OK just to clarify a few points, i'm going to run through some example 'gain trajectories'...
This is for everyone else's benefit, not mine, so hopefully it'll explain exactly what's happening, and what more needs to be done..
First off, let's look at the
shortest theoretical route up the OU ladder..
You'll need Notepad, Calculator and the standard KE formula ½mV² (use this handy online calc
here - i do!):
• Take two 1 kg masses. (They could be 1 kg-m² angular inertias - different dimensions, but same amount of inertia either way)
• Unhanging any portraits of sir Isaac you may have on display, and resting them face down whilst saying three
hail Mary's,
apply a unilateral force to one mass, accelerating it by 1 meter / sec.
So no drama here - just assume you have 1 kg moving at 1 m/s (or 1 kg-m² at 1 rad/s - same deal). How it got to that speed is incidental at this stage..
You should find that you currently have ½ Joule of KE.. and of course, that's
also how much PE you've spent.
The second mass however is still stationary, so:
• inelastically
splat the moving mass into the stationary one.
Bang! Now both inertias are moving or rotating at equal speed.
You
had 1 kg-m/s of momentum, embodied on one 1 kg inertia, and now, post-collision, you
still have that same 1 kg-m/s of momentum... only now it's distributed between
two 1 kg inertias.
So, this means each mass is now moving at ½ m/s, right?
Applying the KE formula, 2 kg moving at 0.5 m/s has 0.25 J of KE..
..but we
spent half a Joule, so 50% of our input energy has been dissipated by the collision. We've lost half our input energy. Bad start for a so-called 'gain-trajectory', eh?
What kind of idiot is gonna repeat the same mistake, expecting a different outcome..?
Exactly!
Our kind! That's
just what we do! It's
what we're built for, derp!
So give it another whirl:
• add
another 1 m/s of velocity to either mass
Again, this is a
unilateral force we're somehow applying - not pushing or pulling against some
other mass - and furthermore,
in spite of that concession, we
still only want to pay the same price for it, of half a Joule.
So now it's moving at 1.5 m/s, or 1.5 rad/s if you chose angular.
Current net input energy is two * ½ J accelerations, so 1 J total.
Again, crunch the accelerated mass into the non-accelerated one:
- so, without any bounce (crucial!), 1 kg moving at 1.5 m/s strikes 1 kg moving at 0.5 m/s; result? You should now find you have two 1 kg inertias moving at 1 m/s..
So how much KE do we have, according to the KE equation?
- Two 1 kg masses at 1 m/s have a total of 1 J of KE..
..and we've
spent 2 * ½ J, so at the second cycle, we've gone from 50% down, to hitting unity! 1 J in for 1 J out..
So the efficeincy of the system is
developing - we're not getting the same result each time! Who's the idjit now eh, mr so-called 'Einstein'..?
So we
have to try a third cycle, to confirm where this is going..
• Again, accelerate one mass by another 1 m/s, for half a Joule..
• ..and
again, prang it into the un-accelerated mass..
So now you should have two 1 kg masses moving at 1.5 m/s.
You've spent 3 * ½ J = 1.5 J.
How much KE does 2 kg moving at 1.5 m/s have?
It's 2.25 J, isn't it?
The ratio of 2.25 / 1.5 = 1.5.
That is, we're now at 150% of unity..
Enthused? Try adding further cycles, and watch in awe as your efficiency converges towards infinity..
However, Bessler does NOT indicate this interaction. His method reaches unity at
four cycles, not two..
Why might that be? Obviously, the requisite conditions here of applying a unilateral force to cause the same acceleration each cycle, invariant of rising net speed, is a total sacrilege upon Newton's 3rd law.
And remember, N3
really IS inviolable, because of mass constancy. A 'genuine' N3 break is literally impossible..
..yet Bessler's wheels were statorless!
..but yet
again, gaining height on a park swing is trivial! And
that requires no stator either! Coincidence? No chance. No other alternative!
So the best we can hope for - and all Bessler managed - was an
effective N3 break - something that achieved the same ends, albeit by alternate means...
..and thus we may conclude that
this trick, of circumventing N3 by
whatever devices, is
inherently a
"4 cycles to unity" optimum outcome - for some reason, it is
intrinsic to the nature of the exploit - a built-in constraint.
Because otherwise, OU in three cycles would be possible, as Bessler would have quickly discovered! Yet evidently, it
isn't.. Bessler's
ideal solution (as portrayed on the Toys page),
necessitates a
"4 cycles to unity" ('4 c/u') trajectory..
So why might that be? Let's frickin'
sim one on the back of a fag packet and find out eh..?
• Let's concede that in order to apply a force to an inertia, it has to be applied
between it and
some other inertia.
