Tri-Force Magnets - Finally shown to be OU?

[sm0ky2]

The increase in energy from A -> B is the difference in the integral force * the distance between an infinite number of points between A and B. i chose to use 13 points for simplicity.

[Omnibus]

The increase in energy from A -> B is the difference in the integral force * the distance between an infinite number of points between A and B. i chose to use 13 points for simplicity.

In the magnetic field the magnetic potential energy decreases from A to B. This is something you never understood.

[sm0ky2]

The increase in energy from A -> B is the difference in the integral force * the distance between an infinite number of points between A and B. i chose to use 13 points for simplicity.

In the magnetic field the magnetic potential energy decreases from A to B.

Yes - this is true in this case.
It is exactly the inverse of the entrance of the Tri-Force (Repulsion - in which it increases).

However - the potential energy of the Ball increases - because the Force is at a vector angle to the gravitational force. vertically below the SMOT gravity is dominant.
Perpindicular to Gravity SMOT is dominant in the horizontal plane.
Movement in the horizontal plays no role on gravitational potential. so once the ball is at the horizontal, potential energy is at Max. It would be more efficient to start the ball to the left of Point B, to obtain more of the magnetic potential of the field. So you are correct in saying this.

But when you are at a vector angle to another force, the Potential Energy of the force*distance is not the full potential of the field. Its the vector resultant. Determined by the location of the ball, with respect to the 'edge' of the field. (Point C).

Examining the full field in this device is irrelevant. You can only examine the observable energy imparted on the ball. Input into the system, and Output of the System (whichever points you choose for your input/output).

If you examine the full field, then you must utilize the full field - by lifting the ball mgh, outside of the field, and placing point B, outside of the field. Then account for the energy from B to the start of the field, where the ball is attracted in. This would be the only applicable set-up, where the full magnetic potential could be used.

In the SMOT you presented this is not the case.  The potential increases as you lift the ball, in fact you can FEEL the ball trying to pull towards point C, you have to hold it back to position the ball at Point B.







     

[Omnibus]

Magnetic potential energy at A in the magnetic propulsor diagram is maximum and this you never understood. Period. Stop with this confused, non-scientific blabber.

[sm0ky2]

Magnetic potential energy at A in the magnetic propulsor diagram is maximum and this you never understood. Period. Stop with this confused, non-scientific blabber.

classical Magnetic potential does not translate to Potential Energy on the ball at that vector angle,
it is at the minimum point in the cycle. potential energy at this point is effectively 0. the ball stops there.

Why don;'t you understand this.?? it is simple.

at Point B, there is more potential energy in the ball, than at Point A. This is why the ball moves to point C.  For someone who claims to be a SMOT "expert" you sure dont know a lot about this Device...


If you lower the SMOT, and eliminate Point A. Start the Ball at Point B, put the entire device on the same horizontal plane - then what you are saying would hold true.  But in the set-up you have here, gravity overcomes the magnetic force at Point A, thus the magnetic force imparts no (net) potential energy onto the ball.