Plying CF as pseudo-inertia to scam N3

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Furcurequs
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Post by Furcurequs »

MrVibrating wrote:
Doubtless it escaped your attention, but i was actually making progress on designing such an system when you and Dwayne barged in, elbows out, trying to read the thread backwards and accusing me of talking gibberish.
...but you have been and you still are talking gibberish.
MrVibrating wrote:Listen chum, i spend all day chucking a big bike through heavy traffic. I get a few hours in the evening to do research, and between you and Dwayne you've wasted the last week insisting i repeat everything for you again and again.
Sorry if you had a rough day, but actually I asked that you not repeat yourself, please.

You've made some false assumptions and have some mistaken math, and so I've been trying to help you see where your errors are.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by ME »

some pages ago (I guess page 16), MrVibrating wrote:
ME wrote:
mrV wrote:Please, ANYONE, show me i'm wrong.??? :(
I tried, but now I have no idea where to begin. I simply don't get it. And you are so convinced...
mrV wrote:You obviously all get the principle by now
I get your search, but obviouslly not your solution.
I was already lost when you asked:
  • Everyone follow that so far?
I think you mistreated kinetic energy when you asked that, with (I think obvious) weird results... but you already concluded things.
Tracking back to your linear-example didn't work, and you're already way paste that slinky...

And with:
  • So input and output energy terms have different, mutually-incompatible dimensions - their line integrals intersect at precisely 4 reactionless accelerations in series.
I have no idea where to start and what to ask.
MrV wrote:There's no free lunch, no magic... no actions without consequences!
...so I guess there's still no N3-break.
Unless ...
mrV wrote:What more could i possibly say to add further clarity?
If you are really convinced then you better level it up to a mechanical exercise.
Good luck.
LOL so you don't understand why CoE depends upon N3?

You don't see that all of the motion imparted by the internal interaction is applied to one mass only (well, the Earth too, but that's a whole 'nother can o' worms)? Instead of imparting positive and negative momenta, we end up with just one sign?

I think you're pulling our leg mate..
Ehhm... are you? You always quote-copy the whole thing, but not actually replying. But it's a "nope" on each sentence.

I could wish I was pulling legs, except such is a very weird thing to wish for, but I'm actually serious. I don't understand any of that.
It shouldn't be a surprise either as I already informed you way back since page 2, and occasionally after.

I see CoE as a byproduct of N2, not N3.
But we can all deduce from that classical stuff that "energy" is just a forced illusion: hence I actually don't understand any of your explanatory "water-tight" math-magic throughout this whole tread where you ]redistribute energy and momentum like a messiah, and be surprised you got more than you started out with.
Therefore my opinion that "your energy values are useless", .... or divine.

It may be possible I overlook the benefits of looking at this differently; so I guess that I (and maybe many more) apparently need a better explanation on how you exactly divide KE between two masses. You assume it's clear in one half word and then rush off with your conclusion. I would like a little more time on that (we are all slow sometimes).
Explain how that works until understanding kicks in and likely that many more too will automatically understand how you get that gain [+25% per interaction], and when agreed then as bonus will also share the rest of your enthusiasm: because such gain would indeed, as understatement, be super-cool.
Marchello E.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by Furcurequs »

rlortie wrote:IIRC, it was Ovyyus who once posted an item about a fly in a capped jar sitting on a scale. The statement goes something like this"

The weight of the fly is felt when it is sitting on the bottom, stuck upside down on the lid or in flight. The only time the weight of the fly does not register is if it is dead and falling.

Ralph
Did they zap the fly with a laser or something while it was in the air? ...lol

That is interesting to think about, though. It does make sense.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

agor95 wrote:@MrVibrating

Best to focus on your thread. Some are reading and learning.

P.S. These roasted hot potatoes are being bounced around and wasting time.
Lets drop them and progress.

Regards
Well i really appreciate the goodwill mate and glad someone's getting something out of it.

But i'm desperate for corroboration of the maths and general principle. Yes, i know time and testing will tell, predictions will either stand or fall, it works or doesn't... but i think it has to, but also has to be done safely.

