Plying CF as pseudo-inertia to scam N3

[cloud camper]

Here's my donkey - actually affectionately known is Miss Piggy.

Bought new in 93, XR650L, very reliable, simple as a stone. Honda has not changed a bolt on this design in 24 years.

Can rebuild the carb by the side of the road in an hour if I had to.



[/img]

[MrVibrating]

MrV. Sorry for my questions, but who would have guessed: Conspiracies 'r us !

Don't let me stop your idea of progress.
I just ask questions.
-- If you can answer them: fine.
-- If you can't: then at least you thought about it.
-- If you change your mind: then I may or may not had some influence...

When I look at the presented formulas, then I see it dangling full of holes - perhaps we can save the good stuff, when it's there.
Pointing out the holes should be the least of anyone's concern when it's correct or has some hidden validity.
The answer would be easy and the question simply a useful helping hand or a tool to describe that idea even better.
Perhaps the theory is in need of a small footnote here and there, that's all I try here and nothing more.
When the answers get more complicated, start to deviate from the issue and even become personal, then that's usually enough indication it's based on wrong assumptions or simply fraudulent.
The conspiracy is one's own!

-- But conspiracy-wise: don't take anyone's word for it, but everyone can check the units on mrV's changing N²-math, and they are wrong in every variant.... all without even touching the physical-validity. (That's why I usually write out those freak'n formula's: so every one is able to check)
-- It's entirely possible I don't understand a concept. In that case: who cares what I think and good luck. I can still ask questions though.

There's only one way to check for gained energy: build that damn thing !
A gain should be translatable into a continuous stream of potential energy.
Not via some formula like Ep=Ek, but the reality of raising weights in the inertial Earth frame.
Water or marbles are perfect, as long as it's external to the mechanism.
Then you'll know it is creating energy somehow.

Marcello everyone can see you're hopelessly confused.

I personally don't see that as problematic. But it's good we finally cleared that one up....

V, I'm not against you... It seems I just have a different perspective.
Can't promise but, I'll try to stop bugging you.

Mate, again, i'm applying a force between two inertias that only accelerates one of them. Net system momentum is usually scalar, but here, it's now a vector, and can be accumulated over successive cycles.

It depends upon countering gravity's acceleration, which is why the cost of a one-second impulse generating 9.81 P is 9.81^2 J.

This obviously invites generalising the term to suit any applied force, so i've used the basic principal of algebra to assign a variable, where N (for 'number') = force / mass, and p=mV, (N^2 J / N P) is accurate, and more readable than using F/m in place of 'N'.

If instead we tried to apply (e^2 / p), without explicitly setting their numerical equivalence, then any reader would be entitled to assign any arbitrary base values to each, and their resulting calculations would not provide the correct results. (e^2 / p) accurately describes the type of relationship, the fields and their respective dimensions, but obviously does not convey their numerical derivation from F/m=A. For instance if the reader chooses a base rate of say (5^2 e / 10 p), then we lose this /4 axis which is a defining feature of the results.

From Dwayne's response, to Eccentrically1's to yours, what all share in common is an apparent under-appreciation for the importance of momentum symmetry in enforcing CoM and CoE.

I say "Einstein" you say "e=mC^2!"... I say "N3" you say "equal opposite reactions!"... but is it just pub-quiz trivia, or really integrated in a comprehensive way? It's always fun realising new ways to apply e=mC^2 to everyday phenomena, but we mostly don't need to know all that stuff. With N3 however it's really critical to our objectives here. If you only knew how crucial, you'd instantly recognise the potential of an effective N3 break.

But no one here appears to. Dwayne used a whole page of calcs to show that momentum's constant if force is halved and mass doubled. Eccentrically1 thinks it's not really an effective N3 break because the counter-momentum is gravity. And you neither seem to see the N3 break, nor the potential value of one.

Yet all these demonstrations - not to mention basic logic - prove that we can reliably produce single-signed momenta, simply by generating and then repaying an equal quantity of momentum from GPE.

