Toad Elevating Moment
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Re: re: Toad Elevating Moment
@rlortie & justsomeone
cheers guys but my quality/quantity ratios are highly non-linear and liable to substantial drift, so if i'm not ranting gibberish yet it probably won't be long.. ;)
cheers guys but my quality/quantity ratios are highly non-linear and liable to substantial drift, so if i'm not ranting gibberish yet it probably won't be long.. ;)
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@Fletcher
BTW thanks for the formulas, i'll need 'em for detailed calcs, but for now i'm still at the 'ready reckoning' stage, and it looks like 'gains' from CF-subsidised lift are offset by losses from non-reciprocal CF workloads - ie. input CF workload = output CF workload, unless we use some of the latter to mitigate our G workload - so what we gain from G we lose to CF, probably...
Obviously any easy discrepancy here would've been found already...
Needs another element in the equation.. a time dependency, counter-balance or something, but who knows..
BTW thanks for the formulas, i'll need 'em for detailed calcs, but for now i'm still at the 'ready reckoning' stage, and it looks like 'gains' from CF-subsidised lift are offset by losses from non-reciprocal CF workloads - ie. input CF workload = output CF workload, unless we use some of the latter to mitigate our G workload - so what we gain from G we lose to CF, probably...
Obviously any easy discrepancy here would've been found already...
Needs another element in the equation.. a time dependency, counter-balance or something, but who knows..
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re: Toad Elevating Moment
No major progress last couple of days, but have been thinking more about potential symmetry breaks...
The most obvious way to try shape an asymmetry is by modulating the values of the forces being used. However if we're just using relative balance to do this, we come up against basic spatial symmetries.
However what if the input and output energies are in different fields, having different dimensions?
So for instance say our input energy was a given mass falling a given height. That's a fixed value per cycle and doesn't vary with speed (at least within mechanical limits).
But for our output energy, suppose we use inertial forces - say CF. Now, the per-cycle output is a function of speed, ie. time, and the faster we spin the more output energy we have relative to our input energy.
Thus even if the input GMH / GPE is higher than that of the initial CF outputs, as speed increases we quickly pass a threshold RPM where I < O. Each mechanism per cycle must be independently over-unity, and sum to sufficient gains to overcome friction in the first cycle.
We can roughly suppose that Bessler's wheels might have lost 10-20% of their efficiency to friction, hence if a single 'crossbar' mechanism suffices to self-loop (albeit barely), then each one must be roughly this much OU.
Hence for 4-8 interactions per cycle the efficiency could be anywhere from 40 to 160% OU per cycle at peak performance.
Of course it is trivial to arrange for unequal energies in either fields; the real trick is in sidestepping (or, perhaps, subverting) these spatial symmetries, allowing a cycle to repeat.
So i'm now looking for ways to vary CF to allow this cycling (varying gravitational weight alone is, by all accounts, a fool's errand). Obviously if CF > G then we can't use input GPE alone to reset. Evidently, then, if Orffyreus was using such an asymmetry (and it seems consistent with the outwards mechanical properties of his wheels) then there is a way to attenuate or 'switch off' CF on the cheap.
Possibilities i'm considering include:
- using the alternating G vector thru a rotation (as described earlier)
- using springs and counter balances (against the CF)
- moving CF loads through the axle and center-of-rotation (this is interesting - doing this for gravitational loads seems a dead end, but i haven't fully considered the case for CF loads)
- flinging (or flailing) a CF workload by pivoting its armature to the hub in the axial plane, and thus extending and retracting the armature off-radially, in the x/y plane
- in another attempt at brainstorming the Apologia wheel, an asymmetric (ie. 1-down, n-up) scissorjack, affixed off the wheel's central axis, whereby the weight attached to the shorter end is n times the mass at the longer end, hence balanced at full extension, but underbalanced upon retraction. This system's interesting because the longer, tail end extends further out and is thus subject to greater CF. Perhaps placing the fixing point 'behind' the axle (as the glyph seems to illustrate) gives us another leverage point to apply torque from (ie. if the jack was fixed to the wheel's exact center, the CF at each end would cancel; small mass in a tight orbit, vs say a 6x smaller mass 6x further out, net force sums to zero). Hence a slightly off-axis fixture point gives us another curve to play with...
