Plying CF as pseudo-inertia to scam N3

[MrVibrating]

MrVibrating,

After having spent a fair amount of time trudging through your thick and heavily made up jargon,

What "jargon" - give an example and show us how you would instead phrase it.

I actually was able to finally figure out what you were trying to do. So, please don't bother repeating yourself.

You mean please don't bother correcting your stupid errors?

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.

Dude, your grasp of what i'm doing is revealed below, and you've no idea what you're even doing..

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...

Energy is conserved, so obviously the system will have 96 J less work done upon it. What is that work, then, if the net momentum is not increased? Where'd that energy go, and why doesn't it have a momentum rise proportionate to its KE?

This should be a laugh..

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

Yes, it would be 865, but close enough

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

Yes.

...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

OK...

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!

Yes! So where did that extra 96 J go then? What has it been spent on, and why are you just ignoring this? Not very thorough are you?

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

Are you really this stupid?

Dwayne, in your above example, all of the momentum came specifically from dropping the two masses.

It all came from GPE.

When we re-lift the masses, what will happen to all that momentum?

That's right - the KE it is part of will be converted back to GPE, leaving a remaining momentum of zero

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...

Yes Dwayne! Yes it is.

Because when we do this, the net system momentum now has two sources - GPE, plus the 96 J impulse!

So when we re-lift the GPE, returning the 9.81 P momentum of its associated KE, we still have the other 9.81 P momentum input via the 96 J.

And it is of one sign only.

So the GPE game is a zero sum deal, for both energy and momentum.

But the work done by our 96 J impulse is conserved.

And specifically, it is a unidirectional rise in momentum that was raised exclusively by the internal expenditure of work, at no cost to the GPE.

So it doesn't self-cancel.

We can equalise this between the two masses and repeat that cycle, in principle indefinitely.

No matter how fast our ambient velocity rises, each 96 J impulse buys another 9.81 P increase in net momentum.

And as you dementedly insist i repeat ad nauseam, after 5 such cycles we open a white hole and flesh-eating ogres start to pour out at an exponentially accelerating rate. I just can't understand what's so complicated, you're just adamantly refusing to follow process, so if you're not just trolling, please address this point directly in your reply, as this is the gain, a bankable net rise in momentum after a zero-sum GPE interaction...

Not a greater net magnitude of momentum!

So no, we don't gain more momentum than if we'd simply dropped the masses passively! You're absolutely correct, but this has absolutely nothing whatsoever to do with the gain principle!?

My god, the effort..

You don't wanna be Prometheus, much less Atlas, it's waay too much responsibility. I get it.

You want Bessler's wheel to be safe, clean OU. Totally understand your mental block.

Or maybe you and Marchello are busy co-writing a patent application of the current art, whatever.

We're not gaining more momentum than we could generate from GPE... we're simply gaining momentum.

And accumulating it.

At a fixed cost, unaffected by rising speed.

Five purchases of 9.81 P at 96 J a pop is 125% over-unity.

You've built a straw man and completely avoided addressing or even acknowledging this central issue. Whether this is intellectual dishonesty or just basic human density, i could've trained a friggin' parrot this in less time.

[MrVibrating]

Here's a more detailed overview of the symmetry break:

Code: Select all

# Cycles:	Energy cost of momentum:	Equiv. Net Value:	Efficiency:
_________	________________________	_________________	___________     


	1		*	(N^2 J / N kg-m/s)		=	1/8mV^2		=	75% Under-Unity

	2		*	(N^2 J / N kg-m/s)		=	1/4mV^2		=	50% UU

	3		*	(N^2 J / N kg-m/s)		=	3/8mV^2		=	25% UU

	4		*	(N^2 J / N kg-m/s)		=	1/2mV^2		=	Unity

	5		*	(N^2 J / N kg-m/s)		=	5/8mV^2		=	125% Over-unity

	6		*	(N^2 J / N kg-m/s)		=	3/4mV^2		=	150% OU

	7		*	(N^2 J / N kg-m/s)		=	7/8mV^2		=	175% OU

	8		*	(N^2 J / N kg-m/s)		=	mV^2		=	200% OU

	9		*	(N^2 J / N kg-m/s)		=	9/8mV^2		=	225% OU

	10		*	(N^2 J / N kg-m/s)		=	5/4mV^2		=	250% OU

	11		*	(N^2 J / N kg-m/s)		=	11/8mV^2	=	275% OU

	12		*	(N^2 J / N kg-m/s)		=	3/2mV^2		=	300% OU

	16		*	(N^2 J / N kg-m/s)		=	2mV^2		=	400% OU

	32		*	(N^2 J / N kg-m/s)		=	4mV^2		=	800% OU

	64		*	(N^2 J / N kg-m/s)		=	8mV^2		=	1,600% OU

_______________________________________________________________________________________


N = m/F - ie. the arbitrary constant N is equal to mass divided by force (ie. acceleration) of the uniform static field, so 9.81 for gravity.