So to me, seeing an inertial interaction in which only one mass accelerates, is incredibly exciting. Everyone else either doesn't see its significance, or just doens't even see it in the first place. To me, calculating and subtracting the KE and momentum from GPE is as trivial as dunking a photo print in developing fluid. The single-signed momentum stands out in stark relief, impossible to mistake, self-evident in every way..

But even if you can see it, that's not enough to understand its value. That value is its speed-invariance; the same input energy produces the same momentum yield, irrespective of rising velocity.

The reason for this is because inertia does not rise with speed, and there's no such thing as terminal velocity - thus no matter how fast the gravitating mass is already moving as it descends, gravity will always accelerate it by another 9.81 m/s^2 if there's nothing else slowing it down. Which means we can always resist that acceleration and use it to generate more single-signed momentum.

I agree that i could probably add more footnotes, but look how long these posts are already - i'm trying to strike some kind of balance between condescending any readers still here, vs putting up a wall of impenetrable assumptions. I'm happy to provide further clarifications on request, but the reservations people are having are due to this presumption that everyone must already realise the centrality of N3 in enforcing mechanical CoE, but also, simply failing to note the key details i've been laying out - such as Dwayne's misconception that i was claiming a greater magnitude of induced momentum, rather than an asymmetric distribution of the same quantity of momentum, which in turn sums to a net gain purely because the balance isn't equal and opposite..

And so armed with that misconception, he got snidey and personal, so i got defensive and frustrated.

And you too are doing the same thing. I can take criticism of an argument on its merits, but when it's based on misconceptions that don't even evaluate the findings, coupled with scathing derision, that's a pressure keg of frustration..

You're criticising my use of a single algebraic parameter to describe a conserved relationship between e & p that you don't even acknowledge as existing yet. You don't have to recognise the significance of a net momentum vector in order to simply be able to see it, but i don't think you've even seen it yet, and yet you're insisting i've mis-described the thing you cannot see...?

Again, the energy value of momentum is not constant, squaring with velocity. Thus a constant cost of momentum, that isn't a function of velocity, is a free meal ticket, a business opportunity. A frikkin' unpatched zero-day exploit.

The significance in all of this is that we have a constant cost of momentum, vs a V^2 value.

But first you have to be able to see this input constant. That this much energy buys this much single-signed momentum, for these reasons.

Then we can argue about the most succinct formal expressions - again tho, that's entirely academic to the mechanical solution.

So of course i'm going to continue trying to deduce a simple design, but in the meantime simply communicating the solution to anyone who'll listen remains a top priority, for me at least. Accordingly, trying to correct or resolve misgivings is a part of that.

You're all intelligent people here. We're all Bessler believers who want, consciously at least, to see a just resolution and vindication. With i'm sure, very little effort, everyone here could see what i'm seeing... and that only adds to the frustration.

[MrVibrating]

Here's my donkey - actually affectionately known is Miss Piggy.

Bought new in 93, XR650L, very reliable, simple as a stone. Honda has not changed a bolt on this design in 24 years.

Can rebuild the carb by the side of the road in an hour if I had to.



[/img]

Fantastic, and near-mint condition! But yeah carbs, man! Carbs, plus gravity. Never had a gravity failure from a loose connection. A fuel pump ain't the kind of thing you can just poke with a screwdriver.

They're also quite efficient for an old design, so were very popular with despatch riders back in the 90's and noughties.

I've still never owned a big thumper - the number of times i've limped home only firing on one or two pots, whereas if a single goes you've no choice but getting your hands dirty, not that i've ever heard of that being a problem with the range.

The XR 125 is still popular with learner couriers, a much more capable package than the CG 125 half its parts are from, better riding position, curb clearance, suspension and running gear.. but yeah, legendary, solid range, and with the way things are going with all these emissions standards and electronic jiggery pokery, old classics like this are keepers..

[Gregory]

Ok, here is an experiment I constructed based on the original idea.

Consider the following thought experiment:

1/ A pair of green masses (identical 1 kg) standing still in outer space, and there is a black box device (for example a constant force spring) between them. After a push of a button the black box device suddenly applies 2 * 9.81 N of force between the green masses for 1 second.