More later...
The most obvious way to try shape an asymmetry is by modulating the values of the forces being used. However if we're just using relative balance to do this, we come up against basic spatial symmetries.
However what if the input and output energies are in different fields, having different dimensions?
So for instance say our input energy was a given mass falling a given height. That's a fixed value per cycle and doesn't vary with speed (at least within mechanical limits).
But for our output energy, suppose we use inertial forces - say CF. Now, the per-cycle output is a function of speed, ie. time, and the faster we spin the more output energy we have relative to our input energy.
Thus even if the input GMH / GPE is higher than that of the initial CF outputs, as speed increases we quickly pass a threshold RPM where I < O. Each mechanism per cycle must be independently over-unity, and sum to sufficient gains to overcome friction in the first cycle.
We can roughly suppose that Bessler's wheels might have lost 10-20% of their efficiency to friction, hence if a single 'crossbar' mechanism suffices to self-loop (albeit barely), then each one must be roughly this much OU.
Hence for 4-8 interactions per cycle the efficiency could be anywhere from 40 to 160% OU per cycle at peak performance.
Of course it is trivial to arrange for unequal energies in either fields; the real trick is in sidestepping (or, perhaps, subverting) these spatial symmetries, allowing a cycle to repeat.
So i'm now looking for ways to vary CF to allow this cycling (varying gravitational weight alone is, by all accounts, a fool's errand). Obviously if CF > G then we can't use input GPE alone to reset. Evidently, then, if Orffyreus was using such an asymmetry (and it seems consistent with the outwards mechanical properties of his wheels) then there is a way to attenuate or 'switch off' CF on the cheap.
Possibilities i'm considering include:
- using the alternating G vector thru a rotation (as described earlier)
- using springs and counter balances (against the CF)
- moving CF loads through the axle and center-of-rotation (this is interesting - doing this for gravitational loads seems a dead end, but i haven't fully considered the case for CF loads)
- flinging (or flailing) a CF workload by pivoting its armature to the hub in the axial plane, and thus extending and retracting the armature off-radially, in the x/y plane
- in another attempt at brainstorming the Apologia wheel, an asymmetric (ie. 1-down, n-up) scissorjack, affixed off the wheel's central axis, whereby the weight attached to the shorter end is n times the mass at the longer end, hence balanced at full extension, but underbalanced upon retraction. This system's interesting because the longer, tail end extends further out and is thus subject to greater CF. Perhaps placing the fixing point 'behind' the axle (as the glyph seems to illustrate) gives us another leverage point to apply torque from (ie. if the jack was fixed to the wheel's exact center, the CF at each end would cancel; small mass in a tight orbit, vs say a 6x smaller mass 6x further out, net force sums to zero). Hence a slightly off-axis fixture point gives us another curve to play with...
More later...
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Hmm... just noticed, the Merseburg illustrations can also be read as hieroglyphs (in the pre-rosetta pictographic sense), from left to right, as a gain cycle...
Using the interaction i describe above!
The reason it/they show multiple pendulums is because the instruction is laid out in a left-to-right sequence;
- we start with an input of CF work to a wheel in profile view (although it's only "input work" for the first cycle - it then becomes purely output work, looped back) - the pendulum.. rather than even considering this as 'input work', it's rather a kind of primer, or could be replaced with a starter mechanism
- then we have the 'real' input to the interaction, the stampers, ie. GMH, AKA GPE
- followed by another pendulum 'output', feeding back into another wheel (this one in plan view)
So we have two force deltas, one a function of space, the other of time, forming a closed-loop cycle. This is what it's really intended to illustrate.
As i reasoned previously, the GMH work is close to the axle, and the CF work out by the rim.
Presumably the distance, size/mass relations indicated by the pendulum are worth further scrutiny (ie. sussing their respective forces and energies) - again, the CF workload here is connected via an off-axis pivot in the axial plane, as i suspected from the Apologia wheel.
Also another checkbox in that the smaller weight is much further away, hinting at a Kepler-esque balance, presumably thrown akilter by the off-axis center of rotation.
Also another in that the CF mechanism has a direct-drive transmission to the axle.
Ditto the GMH mechanism (although as noted earlier, these last two details may be superfluous to the central dynamic, which concerns the form of the asymmetry).
