The ratio of the diametric lever weight's MoI to that of the net system is vitally important, so let's take a closer look at a simple example:
• I've chosen an arbitrary system radius of 2 meters
• Each lever arm is 3.855 meters in length
• Using the standard MoI formulas (ref.
here), the lever arms' MoI are given by 1/3 their mass times their length squared, and the end bobs, by their mass times their radius squared (relative to their pivots,
not the wheel axis); so let's assume a mass of 1 kg for both:
• Arms (each) = (1/3) * 1 * (3.855^2) = 4.953675 kg-m²
• Bobs (each) = 1 * 3.855^2 = 14.861025 kg-m²
..so the MoI of each diametric lever weight is the sum of those two calcs:
• 4.953675 + 14.861025 = 19.8147 kg-m²
OK, so now we want to know their
axial MoI, about the wheel axis. Deriving this using the standard MoI calcs would necessarilly get more involved, so let's use the sim to take a shortcut:
• We know that rotational KE is given by half the MoI times angular velocity squared:
• RotKE = ½Iw²
So we can use the 'kinetic()' function of the sim to take the net energy of the system - all of which is going to be
rotational KE - and then simply
invert that output to derive its MoI component:
• I = rotKE / sqrt(w) * 2
..if we assign an arbitrary value of '1 kg-m²' for the MoI of the wheel body itself, then we can simply deduct that from the above MoI derivation, to leave just the
axial MoI of the diametric weight levers:
..and so the net MoI of both levers about the wheel axis - but not including that of the wheel itself, nor any other mass - is 9.6838 kg-m².
So now we can compare the ratio of the two MoI's:
• 19.8147 / 9.6838 = 2.0461
Thus, each diametric lever weight, individually, has an MoI that is equal to
twice that of
a pair of such levers, rotating about a central axis!
So what is this good for?
What does it all mean?
Recall that the shortest route up the OU ladder is a pair of
equal masses / inertias in a series of reactionless accelerate-and-brake cycles ('braking' = 'inelastic collision').
The number of such cycles required to reach unity, and then, over-unity, is equal to the sum of the ratio of the two interacting inertias in question; thus, with a 2:1 ratio as found above, the unity threshold would be
three cycles, and
four to reach over-unity, at 133%.
However, as noted previously, the Toys page indicates
five cycles to OU, hence either invoking a 3:1 inertia ratio, or else, a 1:1 ratio, but losing half the generated momentum each cycle.
As also noted already, the first option seems the less likely - why intentionally hobble the interaction with an unnecessarily sub-optimal inertia ratio? Thus the latter option must be the correct one; that the effective N3 break involves sinking our counter-momentum (and thus 50% of the total momentum per-cycle) to gravity!
So we ideally want a 1:1 ratio between the interacting inertias.
To put this back into more familiar terms, we want each 'bang' from the descending side of the wheel to cause the maximal rise in system momentum each cycle, hence we want to
increase the MoI of the wheel about its central axis, by adding more mass...
So, what further mass do we need to add?
We're obviously going to need assign some mass to the wheel body itself:
• Bessler noted that his wheel body mass was quite lightweight, using a 'framework'
• The oilskin covers would've added some more mass
• We also know that we need an input GPE, to operate our levers
• This radial GPE may connect via a scissorjack (or gears / pulleys / power conversion generally)
• There may also be a requirement for sprung PE, and thus, springs
So in summary, we can make up that 'target MoI' of the wheel about its axis in any way we like; obviously the wheel body / framework will account for much of it, and the input GPE much of the remainder...
...should it transpire that these components needn't sum to that target MoI, we could just add in some further 'dead mass' purely for the purposes of bumping-up the axial MoI to match that of one weight lever!
So the current 2:1 ratio we're measuring provides us with 'margin' of additional angular inertia for all the other components we're going to need. We already have
half of the total axial MoI we want, and we can make up the
other half using as much input GPE and 'wheel body / framework' mass as we like!
There's further telling in these figures besides; for instance, we know that however lightweight the wheel body / framework, it can't be too weak and flexible; it has to be sufficiently rigid and strong, hence it's inevitably going to require a fair bit of mass...
..we can try design it in such a way as to keep its radius as low as possible, but inevitably, MoI = mr², hence a lil' bit of framework / wheel body mass is going to add
quite a lot of additional MoI..
..the point being, that this would imply that the proportion of additional MoI contributed by the input GPE is inevitably going to be somewhat limited;
in other words, not a lot of input GPE is required for success! This is encouraging, but also informative; remember that our total mass for each diametric lever is currently 2 kg, implying that the input GPE is probably
less than this, which in turn implies that the input GPE either
does not lift these levers, instead impelling them downwards, adding to their gravitation (as depicted in the last MT40-ish doodle), or else, if it
is supposed to lift them, then it must do so over a reduced displacement; IE. 1 kg could raise
more than 1 kg, albeit over a smaller displacement.. to wit, consistent with a need for power conversion such as via scissorjacks...
Now this seems like better progress, huh? We're deducing functional properties from first principles, and it seems to be falling together with some degree of self-consistency..
So what we need now, to guide us across the finishing line, is the N3 break!
How in the hell are these levers applied in order to crack N3? Obvioushly, we need them to undergo a reactionless acceleration prior to their impacts..
So what is the nature of this reactionless acceleration? Do they simply drop, under their own gravity? Is this drop further accelerated by the radial GPE drop? Perhaps also assisted by the scissorjacks?
Or are they lifted by the radial GPE, again, via the scissorjacks?
Seems like it has to be one or other option, and only one of them is going to provide a consistent logical solution..
Open house here folks - we need to somehow apply an acceleration to these levers, such that they fall on the descending side and impact a rimstop or whatever, but this acceleration
mustn't apply counter-torque / counter-momentum back to the wheel axis.
Only their impact must. A radial GPE is apparently part of the key to this feat, possibly connecting via a scissorjack..