2/ There are a pair of gray masses (identical 1 kg) standing still in a gravitational field which only exist for 1 second and produces 9.81 m/s^2 of acceleration. Also here is the same black box device with 2 * 9.81 N of force applied between the gray masses for 1 second when the same button is pressed.

Now, consider Alice is watching and analysing the motion of both pair of masses on a monitor she sits before. She checks and finds that everything is ok, the experiment can be started, so she pushes on the button. Both black box device fires, and applies the same force between the two mass pairs.

The green masses in outer space accelerate equally ending up both having 9.81 kg*m/s momentum, and 48.085 J kinetic energy. Summed up in an absolute sense they have 19.61 kg*m/s momentum and 96.17 J kinetic energy.

The gray masses inside the gravitational field accelerate unequally as the force of the gravity field acting on the upper mass cancels out with the force of the black box device acting on that same mass.
The upper mass ends up with its momentum and KE unchanged while the lower mass ends up having 19.61 kg*m/s momentum, and 192.34 J kinetic energy. Summed up in an absolute sense the gray masses have 19.61 kg*m/s momentum and 192.34 J kinetic energy.

Analyzing the results Alice finds out that at the end of the experiment both the green and the gray pair of masses have the same amount of momentum in an absolute sense, but different kinetic energies as the gray masses now have twice as much KE as the green masses. But that's to be expected because of the V^2 term.

Alice concludes that although the force exerted by the black box device distributes evenly in both cases, in the case of the gray masses one half of this force cancels out against gravity leaving the motion of the upper mass unchanged, while the motion of the lower mass is accelerated twice as much, one half because of the gravity field, and the other half because of the force of the black box device.

Alice thinks this is interesting, because although mathematically the force of the black box device distributes evenly between the masses, but combined with the gravity field the case of the gray masses looks like as an uneven distribution as the motion of the upper mass remains unchanged in the process.

Hhm, is there any way to make use of the increased kinetic energy of the gray masses? - Asks Alice for herself.


Hope I didn't wrote any mistakes into the tale.
Start / Stop screenshots and Wm2d file attached.

Blue arrows: Force of the black box device
Red arrows: Force of gravity

Cheers!

[daxwc]

Gregory:

Analyzing the results Alice finds out that at the end of the experiment both the green and the gray pair of masses have the same amount of momentum in an absolute sense, but different kinetic energies as the gray masses now have twice as much KE as the green masses. But that's to be expected because of the V^2 term.

MrVibrating :

Again, the energy value of momentum is not constant, squaring with velocity. Thus a constant cost of momentum, that isn't a function of velocity, is a free meal ticket, a business opportunity. A frikkin' unpatched zero-day exploit.
The significance in all of this is that we have a constant cost of momentum, vs a V^2 value.
But first you have to be able to see this input constant. That this much energy buys this much single-signed momentum, for these reasons.

Again unless I am totally out to lunch to as to what MrVibrating is trying to convey here the enigma is our definition of energy. That is because energy involves distance squared where momentum does not.
Velocity is distance divided by the time so; KE=1/2mv^2 can be rewritten as KE=1/2 m *(d/t)^2
Momentum rewritten is m*d/t
So the only thing that is changed relative to one another is the distance travelled in MrVibrating Sim.

I my opinion the top mass is a red herring. One may as well say one mass starts with an impulse of x amount of force over x amount of distance and how much difference in energy is calculated due to varying amounts of impulse through the gravitational field.
What will the top mass be doing inside a wheel it is thrown up and becomes weightless to the wheel anyway.

I am sure MrVibrating will set me straight but please do so in the first paragraph. My eyes tend to water over ... yep sort of like the pot calling the kettle black.

[Furcurequs]

Though not my potato cannon, Bruce Yeany's latest video just happens to be demonstrating tennis ball cannons. He doesn't get much into the math, I don't think, but he does show how they are measuring the different speeds of the traveling balls and loose cannons.

https://www.youtube.com/watch?v=UFrLMZqIqlk

His previous video show lots of other Newton's 3rd law demonstrations, also.

https://www.youtube.com/watch?v=24JOUNUzd7U

[Furcurequs]

MrVibrating,

I hope you are catching up on your sleep and haven't had an accident or anything due to lack of sleep.