Finally i suspect that the diagonal delineated by the rope intersecting the plan wheel perhaps denotes another significant clue - maybe one that would tally with the functioning of the CF mechanism, once that becomes clearer...
Thinks, dudes? Compelling or smelling..? :)
EDIT: ...or maybe there's a suggestion here of two CF loads, in alternate phase? Recall the so-called Nice model of planetary dynamics, wherein larger-mass cycles in a factor-of-two relationship combine their forces as they overlap..? Could a similar effect via CF provide a freely-alternating 'overdrive' force to reset another load otherwise trapped out at the rim by a nominally peak CF force?
Again, this last point concerns resetting, not gaining - we already have our gain (a CF / G work asymmetry is trivial), but we need to reset the CF load against what is otherwise the system's maximum force (NB this isn't quite the same as say magnetic sticky spots, which normally prevent a gain - here we have the gain, but precisely because CF > G, we're now stuck.. unless CF workloads have some kind of get-out clause in the small print..! Presumably, there is, and Orffyreus found it.
Somehow, those Merseburg pendulums describe the CF cycle.
Using the interaction i describe above!
The reason it/they show multiple pendulums is because the instruction is laid out in a left-to-right sequence;
- we start with an input of CF work to a wheel in profile view (although it's only "input work" for the first cycle - it then becomes purely output work, looped back) - the pendulum.. rather than even considering this as 'input work', it's rather a kind of primer, or could be replaced with a starter mechanism
- then we have the 'real' input to the interaction, the stampers, ie. GMH, AKA GPE
- followed by another pendulum 'output', feeding back into another wheel (this one in plan view)
So we have two force deltas, one a function of space, the other of time, forming a closed-loop cycle. This is what it's really intended to illustrate.
As i reasoned previously, the GMH work is close to the axle, and the CF work out by the rim.
Presumably the distance, size/mass relations indicated by the pendulum are worth further scrutiny (ie. sussing their respective forces and energies) - again, the CF workload here is connected via an off-axis pivot in the axial plane, as i suspected from the Apologia wheel.
Also another checkbox in that the smaller weight is much further away, hinting at a Kepler-esque balance, presumably thrown akilter by the off-axis center of rotation.
Also another in that the CF mechanism has a direct-drive transmission to the axle.
Ditto the GMH mechanism (although as noted earlier, these last two details may be superfluous to the central dynamic, which concerns the form of the asymmetry).
Finally i suspect that the diagonal delineated by the rope intersecting the plan wheel perhaps denotes another significant clue - maybe one that would tally with the functioning of the CF mechanism, once that becomes clearer...
Thinks, dudes? Compelling or smelling..? :)
EDIT: ...or maybe there's a suggestion here of two CF loads, in alternate phase? Recall the so-called Nice model of planetary dynamics, wherein larger-mass cycles in a factor-of-two relationship combine their forces as they overlap..? Could a similar effect via CF provide a freely-alternating 'overdrive' force to reset another load otherwise trapped out at the rim by a nominally peak CF force?
Again, this last point concerns resetting, not gaining - we already have our gain (a CF / G work asymmetry is trivial), but we need to reset the CF load against what is otherwise the system's maximum force (NB this isn't quite the same as say magnetic sticky spots, which normally prevent a gain - here we have the gain, but precisely because CF > G, we're now stuck.. unless CF workloads have some kind of get-out clause in the small print..! Presumably, there is, and Orffyreus found it.
Somehow, those Merseburg pendulums describe the CF cycle.
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re: Toad Elevating Moment
Ditto the toys page; CF workload (the lower snap-figures at D) vs GPE workload above at C (the smithy and scholar).
I'd previously assumed that D indicated a single, sprung, scissor-cell. However, given that the whistling top is directly below D - and so close as to perhaps be part of it - then taken together the figures' twists more likely indicate "spin force".
Two spin force loads, in alternate phase with a pair of GPE loads.
And D is depicted larger than C (geddit?)
There's 4 sequences possible (138-141) and one of em's a gain.
Presumably, another's a loss, and the other two symmetrical.
The scissorjack E denotes the direction of the CF-to-G translation gradient, from greater CF work to lower GMH work.
The chain links and shaft connections indicated between A and B imply a cyclical pairing or combining / offsetting of these two work loads.