Input efficiency constant is N J/kg-m/s - ie. the speed-invariant energy cost of momentum per cycle is N Joules per kg-m/s.

Note that the number of cycles per revolution of a wheel is arbitrary, so 1600% could be attained after say 8 revolutions of an 8-mechanism wheel, or 16 revs of a 4-mech wheel etc., assuming such sustained accelerations were mechanically practical.

These figures relate to the "2-stroke" version, its efficiency growing as a function of rising velocity. The "4-stroke" implementation would have a fixed efficiency of 125% per cycle, regardless of the number of cycles elapsed.

Since this summary pays no heed to mechanical speed constraints, the "4-stroke" may be the more practical design; as cool as a "2-stroke" version accelerating to failure might sound, synchronisation of the mechs WRT gravity is likely to limit speed.



So here's the design brief again:

---------------------------------------------------

- apply a force between two equal inertias, mutually accelerating one another against each other's inertia, while at least one of them is gravitating

- this introduces an asymmetric distribution of momentum, that we're going to sum between our inertias and keep

- optimally, this force should prevent any acceleration of the gravitating mass, keeping its current speed constant, and only accelerating the other one

- if you're envisioning long protracted accelerations like the ones i've been showing, think smaller - the optimum range being 90° from vertical to horizontal

- then we apply a mutual deceleration between them, without interference from gravity, and so equalising a gain in velocity and momentum

- it doesn't make any substantive difference to the net result whether the lower mass also gravitates or not, since GPE-in = GPE-out

- angular-angular inertias seem simplest to work with, or maybe angular-linear too; Linear-linear is possible, but inelegant

- 2-stroke either needs a bump-start if direct-drive, or else a PE reservoir to get up past unity, but potentially reaching higher speeds and efficiencies

- 4-stroke substitutes the first 4 cycles for a single conventional 1/2mV^2 acceleration of the same value, so achieves a fixed 125% efficiency every cycle

- in either case, the RKE gain can be harvested via CF or GPE; both are absolute, and independent of the internal cost of momentum

- without the maths you are cargo-cult engineering; you must design the momenta being generated, their energy cost, and resulting value!

---------------------------------------------------


The first design to fulfill these criteria will definitely and definitively work. Your optimum cost of momentum-from-gravity is 9.81 J per kg-m/s, in principle across any speed range your mechanism can sustain sync thru, but however much your mech has to pay for it, provided it's constant across some useful speed range, there is always some threshold speed beyond which RKE > net input energy. Obviously tho, if you find you're paying much over ~10 J/kg-m/s then you can probably make some refinements somewhere.

And do remember, "success" here is basically just grabbing a toe-hold onto the ladder - so for instance that first 75% under-unity cycle is actually a significant win.

A second full cycle on top of the first - now only losing 50% energy - would be no less an outstanding triumph than the 125% gain from the fourth cycle to the fifth - different rungs, same ladder, and the only way is up...

So we're talking low-hanging fruit, in nice easy steps, beginning with an interaction that's going to irreversibly waste most of the energy we spend on it.. but to us, that's going to be a Good Thing. Hugely encouraging, in fact..

So, for the sole benefit of anyone who is able to understand the above jargon-laden gibberish, if not now then in future.. you can use the above table and formulas to design any wheel you like, using any ambient force field you like, purely from first principles, knowing precisely how much energy it will gain.

Last one's a cissy. :P

[MrVibrating]

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!

Ralph

I'm recording my research, not trying to sell you anything.

Nothing i have presented is beyond your comprehension. We can buy momentum for less energy than it is worth, so it doesn't exactly take a business mastermind...

It's simple:

Phase 1: Collect underpants

Phase 2: ?

Phase 3: Profit

Get it?

[MrVibrating]

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...

WHY do you think this? I've said no such thing, that was Dwayne's error. I'm calculating KE from 1/2mV^2, in simple annotated steps..?

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:

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.