You need to be alert on that motorcycle. I know. I spent a year using one (a 1981 Suzuki GS550E) as my primary means of transportation, and it's dangerous out there. It's probably even worse now than it was when I was doing my riding. I've had my motorcycle parked for two decades due to my health issues.

Here's a photo of someone's 1979 GS550E that looks just like my 1981 (well, when it;s all together and polished up, at least):

https://i0.wp.com/www.bike-urious.com/w ... r-Left.jpg

Anyway, here are some Spacex rocket landings to show a nice steady descent in a gravitational field:

"SpaceX CRS-11 - Falcon 9 landing (close-up), 3 June 2017"

https://www.youtube.com/watch?v=hulMgWJV3e8


"CRS-10 _ Falcon 9 First Stage Landing"

https://www.youtube.com/watch?v=glEvogjdEVY


"SpaceX Falcon 9 - Successful Drone Ship Landing - 8th April 2016"

https://www.youtube.com/watch?v=RPGUQySBikQ


"SpaceX landing compilation _ best landing montage"

https://www.youtube.com/watch?v=oa_mtakPlfw

[WaltzCee]

I think I figured out how to
derive Bessler's Constant.

[MrVibrating]

Ok, here is an experiment I constructed based on the original idea.

Consider the following thought experiment:

1/ A pair of green masses (identical 1 kg) standing still in outer space, and there is a black box device (for example a constant force spring) between them. After a push of a button the black box device suddenly applies 2 * 9.81 N of force between the green masses for 1 second.

2/ There are a pair of gray masses (identical 1 kg) standing still in a gravitational field which only exist for 1 second and produces 9.81 m/s^2 of acceleration. Also here is the same black box device with 2 * 9.81 N of force applied between the gray masses for 1 second when the same button is pressed.

Now, consider Alice is watching and analysing the motion of both pair of masses on a monitor she sits before. She checks and finds that everything is ok, the experiment can be started, so she pushes on the button. Both black box device fires, and applies the same force between the two mass pairs.

The green masses in outer space accelerate equally ending up both having 9.81 kg*m/s momentum, and 48.085 J kinetic energy. Summed up in an absolute sense they have 19.61 kg*m/s momentum and 96.17 J kinetic energy.

The gray masses inside the gravitational field accelerate unequally as the force of the gravity field acting on the upper mass cancels out with the force of the black box device acting on that same mass.
The upper mass ends up with its momentum and KE unchanged while the lower mass ends up having 19.61 kg*m/s momentum, and 192.34 J kinetic energy. Summed up in an absolute sense the gray masses have 19.61 kg*m/s momentum and 192.34 J kinetic energy.

Analyzing the results Alice finds out that at the end of the experiment both the green and the gray pair of masses have the same amount of momentum in an absolute sense, but different kinetic energies as the gray masses now have twice as much KE as the green masses. But that's to be expected because of the V^2 term.

Alice concludes that although the force exerted by the black box device distributes evenly in both cases, in the case of the gray masses one half of this force cancels out against gravity leaving the motion of the upper mass unchanged, while the motion of the lower mass is accelerated twice as much, one half because of the gravity field, and the other half because of the force of the black box device.

Alice thinks this is interesting, because although mathematically the force of the black box device distributes evenly between the masses, but combined with the gravity field the case of the gray masses looks like as an uneven distribution as the motion of the upper mass remains unchanged in the process.

Hhm, is there any way to make use of the increased kinetic energy of the gray masses? - Asks Alice for herself.


Hope I didn't wrote any mistakes into the tale.
Start / Stop screenshots and Wm2d file attached.

Blue arrows: Force of the black box device
Red arrows: Force of gravity

Cheers!

Sorry for my abrupt absence the last week, you can appreciate i felt it was time to back slowly away from the keyboard, if only for a while...

Gregory, thank you for your excellent summary!