There
may be examples of 'effectively' unilateral forces under certain given conditions (inertial torques, OB torques or whatever), but let's just assume we
have to incur an equal opposite counter-momentum, as per normal, for this current example...
so to accelerate 1 kg by 1 m/s in one direction, we
also have to accelerate the
other 1 kg in the o
opposite direction..
..so we now have 1 kg at 1 m/s with ½ J going on one direction, and then
another, equal mass, speed and KE going the other way..
..so we've initially spent 1 J, with zero net momentum to show for it..
• Now kill that counter-momentum. Destroy, annihilate, liquidate that mofo. However. You work it out. Just, make it happen..
•
Now do the collision.
Rinse, repeat and calculate...
If you actually
do that calc, it's precisely the same interaction as before, except half the net momentum and input energy is lost each cycle.
So, 1 kg at 1 m/s meets 1 kg at 0 m/s, resulting in 2 kg at ½ m/s, and 0.25 J net KE, from 1 J of spent PE; ie. 25% efficiency, 75% loss.
Then, 2nd cyc, 1 kg at 1.5 m/s strikes 1 kg at ½ m/s, resulting in 2kg at 1 m/s and 1 J net KE, for 2 J net input PE, so, 50% efficiency at this stage.
3rd cyc: 1 kg @ 2 m/s hits 1 kg @ 1 m/s - resulting in 2 kg @ 1.5 m/s, so 2.25 J of KE, for 3 J total input PE.
2.25 J out / 3 J in = 0.75% efficiency, so far..
As you can see, net efficiency is increasing by a steady 25% per cycle...
At the fourth cycle, 1 kg @ 2.5 m/s tangles with 1 kg @ 1.5 m/s, leaving both at 2 m/s, and so posessing a
net kinetic energy of 4 J, my idjit friends..
We've spent 4 J too, so it's a '4 c/u' (four cycles to unity) result.
Let's just spend
one more Joule, to witness first-hand the
true miracle of back-of-fag-packet OU for ourselves:
1 kg @ 3 m/s snags 1 kg @ 2 m/s; result = 2 kg @ 2.5 m/s = 6.25 J... but we've only
spent 5 J in total..
6.25 / 5 = 1.25, or 125% of unity
So, that's the gig.
The 2 c/u possibility is apparently too good to be true, but the
4 c/u is friggin' spelled out
in pictures on the Toys page, from 300 years ago..
..so it's totally
game on, people!
There IS a physical mechanism possible, that embodies the 4 c/u trajectory above... and the Toys page, in conjuntion with MT's 30 thru 42 and 133 - 134 etc., describe the nature of the interaction that accomplishes this result.
Thus, each
'bang!' emanating from his wheels was dissipating half the per-cycle input energy..
..his wheels contained sufficient PE to cover the first 5 cycles.. (which, with multiple mechs per wheel could be half a turn or less, it's quite arbitrary, build-time aside)..
..Bessler's unit energy cost of momentum was effectively 1 J per kg-m/s or kg-m²-rad/s.. within a limited RPM range anyway.. (efficacy of the exploit was evidently still somewhat RPM-dependent, since a) his wheels didn't explode, and b) he explicitly wrote that they
"gained further advantage from applied loads" - implying that their peak efficiency was at a lower speed than that which they were able to coast at, but with efficiency
decreasing - the unit energy cost of momentum nonetheless
rising - with increasing RPM.
So, the surgically-clean
theoretical 4 c/u trajectory evidently comes with certain constraints when realised practically... the take-home being that there's a 'sweet spot' RPM margin for optimal momentum gains / counter-momentum destruction.
Note also that simply inverting the sign of counter-momentum - such as by aiming it upwards, to be returned by gravity, is NOT the exploit, since that would conserve all of the momentum and so yield a 2 c/u result instead. The 4 c/u outcome can only arise under two possible conditions: destruction / sinking / expunging the system of counter-momentum on a per-cycle basis.. and also, one further set of circumstances:
• Try repeating the first, 2 c/u example, only this time, make the non-accelerated mass 3 kg. Keep the accelerted one at 1 kg
• You should now find that, despite conserving all momentum and counter-momentum within the system, it's now been drawn out to a 4 c/u trajectory! You get the 25% accumulator again..
In fact, there's a simple back-of-envelope calc you can do to predict the gain trajectory / no. of cycles to unity / OU:
• simply add the ratio of the two interacting inertias together!
So with two equal inertias, 1 + 1 = 2 c/u (their actual masses or MoI's are irrelevant).
If one mass is 1 kg, and the other's 2 kg, then 1 + 2 = 3, so the system will reach unity after three cycles, and 130% at the fourth!
However since we know already that the '2 c/u' possibility is simply not on the cards - the exploit necessitates cancellation or destruction of counter-torque or counter-momentum - we can just double that sum - so 1 Kg playing kiss-chase with 2 kg will reach unity at 6 cycles and be making ever-fatter OU babies from cycle #7 onwards.
