Decoupling Per-Cycle Momemtum Yields From RPM

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MrVibrating
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Post by MrVibrating »

..the obvious issue - which i kind of saw this morning, but in that foggy morning way - is that as the weight radius increases, so does the negative G-time, as it goes back up.. it'll be somewhat accelerated by the momentum dumped from the second rotor via the brakes, so negative G-time will always be less than positive G-time per-cycle..

..however if p/c momentum yields from the s&b cycles are constant, but negative G-time p/c is inevitably increasing, then surely the assumption has to be that this is where unity is going to be enforced - that the amount of momentum necessarily lost to gravity by swinging the weight back upwards at ever-greater radius will eat into those 'constant' p/c yields such that what's left is diminishing by the inverse square of RPM regardless..

Dunno... if i knew diff. calc this could prolly be solved on the back of a fag packet.. is this just another hiding to nowhere?
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Post by MrVibrating »

I see the mistake now - increasing radius doesn't increase 'G-time', since all points on a radial line are still rotating at the same speed; it'd simply be raising the GPE load, which would allow a constant per-cycle momentum yield, but at cost of increasing input GPE with RPM, and thus the 'OU efficiency' of the motor's workload would remain equal to the net input GPE / forsaken output GKE.


Thus it still seems intrinsically impossible to raise G-time to compensate RPM..
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Post by MrVibrating »

Update.

Potentially, quite a significant one..

What if spin and brake phases were in different planes / axes?

For instance consider a tubular weight on a vertical axis; you could thrust this outwards horizontally from a wheels' axis into an OB position using a linear actuator or pulley system, and this same action could pull a ripcord, spinning up the weight as it moves outwards..

..just to throw some numbers on it, suppose the weight has an MoI of '1' in that vertical axis, and is spun up to 1 rad/s, so 1 kg-m²-rad/s of angular momentum, for ½ Joule of work..

The counter-momentum from that spin-up is earthed via the wheels support posts - if the spin-up was clockwise then the room + world was torqued counter-clockwise (in the horizontal plane).

Now, in the extended / OB position, with the weight still spinning in the vertical axis, rotate its axis 90° in the radial plane, from vertical into a horizontal axis; it's now in the same axis as the wheel / net system, so, brake it, feeding that fresh 1 kg-m²-rad/s of prime angular momentum straight into the system axis.

This last vertical to horizontal rotation also applies another counter-torque to Earth, this time in the 'roll' axis, relative to the ground.


And that's it; just retract the now-stopped weight back into the center, rewinding the ripcord on the way in, and just keep repeating that cycle; generating the momentum in the orthogonal axis and then rotating it and dumping it into the system axis.

Each cycle thus raises the net system momentum by 1 kg-m²-rad/s, for ½ J / cycle, irrespective of the rising RPM.

And that's basic over-unity, since the KE value of those 1 kg-m²-rad/s momenta is squaring with RPM.

Use the gain to pay for the OB system and spin-ups.



The only snag i can see is that we currently only have two dimensions of space, and one of time. What we need is, almost quite literally, some kind of '3rd' dimension, if you will, in addition to the familiar x & y planes, in which to mindlessly fly-tip our unwanted-but-permanently-enduring counter-momenta, in such a way that we can at least forget about it temporarily, or just, long enough to gain some small recognition for having appeared to have done something actually useful for once, even though it'll obviously be dwarfed by and long forgotten in the inevitable and personally-devastating blowback.

Aside from that, however, it looks pretty foolproof..?

I need to investigate possibilities for some kind of 3D testing environment, but it has to be cheap or free.. either that, or just resort to mathematical proofs... but you can already see the maths - 10 cycles at half a Joule each would cost 5 J, but if the system axis was also say 1 kg-m² (which in principle is physically feasible) then it would be accelerated to 10 rad/s and 50 J of KE, so 10x OU.