You are very close to seeing what i'm seeing, but not quite there yet:


- you have calculated the energy of the grey system per 1/2mV^2, and produced the 'right' answer insofar as 1 kg at 19.61 m/s has 192 J.


- likewise, you've calculated the energy of the green system as 96 J, also correctly.

- however, you seem to have assumed that this energy difference is the object of interest... it is not!




Specifically, what should interest us is this:

- consider that this linear system could be a tangential (angular) trajectory, at arbitrary radius from its axis

- alternatively, keep it linear, but let it bounce off of a fully-conservative spring

-- either way, rebounding back upwards to its initial drop height, and so returning that portion of its momentum and KE that came from GPE



So now, both green and grey systems have 96.2 J of energy and 19.6 P of momentum...


Yet there is still a key difference between them, and this is absolutely our object of interest...!

The green KE and p is evenly distributed between the two masses, so the system's net p remains zero.


So the green system is fully consistent with Newton's 3rd, and 1st laws; the induced momenta are equal and opposite, and thus we've been unable to change the net system momentum via the application of internal force!


But now look at the grey system - all of the momentum (and thus input energy) has been applied to one mass only!


We have repaid (or just mathematically subtracted) all of the KE and momentum that was sourced from gravity, and so we're left with the KE and momentum input via the black box!

The only reason it was possible to input this internally-applied momentum was precisely because of the mass's mutual resistance to acceleration - the lower mass was accelerated against the inertia of the upper mass, just as in the green system...

...however in the grey system, due to gravity's influence, we performed this inertial interaction such that only one mass was caused to accelerate!

So we have 'broken' N3, and thus with it, N1. An internally-applied force has nonetheless accelerated only one mass, causing a fully-asymmetric distribution of momentum...

...and thus braking that accelerated mass against its non-accelerated partner, distributes that 9.81 p of momentum evenly between them, leaving each with 4.905 p, of equal sign..!!!

So after the GPE interaction is closed, the net system has self-accelerated from the internally-applied inertial interaction, breaking N1 via the effective breaking of N3.





The final point, then, is that we have a means of buying 9.81 p of unidirectional momentum, for 96 J.

It is speed-invariant - the same energy buys the same rise in momentum, no matter how fast we're already going.






This provides us with an alternative exchange rate for energy / momentum, running parallel to the 1/2mV^2 standard..!!!



Two 1 kg masses at a final velocity of 4.905 J each, have 12 J each.

So the final net energy of the grey system after the GPE interaction was fully closed is 24 J.

But the energy cost of the inertial interaction is 96 J.

96 / 24 = 4

So the complete interaction lost 3/4 of its energy.

initially this seems like a bad deal - however it so isn't...


...because this rise in momentum accumulates over successive cycles, at constant cost per cycle, invariant of rising velocity!


So a second such cycle adds another 4.905 P to each mass (capital 'P' now 'cos it's a vector!), and then a third, and so on, and by the time we hit 4 cycles, we start to benefit from the standard V^2 escalator...





This transition from initial under-unity performance, converging with 1/2mV^2 at precisely 4 cycles, and then punching straight up into truly monstrous over-unity, can be simply plotted on a graph, like this:

click me



...as i've always said, in the fantasy menagerie of classical symmetry breaks, an effective N3 violation is the frickin' mother lode.


We're talking hard hats and ear plugs, power densities that would cause "OB" proponents to wake up screaming.


And yet, even amongst the ranks of lifelong OU seekers, the role of N3 in enforcing N1 and thus CoM / CoE has always been vastly under-appreciated. So neglected, that even after explaining it ninety times, i still haven't received a single solitary glimmer of corroboration from anyone else here...

But you're almost there mate... it's all about the momentum, first and foremost...

Inducing, entraining, and building upon (accumulating) momentum.

Buying at the (9.81^2 J / 9.81 P) rate.

Accumulating.

And then selling at the 1/2mV^2 rate.


This is the sole raison d'etre of everything i'm doing here.

