Plying CF as pseudo-inertia to scam N3

[MrVibrating]

Sorry to hear you've thrown in the towel. It was a nice try.

I only meant i'm moderating my expectations of others...


Just given the general level of enthusiasm and all..



As for the concept - buy a PMM, get a free warp drive - what could possibly be the catch?

[eccentrically1]

My replies in blue again.

CC, it isn't reaction less, earth's gravity provides the reaction force. See my post above the last one.

Precisely.

In essence, the reference frame of the on-board inertia the wheel is being torqued against is that of gravity's free-fall reference frame.

As far as a Bessler's wheel is concerned, it's literally as if the ground beneath it were rushing upwards at 9.81 m/s^2.

And rushing away on the ascending side!

In other words the internal inertial frame isn't simply diverging to infinity, but specifically to the 'infinite displacement' of the effective (ie. 'virtual') acceleration of gravity... in which there is no actual net displacement.

Everyone needs to remember tho that the whole Earth is still accelerating upwards towards a weight that isn't accelerating back down...

But it's not accelerating toward the weight. In essence, the preloaded force is preventing the earth from accelerating toward that weight. Sorry.

..so even though a 9.81 kg-m/s momentum divided by the mass of Earth produces a vanishingly small net acceleration, pretty much any acceleration is non-trivial, especially when extrapolated en masse as a global alternative to conventionals.


We're bickering here like savages who've just dug up a radium crystal, arguing over who saw the shiny first and what it all means.. all i know is, i don't wanna be stashing it under my pillow..

We're discussing the merits of your concept. Would you care to address the remarks in my 10:49 a.m. post?

Anyone seen the news reports saying we're in for a quake-tastic 2018?
<snip>

[sleepy]

Mr. V,
on page 25 you said something to the effect of,"get rid of the lower weight and just thrust the upper weight directly against the wheel."

On page 20 I said,
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.

How can what I said on page 20 be dismissed, and then on page 25 when you repeat what I said,it's part of the magic solution? So,can anyone sim this new approach? It should be much easier than simming the original.

[MrVibrating]

And rushing away on the ascending side!

So far i've only modeled an asymmetric inertial interaction on the descending side, so the ascending side remains unaffected. Both energy and momentum from conventional GPE are a zero-sum deal.

But it's not accelerating toward the weight. In essence, the preloaded force is preventing the earth from accelerating toward that weight. Sorry.

Don't be sorry, i've already considered this anyway, and one miracle's enough. But, call me cynical, that's still far too good to be true, so i still have doubts.. especially the linear-linear version.

Have you considered the implications in that instance?

We're discussing the merits of your concept. Would you care to address the remarks in my 10:49 a.m. post?

If i haven't already, then i will if you start replying properly using quote tags. Any colour you like, long as it's readable.

[MrVibrating]

Mr. V,
on page 25 you said something to the effect of,"get rid of the lower weight and just thrust the upper weight directly against the wheel."

On page 20 I said,
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.

How can what I said on page 20 be dismissed, and then on page 25 when you repeat what I said,it's part of the magic solution? So,can anyone sim this new approach? It should be much easier than simming the original.

I'm sorry i missed that the first time, the mental image didn't sink in.

I seriously do not want to be the arbiter of what constitutes 'the magic'.

Either you understand the gain principle or you don't. The maths follow a causal sequence of conditions that result in alternate yet concurrent energy values of momentum within the same system, so able to be applied as source and sink.

If you do not see that, then you obviously can't take anyone else's word on such a claim, and should reject it.

Equally, if you do see it, then please correct me where i'm wrong cos i'm as new to this as you are.

But yes, we have an interesting situation with the inertially-decelerated gravitating weight.

I think i've misinterpreted it multiple times already in this thread...

Is it outputting GPE if it's not accelerating? If it's still getting lower in the field (ie. carrying downwards ambient velocity but not accelerating), then i would expect it has nonetheless fulfilled the key criteria for a GPE to have performed work (it has to get lower in the field, so for instance we cannot extract work from lowering the point of application of a suspended weight, rather than the weight itself).