For example we could mount a pair of Bessler wheels opposite one another to the inside walls of a horizontally-rotating drum, using centrifugal force instead of gravity.

This doesn't seem to cause any stray unbalanced forces. But of course, that's still a far cry from any assurance of safety.

Magnetic systems are equally viable, and can likewise be counter-balanced to isolate stray momenta.

But if the first thing i design is a crude 'gravity wheel' like JB's, that's what's going wild on teh interwebz. That genie's out of the bottle, and no one is going to listen or care about where the other 9.81 kg-m/s our input energy is buying is being induced.

Doing so to profit would be pure wickedness. Doing it to 'win' an internet forum would be just as twisted, but even more stupid.

Yet so far, nobody else here has even corroborated the stray force the theory predicts! So is this really the best place to be unleashing a force the planet's never been subjected to before? have i really exhausted all better options? Am i really doing it for the right reasons? If i think it can be done safely, don't i have a responsibility to ensure that? If no one else is willing or able to reproduce the predictions independently then must i bear sole responsibility in these considerations? If it can be done safely, wouldn't i be stupid not to try and quit the shitty day job? If i turned up at a meeting with techies at a big corp or govt. office and sketched all this on a whiteboard, would i get a similar reception to here? A roomful of Silvertigers and Dwaynes? Would James Dyson just shake his head wearily and tell me i forgot to include a vortex?

Sorry i'm tired, ain't slept more than a few hours last few days as i've been too busy repeating myself all night on teh interwebz..

We need sensible discourse about safety, and that can't happen if i'm the only one who can see any risk, and nobody else will see the risk without following the maths and asking why the fixed-price cost of 96.23 J is precisely twice the value of the 9.81 kg-m/s momentum it appears to buy, and then realising that it's opposite is being transmitted to Earth via the negative GPE of the non-accelerating weight, which is a net upwards linear acceleration of 9.81 kg-m/s being applied to Earth with every single "reactionless" acceleration of the 9.81 kg-m/s raised internally..

Maybe only the linear-linear example is dangerous, and the angular-angular version exerts an equal opposite downforce via the axle and stand? So the inertia of the non-accelerating mass is being used to push the Earth back down, against its upwards attraction to it, both forces equal and opposite?

But in the linear-linear version, with both masses in free-fall when the force is applied between them, there's no avoiding pulling the Earth upwards against the inertia of the lower mass, and the resulting downwards momentum is converted to angular momentum and RKE instead of being transmitted to ground.

I need to sleep on it i guess..

A wanna pick back up where i left off - priming the system with conventionally-raised momentum and KE up to its unity threshold, and so punching straight thru to 125% OU in a single strike..

..and then collecting the resulting momentum gain. That would be good logical progress towards a mechanism that could be segmented around a wheel.

But also, only one step away from releasing something destructive. All it would need then is the CF / gravitational workload harvesting the RKE to power the internal counter-torque pulses. No "PE management" system is required since output can directly drive input from a standing start.

This must be the approach Bessler was using, since without it, a bump-start and PE management system is necessary. Below it's threshold unity energy speed, it would be acting as a non-dissipative brake, effectively destroying the energy you were trying to accelerate it with. Then, as it reached its unity threshold it would stop resisting acceleration and start to coast freely. Any faster and it would start to take off, accelerating hard until the inertial interactions could no longer maintain sync WRT gravity. Or else, catastrophic structural failure.

Not especially desirable properties either way. But also we have no records of such properties observed in JB's wheels.

The single-strike-to-OU approach does seem much more consistent, with each inertial interaction producing an OU result, with multiple such interactions per cycle, outputting constant energy and momentum per interaction, but power simply scaling as a function of the rising velocity (more energy/time).

Anyway, only a couple of days to the weekend..
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

ME wrote: It may be possible I overlook the benefits of looking at this differently; so I guess that I (and maybe many more) apparently need a better explanation on how you exactly divide KE between two masses. You assume it's clear in one half word and then rush off with your conclusion. I would like a little more time on that (we are all slow sometimes).
Explain how that works until understanding kicks in and likely that many more too will automatically understand how you get that gain [+25% per interaction], and when agreed then as bonus will also share the rest of your enthusiasm: because such gain would indeed, as understatement, be super-cool.
To be clear, you're actually asking me F=mA?