[MrVibrating]

Gregory:

Analyzing the results Alice finds out that at the end of the experiment both the green and the gray pair of masses have the same amount of momentum in an absolute sense, but different kinetic energies as the gray masses now have twice as much KE as the green masses. But that's to be expected because of the V^2 term.

MrVibrating :

Again, the energy value of momentum is not constant, squaring with velocity. Thus a constant cost of momentum, that isn't a function of velocity, is a free meal ticket, a business opportunity. A frikkin' unpatched zero-day exploit.
The significance in all of this is that we have a constant cost of momentum, vs a V^2 value.
But first you have to be able to see this input constant. That this much energy buys this much single-signed momentum, for these reasons.

Again unless I am totally out to lunch to as to what MrVibrating is trying to convey here the enigma is our definition of energy. That is because energy involves distance squared where momentum does not.
Velocity is distance divided by the time so; KE=1/2mv^2 can be rewritten as KE=1/2 m *(d/t)^2
Momentum rewritten is m*d/t
So the only thing that is changed relative to one another is the distance travelled in MrVibrating Sim.

I my opinion the top mass is a red herring. One may as well say one mass starts with an impulse of x amount of force over x amount of distance and how much difference in energy is calculated due to varying amounts of impulse through the gravitational field.
What will the top mass be doing inside a wheel it is thrown up and becomes weightless to the wheel anyway.

I am sure MrVibrating will set me straight but please do so in the first paragraph. My eyes tend to water over ... yep sort of like the pot calling the kettle black.

Again, my friend, i'm sorry but you've jumped onto the same initial presumption as almost everyone else here - that i'm trying to leverage some kind of instantaneous difference in absolute energies between the gravitating vs non-gravitating examples. This is not the case.

The advantage is derived purely from the asymmetric distribution of the internally-applied momentum, and its consequent non-zero (ie. non-cancelling) sum.

So, after repaying the momentum and KE that was sourced from GPE, we're left with a single-signed momentum, nonetheless generated against the upper mass's inertia.

Subsequently, equalising the velocity difference between them (such as by a mutual deceleration) leaves both with a net rise in momentum.

This has a KE value of just 1/4 our input energy.

But a second identical cycle only loses 1/2 its input energy.

A third returns 3/4 of our input energy, as KE on the two masses.

A fourth is fully conservative - so four successive purchases of 9.81 P of single-signed momentum, shared between two 1 kg masses, at 96 J a pop, results in exactly the same final momentum and KE as any conventional form of acceleration; we input 96 J, and the masses have 96 J.

Accordingly, a fifth purchase takes us up to 5/4 unity.

In principle, we can stack n quarters, ie. where n = 1 to infinity, in an infinite loop.

Obviously, mechanical limits will apply in practice, but the basic game is simply buying cheap momentum. Or fixed-rate, anyway, whilst its regular value squares with velocity per 1/2mV^2 or 1/2MoI*RPM^2 etc.


Also check my answer to Gregory above, which adds more meat to the same bones.



As for why i insist on using integrals instead of diffs, it's a point i've been making here for years; it's the only way you're going to be able to see OU.


If you're only trying to think in terms of differentials then unless you're some kind of wunderkind, you're blinkering your horizons.

This is a seemingly-complex discussion in its own right, but boils down to very simple rationale - it is axiomatic, from first principles, that an OU system by definition implies an input / output energy / time differential.

However, we do not think in the time plane! We don't think logarithmically, and have to solve for actual instantaneous values of force / time / displacement to derive any potentially-interesting force / time deltas.

Our hardware's wired more for spatial, rather than temporal 'visual logic'.

So while i appreciate that physics is more usually presented in diffs, in the first instance what concerns us in our quest for OU is integrals.

Specifically, we need to be thinking in terms of discrete input and output integrals. Obviously, closed-loop trajectories through static fields yield zero sum deals, and so we do need some kind of force / time delta if OU is to be a possibility (ie. more area under the output line integral).

But obviously, the only way out of that bind is a system in which input and output energy terms are subject to different, mutually incompatible force / time deltas.

So if you're actually more used to thinking in terms of differentials already, that may be some advantage.. provided you have that fluid interchangability with the integral view.