KE is calculated from 1/2mV^2 accurately. We have 1 kg masses, so at say 5 kg-m/s they're moving at 5 m/s, and have 12.5 J.

Scanning back through the thread, you've quoted a line out of context from page 11 where i wrote:

We can make 9.81 kg-m/s of momentum for the fixed, speed-invariant price of 96.23 Joules, which equivalently, works out to 9.81 J per kg-m/s.

Divided between two 1 kg masses, each has 4.905 kg-m/s and 48.11 J.

But per KE=1/2mV^2, at 4.905 m/s 1 kg has just 12.029 J...

So on the first cycle, our very first kg-m/s of manufactured momentum costs exactly four times more to make, than it is worth. To put it another way, it is 4x under-unity.

Everyone follow that so far?

So yes, that bit in red is a mistake, immediately corrected in the following sentence, and irrelevant since it is not used for anything. I'm not calculating anything with it. The conclusion, in bold, is valid.

I am fallible, i've been saying throughout i have and will make mistakes. The maths will need tidying, but the broad strokes are there.

Neither you, ST or Dwayne have found any fatal flaws so far, and your motivations do not seem sincere much less objective.

You have a wide open symmetry break in front of you, and all you can do is indulge in petty ego wars and pedantry?

You freely admit you don't even grasp why CoE depends upon N3 in the first place, yet you aren't compelled to try to understand that proposition? Since the conditions here depend upon it but also, perfectly demonstrate it, how can you possibly weigh in meaningfully otherwise?

Energy has to be quantified in relation to inertial frames because motion is relative, Marchello... So N3 enforces mechanical CoE.

You're on an OU forum flaunting some degree of maths skills - how can you possibly not know this?

Obviously, all three laws are essential to CoE, but i'm not attacking the first two, so if you simply cannot grasp why CoE depends upon N3, even with all this help, then you're never going to be able to understand the battle plan for Operation Over-Unity, since the breach is the cost/benefit ratio of accumulating reactionless momenta.

You're deliberately ignoring all the accurate maths, trawling back 20 pages for irrelevant errors that have no impact whatsoever on the symmetry break. What's next, punctuation errors? Are you gonna claim you don't believe inertia's constant, or that KE=1/2mV^2?

Are you just another pseudo-skeptic troll?




PS. And i'm sorry if i offended your sky fairies. English nerds like Python humour, and tend to regard religion as codified contempt for God and creation.

[eccentrically1]

On page 2, I replied that once the system is started, the upper mass that is acting as a push-off for the wheel will be the negative angular momentum as soon as the mechanism reaches its end of travel. The wheel won't feel its braking effect until then. The upper mass is going to be at rest, and when the delayed reaction kicks in it will have to be accelerated by the wheel. I don't see that this issue has been addressed but maybe you did, sorry in advance if I missed it, in which case you can ignore me too.

[MrVibrating]

On page 2, I replied that once the system is started, the upper mass that is acting as a push-off for the wheel will be the negative angular momentum as soon as the mechanism reaches its end of travel. The wheel won't feel its braking effect until then. The upper mass is going to be at rest, and when the delayed reaction kicks in it will have to be accelerated by the wheel. I don't see that this issue has been addressed but maybe you did, sorry in advance if I missed it, in which case you can ignore me too.

Perfectly reasonable request for clarification.

(And thanks for being so civil! Sorry if i didn't address this issue sufficiently.)

Again, the masses are mutually accelerated against one another's inertia, whilst one or both of them are gravitating.

Then they're mutually-decelerated against one another, when neither are gravitating.

This rectifies momentum from gravity's acceleration.

Because we're applying a mutual force between two inertias, but only one of them is actually accelerating, we're inputting momentum of one sign only.

So when we then come to mutually-decelerate them, instead of having induced positive and negative momenta that sum back to zero, we end up with two positive momenta of equal value.

All of this momentum has been paid for - none of it is 'excess'. At no point do we depend upon 'extra' momentum that hasn't been produced by an internal expenditure of work.

So it's not "free momentum from gravity" - rather, gravity passively assists in causing a momentum asymmetry between the two inertias, all of which has always been paid for, but which sums to a step-wise net increase instead of mutually cancelling.

As we repeat that cycle, each new one begins at the net ambient velocity the last one left off at, but the ever-rising speed has no effect on the energy cost of the momentum we're adding each cycle - it's a constant 9.81 J/kg-m/s.

Thus because KE=(1/2mV^2) or (1/2MoI*RPM^2), input and output energies are completely thermodynamically decoupled, only converging - almost incidentally - at 4 interactions.