A force is applied between two masses. Why do you want to divide KE between them? Where are you getting this from?

Marcello, we're buying momentum at a fixed, speed-invariant cost.

We can accumulate it, always adding the same amount of momentum, for the same amount of energy, regardless of rising net speed.

This is a disunity, because the energy value of momentum is a function of V^2.

Buy 1 lot and we end up with 75% less energy than we've spent.

Buy 2, and now we've only lost 50% of our energy.

Buy 3, and only 25% of our input energy disappeared to nowhere! We get to keep the other 75%..

In all three cases, calorimetry would not be able to account for those losses. They're "non-dissipative".

Buy a fourth, and we finally hit unity. We now have the right amount of KE for our current momentum, per 1/2mV^2.

But buy a fifth, and we now have 125% too much energy.

With a sixth, we have 150% excess.

7th = 175%

8th = 200%

12th = 300%

16th = 400%

Etc.

The basis of all this is that we can buy momentum at a constant rate of 96.23 J per 9.81 kg-m/s, regardless of rising speed.

So multiply that up and you hit unity with KE=1/2mV^2 after four purchases, and 125% OU at five. I've already written out these sums multiple times, please don't make me do it again?
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Post by MrVibrating »

I'd noticed previously that 96.23 = 9.81 * 9.81, so equivalent to 9.81 J/kg-m/s, but just now realise that it's basically 9.81^2, then. Duh.

So 9.81^2 J per 9.81 kg-m/s.


N^2 J / N kg-m/s.

Against which, KE=1/2mV^2.

There's your basic cost / benefit analysis.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by ME »

MrVibrating wrote:To be clear, you're actually asking me F=mA?
A force is applied between two masses. Why do you want to divide KE between them? Where are you getting this from?
I don't know, did I?
I asked you how you divide your kinetic energy, because I thought that's what you do...
I already explained how I see your linear example and there's nothing mysterious to discover. Even your simulator does not complain. So you either do something strange or you conclude something strange just as you asked Dwayne to redo his calculus: but what needs to be changed?
You slippery-slope over the issue. So I don't know what you're doing that makes you so exited. I get it that you are exited once you found some gain.

But if you didn't notice, I did put links in my text.
In this case the KE-division should hint upwards to: "redistribute energy and momentum".
That links to your text where you ask:
MrVibrating wrote:Everyone follow that so far?
My answer: "no".

I get the notion of "division" from that post, and this is how I read it:
  • Kinetic energy induces a change in momentum, check.
    Then [J per kg-m/s]. I think: [J per kg-m/s] that equals [m/s], or actually Ek/p = v/2
    Then you split that and your energy in two... and you add them up again. And compare this new momentum with its energy-level versus ground to this other energy and notice a difference...
    Or at least I think that's what you're doing there. I think: "But why? What is he doing?"
And I stop, scribble something on a piece of paper, gets more confused: let's ask. Get's more confusing answers plus what you can do with all that stuff, and its implications/

I may indeed ask for the basics, sorry. Seems important.
Marchello E.
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re: Plying CF as pseudo-inertia to scam N3

Post by sleepy »

Mr. V,
I cannot corroborate your maths. But I am trying to do something possibly more important by using repeatable real world experiments to see if your theory can actually be put to use. I'm guessing that you haven't done many builds of your ideas,but you have to start with the basics and work your way in. Now during these experiments,I have come across some interesting things.Firstly,there can be no additional momentum transferred to the lower weight unless the mech is free to move before launching the lower weight.So I'm imagining now a double scissor jack where the top and bottom halves point away from each other and can operate independently. The point of connection to the wheel is a rod attached to the middle joint of the upper jack,leaving the jack and upper weight free to move.Jack fires,upper weight moves up (appearing stationary)jack accelerates down and then fires the lower weight, transferring it's extra momentum.Lower weight collides with wheel.Somehow,both weights are retracted when horizontal and the whole process begins again.I may have found an improvement to your concept while doing some experiments. If you have a jack whose base is attached to the wheel and has just a top weight,when it fires on the down side,it can gain some force from lifting the weight and transfer that directly to the wheel,but it has the beneficial side effect of having that weight put it's force on the wheel from 1 oclock to 4 oclock then fires upward and once again places it's force on the wheel from say 2 oclock to 5 oclock.That weight has now acted on the wheel for almost twice the distance.That's gotta be good,right?In simpler terms,the weight has fallen twice without ever removing it's mass from the wheel.
Trying to turn the spinning in my brain into something useful before moving on to the next life.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by Furcurequs »

