# Re: [Vo]:Successful Mechanical OU

```>
> I don't think so.  The earth has experienced an unbalanced attraction to 2
> Kg masses in free-fall near its surface - so it will have accelerated
> upwards slightly to meet these masses (just as it accelerates upwards to
> meet the moon when the moon is overhead).```
```

Precisely!  If we cycle this interaction using the angular - linear
example, then we're inputting momentum, to Earth, from gravity, on each
full cycle.. thus the "center of momentum" reference frame between Earth
and the weight has been shifted upwards on each cycle!  We're applying
reactionless momentum to Earth, from its own gravity field!  The net system
of Earth-plus-weight has thus become an effective warp drive - gaining
unilateral momentum sourced from its own gravity well / 'warp field'..

However there's nothing special about Earth or gravity in this interaction
either - force is force, F=mA, A=F/m, m=F/A, regardless of its provenance
or rest frame!  The 'earth' (or whatever stands in for it) is thus
momentarily accelerating upwards towards an inertially-suspended mass that
is not accelerating back down in reciprocation!

(I imagine you intended 19.62 kg-m^2-rad/sec^2 here?)
>

Beg your pardon, yes - it's just a 1 second acceleration though, so still
the same value.

That value IS 9.81 however, not 19.62 - remember we're now dealing with an
angular inertia, not a gravitating one, so the only force being applied
between it and the weight is 9.81 N / 9.81 N-m.

Agreed but you haven't specified at what radius the ripcord is being
> applied to?  The moment of inertia is one thing (and it seems you are
> trying to keep it constant), but the radius at which you apply the force
> (via a ripcord or whatever) to produce torque, and spin-up the wheel is a
> separate parameter that you haven't discussed?
>

For 1 kg at 1 m radius assume an equal spool radius, so we're on the same
page..   the applied 9.81 N force pulls the cord whilst pushing the weight
up.  The weight thus hovers in mid air under no net vertical force, whilst
the rotor gains 9.81 p/s.  This is effectively 'reactionless' momentum with
regards to the 'closed system' of two interacting inertias - only one of
which was accelerated, but which we also wouldn't have been able to
accelerate without the other one to apply a the force against.  You're
right however that it isn't really 'closed' since it encompasses Earth, via
gravity.

Again, this is not the interaction i'm using to create energy.  But it is
an interesting example of conventional assumptions being challenged.

If you change the moment of inertia of something that is already spinning
> then its spin-rate and stored energy changes.  This is the well known
> effect that occurs when a spinning skater pulls in her arms.  Her moment of
> inertia decreases which means that spin rate must increase (to keep angular
> momentum the same), and likewise the energy stored in the spin must
> increase (she supplied this energy by pulling her arms in against
> centrifugal force).
>

I meant, if we change the mass of the rotor, keeping all else equal but
changing its MoI, then the same 9.81 N force will cause the 1 kg weight to
rise or fall, instead of just sitting there vertically balanced.  Wasn't
referring to inertial torques from changing MoI on the fly.  My point was
simply to provide graphic demonstration of the equality of the two
inertias, despite having different dimensions.

Fundamentally 'inertia' is a function of how much mass has been accelerated
through how much space in how much time - this resolves angular and linear
terms.

Gravity is a 9.81 N force, so the time rate of change in momentum of a
gravitating system is 9.81 p/s, regardless of the relative nature of
radians as units of angular displacement.  Again, consider the weight
tugging the horizontally-sliding mass - no matter how heavy that
non-gravitating inertia, the rate of change in p is 9.81 p/s/kg of
gravitating mass.  This does not change when we switch the non-gravitating
mass to an angular inertia instead, irrespective of the relative units we
use to describe its velocity!

This change in momentum value as a function of time accelerating is usually
symmetrical to the momentum value as a function of GPE / GMH - so we would
calculate the same potential change in net system momentum from either the
time or height references.

But add your 'ice skater' effect into that mix and that symmetry gets
broken...

Sorry but I don't see your argument.  I am not sure what you mean by
> "absolute" inertia?  Are you speaking of inertia (=mass) or moment of
> inertia (=mass x radius^2) or maybe momentum (m v) or maybe angular
> momentum (m r^2 rad/s)?  They are all very different quantities with
> different units and different dimensions and cannot be added or compared in
> magnitude.