Thus an OU result is just as easily obtained as a unity result from the same process... it's just one more cycle of precisely the same interaction.

As i say, if you can reach the second rung on the ladder, you can surely reach the fifth..



PS. on page 6, look again at the green disc demonstration; this is the basic effect, albeit exaggerated to a functionally-useless degree since no further such accelerations are possible after the first.

On page 8 i made a small start on considering mechanisms that might be applied to produce and control the process.

Obviously, all earlier findings are subject to later correction - this is research & discovery not gnosis - but the core exploit is simply a mutual deceleration, following a not-so-mutual acceleration. Any system that can accumulate the resulting gain over four consecutive cycles has no more energy or momentum than it should have, therefore we're all saved and/or doomed.

[daanopperman]

Hi Mr Vibrating ,

Relative to the wheel rim , how fast must the 2 weights part to have your desired effect ?

[sleepy]

Ok,so here is what seems to be the problem.It's not in the math,it's in the transfer of force from the top weight to the bottom weight.Since there must be a mechanism between the two weights,if any part of this reaction is attached to a wheel,then the mechanism absorbs any downward force from lifting the upper weight. There is no transfer of any energy from the upper weight to the lower weight.However,if all 3 items are falling through gravity, then when the jack fires,the extra force of pushing the upper weight against gravity will act on the jack and the lower weight.accelerating the lower weight away from the other two pieces.So the math seems correct,there's just no way to harness the gain into any kind of cycle.So were all correct and we can all be friends again! Unfortunately,we have not solved the problem of PM.

[daanopperman]

After the fireing , will one weight have a 0 or a - value .

[MrVibrating]

Hi Mr Vibrating ,

Relative to the wheel rim , how fast must the 2 weights part to have your desired effect ?

It's completely arbitrary.

Whatever seems manageable. Just don't bite off more than you can chew..

[MrVibrating]

Ok,so here is what seems to be the problem.It's not in the math,it's in the transfer of force from the top weight to the bottom weight.Since there must be a mechanism between the two weights,if any part of this reaction is attached to a wheel,then the mechanism absorbs any downward force from lifting the upper weight. There is no transfer of any energy from the upper weight to the lower weight.However,if all 3 items are falling through gravity, then when the jack fires,the extra force of pushing the upper weight against gravity will act on the jack and the lower weight.accelerating the lower weight away from the other two pieces.So the math seems correct,there's just no way to harness the gain into any kind of cycle.So were all correct and we can all be friends again! Unfortunately,we have not solved the problem of PM.

I'm sorry and not saying this to be combative, but everything you've concluded here is just factually incorrect. You have not paid attention - dropping a spring-loaded scissorjack does manifest the effect, but in a clunky and awkward manner. Page 8 makes a start on potential applications, including the linear-linear example, which i have repeatedly described as an oafish interpretation of the maths. Angular-angular interactions are much more elegant, obviating the need for any intervening mass besides the control mechanisms.

Again you must understand that perfect cancellation of counter-momentum is unnecessary anyway - even if you weren't referring to the idealised demonstrations - the gain is simply inversely proportionate to the degree of successful cancellation.

So even if we came up with a very clumsy design with a 20 J/kg-m/s yield, we'd still hit OU after 10 cycles. At that rate, we'd be paying more than twice as much for our momentum as we need to, yet this simply has no effect whatsoever upon the respective dimensions of the input / output energy fields. One squares with velocity, the other's constant.

If you simply cannot think outside the box of a linear scissorjack example, that's fine - it makes zero difference that the jack gravitates; you can still make the upper mass hover momentarily or even rise, but any upwards-applied force applied to the gravitating mass will cause a corresponding drop in its rate of gravitational acceleration, transferring that acceleration instead to the lower mass / inertia and so causing an associated asymmetry in the resulting momentum distribution.

Announcing that you've somehow disproven these facts with a premature misapprehension is just silly mate - the actuator could be 10,000x the mass of the two weights on its ends, and fire them both up and down like bullets... or anything in-between.. but the effect persists.


LOL hopefully at some point the scale will tip, and people will start to wanna score points from showing understanding of the gain, rather than trying to make the abomination go away..

..if you truly cannot yet see the hefalump decloaking in the corner - you need yer bloomin' eyes tested - but stick with it, think it through carefully. As ever, i apologise if i have neglected to previously clarify these details.

[MrVibrating]

After the fireing , will one weight have a 0 or a - value .

