Mate, i haven't made a spurious OU claim in months. A good few weeks anyway. 'Careful' personified, me. ;)
OK i've had some kip & cofee, let's run thru it one more time: • give both axial and orbital MoI's a value of 1 kg-m² • apply 1 N-m of torque for 1 second each cycle • this raises 1 kg-m²-rad/s of momentum in each direction, at ½ J each, 1 J total • rotate the axial axis 90° to perpendicular; the precessional torque is earthed • brake the axial rotor back to stationary, earthing its momentum • rotate the axial axis back to parallel with the orbital axis, and repeat the cycle This last step, of rotating the axial axis back into parallel with the orbital axis, must be where the magic's occurring. To achieve equal MoI's in each axis requires that the 'turntable' is essentially made from aerogel; since most of the surface area of such a disc would be redundant, it might as well be substituted with a carbon fibre rod, mounted to a pivot at one end with the axial rotor and motor mounted at the outer end. Thus the axial rotor constitutes most of the system mass, for both the axial and orbital MoI's. If the 'axial rotor' is considered a 1 kg point mass at 1 meter radius, and the orbital radius is also 1 meter, then the system has ~1 kg-m² of both axial and orbital inertia. So, while an unusual indulgence, this isn't non-physical. I've knocked up a brief sim of such an impulse being applied, here: https://i.ibb.co/PQv8SnJ/Axial-vs-Orbital-3.gif ..this seems to confirm the initial predicates; applying a 1 N-m torque for 1 second is going to cause a 1 kg-m²-rad/s change in momentum in each axis and direction, regardless of the current system speed.. So let's try it when starting with an initial system speed of 2 rad/s: Relative FoR: https://i.ibb.co/PNdhfPr/Axial-vs-Orbital-31.gif Absolute FoR: https://i.ibb.co/bzkyLsF/Axial-vs-Orbital-32.gif ..eek, that seems to solve it, eh? The T*a is speed-invariant, but once already at speed it's simply transferring momentum between axes, decelerating the axial momentum.. ..then, rotating the axial axis perpendicular to the orbital axis and braking it would be earthing momentum of the sign we wanna keep, not discard.. ..and rotating its axis back to parallel, with zero axial speed relative to the orbital axis, will simply transfer orbital momentum back to axial - so if there were 3 kg-m²-rad/s of system momentum remaining prior to realigning the axial axis, there'd be 1.5 kg-m²-rad/s left on each axis afterwards.. TL;DR - after the first cycle, it ceases earthing counter-momentum and begins earthing positive momentum instead. The 'relative vs absolute' FoR's / momenta don't survive rotation into perpendicular planes, and i'm still an idiot.. On Thu, Jun 6, 2019 at 10:44 AM Frank Grimer <[email protected]> wrote: > Ride carefully. 😉 > > On Thu, 6 Jun 2019 at 06:58, Vibrator ! <[email protected]> wrote: > >> " a 50% accumulator? So 2-cycs to unity, 3 to 133%." >> >> eek i meant "3 to 150%", duh, need slepp.. >> >> On Thu, Jun 6, 2019 at 6:55 AM Vibrator ! <[email protected]> wrote: >> >>> ..on 2nd thoughts, isn't it a 50% accumulator? So 2-cycs to unity, 3 to >>> 133%.. >>> >>> And MoI's obviously supposed to be kg/m² (kg-m²-rad/s is momentum). >>> >>> Whatevs. >>> >>> The same input workload buys the same amount of momentum for the same >>> energy each cycle, in spite of rising RPM's, so plotting that flat trace >>> across RPM's, the rotKE is inevitably going to intersect it after n cycles, >>> and keep on climbing.. >>> >>> The no. of cycles to unity appears to be a function of the sum of the >>> MoI ratio, so for 1:1 = 2 cycs, with a 50% per-cycle efficiency >>> accumulator, for 2:1 = 3 cycs @ 33%, 3:1 = 4 cycs @ 25% etc. >>> >>> Suffice to say if real, it ain't dolphin-friendly.. but does it even >>> work? >>> >>> On Thu, Jun 6, 2019 at 6:12 AM Vibrator ! <[email protected]> wrote: >>> >>>> Magic Roundabout >>>> >>>> >>>> You're standing on the edge of a turntable, holding a heavy flywheel in >>>> your hands. >>>> >>>> Beginning with both axes parallel, spin that baby up.. >>>> >>>> ..then rotate its axis 90° into the perpendicular plane. This exerts a >>>> precessional torque, which is earthed through the turntable's rigid axis, >>>> having no effect upon its current balance of momentum.. >>>> >>>> ..now brake away that counter-momentum in your hands, earthing the lot.. >>>> >>>> ..you're now stood on the edge of a rotating turntable, holding a >>>> stationary flywheel.. >>>> >>>> ..flip it back to parallel and repeat the cycle.. >>>> >>>> >>>> Simplifying, assume equal MoI's for both axes (ie. 1 kg-m²-rad/s each). >>>> >>>> Both the per-cycle input torque * angle and the resulting momentum >>>> yield appear to be RPM-invariant; that is, the input energy cost of >>>> momentum appears to be constant / invariant to system speed, whereas its >>>> rotational KE is obviously squaring up.. >>>> >>>> For example, 10 purchases of 1 kg-m²-rad/s at 1 J each costs a total of >>>> 10 J (this includes dissipating half the input energy per cycle). Yet 10 >>>> kg-m²-rad/s divided by two 1 kg-m² MoI's gives them 5 rad/s each, and so >>>> 12.5 J each, 25 J total. >>>> >>>> Using two equal MoI's, we find a 75% net loss after the first cycle, >>>> 50% following the second cycle, 25% at the third.. we hit unity at the >>>> fourth cycle, and 125% of unity at the fifth.. and efficiency keep rising >>>> by that same 25% per cycle as we accumulate ever-more 'unilateral' >>>> momentum, at fixed cost, its KE value squaring with rising velocity.. >>>> >>>> >>>> Only thought this up 24 hrs ago but barely slept since.. where am i >>>> going wrong? It's too simple! >>>> >>>>