Incredibly simple exploit, no? Am i overlooking something?
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

..here's a rough graphic of what i'm trying to describe:

Image


So basically the GPE interaction - radial lifts w/ angular drops - is a zero-sum, but the spin'n'brake cycles are OU; the rotational KE value of the angular momentum being dumped into the main system axis quickly outgrows the PE cost of spinning up the weights, once a sufficient number of cycles have elapsed (determined by the sum of the MoI ratio between the weight axes and system axis).

Essentially we're generating angular momentum for fixed unit energy cost per cycle, but escalating KE value.

So long as the work / energy required to keep spinning up the weight doesn't increase with the wheel's RPM - and there's no apparent reason it should - the input energy's rising linearly whilst the output rotKE's squaring with RPM..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Fletcher »

Still digesting.

Thoughts :

Two side by side experiments.

1. a cylinder weight on an ice sheet. Apply a force to its middle to move it horizontally to a certain translational speed in m/s. At that speed apply a rotational force to flip it 90 degrees.

It translates with no rotation (assume zero frictions). The force only has to overcome its inertia. Before the flip it has translational KE in joules which is equal to the force applied x displacement. When at speed more work is done to flip the weight altho its translational velocity is the same. The sum of the Work Done in joules equals the TKE plus a wee bit for the flip.

2. the cylinder runs in a rack and pinion arrangement. Apply the same translational force over the same horizontal distance. The rack and pinion insures the cylinder gains translational velocity plus rotational velocity. IOW's Total KE = TKE + RKE. Then it is flipped as before.

IINM the theory is the translational velocity in case 2. will be less than in case 1.
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Post by MrVibrating »

There's an even simpler way to explain it, sans gravity (since it's irrelevant):

• You're standing, stationary, upon a small turntable, on excellent bearings. In your hands is a hefty gyroscope, spun up by a motor or ripcord or whatever you like.

• If the gyroscope's oriented in the same plane as the turntable, then imparting say 1 kg-m²-rad/s of momentum to it will likewise apply 1 kg-m²-rad/s of momentum to yourself, about that vertical axis.

• If the axis of the still-spinning gyroscope is then rotated 90° into the horizontal plane, this applies a twisting torque to the planet, in the 'roll' axis (this same precessional counter-torque is used by satellites and spacecraft to control their attitude).

• You still have 1 kg-m²-rad/s in your hands, which remains conserved and unaffected by the rotation of its axis from vertical to horizontal. So, now brake the gyro back to a halt, in that horizontal axis; this dumps its angular momentum to earth, in the 'pitch' axis.

• You and the turntable still have the 1 kg-m²-rad/s of counter-angular-momentum in the vertical axis..

• ..and you can keep repeating that cycle, gaining 1 kg-m²-rad/s each cycle, for the same half-joule work / input energy (assuming a 1 kg-m² gyro at 1 rad/s per cycle).


Alternatively, instead of grounding the 'primary' momentum and accumulating their counter-momenta, you could do it the other way round; accelerating the gyro in the vertical plane and decelerating it in the horizontal..

..either way, we're simply generating momentum in one plane, and then simply rotating it into another plane and dumping it there, repeatedly.

The spin-up work / energy in the vertical plane is constant per cycle - the same angular inertia is being accelerated to the same angular velocity each cycle, irrespective of the RPM rise in the horizontal plane.

The work involved in rotating the axis of the gyro / flywheel into an orthogonal plane is also invariant of the rising RPM - the precessional inertia is a function of the gyro's speed about its own axis, not its 'orbital' axis.



And so the proposition is extremely simple: that we might simply spin up some internal momentum in an axis to which the system cannot yield (no freedom of movement in that plane, unless one end of the axle were able to rise up off its post), and then fully-conservatively rotate the axis of that internal momentum into alignment with the system axis, before braking it back to stationary relative to the wheel / net system axis, thus emptying its momentum into the main system axis.