But the only differential of interest is the input energy / time vs output energy / time.

Ultimately, it all comes back down to the energy under the respective curves.....

That integral is the picture that a differential equation can evoke no better than a thousand words... it really is the 'bottom line' summary..

[MrVibrating]

Though not my potato cannon, Bruce Yeany's latest video just happens to be demonstrating tennis ball cannons. He doesn't get much into the math, I don't think, but he does show how they are measuring the different speeds of the traveling balls and loose cannons.

https://www.youtube.com/watch?v=UFrLMZqIqlk

His previous video show lots of other Newton's 3rd law demonstrations, also.

https://www.youtube.com/watch?v=24JOUNUzd7U

Cool vids!

Imagine if the gun didn't recoil tho, and the potato was connected to the gun via a slack line....

...the net system would've been accelerated by an internal expenditure of work!

Instead of self-cancelling, the net system momentum would've risen.

If the ball were wound back into the cannon, consistently with N3, it could be re-fired using the same trick, adding the same net rise in momentum again..!

The first time we do this, the final system KE of that un-cancelled momentum will be precisely 1/4 our input energy.

The second time will leave us with 1/2 our input energy.

The fourth will result in the net system of cannon plus ball, having the same net KE as the energy spent inside the cannon.

The fifth will result in a final KE that is 5/4 more than the energy spent by the cannon.

The 6th, six-quarters, and so on...



This could be achieved simply by aligning the cannon to gravity, and letting it passively invert the sign of some or all of the counter-momentum (depending upon the magnitude of the mutual acceleration between the two masses - obviously, 2 G equals full cancellation / inversion).

[MrVibrating]

MrVibrating,

I hope you are catching up on your sleep and haven't had an accident or anything due to lack of sleep.

You need to be alert on that motorcycle. I know. I spent a year using one (a 1981 Suzuki GS550E) as my primary means of transportation, and it's dangerous out there. It's probably even worse now than it was when I was doing my riding. I've had my motorcycle parked for two decades due to my health issues.

Here's a photo of someone's 1979 GS550E that looks just like my 1981 (well, when it;s all together and polished up, at least):

https://i0.wp.com/www.bike-urious.com/w ... r-Left.jpg

Anyway, here are some Spacex rocket landings to show a nice steady descent in a gravitational field:

"SpaceX CRS-11 - Falcon 9 landing (close-up), 3 June 2017"

https://www.youtube.com/watch?v=hulMgWJV3e8


"CRS-10 _ Falcon 9 First Stage Landing"

https://www.youtube.com/watch?v=glEvogjdEVY


"SpaceX Falcon 9 - Successful Drone Ship Landing - 8th April 2016"

https://www.youtube.com/watch?v=RPGUQySBikQ


"SpaceX landing compilation _ best landing montage"

https://www.youtube.com/watch?v=oa_mtakPlfw

Cheers mate, fantastic what Musk is doing (wait til he gets a load of this tho). And thanks for your kind words, i'm so sorry i got into a rant at you, and just the fact that you'd raised any objection at all did give me some degree of relief for a day or two - i honestly hoped you were correct, and had found a weakness...

I've ridden a few GS 500's back in the day, they were also a solid workhorse. My fave Suzi's tho are the mk1 bandits, esp. the 1200. Mine's also currently in pieces tho - ran one cylinder too lean for too long, burned it out so now need to take the block in for a re-sleeve and re-bore.. putting it off til next summer as i don't have a garage to work in and it's grim banging the metal in winter..

[MrVibrating]

MT 134, people:





...noting that the preceding image, MT 133, appears to suggest a correlation between the shape of the 'hammers' with those of the lower hammer toy on the 'toys page', and also that the subsequent MT 135 shows a simple axle pass-thru principle, i believe MT 134 is showing us a viable mechanical approach.


A "cross bar" would thus be two identical pairs of mechanisms at orthogonal angles.

At vertical, under a combination of gravitational weight and CF (a rotating system wants to maximise its MoI to mimimise its RKE for its given angular momentum), this pair of masses and the beam between them drops down thru the axle, applying an upwards force against the two horizontal masses.

