New Perm Mag Engine Design 1.5 : 1 Ratio (work from magnets)

[Floor]

@conradelektro

[lumen]

@Floor
Interesting results!

If you get time to make some changes in testing a similar operation, I would like to suggest a change that may result in a larger work result.

1: Change the rotating magnet to a diametrically magnetized cylinder. The rotation axis will need to change also, so the poles are flipped on 180 degree rotation.

2: Change the slide magnet to a ring magnet with a hole diameter about 1/2 the length of the cylinder magnet, and an outside diameter of twice the length of the cylinder magnet.

3: Bring the two together in attraction until the cylinder stops attracting and use this point as the end of the slide action.

4: Rotate the cylinder 180 into repel and do all the same fine testing as you have already done!

Just a suggestion. Keep up the good work!

[conradelektro]

@Floor:

Thank you for the very detailed explanations. It helped me to understand what you measured.

My thoughts:

Situation 1): rotating magnet and sliding magnet are parallel, repelling force = force to move the magnets into close proximity:

The 750 mL of water (750 grams) are necessary at the end of the movement of the sliding magnet to bring it into close proximity to the rotating magnet. But the repelling force depends on the distance of the magnets. So, when the magnets are farther apart one needs less than 750 mL (750 grams) to move the sliding magnet. The force is not constant, but increases the nearer the magnets are. Conclusion: the force necessary to move the magnets together (or the repelling force) is not constant over the whole way, therefore the simplified work calculation (constant Forth x Distance) is wrong.

Situation 2): rotating magnet and sliding magnet are at 90°, forth to turn the rotating magnet 90° into the parallel position:

The 500 mL of water (500 grams) are necessary at the end of the turn once the rotating magnet and the sliding magnet are almost parallel. At the beginning of the turn (when the rotating Magnet and the sliding magnet are at 90°) the force is much smaller. So, in this situation the force depends on the angle between the rotating  magnet and the sliding magnet and is never constant. Conclusion: the force to rotate the magnet is not constant over the 90° turn, therefore the simplified work calculation (constant Forth x Distance) is wrong.

Remarks:

It is rather complicated to do a correct forth and work calculation because the changing field strength of the magnets has to be known at every distance (or angle). The assumption that the forth is constant over the whole way (or the whole 90°) is wrong.

The change of field strength is different in situation 2) and situation 1). It can be assumed that the turning magnet experiences a stronger field on average because its distance to the sliding magnet is always small. On the other hand, the field strength  between the sliding and the rotating magnet diminishes rapidly as the magnets move apart.

Said in short:

The forces are not constant, therefore the simplified work calculation (constant Forth x Distance) is wrong.

The measurements only told the maximum force near the end of each movement (rotation or sliding).

It would be necessary to measure the changing forth in very small steps between the beginning and the end of the movement (rotation or sliding) along the changing field. But in practice this would be difficult if not impossible.

When done in e.g. three equal steps the values of the work in situation 1) and 2) will be much closer than in the wrong calculation (which assumes a constant force). The calculation of the approximation would be: (F1 + F2 + F3)/3 , F1 is the force at a third of the way, F2 is the force at two thirds of the way and F3 is the force at the end of the movement (at three thirds of  the way, F3 is the value used in the wrong calculation).

Speaking in mathematical terms:

The changing magnetic field between the magnets is described by a "differential equation" and the factors in this equation depend on the physical properties of the magnets (which are hard to know).

To calculate the work one "integrates" the force over a "path" through the changing magnetic field.

Conventional theory says that the work in situation 1) and situation 2) is equal (not accounting for friction losses).

Greetings, Conrad

[conradelektro]

Because the contraption is beautifully built and very instructive I do a more complete analysis:

In a world without friction

W1 = work necessary to move the "sliding magnet" towards the "rotating magnet", both magnets are in the horizontal position, movement starts from the point where the two magnets are farthest apart and ends when the magnets almost touch

W2 = work necessary to turn the "rotating magnet" from a vertical position to a horizontal position, the "sliding magnet" is very close to the "rotating magnet", the movement is a 90° turn of the "rotating magnet"

W3 = work necessary to move the "sliding magnet" towards the "rotating magnet", the "rotating magnet" is in the vertical position and the "sliding magnet" is in the horizontal position, movement starts from the point where the two magnets are farthest apart and ends when the magnets almost touch

W4 = work necessary to turn the "rotating magnet" from a horizontal position to a vertical position, the "sliding magnet" is farthest apart from the "rotating magnet", the movement is a 90° turn of the "rotating magnet"

(power stroke) W1 = (recovery stroke) W2 + W3 + W4 ;  W3 and W4 are rather small (in a world without friction) because the magnetic field between the magnets is rather weak


In the real world

W3 and W4 have to be done essentially to overcome friction (the magnetic fields are weak).

W3 (in essence against friction) also has to be done while doing W1 (in essence against magnetic filed).

W4 (in essence against friction) also has to be done while doing W2 (in essence against magnetic filed).

W3 and W4 are much bigger in the real world (against friction) than in the ideal world without friction (only against weak magnetic field).

Power stroke: W1 + W3

Recovery stroke: W4 + W3 + W2 + W4


Conclusion

In the real world the "recovery stroke" costs more work than the "power stroke" because of friction.

In an ideal world (without friction) the "recovery stroke" and the "power stroke" would need the same work.


Remarks

All my explanations (in my previous post) about the changing forces in the changing magnet field still hold. It is therefore rather difficult to calculate W1, W2, W3 and W4.

In conventional theory:

Moving two magnets together costs work.

In a world without friction it does not matter along which path the magnets are moved together, it will always cost the same work (always starting from the same start positions).

In the real world the more complicated path will cost more work because of friction losses.


Note

Work = Force x Distance (in case the force remains constant over the whole distance)

Work = "average force" x Distance (in case the force changes along the distance in a linear way, which is not the case with magnets, but could be used to get an approximation)

Work = "force integrated along a path through the changing magnetic field" (this is the case in this machine)


I hope this helps, greetings, Conrad

[Floor]

@lumen

Hello Lumen

I was not totally able understand  the descriptions and explanations given in your last post. Can you clarify them.

Please find attached the file TD floor 2 Lumen 2