And that's it - just basically shovelling momentum from pile 'A' into pile 'B' - its shovelling-cost per bucket-load from pile 'A' is constant, but its KE value to the main system axis is squaring with RPM.


Because the exploit isn't dependent on gravity * time in the first place for its momentum yield, it's RPM-invariant - the same input energy / work buys the same quantity of angular momentum each cycle, for an ever-escalating KE value on the main system axis: just consider that the net MoI of the main system axis is constant over each full cycle (not progressively increasing or decreasing with RPM), and the same amount of angular momentum is being dumped into it each cycle; thus the main system axis undergoes equal angular acceleration from each cycle, always rising by say 1 rad/s assuming the main axis were also 1 kg-m², or 0.5 rad/s if it were 2 kg-m² etc. The KE value of a succession of equal but accumulating accelerations squares with rising net velocity... even though the tally of input work / expended PE is increasing linearly over successive cycles.

Likewise, if the axial to orbital MoI ratio were, say, 1:4, then each 1 kg-m²-rad/s spin-and-brake cycle would cause a 0.25 rad/s acceleration of the net system axis; 4 such cycles would have a net cost of 4 * ½ J = 2 J, and a KE value (from a standing start) of [4 * 0.25 rad/s accelerations of 4 kg-m²] = 1 rad/s final velocity and thus also 2 J, but a fifth such cycle takes the net input energy up to 2.5 J, but the net output energy is 4 kg-m² @ 1.25 rad/s = 3.125 J.

And it just gets even-more over-unitier the faster it gets - after a sixth cycle the main system axis of 4 kg-m² is at 1.5 rad/s, for a net energy cost of 3 J, but a total KE of 4.5 J.


After just ten such cycles, we'll've spent 10 * 0.5 J = 5 Joules, but we'll have 4 kg-m² at 2.5 rad/s, so 12.5 J of rotational KE. But 'ten cycles' is just an arbitrary waypoint - it just gets even more efficient, the more momentum is banked this way.


Honestly, i've just spent an hour in the bath playing with a little gyroscope and trying to work out what i might have overlooked, but i can't yet see it; all it seems to rely on is conservation of angular momentum - on the spun-up momentum being conserved through the rotation of its axis into the orthogonal plane. And that's about it! It appears to be an inherently-OU process of generating momentum in one plane and then simply rotating and accumulating it into another..

Wish there was a 3D version of WM - Autocad looks daunting, if it can even produce reliable precessional torque * angle data in the first place. A simple test rig might suffice for testing the key principles and assumption, but the efficiency measurements are going to be fully manual..

..if the prospects do pan out then a simple motorised build would seem in order.. pickup coils on the main axis charging the cells driving the spin-up motor, and just dissipating the braking losses (we're mechanically OU even without 'em)..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Fletcher wrote:Still digesting.

Thoughts :

Two side by side experiments.

1. a cylinder weight on an ice sheet. Apply a force to its middle to move it horizontally to a certain translational speed in m/s. At that speed apply a rotational force to flip it 90 degrees.

It translates with no rotation (assume zero frictions). The force only has to overcome its inertia. Before the flip it has translational KE in joules which is equal to the force applied x displacement. When at speed more work is done to flip the weight altho its translational velocity is the same. The sum of the Work Done in joules equals the TKE plus a wee bit for the flip.

2. the cylinder runs in a rack and pinion arrangement. Apply the same translational force over the same horizontal distance. The rack and pinion insures the cylinder gains translational velocity plus rotational velocity. IOW's Total KE = TKE + RKE. Then it is flipped as before.

IINM the theory is the translational velocity in case 2. will be less than in case 1.
I initially over-complicated it with the GPE interaction, which is irrelevant - no translational motion required; just rotating the spin axis 90° into another plane, and braking it there.. if the system has freedom of rotation in one plane but not the other then we accumulate momentum or counter-momentum, depending on whether the 'spin' or 'brake' axis is free or grounded..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Fletcher »

..if the prospects do pan out then a simple motorised build would seem in order.. pickup coils on the main axis charging the cells driving the spin-up motor, and just dissipating the braking losses (we're mechanically OU even without 'em)..
Sounds like you got a POP to build :7)
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Tarsier79 »

I tried something sort of relative, using a change in inertia to transfer energy during braking, but could not get a complete transference.

