The Paradox Engine

[Tusk]

I've estimated that for the disk speed of 100m/sec with the disk's diameter of 30 cm,
it will need to rotate at 6000 rpm.
What would be the rpm of the arm in this case?
How much energy will it produce by cycling between , say, 6000 to 7000 rpm
for the disk?

Q1. As I've often said telecom, the twin disk setup is probably optimal; also as near as possible to a ring mass rather than a disk, and until recently I've preferred them mounted on a lightweight rotor arm such that we can virtually ignore it's mass. With this latest idea (multiple cycling of the disks to achieve OU with the rotor arm) we may be looking at a deliberately massive arm. But staying with the earlier setup for now, in that configuration the rotor arm RPM matches the disks/rings.

Q2. That would depend on the mass. But with the ring mass you can treat rotation velocity (I.E. the velocity of the mass around the disk/ring axis) much like linear velocity, and if you study the layout of the device you will notice that the radius of the disks/rings is similar to the radius of the rotor arm provided you don't extend it too far beyond the axes of the disks. And you can treat the rotor arm configured in this way like another ring mass, since the disk axes are at their cm and are mounted at the ends of the rotor arm. So popping in some values and calculating the outcome is fairly straightforward.

Do You understant what mean difference between paralel and perpendicular situation

Was your intent to make my answer quick and simple tesla2, or does friendly scientific inquiry in your first language translate into aggressive insults in English?

I'll make some allowance for the language thing. This is how it works; you find something that doesn't seem 'quite right' in the information I provide, and either pop in a quote of the offending material with your question or rebuttal, or at least mention it as a starting point for your question or rebuttal.

As it stands I have no idea what got stuck in your windpipe, and the very last thing I need right now is a lecture on FoR and Relativity. But just putting myself in your shoes for a minute, considering I got this far it seems a fair bet that I do know the difference between parallel and perpendicular. I'm just taking a wild guess here but are you by any chance referring to my earlier statement re secondary reaction with that question? If so then it's also a fair bet that you haven't even got beyond accepting it, which is a tough one I agree but there you have it. Either provide an alternative or dismiss it as just one more vague and elusive unknown in your life, move on and leave the crazy old guy to his unintelligible raving and ranting.

[CANGAS]

Tusk, we have definitely suffered from a failure to communicate. Just like the Georgia cracker warden said to Luke. The failure to communicate has been all yours, dear fellow.

A little while, a few days, ago I responded to this specific post. For once, please focus your communication faculties upon the mentions of "squared" and "1/2" in the specific post copied and quoted below.....

This is a really good thread!  Keep up the good work Tusk!  Below is a copy and paste summary of a publication on the kinetic energy equation, by Miles Mathis (reference link provided below).

Why is the velocity squared in the kinetic energy equation, E = ½mv²?  Why should the energy depend on the square of the velocity? We have the same question with the equation E = mc².  Why square the speed of light? Why should the energy depend on c²?  Or, to extend the question, why should the energy of any moving object, moving with a constant velocity, depend on the square of that velocity?

In Miles Mathis' paper on photon motion, he showed how the measured wavelength and the real wavelength of the photon differ by a factor of c². This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c². Energy, like velocity, is a relative measurement. A quantum with a certain energy has that energy only relative to us, since it has its velocity only relative to us. If the wavelength has to be multiplied by c² in order to match it to our measurements, then the mass or mass equivalence will also. Hence the equation E = mc². In this way, c² is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.

The same analysis can be applied to any object. The energy of any object is determined by summing the energies of its constituent atomic and quantum particles, and all these particles also have spins. The quanta will impart this spin energy in collision, so this spin energy must be included in the total kinetic energy.  So the short answer is that the kinetic energy equation, like the equation E = mc², always included the spin energy; but no one recognized that.  Just as with the photon, all matter has a wavelength (see de Broglie), and the wavelength is determined by spin. The spin has a radius, and this radius is the local wavelength. Any linear velocity of the spinning particle will stretch our measurement of this wavelength, in a simple mechanical manner, as Mathis showed in the photon paper. As the linear velocity increases, the spin velocity relative to the linear velocity decreases, by a factor of 1/v. This makes the difference between the linear velocity and the spin velocity v². The term v² transforms the local wavelength into the measured wavelength. This is why we find the term in the energy equation.

The only question remaining is why we have the term ½ in the kinetic energy equation. The reason is simple. We are basically multiplying a wavelength transform by a mass, in order to calculate an energy.  So we have to look at how the mass and the wavelength interact.  Mathis has shown that the wavelength is caused by stacking several spins (at least two spins), so what we have is a material particle spinning end-over-end. If we look at this spin over any extended time interval, we find that half the time the material particle is moving in the reverse direction of the linear motion. Circular motion cannot follow linear motion, of course, and if we average the circular motion over time, only half the circular motion will match the linear vector. This means that half the effective mass will be lost, hence the equation we have.

Reference:  The kinetic Energy equation, by Miles Mathis

Additional Resources:  Angular Velocity and Angular Momentum, by Miles Mathis (Both current equations are shown to be false)

Gravock

Upon reading this garbage, I could not believe my eyes and asked the poster if he was serious, if he REALLY did not understand where the "squared" and the "1/2" came from in the standard Newtonian derivation of the formula for Kinetic Energy. Do you remember any of this?

