It was late and i didn't read it back.. i meant that the roles of both Newton's 3rd law and Lenz's law in shoring up conservation of energy and momentum is often under-appreciated - as evinced by the previous replies to this thread, which prompted me to delurk after so many years..
Interia is velocity independent due to mass constancy - a body's inertia (its resistance to a given acceleration) is constant, and not a function of its current velocity. But kinetic energy IS a function of velocity - it costs more energy to further accelerate a mass, the faster it goes - for a 1 kg mass, the special introductory one-time-only offer is just 1/2 a Joule for the first meter per second of acceleration. Bargain basement. But then each additional m/s of acceleration costs more and more, escalating as half the square of the rising velocity. If our target speed is 100 m/s, then that final m/s of acceleration from 99 m/s costs a whopping 99.5 Joules - almost 200 times the cost of the first meter / sec. So my point was simply that an N3 violation would allow us to maintain that 1/2 J/kg/m/s deal indefinitely - we'd always begin stationary relative to our reaction mass, regardless of the rising system velocity. So our input energy would scale following the same dimensions as momentum (P=MV). Since KE = 1/2 MV^2, P and KE both converge at a value of two anyway (since half two squared is still two) - so in principle their units and dimensions would become identical.. at which point the question of whether we're inputting "energy" or "momentum" becomes entirely academic - if the system momentum gained 1/2 a meter per second per Joule, it would sustain that ratio independently of rising net system velocity. And yes, a Lenzless motor presents no load upon the power supply. To elaborate, in a conventional motor we could monitor the rotor's activity via a scope attached to the power supply - accelerations of the rotor cause back EMFs - if a rotor is pushed one way, the voltage supplying the current is pushed back... equal and opposite reactions, and so active feedback to the power supply. But a Lenzless motor accelerates its rotor without inducing counter-EMFs - so there is no feedback on the power supply. The rotor is pushed one way, but the current and voltage are completely unfazed by its acceleration. Hence the only dissipative workload on such a solenoid would be resistance losses - buy a heater, get a free blender. The free RKE will dissipate to heat and so register in a calorimeter as a gain over and above Joule heating. As for the power of ten thing, as already explained, the baseline minimum cost of a 1 kg/m/s of acceleration is 1/2 a Joule (because KE=1/2MV^2). And an N3 violation would enable us to keep taking that introductory offer, repeatedly - basically re-joining the back of the queue in a ropey disguise. Which would mean we could accelerate a 1 kg mass, say, up to 100 m/s, say, using 100 discrete 1/2 J per 1 kg/m/s thrusts, for a total expenditure of 50 Joules. Yet at 100 m/s, from a stationary reference frame, a 1 kg mass has 5 kJ of energy. 50 J in for 5,000 J out is actually two powers of ten, so my bad. And i also mis-spelled "motherlode"... but there it is. For such an oft-overlooked energy asymmetry, it's a particularly bountiful one.. But EM or mechanical, it's the same fundmanetal asymmetry, creating energy by not conserving momentum. Furthermore the gain in energy is not apportioned from the input energy - all of which is always accounted for, internally, in the form of mass displacements, but rather from its asymmetric distribution (specifically its momentum component) relative to a third (ie. static) reference frame. PS. On reflecton, it seems the powers of ten are just waypoints - at 500 J input we'd have 50,000 J output, so three powers of ten gain margin. The peak efficiency is thus arbitrary and simply equal to the standard divergence between P and KE for their respective linear vs exponential dimensions. On Tue, Feb 9, 2016 at 4:31 AM, Eric Walker <eric.wal...@gmail.com> wrote: > I wrote: > > I followed your presentation, ... >> > > The presentation was a little more opaque than I at first appreciated. > > "As with Newton's 3rd law, many people miss why conservation of energy > should be dependent upon equal and opposite reactions." > > "Inertia is velocity-independent." > > "Essentially, accelerating a mass Lenzlessly would present no load upon > the power supply - only usual resistance losses remain, following Joule's > 2nd law for heat (Q=I^2RT where Q = J and I^2RT is current squared times > resistance times time)... calorimetry would thus show gains" > > "EM or mechanical, the peak efficiency of an N3 break is a power of ten of > whatever's input." > > I take it back. I didn't really understand it. > > Eric > >