And yet, in the initial interaction from a standing start, all the same forces and workloads are there, as when it's already rotating... except there's no change in GPE during the first stroke... At least, in the angular-angular or angular-linear case.

So surely for consistency, if no work is output by GPE on the first stroke, then it cannot be contributing anything on subsequent strokes, either..? Even though it is descending in the field..!?

I ran through an angular-linear example in the last analysis. All input work was taken as internally-applied, with zero borrowed from or repayed to gravity.

And everything seemed to check out - it all squared up without any apparent ambiguities. From my end, anyway.

So there seems to be this paradoxical implication that a weight can descend without performing work..?

There has to be a way to resolve this?

[MrVibrating]

@eccentrically1

We must be in different time zones, so i don't know the post you mean or which question i've left unanswered.. just ask it again, whassup?

[eccentrically1]

Edit: New quote boxes!

This one. The entire post was quoted. That saves on having a separate quote box for each reply in the post. It's done occasionally here.

Edit - My responses to MrV are in blue...

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 is the magic step/mechanism , no one has found a way around it. Everyone's design stalls as this point. Why would a preloaded force (spring, jack, etc) between the upper mass and the wheel all of a sudden work now?

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.

This is the reactionless acceleration you've talked about, but the reaction force for the acceleration is hiding there in plain sight- Earth's gravity! Similarly, the reaction force for Cp acceleration is inertial.

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.

That's not the way you described it earlier or the way I see it. The preloaded force is divided into a positive momentum (the wheel) and a zero momentum {the upper mass providing the inertial (fictitious) force}.

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.

Yes, the momentum is preloaded in the mech for startup. No one has been able to design a mech that internally produces that needed extra momentum to keep things going.

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.

The momentum is asymmetrically distributed, but not increased. Lots of interactions have asymmetric results, but in the final analysis, all of the different system properties (different types of energies, linear and angular momentums, different types of forces,) sum to zero. If they don't, we haven't included all of the contributing properties in that final analysis and our results are skewed.

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.

We haven't repeated a cycle yet, or I missed that page. If I didn't, I'd like to see you (or anyone with 2d!) simulate a design, the simpler the better. Just a 1G preloaded spring (not a massless spring please) with a 1 kg mass on one end, one mech on each side for balance, or whatever the force/mass match needs to be for the concept to work since the wheel and balancing mech will also have mass and some startup friction, attached to the wheel spokes, turn on air resistance, fire it off, and see if it gains speed, even after the opposing mech is fired. Did I miss that?

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.

The input energy is GPE and elastic potential, of external origin to the wheel system. The springs didn't load themselves, or attach their selves up on the wheel spokes. When you include the wheel environment, the input and output are thermodynamically coupled.

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.

Well, I agree the green disc on page 6 is useless.

The mech on page 8 is a small start, it seems to have fallen by the wayside.
Looking forward to any results from your concept, from anyone.

[MrVibrating]

OK i remember writing something before along the lines "the GPE is transmitted to the wheel instead".. or at least thinking it... But if that's not true from a standing start then how can it be true while rotating?

These seem like two mutually-contradictory conditions - we cannot have extracted work from a GPE that hasn't changed height, right?

Equally, if it has changed height, then surely work has been performed?

So this seems to be where some concept of "negative GPE" would creep in...

Because we would be extracting work from GPE if the planet was accelerating towards the suspended mass, regardless of whether it was actually descending or not!

It goes without saying this whole 'negative GPE' thing sounds like a bad sci-fi plot line...

..yet it would seem to be the simple solution to the above paradox..

[MrVibrating]

@eccentrically1

How is anyone supposed to wade through all your missing quote tags!

It'd take half an hour just to re-format so i didn't end up trolling myself (you wanna see a real flame war?)

Please just ask your questions in the normal format - as much so i notice them in the first place, as to cut the unnecessary work involved in answering them. I didn't ignore your questions previously, i just didn't notice them due to the absence of normal quote-formatting.

[eccentrically1]

There you go. Is that better?

[MrVibrating]

Thanks.