I suspect, even if you can convert all the rotational KE to the wheel, you will be doing just that. Transferring energy, not speed.

I hope I am wrong though. All the best!
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Post by MrVibrating »

I think i found the potential answer:

• suppose you're stood on the turntable, holding the static gyro's axis horizontally

• spinning it up in that plane grounds the counter-momentum, all well and good

• but when we then rotate its axis into the vertical plane - before we have a chance to brake it - does it not apply counter-momentum to the turntable during that rotation of its axis?

So in other words, does the act itself of rotating the spinning gyro's axis into the same plane as the turntable, also cause the turntable to start rotating.. in the opposite direction? Thus, when we brake the gyro, instead of imparting its momentum to the turntable, it just cancels out the momentum induced by rotating its axis?


So the problem boils down to a very simple question:

• Could you, in principle, add momentum to a kid's carousel / merry-go-round (allowing for nearo-zero friction), using an onboard gimballed gyro / flywheel with a motor and brake? Spin up in one plane, brake in the other..


Thus the question of whether it can be accumulated at OU efficiency is ultimately the same question of whether the very first cycle #1 will actually impart any net momentum. If that does, then all subsequent ones should too and the maffs of OU should kick in..

..but if OTOH we impart angular momentum to the turntable by the very act of changing the gyro's plane of orientation - quite aside from the counter-momenta caused by changing the gyro's speed - then the presumption has to be that all momenta cancel out over a full cycle.

So, you're stood on a turntable, with an already-spinning gyro in your hands, with its axis held horizontally. You then lower one arm while raising the other, tilting the gyro's axis into the vertical plane - but without touching its speed: Do you thus start rotating on the turntable, in the opposite direction to the gyro?

Cos if the answer's 'yes', then there's my error.

If 'no' then we may have a winner..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Fletcher »

You better get out of the bath and head down to the local playground and kick the kids off the carousel ;7)
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Post by MrVibrating »

LOL no i'd go in the middle of the night, obviously.. be weird otherwise.


Still - if you apply counter-momentum to earth in spinning up the gyro horizontally, but then also induce a second counter-momenta to yourself on the turntable when rotating the gyro into the upright position, then how's that conserving net momentum? I mean, presuming that this 2nd counter-momenta was indeed cancelled out by braking the gyro - thus scuppering my plans - what of the 1st counter-momenta induced to earth? That can't be right eh?

So, what if that 2nd momenta induced to the platform is actually just the first one, but also rotated along with the gyro's axis? IE. so when we spin it up we induce a counter momentum to earth, but then when we rotate the axis into vertical alignment, this action, at the same time, is also rotating the counter-momenta into the other plane? But then again you'd run into the same paradox; the initial counter-momenta was induced to earth, not the turntable, hence if rotating the gyro's axis also rotated the plane of that counter-momentum then it would still be embodied on the earth.. it can't 'leap' into the turntable's axis, not least because it's on low-friction bearings..

TL;DR - if pitching up the spinning gyro's axis into the vertical plane - the same plane as the turntable you're standing on - but without changing the gyro's speed nonetheless induces a counter-momentum to yourself about the turntable's axis, in the opposite direction to the gyro and thus perfectly cancelling its momentum when braked, then we'd seem to be invoking a 2nd counter-momentum from nowhere, whilst discarding the counter-momentum already induced to earth when we spun up the gyro. It would seem to violate CoM.