The fork atop shaft B indicates two additional, off-axis centers of rotation, in the axial plane.
- presumably the apparent eye formed by the hole in the whistling top is an instruction to stop reading whatever i want into everything, and go to bed..? Am i on the "right path" yet..?
I'd previously assumed that D indicated a single, sprung, scissor-cell. However, given that the whistling top is directly below D - and so close as to perhaps be part of it - then taken together the figures' twists more likely indicate "spin force".
Two spin force loads, in alternate phase with a pair of GPE loads.
And D is depicted larger than C (geddit?)
There's 4 sequences possible (138-141) and one of em's a gain.
Presumably, another's a loss, and the other two symmetrical.
The scissorjack E denotes the direction of the CF-to-G translation gradient, from greater CF work to lower GMH work.
The chain links and shaft connections indicated between A and B imply a cyclical pairing or combining / offsetting of these two work loads.
The fork atop shaft B indicates two additional, off-axis centers of rotation, in the axial plane.
- presumably the apparent eye formed by the hole in the whistling top is an instruction to stop reading whatever i want into everything, and go to bed..? Am i on the "right path" yet..?
Last edited by MrVibrating on Tue Jul 23, 2013 3:21 am, edited 2 times in total.
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Nb. sorry just jumped outa bed and fired up the PC for one last strike (and if this was warm yesterday it's scorching today); ladies and gents, i give you the hieroglyphic version of MT41:
- again, there's two CF workloads, off the central axis but in the axial plane
- they're in direct-drive transmission with the axle
- the two lower weights close to the axle at B are a gravitational load, counter-balanced by equal and opposite CF loads
- the upper two at A are the CF load, out by the rim, and the equal-and-opposite CF forces are cancelling eachother (or perhaps combining additively)
These elaborate hieroglyphs are saying the same thing, in different ways, and it's all about the interplays of CF and G forces..!!
(i'd always wondered why A and B are labelled as distinct pairs (it's not his usual convention))
- again, there's two CF workloads, off the central axis but in the axial plane
- they're in direct-drive transmission with the axle
- the two lower weights close to the axle at B are a gravitational load, counter-balanced by equal and opposite CF loads
- the upper two at A are the CF load, out by the rim, and the equal-and-opposite CF forces are cancelling eachother (or perhaps combining additively)
These elaborate hieroglyphs are saying the same thing, in different ways, and it's all about the interplays of CF and G forces..!!
(i'd always wondered why A and B are labelled as distinct pairs (it's not his usual convention))
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Wow, today i realised a cunningly simple way to reset the CF load.
We don't need to yank it back up the CF gradient at all. Not an inch.
We can just let it go. Let it fly off the edge of the rail (or whatever it travels on).
Of course, this would mean it will impact with the side of the wheel - this gives us our 'thumps', as the flung and released CF weights collide with the downwards-travelling outside edge of the wheel.
Furthermore, at this point, freed from the bounds of the CF mechanism, the weight is now a gravitational load - it can perform more work as it falls.
Hence all that remains to be discovered is a mechanism to convert a previously gravitating weight into a CF weight, and vice versa. Then we can use identical weights for both workloads and cycle them between outer and inner loops, each in turn.
The thermodynamic cycle seems complete, and the mechanics, eminently feasible. Weights are flung upwards by CF, using that work to re-lift a previously-fallen gravitating weight, and then dropped from the end of the 'flail' arm (whatever comprises that), falling back down towards the axle while performing overbalancing work (RKE), before being re-lifted by another CF impulse, and then re-loaded into the CF mechanism to be re-flung themselves.
Simply dropping the CF load from the point of max excursion is a perfect solution, the ultimate get-out-jail-free card, with a new identity and job to boot.
If anyone's got suggestions for a mechanism to exploit this cycle, do pipe up..!
We don't need to yank it back up the CF gradient at all. Not an inch.
We can just let it go. Let it fly off the edge of the rail (or whatever it travels on).
Of course, this would mean it will impact with the side of the wheel - this gives us our 'thumps', as the flung and released CF weights collide with the downwards-travelling outside edge of the wheel.
Furthermore, at this point, freed from the bounds of the CF mechanism, the weight is now a gravitational load - it can perform more work as it falls.