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

This is the magic step/mechanism , no one has found a way around it. Everyone's design stalls as this point. Why would a preloaded force (spring, jack, etc) between the upper mass and the wheel all of a sudden work now?

All this step depends upon is Newton's 3rd law. One inertia is moving faster than the other. If they're angular then they're always side-by-side, so just brake them against one another. It's a passive step, not active.

This is the reactionless acceleration, but the reaction force for the acceleration is hiding there in plain sight- Earth's gravity! Similarly, the reaction force for Cp acceleration is inertial.

Well yes, the whole concept is basically using gravity's effective 1 g downwards acceleration to cancel the counter-momentum of a vertical 2 G mutual acceleration, all of which is commuted to the wheel as one sign of momentum only.

I suspect we should be expecting a stray vertical component of work here, if only because non-polluting free-energy from nowhere sounds like giddy naivete. But until we have testable designs, we might as well be panicking about inadvertently summoning Imperial Star Destroyers into low-earth orbit. Hopefully simulation will flag any such risks. Maybe hi-res scale tests, too, tho probably at sensitivities beyond our means.

That's not the way you described it earlier or the way I see it. The preloaded force is divided into a positive momentum (the wheel) and a zero momentum {the upper mass providing the inertial (fictitious) force}.

It seems a valid-enough point, yes. But the net effect is that a force was applied between these two inertias, whilst only accelerating one of them. The resulting single-signed, and thus non-self-cancelling, momentum, is the baby in the bathwater..

Yes, the momentum is preloaded in the mech for startup. No one has been able to design a mech that internally produces that needed extra momentum.

PE is preloaded for startup in the "2-stroke" implementation, in which we begin at the bottom rung of the (e^2 / P) ladder. The first three interactions effectively destroy energy, in exactly the same manner it is later created. Hence some kind of bump-start is required to get it up past 5 'reactionless' momentum rises.

But if you just look at the energy tables - the 'ladder', in progressive +25% steps - it becomes clear that we can jump straight to a 125% gain in a single bound by unleashing that same amount of PE upon every single cycle. This is what i've dubbed the "4-stroke" approach - so the PE would need to be repaid immediately, but at least ample supply is available from the get-go.

Obviously, the gain takes the form of the speed-invariant cost of momentum bought this way, versus its usual value squaring with rising velocity, so we're probably best off looking to harvest the KE gains via CF workloads. No work has been done on this yet, aside from establishing that the discount is actually there and viable. The "gain" however is not an excess of energy per se, only the usual value of 1/2mV^2 or 1/2MoI*RPM^2 etc. The advantage is the discount cost of purchase of the momentum that becomes that RKE.

The momentum is asymmetrically distributed, but not increased. Lots of interactions have asymmetric results, but in the final analysis, all of the different system properties (different types of energies, linear and angular momentums, different types of forces,) sum to zero. If they don't, we haven't included all of the contributing properties in that final analysis and our results are skewed.

Absolutely, i salute your intelligence, this is a point that will undoubtedly be lost on most folks. If this symmetry break between apparent input and output energy fields pans out the way it seems to be going, then by definition we do not have a closed thermodynamic system, period.

What fields exactly it remains in open-circuit with is arguably a more important concern then who ends up with the biggest pile of candy..

Remember Leibniz and Wolffe's conclusions - both effectively seemed resigned to the same implications as us - basically, some kind of OU, but evidence of creation ex nihilo is intrinsically impossible; just because we haven't identified the source doesn't mean it ain't there, and ignorance of the law is no defence.

IMHO we absolutely should not be trusting our testosterone-fueled monkey-brains with the potential risks here, if this scheme checks out. But advocating we go into lock-down mode over a sci-fi fantasy fueled by a stupid mistake isn't gonna win me much credibility either.. most here prolly think i've gone off a cliff already. And they're probably right.

Either way, N3 is immutable, because mass constancy. A genuine N3 break remains sci-fi.

We haven't repeated a cycle yet, or I missed that page. If I didn't, I'd like to see you (or anyone with 2d!) simulate a design, the simpler the better. Just a 1G preloaded spring (not a massless spring please) with a 1 kg mass on one end, one mech on each side for balance, or whatever the force/mass match needs to be for the concept to work since the wheel and balancing mech will also have mass and some startup friction, attached to the wheel spokes, turn on air resistance, fire it off, and see if it gains speed, even after the opposing mech is fired. Did I miss that?

