# Re: [Vo]:Successful Mechanical OU

```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.

- 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

- 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.

- 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.

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.

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.

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)

On Mon, Jun 4, 2018 at 7:18 PM, John Shop <quack...@outlook.com> wrote:

> On 5/06/2018 12:37 AM, Vibrator ! wrote:
>
> Consider a 1 kg weight, connected by a pulley cord to another mass that
> slides horizontally without friction.  You may verify that the rate of
> change of net system momentum is a constant, invariant of the ratio of
> gravitating to non-gravitating mass - taking gravity as 9.81 N, it is
> precisely thus 9.81 kg-m/s per kg of gravitating mass.
>
> So, the amount of non-gravitating mass could be anything from zero to
> infinity, but regardless of whether the gravitating mass is rising or
> falling, the rate of change of net system momentum is always 9.81 p/s/kg
> (where p=mV).
>
> This is not, as one might suspect, a consequence of Galileo's principle -
> that gravity defies F=mA - but rather a direct manifestation of it.
> Same-same, no matter what force we apply.
>
> Now switch out that linear-sliding mass for an angular inertia instead.
> If we measure its angular inertia in terms of kg/m^2, and given that moment
> of inertia (MoI) is equal to mass times radius squared, we can select a
> mass of 1 kg at 1 meter radius for an MoI of '1'.
>
> If we measure its angular velocity in terms of radians per second, then we
> have numerical parity with its linear equivalent for an equal distribution
> of absolute momentum - that is, if we applied a linear to angular force
> between them of 1 Newton for 1 second, we obtain 1 kg-m/s of linear
> momentum, and also 1 kg-m^2-rad/sec of angular momentum.
>
> Likewise, if we employ a 1 kg drop-weight to torque up that MoI, the
> system gains 9.81 p of net momentum per second.
>
> Since they're equal absolute magnitudes of inertia, albeit in their
> respective dimensions, the net system velocity remains equally-distributed
> between them.
>
> Hence with 9.81 p of net system momentum, we have 4.905 p on each inertia
> - 1 kg dropping at 4.905 m/s, and an MoI of '1' rotating at 4.905 rad/s.
>
> However, since the objective distance 1 radian corresponds to is dependent
> upon the dimensions of the circle in question (it's a relative, not
> absolute, quantity), this same point applies to the 'magnitudes' of angular
> momentum we're measuring for any given angular velocity; so for instance if
> we double the mass radius, then per mr^2 we quadruple the MoI,
>
> All looks OK (even if rather strange language) to here.
>
> but also halve the relative (angular?) velocity compared to the linear
> value wherein inertia is a fixed function of rest mass.  Hence, repeating
> the 1 second, 1 kg drop, we'd again obtain 4.905 p on the weight, but
> '9.81' p on the MoI - for a 'net' total of '14.715' p
>
> This is numerically correct but dimensionally incorrect (which is maybe
> why you use the quotes).  Angular momentum does not have the same
> dimensions as linear momentum and so they really cannot be added in this
> fashion (just as you can't add 4.905 meters to 9.81 square-meters and
> obtain a reasonable result as 14.715 somethings).
>
> ... i'm using scare-quotes there to highlight my point; the objective
> value of the absolute magnitudes of momentum and their distribution remains
> 9.81 p/s for the net system, regardless of how the angular component is
> represented.
>
> .  .  .
>
>
>
```