Hence all that remains to be discovered is a mechanism to convert a previously gravitating weight into a CF weight, and vice versa. Then we can use identical weights for both workloads and cycle them between outer and inner loops, each in turn.
The thermodynamic cycle seems complete, and the mechanics, eminently feasible. Weights are flung upwards by CF, using that work to re-lift a previously-fallen gravitating weight, and then dropped from the end of the 'flail' arm (whatever comprises that), falling back down towards the axle while performing overbalancing work (RKE), before being re-lifted by another CF impulse, and then re-loaded into the CF mechanism to be re-flung themselves.
Simply dropping the CF load from the point of max excursion is a perfect solution, the ultimate get-out-jail-free card, with a new identity and job to boot.
If anyone's got suggestions for a mechanism to exploit this cycle, do pipe up..!
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Interesting; just reading up on the history of centrifugal force in classical mechanics, and its modern formalisation begins around the time of Bessler. Furthermore however it also plays into the spat between Newton and Leibniz - the latter's work trumped by the former.
For anyone who's read JC's Perpetual Motion: an Ancient Mystery Solved?, this particular dispute may have some bearing on Leibniz's interest in Bessler's wheel - indeed, being one of the most foremost authorities on CF of the day, it seems inconceivable that Leibniz wouldn't have strongly suspected - if not at least considered - that CF may be a significant factor in Bessler's mechanism.
Perhaps he initially hoped it might validate his conception of the force, over Newton's, thus perhaps vindicating his own cosmogony over that of his upstart rival...?
https://en.wikipedia.org/wiki/History_o ... tal_forces
For anyone who's read JC's Perpetual Motion: an Ancient Mystery Solved?, this particular dispute may have some bearing on Leibniz's interest in Bessler's wheel - indeed, being one of the most foremost authorities on CF of the day, it seems inconceivable that Leibniz wouldn't have strongly suspected - if not at least considered - that CF may be a significant factor in Bessler's mechanism.
Perhaps he initially hoped it might validate his conception of the force, over Newton's, thus perhaps vindicating his own cosmogony over that of his upstart rival...?
https://en.wikipedia.org/wiki/History_o ... tal_forces
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re: Toad Elevating Moment
As for the relative energies, i found this SI formulation on Wikipedia:
https://en.wikipedia.org/wiki/Uniform_circular_motion
________________________________________________________________
Consider a body of one kilogram, moving in a circle of radius one metre, with an angular velocity of one radian per second.
The speed is one metre per second.
The inward acceleration is one metre per square second[v^2/r]
It is subject to a centripetal force of one kilogram metre per square second, which is one newton.
The momentum of the body is one kg·m·s−1.
The moment of inertia is one kg·m2.
The angular momentum is one kg·m2·s−1.
The kinetic energy is 1/2 joule.
The circumference of the orbit is 2π (~ 6.283) metres.
The period of the motion is 2π seconds per turn.
The frequency is (2π)−1 hertz.
________________________________________________________________
So it seems that we need a centrifugal excursion just over 2x that of the gravitational drop...?
Am i getting that right? Ie. it takes 1J to accelerate 1KG by 1 meter/sec in a straight line, but, as shown above, in a constant-curvature path at 1m radius it only has .5J of CF energy. Thus to restore the spent GMH, the CF workload needs twice its displacement.
I may be mis-reading this so do chime in if that looks suspect.. but the general dynamic here seems sound enough, to me at least; The CF excursion can be pretty much arbitrarily longer than the GMH drop.
Of course, one compact way to double-up the CF excursion would be to fling two weights at once, in equal and opposing 180° arcs...
https://en.wikipedia.org/wiki/Uniform_circular_motion
________________________________________________________________
Consider a body of one kilogram, moving in a circle of radius one metre, with an angular velocity of one radian per second.
The speed is one metre per second.
The inward acceleration is one metre per square second[v^2/r]
It is subject to a centripetal force of one kilogram metre per square second, which is one newton.
The momentum of the body is one kg·m·s−1.
The moment of inertia is one kg·m2.
The angular momentum is one kg·m2·s−1.
The kinetic energy is 1/2 joule.
The circumference of the orbit is 2π (~ 6.283) metres.
The period of the motion is 2π seconds per turn.
The frequency is (2π)−1 hertz.