Not yet, unfortunately. We'll get there, if this don't go pop first.

So far, the best we have is predicted and simmed first interaction from stationary, ending with a net 9.81 kg-m/s vectored downwards, and the upper mass stationary.

I then just notionally apply the mutual deceleration, taking half that lower mass's momentum - 4.905 - and applying it to the upper mass, so each now have a uniform 4.905 m/s velocity, and 24.04 J net KE - 25% - of the 96.23 J paid for that momentum.

I've then performed a second set of predictions and tests, giving the identical system an initial downwards velocity of 10 m/s (rather than just the 4.9 left over from the preceding run). The objective of this test was to ensure that the same 96.23 J applied internally bought the same 9.81 kg-m/s after all KE and momentum from GPE is notionally 're-lifted' (ie. i just calculated and subtracted it). This too was positive.

Then i did a third run, as before, but this time giving the system an initial downwards velocity equal to the predicted result of four successive 'reactionless' acceleration cycles, so jumping straight to 125% unity, instead of just 25% or 50%. This too was a positive outcome.

So for now it's all step-wise solutions, but everything seems to dove-tail without conflict in principle.

Hopefully we may come up with a simple interaction demonstrating the viability of using this KE 'gain' (remember, it's only subjectively a gain, relative to the reduced internal cost) to actually reload a spring or something... that would seem a reasonable intermediary step between here and a full wheel design. Give it a few weeks, see what happens i guess..

The input energy is GPE and elastic potential, of external origin to the wheel system. The springs didn't load themselves, or attach their selves up on the wheel spokes. When you include the wheel environment, the input and output are thermodynamically coupled.

You misunderstand, i'm talking about the respective dimensions of the input and output energy terms.

Input energy has the form (e^2 / P). Or equivalently, (F/m^2 J / kg-m/s). This is a linear function, and constant with respect to speed.

Output energy has the regular form 1/2MoI*RPM^2, so squares with angular velocity.

The symmetry break - the apparently-free-energy-gradient - is between these two incompatible scaling dimensions of the input vs output energy terms.

That is the exploit, and the foundation of our business model. Farming RKE by accumulating momentum at constant (speed-invariant) efficiency.

Well, I agree the green disc on page 6 is useless.

The mech on page 8 is a small start, it seems to have fallen by the wayside.
Looking forward to any results from your concept, from anyone.

Answered in detail already.

[eccentrically1]

Hopefully we may come up with a simple interaction demonstrating the viability of using this KE 'gain' (remember, it's only subjectively a gain, relative to the reduced internal cost) to actually reload a spring or something... that would seem a reasonable intermediary step between here and a full wheel design. Give it a few weeks, see what happens i guess..

That would seem reasonable...

You misunderstand, i'm talking about the respective dimensions of the input and output energy terms.

Input energy has the form (e^2 / P). Or equivalently, (F/m^2 J / kg-m/s). This is a linear function, and constant with respect to speed.

Output energy has the regular form 1/2MoI*RPM^2, so squares with angular velocity.

This is where you're approaching the cliff, for me. Those values for input energy don't ring any bells.

The symmetry break - the apparently-free-energy-gradient - is between these two incompatible scaling dimensions of the input vs output energy terms.

And 'two incompatible scaling dimensions of I-O energy terms' is another one. If over unity depends on anything, it would be more energy out than in.
Is that what you're saying?

[MrVibrating]

I'm sorry mate but this is great, if you don't understand what i'm saying then i am talking word-salad, no matter how painstakingly accurate i'm trying to be.. if the information content's basically nil then it's no answer.

So... let's try clarify precisely what the hell i'm trying to convey:

The standard kinetic energy term is KE = 1/2mV^2. The rotational version's RKE = 1/2 angular inertia * RPM^2.

Both are equivalent in their respective domains.

