ME wrote:Did the last animation really not help? I thought it was the smoking gun.. :|
Sorry, I don't get it either...
Creating momentum is as simple as pulling mass inwards under peak CF, while re-extending it under zero-CF.
I guess you suggested (at least twice) that this would be an effortless action.
If that's possible you don' t need a "Robernoster", only a wheel or an ice skater: Pulling a mass inwards is indeed the winning action (when done without effort). As you've mentioned: you have to work against CF (peaked or not).
In you latest animation (t>5) you can clearly see the drop in all the values when turning the corner and masses extend, I guess you still have to fight against that?
eeh, right?
WE HAVE TO TRY THIS!!!
Please do !
To pull the mass inwards against CF, yes of course, that's input work, and the energy you spend is equal to the gain in KE.
Normally, with pure rotation, letting the mass fling back out under CF would then take away all the KE you'd just input.
Unless we let the mass back out while it
isn't rotating - in which case we can get back out to the rim, without causing the usual deceleration. And when we do this, we have gained momentum.
When the masses extend in the lossy interactions here, inertia causes a deceleration.
This deceleration doesn't change the momentum of the extended weight itself - on the contrary, the weight's RPM is decelerated precisely because its angular momentum was conserved - MoI rises with radius, so RPM must decrease to conserve net AM.
Likewise, when we pull a mass inwards and its orbital velocity rises, it does so because it's conserving angular momentum - MoI goes down with radius, so RPM must increase to conserve net AM.
So CoAM does not change the total angular momentum
of the radially-translating mass itself.
It conserves it, by either accelerating it, or decelerating it, to compensate the change in MoI with radius.
Accelerating anything else that is attached
to it!
IOW, these radial translations, which apply torques only
because they're supposed to conserve momentum, are ALSO being applied to all the other,
non-radially translating mass - whether it's rotating or traveling in a straight line or whatever.
For instance, stick a radially-sliding mass on a flywheel, spin it up and pull the mass inwards, causing an acceleration - the AM of the radially-moved mass remains constant,
but the AM of the flywheel itself does not!!!
In exactly the same way, consider the paternoster, as a chain around two sprockets - we can move mass in and out radially on the sprockets.
We could even attach a giant flywheel to the main axle, and move masses in and out radially, to much higher radii than that of the sprockets turning the chain.
We could even gear up that flywheel, so that one revolution of the axle causes 10 revs of the giant flywheel.
Whatever, when we move the mass in, its momentum remains conserved, but in so doing, the momentum of the flywheel, and sprockets, and chain,
all increase.
The momentum of the whole system increases, because its velocity has increased, and the flywheel's MoI is constant, and the sprocket's MoI is constant, and MoI isn't even a thing as far as the linearly-moving parts on the chain are concerned, and its rest mass certainly isn't variable... therefore where speed increases,
so does momentum UNLESS the inertia in question decreases, which it can't in the linear scenario, being invariant rest mass.
So, in the sim above, when the locks are opened at 5 secs, the masses start to slide out under CF, which causes an angular deceleration to conserve their (and only their) momentum - however all the other mass that
isn't changing radius is also being decelerated - and so draining momentum away.
When the masses come back onto the next straight, they're no longer subject to CF, and so moving them back inwards (via springs, pulleys or whatever)
doesn't cause an angular acceleration, which would otherwise restore the system momentum. Instead, we move back in under zero CF, causing no inertial torque, and are thus ready to be flung out again, compounding the first momentum loss with a second, and so on - each time a mass cycles in and out, it applies negative inertial torque but not equal opposite positive torque.
And again, remember, the very thing
causing these torques that are destroying momentum is CoAM!
And there's nothing stopping us rectifying only the positive torques, and dodging all the negative ones - yes, we'd need on-board PE stores for this, but don't worry about that for now - just supposing
we had sufficient energy, we could keep adding ever-more momentum, side-stepping the reciprocal negative torques from the re-extending masses, for any number of cycles, until we ran out of energy.
The point of course is not that infinite energy could buy infinite momentum, but simply that we could buy
any at all from within a closed system of radially-cycling masses. The point is that it's even on offer and available in the first place.
And this applies equally WRT getting shot of the stuff.
You're seeing it, but not putting it together... if ALL the masses extended at the same time, while rotating only (no linear section), then the system momentum would remain constant! It would not waiver!
If only one mass extended, while the others stayed locked to their poles, rotating only with no linear section, the net system momentum will decrease.
Basically, the momentum of ANY and ALL non-radially-translating mass IS modified precisely BY the action of CoAM on the radially-translating mass, in so doing, changing the momentum up or down of all the other moving mass
connected to the same motion.
Hence by successively only applying inertial torque of one sign, and skipping over the corresponding opposing torque - by changing radius around the zero-CF straight section instead of the rotating section - we progressively lower or raise the momentum of what would normally be regarded as a very much closed system.
You're seeing the wood, you're seeing the trees... i can't be the only one seeing the unicorn?