MrVibrating wrote:
Furcurequs wrote:MrVibrating,

After having spent a fair amount of time trudging through your thick and heavily made up jargon, I actually was able to finally figure out what you were trying to do. So, please don't bother repeating yourself.

Since I do understand what you are trying to do, that is why I'm now trying to show you what you've done wrong with your math.

So, let's first consider what the conditions would have been like if you hadn't added your 96 Joules of energy to the falling system of mass...

We would have had two 1kg masses each with an initial velocity of 19.62 m/s downward that accelerate downward for 1 second in the earth's gravitational field.

We can calculate the distance they both fall:

S = Vi x t + 1/2 x g x t^2

= 19.62 m/s x 1 s + 1/2 x 9.8 m/s^2 x (1 s)^2

= 19.62 m + 4.9 m

= 24.52 m

We can calculate the total work done on them due to the force of gravity:

F x d = m x g x h = Fg x S (from above)

= 2 kg x 9.8 m/s^2 x 24.52 m

= 19.6 x 24.52 J

= 481 Joules

We can calculate their total energy after the 1 s fall:

KEinitial + F x d (from above) = 1/2 mass x Vi^2 + 482 Joules

= 0.5 x 2 kg x (19.62 m/s)^2 + 482 Joules

= 385 Joules + 482 Joules

= 867 Joules

We can calculate their final downward velocity from this:

Vf = (KEf x 2 / mass)^0.5

= (867 Joules x 2 / 2 kg )^0.5

= 29.4 m/s

...and we can, of course, calculate the downward momentum of both masses:

P = mass x velocity

= 2 kg x 29.4 m/s

= 58.8 kg x m/s

MrVibrating, does that number look familiar to you?!

It should!! ...lol

...because it is the same total momentum that you have in your example even with your added 96 Joules of energy! You have bought nothing!

So, do you need me to explain this to you?!!!

Well, let me do that anyway...

Since you added energy internally to the falling 2 mass system, the upward and downward forces sum to zero and so don't change the total system momentum in the vertical direction at all! ...lol

It's a zero sum game. ...as I've already tried to point out.

The momentum you think you bought was only borrowed from the upper mass and so whether you know what you are doing or not, it's going to get paid back in reality.

Your best bet is to just try to hang on to the energy you are inputing.

...sorry...

I'll try to post another explanation later.

Dwayne

Oh, and don't worry about calling me Wayne. Even my own dad sometimes forgot the "D" for some reason.

(Btw, if my dad were still alive, he'd be 3 days shy of 100 years old right now. I just happened to be thinking about that.)
Yeah wow you've totally got it. Spot on! Fantastic. I can see i'd have to get up very early in the morning to get one over on you, eh? You've simply cut straight to the thrust of it, with that rapier-like incisiveness, explicated the very nub of the gist of my so-called thesis and left it in ruinous tatters, like the confused and meaningless word-salad it always was. Good show, take no prisoners! Capital stuff.
Thank you, and you are quite welcome! :)
MrVibrating wrote:No, seriously. You really, totally shot my fox there, i feel such a fool. Well and truly cooked. My.


Goose.




Formidable.
Oh, don't make me blush. :">

Again, though, you are quite welcome. Better luck next time. We can all make mistakes, I guess. Glad you understand.
MrVibrating wrote:
(yawn)
...uh... ...but maybe you should try to stay awake around someone who really can cook your goose! ;P
Last edited by Furcurequs on Thu Nov 16, 2017 5:38 pm, edited 1 time in total.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by Furcurequs »

MrVibrating wrote:
Furcurequs wrote:MrVibrating,

After having spent a fair amount of time trudging through your thick and heavily made up jargon, I actually was able to finally figure out what you were trying to do. So, please don't bother repeating yourself.