In either case, since both are a function of inertia and velocity, both also have corresponding momentums, depending on the particular distributions of inertia and velocity of the system being measured. For instance a 1 kg mass at 1 meter / sec has 1 kg-m/s of momentum, and 1/2 a Joule of KE. We know it's half a Joule because KE=1/2mV^2, so 1/2 * 1 kg * 1 m/s^2 = 1/2 J.

So half a Joule is enough energy to accelerate 1 kg by 1 m/s, and so raising 1 kg-m/s of momentum.

Or at least, it was on that occasion.

But per 1/2mV^2, if we want to add a second kg-m/s of momentum, the price has gone up fourfold to 2 J, since 1/2 * 1 kg * 2 m/s^2 = 2 J.

If we want a third serving, just exactly the same amount of momentum we're buying each time, 1 kg-m/s, then the cost is 1/2 * 1 kg * 3 m/s^2 = 4.5 J.

And so on, the unit energy cost of momentum rising as a function of V^2, per KE=1/2mV^2.

By the time we get up to, say, 50 meters / sec (just to illustrate the point), the 1 kg mass has 1250 Joules. A 51st kg-m/s would take us up to 1300.5 J of KE. Obviously, the difference between 1300.5 - 1250 = 50.5 J. That's how much the cost of purchase of 1 kg-m/s of momentum has grown, due to the accumulating speed.

So the first kg-m/s cost 1/2 a Joule, but the 51st cost 50.5 Joules... same quantity of momentum each time, 100x the energy cost.

So, if we plot input energy as a function of velocity, we get a steepening curve, as the same quantity of momentum costs ever-more energy to buy, the more we have accumulated already.

This rising cost is due to the practicalities of Newton's 3rd law, but i don't want to distract you too much with that for now (it'll become clear in time anyway if you think it over).

What concerns us is this - what if we could somehow maintain that initial, introductory offer of 1/2 J per kg-m/s... and just maintain it, indefinitely, irrespective of accumulating momentum and thus rising velocity?

For instance, say 200 kg-m/s of momentum would only cost us precisely half that in Joules - so 100 J for 200 kg-m/s.

Now, if all that momentum were embodied on a single 1 kg mass, it'd be traveling at 200 m/s, and so have a standard KE of 1/2 * 1 kg * 200 m/s^2 = 20,000 Joules.

But we only paid 100 J for it, in 200 individual 1/2 J/kg-m/s purchases.


Suppose we were to plot the input energy required to keep buying at that rate from a standing start, as a function of rising velocity. So plot input energy on the vertical axis, and velocity on the horizontal. As velocity rises, per-cycle input energy stays the same, hence we get a straight diagonal line that just maintains that angle up to infinite speed.

But if we overlaid that with another plot in another colour for KE as a function of speed, that steep-curving line will punch through the previous one and quickly diverge upwards ballistically..

The area under the first curve is our input energy, and the area under the latter curve is our output energy.

I can't offer you 1/2 J / kg-m/s, but the rate i have got is still a guaranteed earner - 96.23 J per 9.81 kg-m/s.

It might not seem such a bargain at first - the initial 9.81 kg-m/s is only worth 48.11 J according to 1/2mV^2, so 96.23 J is almost twice the going rate...

I mean, if you just went with the regular base rate, that 96.23 J on a 1 kg mass could buy you 13.87 kg-m/s, straight off the bat, no messing around.

Obviously tho, if you then wanted another 13.87 m/s... yeah, no - that's gonna set you back another 288.69 J, because per 1/2mV^2, at 2 * 13.87 = 27.74 m/s, 1 kg has 384.92 J. So minus the 96.23 already spent, that's another 288.69 J we have to pay, for the same amount of momentum we bought first time round. The cost went up from 96.23 J for the first 13.87 m/s to 288.69 J for the second - a threefold increase.

Screw that.

So let's stick with my special rate, and see how it pans out. The only condition is, you have to use a pair of masses. The deal's only valid for couples.

So we make our first purchase. 96.23 J for 9.81 kg-m/s. Divide that momentum between our two 1 kg masses, so now they have 4.905 kg-m/s each, and thus also 4.905 m/s each. At 4.905 m/s 1 kg has 12.02 J, per 1/2mV^2.

