" 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! >> >>