Since I do understand what you are trying to do, that is why I'm now trying to show you what you've done wrong with your math.

So, let's first consider what the conditions would have been like if you hadn't added your 96 Joules of energy to the falling system of mass...

We would have had two 1kg masses each with an initial velocity of 19.62 m/s downward that accelerate downward for 1 second in the earth's gravitational field.

We can calculate the distance they both fall:

S = Vi x t + 1/2 x g x t^2

= 19.62 m/s x 1 s + 1/2 x 9.8 m/s^2 x (1 s)^2

= 19.62 m + 4.9 m

= 24.52 m

We can calculate the total work done on them due to the force of gravity:

F x d = m x g x h = Fg x S (from above)

= 2 kg x 9.8 m/s^2 x 24.52 m

= 19.6 x 24.52 J

= 481 Joules

We can calculate their total energy after the 1 s fall:

KEinitial + F x d (from above) = 1/2 mass x Vi^2 + 482 Joules

= 0.5 x 2 kg x (19.62 m/s)^2 + 482 Joules

= 385 Joules + 482 Joules

= 867 Joules

We can calculate their final downward velocity from this:

Vf = (KEf x 2 / mass)^0.5

= (867 Joules x 2 / 2 kg )^0.5

= 29.4 m/s

...and we can, of course, calculate the downward momentum of both masses:

P = mass x velocity

= 2 kg x 29.4 m/s

= 58.8 kg x m/s

MrVibrating, does that number look familiar to you?!

It should!! ...lol

...because it is the same total momentum that you have in your example even with your added 96 Joules of energy! You have bought nothing!

So, do you need me to explain this to you?!!!

Well, let me do that anyway...

Since you added energy internally to the falling 2 mass system, the upward and downward forces sum to zero and so don't change the total system momentum in the vertical direction at all! ...lol

It's a zero sum game. ...as I've already tried to point out.

The momentum you think you bought was only borrowed from the upper mass and so whether you know what you are doing or not, it's going to get paid back in reality.

Your best bet is to just try to hang on to the energy you are inputing.

...sorry...

I'll try to post another explanation later.

Dwayne

Oh, and don't worry about calling me Wayne. Even my own dad sometimes forgot the "D" for some reason.

(Btw, if my dad were still alive, he'd be 3 days shy of 100 years old right now. I just happened to be thinking about that.)
OK you've obviously earned more than a tart riposte, you're making an effort and i should try and meet you halfway.

The whole concept here however is obtaining a net momentum.


Thanks.

...but as you should be able to see from comparing the results of your simulation to the results of my calculations, the net momentum of the two masses did not change with the forces you were applying in your simulation.

In your simulation, the impulse down on the lower mass is equivalent in magnitude to the impulse upward on the upper mass. Since an impulse equals the change in momentum, we can see that the momentum added to the lower mass is equal to that taken away from the upper mass.

There is no net change in momentum due to your internal forces.
MrVibrating wrote:That's what all the calcs are for. To track the sources, amounts and signs of momentum.
The forces are internal to the two mass system and so, again, there is not net change in the total momentum of the two masses.
MrVibrating wrote:In order to 'see' the results of the inertial interaction, we need to filter out the gravitational interaction, and its associated momentum.
The momentum of the two mass system only changes due to the gravitational interaction since, again, the forces you are applying are internal to the two mass system.
MrVibrating wrote:That's what the calcs did. I took the change in GPE, the RKE it converted to and the corresponding momentum and subtracted them, to leave only the KE and momentum applied between the two masses.
I think you meant KE where you said RKE. The kinetic energy you added is internal to the two mass system and so doesn't change the net momentum of the system.
MrVibrating wrote:All of which is of one sign - only one of the two inertias gets accelerated, so its rise in momentum (or velocity) can be shared with its partner, bringing both to a new equilibrium momentum / velocity.
Maybe instead of talking about "two inertias" you should be speaking of the motion of the two masses so that we don't lose track of what we are talking about. Whereas you have introduced energy, there is no net change in system momentum due to the internal forces.