I think i've made my point here - just because we're measuring, say, 19.62
kg-m^-rad/sec /sec^2 does not mean we're obtaining twice as much angular
momentum as another rotor with half the MoI, or indeed a
horizontally-sliding mass.  The 'objective' or absolute rate of change of
momentum for a gravitating system remains 9.81 p/s, irrespective of the
relative units of the measurement dimensions; the
space/mass/acceleration/time relation is constant, or else we'd likewise be
varying the 9.81 p/s rate along with the amount of non-gravitating mass in
the linear horizontal case.

This is true.  It takes much more energy to accelerate a car from 100mph to
> 110mph than it did to accelerate it from 0 to 10mph.  One reason is obvious
> - you would have to apply a constant accelerating force (acting against the
> earth) over a much greater distance for the period when the car's average
> velocity is 105mph than you had to when its average velocity was only
> 5mph.  And energy = force times distance.
>

..thus a rip-off velocity tax; the detail they're catching you on is the
simply drags along beneath it, then you could keep raising momentum against
it without that velocity and thus distance increasing!  IOW, CoE is
enforced by N3.  Effective violations of the 3rd law 'create' mechanical
energy, by raising momentum on the cheap!  The value of that momentum
however remains a standard function of V^2 in the static frame, hence
collect underpants and profit.

On Tue, Jun 5, 2018 at 6:22 AM, John Shop <quack...@outlook.com> wrote:

> On 5/06/2018 4:32 AM, Vibrator ! wrote:
>
> LOL have i not just clearly delineated the terms of their equivalence?
>
> Allow me to put it more tangibly:
>
>  - Apply a 9.81 N force vertically between two 1 kg masses, the moment
> both are dropped into freefall.
>
>  - We observe a kind of inverted 'slinky drop' effect - the upper mass
> hovers stationary in mid-air, whilst the lower one plummets at 2 G.
>
>  - We've thus input momentum to the system, by applying a force between
> two masses, but which has nonetheless only accelerated one of them.
>
>  - Without the upper mass to push against, we couldn't've applied any
> further acceleration to the lower one, beyond that from gravity.
>
>  - So the lower mass will reach a speed of 19.62 m/s in a one second drop
> time.
>
>  - 1 kg @ 19.62 m/s = 19.62 kg-m/s.
>
>  - Half this momentum came from gravity.
>
>  - The other half came from the internally-applied 9.81 N force.
>
>  - So we've definitely raised some 'reactionless momentum' here - with
> certain caveats of course.
>
> I don't think so.  The earth has experienced an unbalanced attraction to 2
> Kg masses in free-fall near its surface - so it will have accelerated
> upwards slightly to meet these masses (just as it accelerates upwards to
> meet the moon when the moon is overhead).
>
>  - Now let's get rid of the lower mass, and replace it with an angular
> inertia, rotating about a fixed axis.
>
>  - We can apply the 'downwards' end of the linear force to the rim, or
> else the axle of the rotor, such as via a ripcord or whatever.  Forget
> about the mass of the 'actuator' for now, just consider the raw
> distributions of momentum from the applied forces.
>
>  - If we choose an MoI of '1', then as before, the upper 1 kg mass will
> hover stationary, experiencing equal 9.81 m/s accelerations in each
> direction, up as down, whilst the rotor spins up at the rate of 9.81
>
> (I imagine you intended 19.62 kg-m^2-rad/sec^2 here?)
>
>  - That MoI of '1' could be comprised of 1 kg at 1 meter radius...
>
>  - ...or equally, 4 kg at 500 mm radius...
>
>  - ...or 250 grams at 2 meter radius..
>
>  - Or indeed any arbitrary distribution of mass and radius within
> practical limits.
>
> Agreed but you haven't specified at what radius the ripcord is being
> applied to?  The moment of inertia is one thing (and it seems you are
> trying to keep it constant), but the radius at which you apply the force
> (via a ripcord or whatever) to produce torque, and spin-up the wheel is a
> separate parameter that you haven't discussed?
>
>  - However, since 'radians' are a function of diameter of the rotor, the
> actual angular momentum we measure IN those units is by definition
> speed-dependent (kg-m^2-rad per second).  It's a relative measure - and a
> very useful one at that - but it also has an objective magnitude, a scalar
> quantity independent of its actual spatial dimensions!
>
> We have proven this, since changing the MoI whilst maintaining the
> internally-applied 9.81 N force will break this balancing act, and the
> 'suspended' 1 kg weight will instead rise or fall.
>
> If you change the moment of inertia of something that is already spinning
> then its spin-rate and stored energy changes.  This is the well known
> effect that occurs when a spinning skater pulls in her arms.  Her moment of
> inertia decreases which means that spin rate must increase (to keep angular
> momentum the same), and likewise the energy stored in the spin must
> increase (she supplied this energy by pulling her arms in against
> centrifugal force).
>
> Thus the equality of the magnitude of absolute inertia - independent of
> its time-dependent measurement dimensions - has been empirically proven.
> You've just disproven anyone who tries to tell you it's conceptually
> 'impossible' to convert, much less compare, between them.
>
> Sorry but I don't see your argument.  I am not sure what you mean by
> "absolute" inertia?  Are you speaking of inertia (=mass) or moment of
> inertia (=mass x radius^2) or maybe momentum (m v) or maybe angular
> momentum (m r^2 rad/s)?  They are all very different quantities with
> different units and different dimensions and cannot be added or compared in
> magnitude.
>
> Now, if i'd just Googled that question, i'd've come to the same conclusion
> as you and everyone else.
>
> But having worked it out from first principles, i do not need to worry
> what anyone else thinks.  Hypothesise, test, rinse and repeat.  Whatever
> the result, it is what it is.  That's the only kinda 'Googling' that really
> counts.
>
> The upshot of that equivalence, however... is an effective 'reactionless
> acceleration', with no change in GPE.
>
> We've applied gravity to cancel or invert the sign of our
> counter-momentum.  There's actually a few different ways of doing this, but
> the most interesting ones are of course those that enable the accumulation
> of such momentum, thus allowing its constant energy cost of production to
> diverge from its effective value as a function of the accumulating V^2
> multiplier, via the standard KE terms.
>
> Consider the energy cost of 1 kg-m/s of momentum, for a 1 kg mass:
>
>  - per 1/2mV^2, at 1 m/s 1 kg has 1/2 a Joule
>
> However if we then wish to add exactly that same amount of momentum again
> - so raising its speed by a further 1 m/s, we find that we need to input 2
> Joules.  Four times as much energy, for the same rise in momentum.
>
> Thus the cost - and value - of accumulating momentum increases via the V^2
> exponent, the more momentum and thus velocity we already have.
>
> This is true.  It takes much more energy to accelerate a car from 100mph
> to 110mph than it did to accelerate it from 0 to 10mph.  One reason is
> obvious - you would have to apply a constant accelerating force (acting
> against the earth) over a much greater distance for the period when the
> car's average velocity is 105mph than you had to when its average velocity
> was only 5mph.  And energy = force times distance.
>
> Taking lightspeed as 299792458 m/s, 1 kg would require 44937758936840900 J
> to reach C.
>
> 1 m/s less than this is 299792457 m/s, at which speed we have
> 44937758637048400 J.
>
> Therefore while our 1 kg mass's first kg-m/s of momentum only cost 500 mJ,
> the final kg-m/s up to C costs a whacking 44937758936840900 -
> 44937758637048400 = 299792500 J.
>
> Incidentally, we might note that this KE value is precisely equal to the
> speed of light in m/s, minus exactly 42 (299792500 Joules - 299792458 m/s =
>
> I digress; point is, it's a right rip-off innit?  Velocity tax,
> basically.  What if we could maintain that initial 500 mJ / kg-m/s or
> kg-m^2 in the angular case... all the way up to C?
>
> 299792458 * 1/2 = 149896229 J.  That's how much energy it'd cost to get 1
> kg up to lightspeed, using reactionless momentum.
>
> Subtracting that from the standard cost: 44937758936840900 - 149896229 =
> 44937758786944671 J.  That's how much energy we'd save.
>
> Incidentally again, however, if we see how many times that cut-price
> ticket divides into the standard V^2 value: 44937758936840900 m/s /
> 149896229 J = 299792458...
>
> Recognise that number?  Yep that's right - using reactionless momentum to
> get to lightspeed is as many times over-unity as there are meters per
> second in C.
>
> So C is 299792458 m/s, and accelerating 1 kg to that speed sans N3 is
> 299792458x OU.
>
> Obviously i'm not suggesting running Bessler wheels at those speeds... but
> i am positively dribbling at the 1.4 meter shift in the zero-momentum frame
> on each cycle of the gain interaction i'm looking at..
>
> With both effects harnessed, not only can we reach 299792458 m/s , but we
> can do so using 299792458 times less energy than would normally be
> required.  Not that we haven't got infinite energy to play with of course,
> but still, waste not want not..
>
> (before anyone dives overboard the above interaction's not the one i'm
> claiming to have successfully implemented)
>
>
>
```