As mentioned above, any degree of cancellation of the gravitating mass's acceleration pays momentum dividends which can be accumulated.

Perfect cancellation yields an optimum constant input energy of 9.81 J per kg-m/s.

Where P = mV and N is the acceleration of the force (ie. equal to m/F):

N^2 J / N P

So if we're using gravity, 96.23 J buys 9.81 P

However because the output energy field has the standard KE dimensions of 1/2mV^2, while the above input energy term scales linearly per cycle, even a very inefficient cycle still has a break-even speed beyond which gains rise per V^2, minus those losses.

[rlortie]

We can buy momentum for less energy than it is worth, so it doesn't exactly take a business mastermind...

You mean like a sale? You raise the price 20% percent blaming it on inflation. One week later you advertise it on sale for 20% off. Or is it: "Get two free if you buy three at the regular price of five?

It's completely arbitrary.
Whatever seems manageable. Just don't bite off more than you can chew..

Bite of more of what is where I am at a loss. Where is the physical connectivity that puts your scheme to work? Note: I do not call it a design, as a mechanical engineer I have yet to see or imagine any concept of levers, ropes, or pulleys that is going to fulfill your descriptive equations.

Ralph

[MrVibrating]

You mean like a sale? You raise the price 20% percent blaming it on inflation. One week later you advertise it on sale for 20% off. Or is it: "Get two free if you buy three at the regular price of five?

Just buy low, sell high.

Sorry mate but you simply cannot have any real comprehension without the maths.

We have two 1 kg inertias. Could be angular or linear, makes no difference.

We can use them, plus gravity, to buy 9.81 P of momentum, for 96 Joules.

We then divide that momentum evenly between them, so each has 4.905 P.

Since the masses are 1 kg, each is traveling at 4.905 meters / sec.

Per the standard KE term 1/2mV^2, each mass has an energy of 12.02 J.

So our total 'output energy' is 24.04 J.

But we spent 96.23 J, to be exact:

96.23 / 24.04 = 4.

Our output energy is thus only one quarter of our input energy.

The other three-quarters - 72.19 J - is completely wasted. Gone. Brilliant!

But, we've accelerated a 2 kg system by 4.905 m/s for 96.23 J, with an expenditure of work applied between them, and can follow that first 4.905 m/s acceleration with any number of others.

Refer to the previous table to see how this differential evolves as the number of elapsed cycles increases. After four cycles the system's energy and momentum are in full agreement with 1/2mV^2, with absolutely no excess of any kind whatsoever, thus proving the existence of hefalumps.

Bite of more of what is where I am at a loss.

Hefalump. I do wish people would pay attention.

Where is the physical connectivity that puts your scheme to work? Note: I do not call it a design, as a mechanical engineer I have yet to see or imagine any concept of levers, ropes, or pulleys that is going to fulfill your descriptive equations.

I'm sorry where did i claim i had such a machine?

I'm deducing a means of generating apparently infinite free energy using just the basic laws of motion plus gravity, entirely from first principles.

This is happening now, here, in this thread.

It didn't happen somewhere else, previously, like i've just ported it over here to tantalise you.. it's unfolding day by day. I'm not withholding secrets, i'm not asking for anything, instead ensuring that you have as much accurate relevant info as i'm able to determine in order to begin designing viable systems.

To that end, i have reserved for myself absolutely no advantage whatsoever, and you know about as much about prospective implementations as i do - possibly a good deal more, since i'm no mechanical whizz.

You can appreciate tho that i found this by delving into the math (shallow that it is), not by tinkering with builds or sims. I've finally fitted the jigsaw pieces i've been juggling the last few years, so it's a conceptual breakthrough, and not the chance discovery of an anomaly in a build or sim..

So if you wanna design something go ahead. Or else, don't. Presuming no one else beats me to it, i'll get there soon enough..

Obviously, building without knowledge of a viable principle energy source has been the chief problem dogging the field in recent centuries, so i figured that finding one could really help turn things around, type stuff. Sorry it's so general, but that is kind of the whole point.

But if you can understand concepts like force, inertia and gravity then you can design an OU momentum gain using this principle, today. It really is blessedly-easy rocket science..

[daanopperman]

Hi Mr V ,

In a wheel 2m dia , 60 rpm , rim v = 6.2 m/s .
If we can transfers all vel to one of the 2 masses and have the other mass come to a complete stop , without any back torque to the wheel , then the fired mass will have a vel of 12.5 m/s .
Will this velocity be of any use .