Likewise, a 2 kg-m² MoI in a series of 'spin & brake' cycles with a 10 kg-m² MoI has an inertia ratio of 1:5, and 1 + 5 = 6, so allowing for the requisite 50% momentum loss per cycle, the system would hit unity at the twelfth cycle, and 112% at the 13th... Geddit? Piss-easy huh?
So that's full disclosure. MT, and
all the clues, are about the means to harness this KE gain principle.
If you
really think it through (seriously, take a hot bath / sunbathe or whatever to dwell on it) - it's hard to disagree with Bessler's point that in
"all true PMM's, everything must, of necessity, go around together" - in other words, that KE gains from a statorless wheel / effective N3 break
really are the only game in town. There simply
is no other possibility - no margins for doubt, even - anywhere else in mechanics / classical physics.
KE gains / mechanical OU is frickin'
kids stuff - just
fling & bang, repeatedly... flinging and banging, and flinging again.. so long as the 'flings' are
reactionless, somehow (?), you can't
not make energy. :|
So,
obvioushly, gravity-assisted asymmetric inertial interactions are the name of the game. Gravity can be used to cyclically sink counter-momentum, or else, prevent its generation in the first place, such as by cancelling counter-torques. Spin & bang / brake / clutch / splat / crunch / land gracefully upon - basically any collision
except bounces /
boings - and you get a constant per-cycle efficeincy accumulator.
Just find some mechanism that literally articulates the accelerate & brake maths above, and you'll know it's working when it loses 75% input energy the first cycle, but only 50% on the second..
The Toys page and MT are obviously explaining a mechanism/s for achieving this.
The key for us is to understand what the gain principle
is - the only thing it
can be - because this is the context in which to be viewing MT and the Kassel engravings etc.
A pendulum with a wheel is a gravitating inertia with a non-gravitating one.
A wheel biased by a bucket spooling off the axle is a wheel biased to sink counter-torques
applied to that axis, to gravity.
A scissorjack is a frickin' linear lever - it just converts a high-displacement, low force action into a low-displacement, high-force one. MT repeatedly shows us ways this can be used to couple GPE and MoI interactions.
We already know that GPE-GPE interactions are a zero sum, and that inertial interactions are likewise symmetry-bound by N3.
Obviously, then, the symmetry break depends upon the interaction of gravity
and inertia.
Obviously, it's a gravity-assisted asymmetric inertial interaction,
derp.
That's all it
can be.
It's a wheat-and-chaff game, in every sense, but threshing momenta from counter-momenta is what this is all about..
"Driver" vs
"driven", or sources / sinks, or causes / effects might be better fits to Bessler's thought processes than our modern concepts of "inputs" and "outputs". Likewise, he wouldn't have recognised the term "rectification" as we do, but that's precisely what he was doing.
My best bet is "sinking counter-torques to gravity on a per-cycle basis".
The GPE-MoI coupling afforded by the scissorjack is the key to decoupling the momentum yield from RPM - ie. gaining the same momentum, for the same PE cost, in spite of rising RPM, across some useful RPM range.
Obviously, those diametric or radial weighted levers have high MoI, but small displacements.
One or other of the torques or counter-torques involved in their operation is being somehow sunk to gravity.
This yields a constant momentum gain per cycle, which has a relatively small 'velocity' component, but high 'inertia' component.
Its constant, speed-invariant energy cost is the speed-invariant GPE of the
'angular lift & radial drop'.
So that same GPE output buys the same rise in net system momentum per cycle, or at least, at a rate that
doesn't diminish by the square of velocity, across some useful RPM range.
Somehow, the confluence of 'over-balancing' torque, coupled with the sequence of positive-then-negative inertial torques caused by the radial translations,
combined with radial GPE outputs, hustles a per-cycle net momentum rise from those weighted levers.. levers which are,
more specifically,
heavy angular inertias..
Finally, note also that this GPE-MoI coupling is going to transition every 180° - each time it rotates around 180°, the radial GPE drops, applying two opposing torques and their respective counter-torques to those high-MoI armatures..
..yet the Toys page shows two
'impetuses' for 180°, then only one for the next 180°..
..therefore it is evident that the asymmetric inertial interaction depends upon the different ways gravity interacts with the mechanism, when oriented in the two 180° phases..
..so it's not simply the same interaction repeated twice per rotation - rather, the momentum gain is the result of one complete turn of a given mechanism.
Something happens differently when rotating 'under' compared to 'over'.. and the momentum gain
through the full 360° is the result of that difference.
(Apologies to anyone who wanted 50,000 words or less. It's
worse than that - you need a calculator! Eek! But hey precious little jargon, plain-english and just
one equation.. all i currently have is summarised right here.)