The Paradox Engine

[lumen]

I'm not sure how you are driving the wheel but a test should be made on the energy required to spin the disk X RPM when the arm is free, as opposed to the same RPM when the arm is fixed.

It stands to reason that the disk would take less energy to reach X RPM with a fixed arm. The RPM would need to be checked with the arm stationary in both tests since a mark on the disk would pass the center of the arm in only 3/4 turn if the arm rotated 90 degrees before the mark on the disk passed the center of the arm.

It's clear to see that the RPM of the disk's mark passing the center mark would appear the same when the arm is rotating as not rotating, but in the stationary arm case the disk would actually be spinning faster.

So in the end, the energy applied is simply divided between the two rotations.

This is of course only my view and I have not done the testing myself.

[infringer]

This is a rather interesting way to look at it lumen it may in fact take more energy to get to the higher speeds with a lose rig then something that is fixed like a football player playing on artificial turf without cleats the start and the top speed of his run will be far different than if he had cleats.

But it also does make one wonder about stuff like planetary rotation and the forces that keep things going within not only a planetary system but a galaxy as well.

It may just be that this is a more profound discovery than we are giving it credit for.

Either way this theory could easily be proven with a fairly simple but semi expensive measuring device which I assume the builder already has within his toolbox

Now this is collective thinking I am happy for this input to allow the user to take a closer examination this is a fairly simple test simply secure the arm take measurements and then run it loose and take the same very measurements and investigate data for any conflicting information before moving on to power generation and investing more time and money in that.

Thanks a bunch lumen.

[Tusk]

Thanks once again, I appreciate all your input. Regarding the suggestions from infringer and telecom - if I attempt any further experimental work I'll keep those ideas in mind.

Also broli, you may be interested in this:

http://www.overunity.com/13079/the-pendulum-bias-paradox-experiment/#.UrZiTPvcDcw

... and I do believe that if you familiarise yourself with frame of reference manipulation, you will gain a rare clarity of insight on the KE issue. Since you asked for some numbers for the PE apparatus it might be worth running a few for a theoretical idealised example. If we allow a totally frictionless system and distribute the mass for simplicity and greatest effect, we can aim at the best possible result and work backwards from there adding in the usual inefficiencies etc later if need be.

I'll specify a single disk PE device similar to the experimental apparatus already in existence, but with a main rotor arm of mass = 0 (therefore no counterweight). Obviously this would cause some practical problems which we can disregard for the sake of this example. The disk itself has it's total mass concentrated around an outer ring which we will consider as sharing a common radius.

mass of disk = 200gm
circumference of disk = 120cm

We already know that any force applied to the disk at the outer edge will create an equal secondary reactive force in the same direction at the disk axis. We can allow that the total mass of the disk can be considered to act on the rotor arm at the point of the disk axis. Since the disk and the rotor arm have the same radius:

one rotation of the rotor arm = one rotation of the disk

(in terms of mass in motion, whether that be momentum, acceleration or kinetic energy.)

In explanation of the above, both the disk and the rotor arm can be considered as having the same mass rotating around their axis at a radius of (approx) 19cm.
The application of force will be such that the rotor arm accelerates to complete one full revolution; the disk will advance under inertial and geometric advantage thus completing two full revolutions in the same period, since the system is free of any resistance (but subject to inertia which is causal to the additional rate of rotation).

allow the velocity of the disk axis due to rotor arm motion to be  (after one full revolution):

v = 100cm/sec

v²-u² = 2as              (u = 0)

also      F = ma       (m = 200)

therefore (approximating)      F = 8333dyne

When considering work done we need to interpret the displacement of the point of application of force, which in this instance has a direct relationship with the displacement of the mass (the disk);  although not a linear displacement it may be considered as such. So, since 1 erg represents the amount of work done when a force of 1 dyne moves it's point of application a distance of 1 centimetre and the force applied over 120cm:

work done accelerating the rotor arm 1 complete revolution = 999960 ergs

Note that I did not begin with the disk due to the rotation rate being double that normally expected due to the application of the applied force if it were a simple flywheel (i.e. 'rotor arm secure' in my actual experiments) or if you prefer, double the rate of the rotor arm. And as we will be braking the rotor arm motion first, the additional rotation of the disk will be eliminated, reducing to one revolution in the period of time which would have otherwise been one revolution of the rotor arm, which will now be motionless (briefly).