The left mass is thus braked against its acceleration under gravity (the system rotates counter-clockwise), whilst the right mass is accelerated along with all others stacked resting upon it above.

So we have an input energy provided by the vertical mechanism dropping thru the axle, performing output work in the form of a mutual acceleration of the right-side mass and all the other masses above it, against an equal opposite deceleration of the left-side mass, which would otherwise be accelerating downwards under gravity.

This generates single-signed momentum.

I know everyone will already have their own interpretations of these images, but this one really seems to tie everything else up.


MT 134 is obviously not purely 'literal' - it's intended to convey certain points, that are only apparent once you already know the gain principle.

So for example, the 'hammers' in MT 133 have now become 4 right triangles in MT 134.

This indicates that it is what i've been dubbing a "four-stroke" implementation - each interaction is individually 5/4 over-unity, in its own right. The input energy thus available from dropping the vertical mechanism thru the axis is equal to 'four quarters' - ie., to the KE and P products of four single-signed momentum rises, equal to unity at 1/2mV^2.

Thus Mt 134 represents a system in which each individual interaction is 5/4 efficient. Four 'quarters' are input each stroke, and five 'quarters' are output.


Again, a 'quarter' refers to the 25% of returned output KE that accumulates over successive cycles, hence resolving with 'unity' at 1/2mV^2 after four such 25% yields.

So we're seeing a composite of abstract and practical concepts.

The system for accomplishing this momentum distribution is represented in two different ways - in the upper main image, there are two chords, one running an upper loop, the other a mirrored lower loop. If the central vertical column drops down, we get an output motion akin to a 'Y' clapping its hands..

In the lower frame, an alternative means of accomplishing this same motion is depicted - it is not so much a 'Roberval' as a brace, again converting vertical linear motion to opposing angular motions.

Neither approach is necessarily better or worse, both have their pros and cons, and perhaps may even be applied complimentarily. The motion we need to accomplish is relatively straightforwards, however tricky it may seem to coordinate... And it looks to me like Machinen Tractate really does lead us right up to the safe... provided we already have the key..!

[MrVibrating]

...so, this is the kinds of mechs i've been playing with the last week or so. Basically a 'Y' shape, tug its tail and it claps its hands, one side accelerating the net system, the other decelerating an otherwise-gravitating mass.


A 'minimum system' would appear to be two such mechs, oriented orthogonally to one another at 90° angle, such that the vertical one, acting as the 'input' load, is at the same time subject to the rise in net system angular momentum being introduced by the horizontal one it is powering.


90° later, the two mechs trade roles, self-perpetuating a 25% KE gain, twice per full-cycle.


So this is what i'm currently working on, slowly but surely. Will post results as i get them..

[MrVibrating]

I think I figured out how to
derive Bessler's Constant.

Yes, it's just the optimum-possible fixed rate of input energy for accumulating momentum, with a precise value of 96.17 Joules per 9.80665 kg-m/s of unidirectional momentum when using gravity.

Obviously though, the static uniform force could be anything, from a magnetic source to centrifugal / centripetal pseudo-forces.

I like the example of two Bessler wheels opposite one another in a horizontal centrifuge they're also driving, thus super-charging the CF/CP forces and corresponding momentum yield per cycle.


Imagine it - 1 kJ per 100 kg-m/s outputting 2 kJ per cycle at 8 interactions per cycle, from a tiny generator spinning at a few hundred RPM, with a net weight of under a kilogram and no stray external forces or vibration.

Power densities Iron Man could only dream of. Infinite-range lightweight 'jetpacks' using electric moto-gen combos, or just good ol' steampunk mechanical assemblies, whatever the forces applied or their magnitudes, the input rate of energy-for-momentum is fixed across whatever speed range we can manage, at a^2 J per a kg-m/s, yielding a 25%-of-unity output KE that accumulates over successive cycles, from an initial 75% loss upon the first cycle, to a 25% gain at the fifth, 50% at the sixth, and so on up to 200% at the eighth, and beyond..