________________________________________________________________
So it seems that we need a centrifugal excursion just over 2x that of the gravitational drop...?
Am i getting that right? Ie. it takes 1J to accelerate 1KG by 1 meter/sec in a straight line, but, as shown above, in a constant-curvature path at 1m radius it only has .5J of CF energy. Thus to restore the spent GMH, the CF workload needs twice its displacement.
I may be mis-reading this so do chime in if that looks suspect.. but the general dynamic here seems sound enough, to me at least; The CF excursion can be pretty much arbitrarily longer than the GMH drop.
Of course, one compact way to double-up the CF excursion would be to fling two weights at once, in equal and opposing 180° arcs...
re: Toad Elevating Moment
MrVibrating wrote: (edited for brevity)
Daniel Bernoulli, son and student of Johann and student Leonhard Euler. Both of which become involved with fluid mechanics and dynamics. They in their own way found that some laws are written to agree with Newton while others we are left to assume!
If interested: http://www.mathcs.org/analysis/reals/hi ... noull.html
Ralph
Leibniz and Johann Bernoulli did indeed cause havoc creating a delay in Western Europe's acceptance of Newton's Principia. Although Newton triumphed in the end, there was two noted individuals that held out.Interesting; just reading up on the history of centrifugal force in classical mechanics, and its modern formalization begins around the time of Bessler. Furthermore however it also plays into the spat between Newton and Leibniz - the latter's work trumped by the former.
Daniel Bernoulli, son and student of Johann and student Leonhard Euler. Both of which become involved with fluid mechanics and dynamics. They in their own way found that some laws are written to agree with Newton while others we are left to assume!
If interested: http://www.mathcs.org/analysis/reals/hi ... noull.html
Ralph
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Perhaps some relevance here to MT137?
If, as JC supposes, this glyph refers to the circle of fifths, a perfect fifth is of course half an octave. For instance, if 220Hz = A3 and thus 440Hz, A4, if we instead add half their difference (110) to the fundamental of 220, we get a perfect harmonic fifth at 330Hz.
Any two such fifths are thus equidistant by one octave - that is, 660Hz and 165Hz are the next fifths up and down the spectrum, respectively.
In other words MT137 may be drawing attention to the respective displacement relationship between inputs and outputs in the two different fields...
Pretty tenuous i know, but hard to see what else it could signify..
If, as JC supposes, this glyph refers to the circle of fifths, a perfect fifth is of course half an octave. For instance, if 220Hz = A3 and thus 440Hz, A4, if we instead add half their difference (110) to the fundamental of 220, we get a perfect harmonic fifth at 330Hz.
Any two such fifths are thus equidistant by one octave - that is, 660Hz and 165Hz are the next fifths up and down the spectrum, respectively.
In other words MT137 may be drawing attention to the respective displacement relationship between inputs and outputs in the two different fields...
Pretty tenuous i know, but hard to see what else it could signify..
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@rlortie
Interesting cheers. Lol, in my wilder daydreams i've toyed with the idea, should i ever succeed in solving this riddle, of hanging a Bessler wheel around a prominent statue of Newton in central London, in a kind of "guerilla marketing campaign".. How outrageous would that be?
Probably not much.. come to think of it i'd probably fall and wake up in A+E to be promptly arrested, but still, just the thought gives me a wry smile...
Interesting cheers. Lol, in my wilder daydreams i've toyed with the idea, should i ever succeed in solving this riddle, of hanging a Bessler wheel around a prominent statue of Newton in central London, in a kind of "guerilla marketing campaign".. How outrageous would that be?
Probably not much.. come to think of it i'd probably fall and wake up in A+E to be promptly arrested, but still, just the thought gives me a wry smile...
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Another week spent piffling around with sims and half-baked ideas, and i've suddenly done an about face and come back to this 'un.
I'm pretty sure it works.
I dunno why i lost confidence in it - i think tired myself out and got confused, assuming wrongly that the lift against gravity ate all the work gained from the extra work perfomed by CF.
But it doesn't - i've just gone over it again and it's a gain - again. Again, we gain gain by making CF force perform RKE work - so a mass gets flung out from axle to rim, in the process this 'fall' is harvested (say by springs, flywheel or whatever) and fed back into the wheel, accelerating it. If this fall also coincides with the direction of gravity (ie. downwards) then both CF and G vectors sum to a greater net magnitude, and we get a big RKE boost by falling out to the rim in the lower 180° arc.