Soon you jumped in to defend the poster and attacked me even to the extent of describing your hallucination of a Wild West shootout in which you would wield a double barrel shotgun. I had simply asked the poster if he really did not understand the "squared " and the "1/2" or was he joking. Since I have derived the Kinetic Energy equation myself, I have long since been completely satisfied that the logic and math , particularly the "squared" and the "1/2" are perfectly logically and mathematically self consistent within the framework of Newtonian physics. You, by vehemently defending the poster obviously took his side that you also did not understand why or how the "squared" or the "1/2" are in the Newtonian Kinetic Energy formula. Your comments in this regard caused me to waste a substantial amount of time and energy which, in my old age, are increasingly important and which I cannot retrieve and you cannot possibly re-supply to me. Do you remember any of THIS, or are you wandering around in your own imaginary world of double barrel shotguns and fake Kinetic Energy equations?

Thanks a lot for all the BullShit.


CANGAS 78

[Tusk]

Indeed I had little real hope (based on the tone of your posts) that you would recognise an olive branch for what it represents, CANGAS.

Since you have quoted gravlock's apparently 'offensive' post several times I must assume that you read and understood that he was himself quoting from another source:

Below is a copy and paste summary of a publication on the kinetic energy equation, by Miles Mathis (reference link provided below).

It seems fairly self evident that he felt the material he referenced might be of some interest and relevance to this thread, or at least of some general interest; and relevant or not (I have not yet decided) it is certainly an interesting theory, especially taken in context with the vast repository of original work on the source website. So yes, a friendly and informative greeting by gravock and received in kind. I wish there were more like him.

I have explained the playful nature of my wild west 'hallucination' as you call it but you clearly reject that; I have responded to what can only be described as an unpleasant post with the offer of a fresh start and no hard feelings, and you have rejected that. On this occasion, much like yourself I consider my time valuable, so that time spent in reply to your unfortunate hostility has blown out of all proportion to any possible advantage and therefore must be curtailed. Since it's not clear what your true agenda is, and considering the disruption you have caused thus far - to no good purpose other than reinforcing your own sense of superiority and appearing unnecessarily rude - I can only hope that you will make yourself content with a final string of invectives on your way out the door.

[telecom]

Q1. As I've often said telecom, the twin disk setup is probably optimal; also as near as possible to a ring mass rather than a disk, and until recently I've preferred them mounted on a lightweight rotor arm such that we can virtually ignore it's mass. With this latest idea (multiple cycling of the disks to achieve OU with the rotor arm) we may be looking at a deliberately massive arm. But staying with the earlier setup for now, in that configuration the rotor arm RPM matches the disks/rings.

Q2. That would depend on the mass. But with the ring mass you can treat rotation velocity (I.E. the velocity of the mass around the disk/ring axis) much like linear velocity, and if you study the layout of the device you will notice that the radius of the disks/rings is similar to the radius of the rotor arm provided you don't extend it too far beyond the axes of the disks. And you can treat the rotor arm configured in this way like another ring mass, since the disk axes are at their cm and are mounted at the ends of the rotor arm. So popping in some values and calculating the outcome is fairly straightforward.

Hi Tusk,
do you mean that we can use a simple kinetic energy equation E = 1/2 mv^2?
Presuming, that as you said, rpm of the rotary arm is equal the rpm of the disk,
and its mass equals 1 kg,
E will be 5000 J at 6000 rpm and approx 6800 J at 7000 rpm.
So during the cycling we should be able to harvest from the rotor arm 1800 J during
the acceleration, and equal number during the deacceleration at reverse, 3600 J in total.
At the same time we should spend 1800 J to accelerate the disk, of which we should be
getting back 70 % during the deacceleration, with the losses of 540J.
Total surplus should be 1800 + 1800 - 540 = 3060 j.
Considering that the cycle will take 10 seconds, the device should produce 3060 / 10 =
306 W of power.
Is this within the reasonable margin of error or not???

[gravityblock]

Hopefully you will be able to identify the source of any advantage amongst all these dark arts? lol. Will you start another thread or post news/results here?

An afterthought gravock - the point of applied force on the disk/s must be at or near the edge and keeping a small footprint for best effect; not sure how you are going to achieve this with the method you outlined.

I'm going to use the drive wheel instead of the EM drive.  This eliminates all permanent magnets!  Disk A and disk B will be based around the homopolar principals as discussed previously.  Extracting current from disk A will induce a torque from the edge of the disk to the disk's center of mass which will be applied to the outside edge of the rotor arm and will also be in the same direction as the rotor arm, causing it to further accelerate (either a CW spiral or a CCW spiral according to rotation direction of the drive wheel).  The secondary reaction force will be applied at the outer edge of the rotor arm, which is also the disk's center of mass where the force is applied to the outside edge of the rotor arm in this case, which is an inversion of cause and effect.  In addition to this, disk A will be induced with an additional rotation due to the Lorentz force, further accelerating the disk's rotation.  Disk B will be similar to disc A, and each disc will be accelerating both the rotor arm and it's own disk rotation.  In this way, the rotor arm and both disks are utilizing the secondary reaction force to provide a continuous acceleration for both the disks and the rotor arm.

Gravock