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 kg-m^2-rad/sec.
(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 
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 

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 

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 = 42).  Eat 
your heart out Douglas Adams..

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 = 

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)

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