So we've gained energy from a radial translation, but now we need to get it back to the center, for less energy than its already paid out.
My mistake earlier in dismissing this was thinking this subsequent axial / circumferential lift performs an input of work to the CF gradient... which is nonsense!
Rather, we do RKE work by being flung outwards while falling, then we hit the rim of the wheel and can't BE flung any further. Hence no more CF work is performed - only gravitational work is involved in relifting the weight from the bottom of the wheel back to the top - the weight just sits there resting against its rimstop.
Once it's 'upside down' - with CF pinning it to the ceiling, as it were - we yank it back down, against this CF force, towards the axle. This input energy is less than the output energy because gravity was helping us!
Gravity plus CF pulls the weight down from axle to floor. We re-lift it, but only against gravity, CF's contribution is neutral in this action. Then we retract it to the axle, against CF but with the assistance of gravity.
The circumferential / axial lift - the 180° rotation back up to TDC (while still stuck to the rim) - can be counter-balanced by running the lift and drop in parallel, 180° out of phase.
In other words, connect the weights with rigid rods as per MT 41, where the rod length is about equal to the wheel's radius; hence one's at the axle while the other's at the rim.
And again, crudely, if CF equals roughly 1G then we get 2G of output force upon extension, zero input force for retraction and 1G to rotate against. This leaves a gain roughly equal to the GMH of the mass drop.
Anyone else following me here? If i can just keep focused on this, i think it might lead somewhere...
The trick will be turning CF translations into RKE - essentially, turning CF into torque.
Free-sliding radial translations tend to eat RKE, however, as i think i've illustrated with the flywheel-on-a-roundabout thought experiment, tethered one's needn't! The work that can be performed by this outwards-bound force experienced by a body in a rotating frame appears to be thermodynamically decoupled from the RKE of the system.
----
Again, to clarify the thought experiment, suppose we're on a roundabout in space (in zero-G), and there's a big flywheel in the centre of the roundabout, on a low friction bearing. It also has a spool of chord.
If we allow ourselves to be flung off, into the merciless depths of space, without pulling the chord, the net energy of the system is reduced by our departure (from the system, if not this mortal coil).
If however we grab the flailing end of the chord, in a last, desperate-if-futile grasp at survival, say, then as we shoot off into the hungry blackness of eternity we can at least accelerate the flywheel (for all the help that'll be).
The point however is, what's the mass limit on the flywheel...? It's arbitrary, innit? If its inertia's high enough, it might take 10 minutes to spin up, relative to the already-spinning roundabout, before the chord runs off the spool and you're flung unceremoniously to your slow, lonely death by suffocation or whatever.
IOW, the amount of energy you can impart to the wheel in your last desperate acts is a function of the rotational inertia of the system, and, ultimately, the amount of time the rip-chord is subject to torque (the torque you generate as you're ejected off to infinity).
Can WM2D handle ripchords? How else might CF work be converted to RKE? Springs come to mind, but a precise mechanism eludes me for now...
Edit to add:
Re. counterbalancing the axial re-lift - two pairs of MT41 type weights, inverted 180°, zeroes the RKE cost in re-lifting..!
I'm pretty sure it works.
I dunno why i lost confidence in it - i think tired myself out and got confused, assuming wrongly that the lift against gravity ate all the work gained from the extra work perfomed by CF.
But it doesn't - i've just gone over it again and it's a gain - again. Again, we gain gain by making CF force perform RKE work - so a mass gets flung out from axle to rim, in the process this 'fall' is harvested (say by springs, flywheel or whatever) and fed back into the wheel, accelerating it. If this fall also coincides with the direction of gravity (ie. downwards) then both CF and G vectors sum to a greater net magnitude, and we get a big RKE boost by falling out to the rim in the lower 180° arc.
So we've gained energy from a radial translation, but now we need to get it back to the center, for less energy than its already paid out.
My mistake earlier in dismissing this was thinking this subsequent axial / circumferential lift performs an input of work to the CF gradient... which is nonsense!