Alternatively we could simply recover energy from the disk at this point and sacrifice the rotor arm reversal under disk braking (the entire system would cease all motion at the conclusion of disk braking under these conditions). In the frame of reference of the drive unit, disk rotation will have been double as noted earlier, but this must have been achieved with an applied force equal to the secondary reactive force since the two are equal; not unlike accelerating a car downhill the reactive force allows us (in this instance)  a 'two for one' return on our investment of energy. But we can do better than 'two for one'.

So, adopting the preferred method (and allowing 100% efficiency):

the work potential of the rotor arm = 999960 ergs (primary motion)

and the work potential of the disk after recovery of the energy of the rotor arm = 999960 ergs

and since recovery of the energy of the disk creates another secondary reactive force in opposite direction to the first, the rotor arm accelerates again in opposite direction but as before, therefore:

work potential of the rotor arm = 999960 ergs (secondary motion)

999960 x 3 = 2999880 ergs (from the 3 separate motions)

Thus for the ideal 100% efficient theoretical limit on work potential of the PE apparatus as described with an applied force as defined:

work expended = 999,960 ergs

work recovered = 2,999,880 ergs (300%)

Btw lumen there was an earlier post of just such data, I'll add another here though of a typical test run. In fact the disk gains energy significantly faster with the rotor arm free.

[broli]

Tusk SI units are a little easier on the brains . However I seem to agree with your cycle, to sum it up.

1) Put in X units of energy to accelerate disk from the arm's reference point
2) Both disk and arm will be rotating now, in opposite directions
3) Break arm from earth's reference point
4) regain X units of energy from this
4) Disk will be still spinning with X units of energy
5) Break disk from arm's reference point
6) regain X units of energy from this
7) this will cause the arm to start rotating with X units of energy
6) Break arm from earth's reference point
9) regain X units of energy from this
10) No motion in the system left
11) repeat 1)

[Tusk]

Well done broli, if I had thought to put things so succinctly myself perhaps we could have got to this point sooner. One small criticism, a potential clarity issue for anyone reading through your steps without prior knowledge of the device; in steps 3, 5 and 6 I think you meant to say 'brake' rather than 'break'. So if I may repeat with the corrections and some renumbering, since it's such a fine explanation:

1: Put in X units of energy to accelerate disk from the arm's reference point
2: Both disk and arm will be rotating now, in opposite directions
3: Brake arm from earth's reference point
4: regain X units of energy from this
5: Disk will be still spinning with X units of energy
6: Brake disk from arm's reference point
7: regain X units of energy from this
8: this will cause the arm to start rotating with X units of energy
9: Brake arm from earth's reference point
10: regain X units of energy from this
11: No motion in the system left
12: repeat 1)

(with thanks to broli for this analysis)


Outstanding effort 

I guess that makes it somewhat easier for anyone looking to find a hole in the plot, so fire away by all means (but please read the entire thread first since I've covered quite a lot of ground already and don't wish to choke the thread with repetition). The most common areas of concern (i.e. disbelief) are as follows:

1. That a secondary force of equal magnitude and acting in the same direction as the applied force manifests at the axis of the disk.

2. That the location of the EM drive unit mounted at the main rotor arm axis allows both disk and rotor to accelerate with a single applied force, sufficient in convention for one or other motions but not both simultaneously.

3. That a second acceleration of the rotor arm equal and opposite to the first manifests due to the secondary force acting when the disk is subjected to braking.

These three issues have been discussed at length, however if any of these or some other aspect of the material continues to elude acceptance after reading through the thread then I will happily attempt a clearer explanation.