We spent 96.23 J, we now only have 12.02 J on each mass, so 24.04 J KE total. 96.23 / 24.04 = 4, so we only have 1/4 of our initial input energy remaining.

So at first, this seems like a far worse deal than 1/2mV^2. Sure, it quickly turns into a ripoff, but at least it wasn't destroying our input energy. Plus it bought us more momentum!

Round one to 1/2mV^2 then.

But stick with this, let's buy another 9.81 kg-m/s, for another 96.23 J.

So now we've spent 2 * 96.23 J = 192.46 J, for which we've bought 2 * 9.81 = 19.62 kg-m/s.

Dividing that momentum between our two 1 kg masses leaves 9.81 m/s on each, with a 1/2mV^2 value of 48.11 J on each, so 96.22 total (it's really .23 but we let in a .01 rounding error).

So we have precisely half our input energy left after two purchases at this special rate.

So i can probably skip the third if you'll accept that i've done this calc so many times now - we end with 75% of our initial input energy.

Because it's increasing 25% with each purchase, after we get to our fourth round, we're in a perfect tie with 1/2mV^2... they're neck and neck here.

This is the point on the graph plot where the steepening 1/2mV^2 curve punches through the straight diagonal 96.23 J / 9.81 kg-m/s line. From hereon, there's clearly going to be more area under the steepening curve, and moreso the higher we plot the accumulating velocity...

And so from thereon, we surpass unity, and a fifth identical investment of 96.23 Joules of energy in 9.81 kilogram-meters-per-second of momentum takes us to a final energy that is 5/4 more than we started with.

Five times 25 = 125.
Six * 25 = 150.
7 * 25 = 175.
8 * 25 = 200.

These are all over-unity percentages we're talking.

We're over unity, simply because the standard value of KE squares with rising velocity... whereas here, we've found a super-secret special fixed-rate, for special people like us, which we can maintain at any velocity. No matter how much momentum we accumulate, the next 9.81 kg-m/s costs the same flat-rate 96.23 J as all preceding ones.

So, if you can digest this differential between the cost and value of momentum between our special input rate and regular unadulterated output rate, and integrate that with your good grasp of the general mechanical and gravitational principles, then we're basically on the same page.

You can also see though how all conservation laws are met during the exploit - if you're seeing any magic then just ask.

So again you can integrate this with your good grasp of CoM and CoE in the bigger picture..

I say "negative GPE", but obviously that sounds like pure pseudoscience gobbledygook. I mean this daft notion of an unreciprocated fall of the planet towards the descending mass, of course. Sounds dumb, but it's all i've got for now..



ETA: oh and (e^2 / P) is just another way of saying "96.23 J per 9.81 kg-m/s".

since 96.23 = 9.81 * 9.81

we could write:

9.81^2 J / 9.81 kg-m/s

Obviously it's no coincidence that we have the same number on each side, and furthermore, we instantly recognise this number as the value of gravity's acceleration.

Hence we can now generalise the term to apply to any substitute for gravity in this scenario - whatever its acceleration value, the proportions N^2 J / N kg-m/s are evidently constant.

Since F=mA, inversely, A=F/m. Since N clearly derives from gravity's acceleration, we can substitute them:

(F/m^2) / (F/m kg-m/s)

This gives a more readable, if longer expression, but saying precisely the same thing.

Or equivalently, we could just completely ignore the fact that the same number arises on each side of the term, and just express the field and dimensions:

(e^2 / P)

Where e^2 is for energy squared, and P is for momentum, P=mV.

This looks like the simplest shorthand expression for now. However it's written, i've taken to thinking of it as 'Bessler's constant'.

What it is, is a new, parallel going-rate of momentum, running alongside the regular 1/2mV^2 rate, basically. We can trade between the two different exchange rates, apparently freely.

Buy low, sell high.

Source & sink.

Collect underpants & profit.

[Grimer]

Do you read your Private Messages, MrVibrating?

[MrVibrating]

Soz mate, never check 'em, just answered now tho!