MrVibrating wrote:Momentum generated this way - from asymmetric inertial interactions - is special, because the process generating it is speed-invariant - it costs the same energy, doing the same workload, to generate the same rise in net momentum, whatever our current velocity.
Due to Newton's 3rd law in which there are equal magnitude and opposite direction forces in an interaction, there was, again, no net momentum introduced in the interaction. So, it wasn't a special interaction. The momentum gained by the system is thus only due to the external force of gravity. The center of mass of the system behaves the same in the gravitational field as if the two masses were still together.

MrVibrating wrote:But the energy value of that momentum is not constant, instead it squares with velocity.
Your interaction didn't change the net momentum.
MrVibrating wrote:If we buy five or more lots at that fixed price, that V^2 multiplier takes us over-unity.
Since there was no net change in system momentum beyond that introduced by gravity, what you are talking about having "bought" for the lower mass was simply "bowered" from the upper mass for, again, no net change in system momentum.
MrVibrating wrote:The optimal cost of this momentum, using gravity to fully cancel counter momentum between two 1 kg inertias, is 96.23 J per 9.81 kg-m/s.

Four times 9.81 = 39.24.
Four times 96.23 = 384.92

39.24 kg-m/s divided between two 1 kg masses means both are at 19.62 meters / sec

2 * 1 kg masses at 19.62 m/s have 192.47 J each, so 384.94 J total.


So after precisely four reactionless momentum rises, the two masses have exactly the same momentum and energy they would've had, no matter how they were accelerated.

But below four, they have less RKE value than their cost of generation.

And above four, they have more value than their cost of generation.
There were no reactionless interactions. The energy you introduced internally to the two mass system were normal Newton's 3rd law reactions, and the other is the normal gravitational reaction with the earth.

If you want to keep the energy you introduced into the system by bringing the masses back together, we will have to manipulate their direction of motion through further interactions with the earth and thus change their momentum that way.
MrVibrating wrote:Get it?
I get that you are confused.
MrVibrating wrote:Mate... you haven't calculated the momenta!

That was the whole point of the hypothesis, calculations, sims and results analysis, and the context of the resulting discussion of those findings.

Is there or is there not a net momentum, after subtracting (ie. notionally repaying) the KE and momentum generated from the GPE? That's the objective of the analysis.

Not an excess of inertial work - the amount of momentum raised is initially the same, whether symmetrically-induced or not.

But if it's symmetrical, the net change is zero. Any degree of asymmetry results in a proportionately non-zero sum, and here we have a fully 100% asymmetric inertial interaction - one of the inertias is not accelerated at all, despite being used to successfully raise momentum in its partner.

So disproving me (if that's your goal, which is great BTW) requires following the sign of the momentum, specifically the momentum produced by the inertial interaction.

All that concerns you should be three issues:

- can momentum be generated this way?

- can it be accumulated?

- is its cost of generation constant, irrespective of ambient speed?

My case - OU - stands or falls on these predicates.
The recombining of the masses in the way you described would require some interactions with the earth and subsequent changes in momentum but also a loss of much of the energy you input, so you would have been better off having just applied downward forces to both masses to begin with (pushing off something affixed to the earth).

Dwayne
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Post by MrVibrating »

OK i'm gonna have to ignore Marcello, Dwayne, ST and anyone else struggling for now, no more repetitions, but all three need to read up on CoM and CoE.

If you don't even understand how they apply in the first place, you're not even in the game. I find it so incredible that Marcello doesn't grasp CoE's dependence on N3 that i no longer trust his intellectual capacity, honesty or motivations, and the same basically goes for Dwayne. Both are inherently incapable of making any useful assessments whatsoever. I mean if you can't understand how N3 enforces CoE then you can't even measure or calculate anything - none of your velocity, acceleration, force, displacement or momentum or KE maths has any context without an inertial frame, which depends entirely upon N3. It's just absurd, farcical, borderline suspicious even.. so stupid, it's beginning to seem suspect, even by my impressive standards.. :/
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re: Plying CF as pseudo-inertia to scam N3

Post by rlortie »

That's Marchello not Marcello, and if you are going to ignore the three aforementioned members you might as well close this thread. The only other one who has attempted to understand 20 pages of already-called gibberish is sleepy who admits he cannot corroborate your math either.