Rather, we do RKE work by being flung outwards while falling, then we hit the rim of the wheel and can't BE flung any further. Hence no more CF work is performed - only gravitational work is involved in relifting the weight from the bottom of the wheel back to the top - the weight just sits there resting against its rimstop.
Once it's 'upside down' - with CF pinning it to the ceiling, as it were - we yank it back down, against this CF force, towards the axle. This input energy is less than the output energy because gravity was helping us!
Gravity plus CF pulls the weight down from axle to floor. We re-lift it, but only against gravity, CF's contribution is neutral in this action. Then we retract it to the axle, against CF but with the assistance of gravity.
The circumferential / axial lift - the 180° rotation back up to TDC (while still stuck to the rim) - can be counter-balanced by running the lift and drop in parallel, 180° out of phase.
In other words, connect the weights with rigid rods as per MT 41, where the rod length is about equal to the wheel's radius; hence one's at the axle while the other's at the rim.
And again, crudely, if CF equals roughly 1G then we get 2G of output force upon extension, zero input force for retraction and 1G to rotate against. This leaves a gain roughly equal to the GMH of the mass drop.
Anyone else following me here? If i can just keep focused on this, i think it might lead somewhere...
The trick will be turning CF translations into RKE - essentially, turning CF into torque.
Free-sliding radial translations tend to eat RKE, however, as i think i've illustrated with the flywheel-on-a-roundabout thought experiment, tethered one's needn't! The work that can be performed by this outwards-bound force experienced by a body in a rotating frame appears to be thermodynamically decoupled from the RKE of the system.
----
Again, to clarify the thought experiment, suppose we're on a roundabout in space (in zero-G), and there's a big flywheel in the centre of the roundabout, on a low friction bearing. It also has a spool of chord.
If we allow ourselves to be flung off, into the merciless depths of space, without pulling the chord, the net energy of the system is reduced by our departure (from the system, if not this mortal coil).
If however we grab the flailing end of the chord, in a last, desperate-if-futile grasp at survival, say, then as we shoot off into the hungry blackness of eternity we can at least accelerate the flywheel (for all the help that'll be).
The point however is, what's the mass limit on the flywheel...? It's arbitrary, innit? If its inertia's high enough, it might take 10 minutes to spin up, relative to the already-spinning roundabout, before the chord runs off the spool and you're flung unceremoniously to your slow, lonely death by suffocation or whatever.
IOW, the amount of energy you can impart to the wheel in your last desperate acts is a function of the rotational inertia of the system, and, ultimately, the amount of time the rip-chord is subject to torque (the torque you generate as you're ejected off to infinity).
Can WM2D handle ripchords? How else might CF work be converted to RKE? Springs come to mind, but a precise mechanism eludes me for now...
Edit to add:
Re. counterbalancing the axial re-lift - two pairs of MT41 type weights, inverted 180°, zeroes the RKE cost in re-lifting..!
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Another half hour going over this and it still looks good - 4 weights each undergoing one full gravitational and one full CF interaction per cycle, swapping axle/rim positions every 180°. All force/distance integrals sum to zero in the inertial frame (we're dropping and lifting the same weights the same distance against the same given forces). From the non-inertial frame though the specific sequence of interactions yields a non-zero lagrangian - the alternately additive and subtractive vector sum is a free energy gradient, precisely as stated in the OP.
:)
:)
re: Toad Elevating Moment
This may be of interest to you & perhaps be a method you could use or modify for your needs.
It is a sim I once built whilst doing my own Cf studies.
The dual masses change radius in & out simultaneously as the pendulum shaft swings & the translation energy [Cf's ?] of the sliding masses is channeled via the geared "Pull Mech" into Rotational System Energy at the axle.
In short, it was a way to transform linear/radial motion into rotational acceleration at the axle.
Pull the sim apart to see how everything works - I have stripped it to bare essentials so it shouldn't be to hard to follow.
It is a sim I once built whilst doing my own Cf studies.
The dual masses change radius in & out simultaneously as the pendulum shaft swings & the translation energy [Cf's ?] of the sliding masses is channeled via the geared "Pull Mech" into Rotational System Energy at the axle.
In short, it was a way to transform linear/radial motion into rotational acceleration at the axle.
Pull the sim apart to see how everything works - I have stripped it to bare essentials so it shouldn't be to hard to follow.
- Attachments
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- CfTestC4[3].wm2d
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