So from that rationale alone, we might assume that no, rotating the gyro's axis couldn't provoke a 2nd and perfectly-cancelling counter-momentum on the turntable's axis, because that would leave the original spin-up counter-momentum orphaned; each full cycle would then accelerate the planet a little bit.. but where did that momentum come from but the fact that we specualtively invoked it to cancel out a KE gain downstream? We're not even playing with the gravity * time equation here, it's just basic inertial interactions, so all momenta should cancel out, right?


Or maybe the 2nd counter-momentum induced to the turntable by rotating the gyro's axis came at the expense of the gyro - ie. does it get decelerated, despite the absence of friction, by the rotation into another plane, thus donating half its share of momentum to the turntable's axis?

Just getting even more convoluted trying to explain it away.. i think Occam would agree the turntable won't be accelerated until the gyro's braked in the same axis.. and thus that the exploit seems practical..

Give it a few days - i'll either come to my senses or work out a simple test rig; i have gyros and motors etc., expanded foam for to make a floating platform, and a bath, so all i need a way to rotate the axis of a spinny thing that's sat on a floaty thing to check if this spins up the float - if so then does it also spin-down the gyro in direct proportion such that all motion ceases when the gyro's braked against the float while aligned in the same axis..

..and if not, does the per-cycle momentum gain stay constant with rising system RPM - ie. do we see the same amount of acceleration of the main axis each cycle at say 1 rad/s as from a standing start?

There's prolly an Eric Braithwaite video out there that'd answer this, give it a few days to stew on eh..
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Post by MrVibrating »

..sorry but this is doing my nut in - suppose we're floating in space and i hand you a gyroscope that's already been spun up; simply taking possession of it in your hand isn't going to cause you to suddenly start rotating, right?


Now suppose you just change the angle of its axis by 90°, from parallel to perpendicular to your body; this will also cause you to rotate about the gyroscope slightly, but only while you're rotating the gyro's axis - as soon as it stops in the new alignment, so does your motion... but you certainly wouldn't expect to start counter-spinning relative to the giro itself, surely?

Again, this simple thought experiment seems to agree with the hypothesis that the main axis will only be accelerated by the braking counter-momenta, not the rotation of the gyro's axis, thus providing constant per-cycle momentum yields at linearly-accumulating cost and squaring benefit..
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Post by MrVibrating »

A polystyrene disc, and a small motor running off a small battery; the motor's intrinsic MoI is 'the gyro', and the battery ensures it runs at constant speed..

..so all i need is to sit the battery-powered motor on the floating foam disc, switch it on whilst horizontal to ground the spin-up counter-momentum, then let go and allow the running motor's axis to tip or tilt from horizontal into vertical alignment..

Maybe the motor's taped to a horizontal beam that hinges at one end, with the other end suspended by a thread that i can burn thru with a lighter without imparting any torques.. so the motor swings down under gravity into vertical alignment, and either the float does or does not spin up in response.. if not, win, if so it'll need unbaffling all the same..
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Post by MrVibrating »

Might've just busted it - were the hypothesis correct, then surely by the same token it'd be possible to spin up the gyro in the aligned plane, thus accelerating the turntable with counter-momentum.. and then simply flip the axis of the gyro a full 180°, swapping 'heads' for 'tails; now both primary and counter-momenta would be vectored in the same direction of spin, a 100% yield you could repeat every cycle.. and OU would've been discovered centuries ago.

OK, even more centuries ago.

It thus seems logical to presume that rotating the axis of the spinning gyro probably would, somehow, induce counter-rotation to the turntable, such that subsequently braking it in that aligned plane would simply cancel out all motion.


Still needs further investigation tho, since i'm obviously missing something simple..

Thought more about a physical test, and one issue would be that any gyro or motor's going to have some friction, and this will inevitably start imparting momentum to the turntable the instant the axes are aligned - so if the test's sensitive enough (say, using very low-friction bearings rather than a water float), then how to differentiate angular displacement of the base from that gyro / motor bearing friction, from that induced by rotating its axis?

For this reason it might simply be more practical and clear-cut to settle for a better understanding of the theory, rather than an actual experimental result..
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