view) where I was so frequently whacked for cheek and other misdemeanours. .... ;-)

In the sixth form we were allowed out to Gunnersbury Park in the lunch hour.
When I still drove I often used to stop at the northern carpark and have a wander around with my old school friend.

Happy days - in spite of the beatings - which I thoroughly deserved. ... :-)

[MrVibrating]

Looks like some great contributions over the last page and i'll try to get back to them after work this evening.


Meantime just a quick update...

Last night i realised a different approach to the asymmetry, which seems much more robust and easy to follow.

RKE squares simply because the given force must be applied over a greater distance for each successive acceleration, due to the accumulating speed (rate of change of position).

Therefore, a small acceleration in a moving frame can be equal in energy to a much larger one, from a stationary frame.

In other words, the increased displacement over which the force must be applied can be subsidised by the motion of the whole system!

Example: suppose we have a heavy 100 ton platform moving sideways at ten meters per second. Upon it, a 1 kg mass is accelerated by 1 meter per second in the same direction as the platform, using a pre-loaded spring.

Aboard the platform, the rise in energy of the 1kg mass is effectively linear. Even if it is high enough to square up, it's basically academic, compared to the result in the stationary frame.

From the stationary perspective, the pre-existing velocity of the moving system simply adds to the effective distance over which the force was applied - it gives us free extra displacement in the stationary frame... free extra work done. The 1 kg mass's velocity has only increased by 1 m/s, from 10 m/s to 11 m/s, but the half square of this magnitude of velocity for the same mass yields a much higher rise in energy.

Granted, the whole platform has also been decelerated by the same amount of work, although the change in velocity is inversely proportional to its mass, so the trick here would be to use the comparative gain to re-lift a weight or something in the stationary frame.

Not a closed loop yet, but definitely a means of opening a window.

If you have sim software, try it - accelerate a mass on a heavy static platform, measure its energy, then preset the same velocity to platform and projectile and re-run the stationary measurement... the gain is multifold.

The trick is gonna be grabbing it and using it before the system has a chance to equalise it. Presumably gravity will have to provide such a trap..

Back later tonight..

[MrVibrating]

Posting from mobile so briefly, run the platform config with a scissorjack that accelerates equal masses in opposite directions...

Input workload doubles, but N3 is cancelled...

Therefore input energy stays constant, momentum stays constant, but RKE of the accelerated mass still squares up...

[Grimer]

Bravo. You are well on the way to discovering Iterative Hierarchical Mechanics.

[MrVibrating]

The only question that matters here is:

Why does KE square up, and does it have to (as in really, really have to)?

It's as fundamental as doubling the radius and getting 4 times the area.

You're almost certainly correct, in which case the proposed asymmetry can never be physically realised.

However i'm not sure how well the analogy answers to the specifics of the question - the line integral of a circle must by definition have constant curvature, whereas that of an energy integral can be dependent upon the frame of reference from which it is measured.

If you really think through the practicalities of just exactly why RKE does square with velocity, some subtleties become apparent.. as i said before, the main reason is that the accumulating velocity effectively adds its pre-existing displacement to that of each successive acceleration; ie. the given force must be applied over a progressively greater net displacement.

But if we wanted to somehow try and side-step that increasing displacement penalty for each additional burst of acceleration, the first hurdle we'd come up against is Newton's 3rd law.

If it wasn't for that, we'd be able to keep getting the same amount of acceleration for the same amount of input energy.

So the reason i pose the question "must KE square?" is to prompt folks to think through the issue and so better understand my own conclusions.

What i find encouraging is that N3's role in shoring up CoE looks decidedly indirect, conditional and almost incidental.. it's simply a law that says "actions have equal reactions" - its central function in causing KE to square up is so abstract that even after 6 pages of analysis, very few readers seem to be connecting it to the problem.

Although Kirk got it in one in the very first reply to this thread..

KE squares up because of Newton's 3rd law.

But its very unintuitiveness, to me, smells like weakness, so maybe there's a line of attack here.

So then the next question becomes "is there a way around it?"

Assuming N3 is immutable, we're left to consider ways of emulating the results of a violation by other means...

N3 i think is more directly analogous to a geometric principle. But the reason RKE squares - and N3's role in it - is slightly more nuanced.

[MrVibrating]

It's another fundamental: a basic integral.

a=acceleration[m/s^2], h=height[m], m=mass[k], v=velocity[m/s], t=time[s]
Drop a mass, under influence of acceleration
As it makes a 'louder bang' when dropped on some wood, let's call this Energy: Ep=m*a*h
We know v=a*t and the integral: h=(1/2)*a*t^2 ... + vt+h(0), note: v^2=v*v, v=Sqrt(v*v)
With h=(1/2)*a*t^2--> t=Sqrt(2*h/a), note v(0)=0, h is a delta, so h(0)=0
With v=a*t and t=Sqrt(2*h/a) --> v=Sqrt(2*h/a)*a = Sqrt(2*h*a)
So, v= Sqrt(2*h*a) --> v^2=2*a*h --> (1/2)*v^2=a*h
Make it more familiar in the form of energy, and reintroduce mass
With (1/2)*v^2=a*h and m --> (1/2)*m*v^2=m*a*h
or Ek=Ep

Fantastic. I love cross-referencing and this is pretty much the prison blueprints. KE is symmetrical to PE.

And we can extrapolate from it that, by definition, the energies being juggled by Bessler's wheel (or any OU system) must reside in different reference frames.

Furthermore we can suppose that if gravity is constant and so gravitational asymmetries impossible, then at least one (or more) of these alternate reference frames must be inertial... and the "prime mover" must be an effective N3 break.

So on the one hand i accept that KE squares with velocity "because it just does", but when we consider the practical factors acting on the system to drive the rising KE integral - which is what i was really asking - we see that the displacements over which successive acceleration forces will act are rising cummulatively with velocity; work = F*d and if work is increasing but F is constant then d is increasing, and vice versa.

All of the PE/KE equivalencies are reference frame dependent. If GPE is always going to equal GMH then we're left with PE to KE generally, and some kind of inertial exploit.

Some means of either overcoming an inertial reaction, or else turning one to our advantage.

This seems to me the certain implication of the cross-symmetries you've so astutely summarised.. Combine them with the inevitable properties of an OU system - ie. one in which the same field has two different energies depending on how it's measured, and Bessler's asymmetry must've been inertial.

And this is the reason he insisted that in a true PM, everything must go around together - the external RKE was decoupled from the internal energy of the prime mover, because it 'saw' the wheel as stationary - ie. it's energy isn't squaring up symmetrically to the system's RKE.

This also explains why he geared down the demo loads; to prevent the actual form of the gain - RKE rather than voided weight - from spiraling down self-destructively. It is also consistent with his claim that a very slow, but higher power version was possible - a flywheel with a very high MoI (such as a long lead tube) could carry a lot of RKE at lower velocities, so still squaring up to much higher values than the internal energy per cycle, which remains a constant function of the asymmetric distribution of momentum per cycle.


The net KE of an OU system is not a basic integral.

An integral with two alternate reference frames has a third axis, and its net energy is in a kind of superposition, depending on the reference axis. Any OU system must have one at its heart. This is what enables input vs output asymmetries and path dependent interactions.

We have to use what we can see to infer what we can't, reading between the lines and checking the smallprint for loopholes. The most viable conditions for breaking unity must be implicit within your description of it...

To my thinking, RKE squares because cummulative d squares with rising v.

Everything points to B. hacking d via N3.