Sorry but for me this topic has been nothing more than a waste of bandwidth. If you wish to save it, give us something substantial/explanatory like sleepy's last post, something that the average member can gnaw on!

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re: Plying CF as pseudo-inertia to scam N3

Post by ovyyus »

MrV wrote:OK i'm gonna have to ignore Marcello, Dwayne, ST and.... all three need to read up on CoM and CoE.
Your only solution is a proof-of-principle demonstration. Then we all get to learn more about physics and psychology.
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re: Plying CF as pseudo-inertia to scam N3

Post by Grimer »

I've copied below MrVibrating's paragraphs on page 20 of this thread where he refers to inertia.
MrVibrating wrote:Maybe only the linear-linear example is dangerous, and the angular-angular version exerts an equal opposite downforce via the axle and stand? So the inertia of the non-accelerating mass is being used to push the Earth back down, against its upwards attraction to it, both forces equal and opposite?

But in the linear-linear version, with both masses in free-fall when the force is applied between them, there's no avoiding pulling the Earth upwards against the inertia of the lower mass, and the resulting downwards momentum is converted to angular momentum and RKE instead of being transmitted to ground.

This must be the approach Bessler was using, since without it, a bump-start and PE management system is necessary. Below it's threshold unity energy speed, it would be acting as a non-dissipative brake, effectively destroying the energy you were trying to accelerate it with. Then, as it reached its unity threshold it would stop resisting acceleration and start to coast freely. Any faster and it would start to take off, accelerating hard until the inertial interactions could no longer maintain sync WRT gravity. Or else, catastrophic structural failure.

The single-strike-to-OU approach does seem much more consistent, with each inertial interaction producing an OU result, with multiple such interactions per cycle, outputting constant energy and momentum per interaction, but power simply scaling as a function of the rising velocity (more energy/time).

If you don't even understand how they apply in the first place, you're not even in the game. I find it so incredible that Marcello doesn't grasp CoE's dependence on N3 that i no longer trust his intellectual capacity, honesty or motivations, and the same basically goes for Dwayne. Both are inherently incapable of making any useful assessments whatsoever. I mean if you can't understand how N3 enforces CoE then you can't even measure or calculate anything - none of your velocity, acceleration, force, displacement or momentum or KE maths has any context without an inertial frame, which depends entirely upon N3. It's just absurd, farcical, borderline suspicious even.. so stupid, it's beginning to seem suspect, even by my impressive standards.. :/
I think he is absolutely spot on in his attention to this property of inertia.

The falling yo-yo phenomena which will be shown to generate energy is a manifestation of the polarization of inertia, the polarization of mass in other words.

I think MrV might find it conceptually easier to think of gravity as a compression rather than a tension. The two are equivalent.

We are all familiar with the polarization of mass though we don't recognise it for what it is.

For example. Hold a bicycle wheel by an extended axle on either side, the left hand holding the extended axle on the left side and the right hand holding the extended axle on the right.

Try to rotate the wheel in the horizontal and vertical plane. No problem. It takes very little effort.

Now rotate the wheel at some moderate rpm and try again. You will find it is extremely difficult to keep it in either plane. Not only is the difference vast but the damn wheel tries to twist away from the plane.

Why.

Because the rotation of the wheel in one plane has polarised the inertia, polarized the EG mass. The NG mass is unaffected. A rotating wheel weighs the same as a stationary wheel.

If people don't like the term EG mass they can call it centripetal/centrifugal mass if they like - or inertial mass to distinguish it from Newtonian gravitational mass. A rose by any other name .....
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by Grimer »

rlortie wrote:That's Marchello not Marcello ...
Quite right. Thinking of Inspector Morse I've often wondered if the E stood for Endeavor. :-)
Who is she that cometh forth as the morning rising, fair as the moon, bright as the sun, terribilis ut castrorum acies ordinata?
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