Getting energy from asymmetry of the magnetic field experiment

[ayeaye]

One more idea came to my mind. I don't say that it works, but something to try, and it may work. Considering how this experiment is done. There is a box, and the small magnet is on the lid of the box.

The force gauges are expensive. But there are electronic scales, that are very precise, and quite cheap. Of course they assume an equal distribution of weight, and an exactly perpendicular force.

Attach a hard disc, and a rod on it, to the electronic scale. So that the scale is put 90 degrees, and the rod reaches the small magnet. Then don't pull, but push. I don't know all the details, but when carefully done, this may actually enable to measure forces.

Another idea, that someone else, somewhere else told. Now that works by pulling. Attach a string to the magnet, over a pulley, and attach to the end of it a known weight. Then put that weight on the scales. Somehow make the length of the string adjustable. Then the scales will show the weight of the known weight, minus the force to the small magnet.

This will more likely work, but is somewhat more difficult to make. But people have tried, when putting an electronic scale to stand on its edge, it shows some force. This depends on the construction of the scales, some may not. But that with a weight and a pulley always works.

Also, just pulley, and a very exactly adjusted weight. Weight is the force. And the position of the magnet is exactly where the force to it is that great. This requires some good set of weights. And doesn't enable to measure force when moving, which the previous method may somewhat do, when moving the scales down some way.

[ayeaye]

As i see it. When using only a pulley and a weight to measure the force. When there is no friction, there is only one force to measure. And the magnet is always at the position where there is such force.

When there is friction, there are two forces to measure at any position of the magnet, instead of one. One is the minimum weight with which the weight still doesn't move up. This corresponds to the standing force. The other is the maximum weight with which the weight still doesn't move down. This corresponds to the moving force.

The real force is the average of these two forces. The friction can be calculated too, which in this case is the static friction, if the static friction is greater. The friction force is a half of the difference between these two forces. The friction in that case is also the friction of both the magnet and the pulley. Except when the friction of the pulley is very small, and can be disregarded.

What concerns the static friction, in that case what we always deal with, is static friction. To find the moving friction, if necessary, the ratio between the static friction and the moving friction should be measured. There is likely no greater static friction in case of dry slippery surfaces. But static friction is greater, like when there are ball bearings in the pulley.

Hope that this may help.

[ayeaye]

Floor,

"you will find that it requires more work to separate
two magnets in attraction form one another, when they are slid side ways, than the work done when they
are pulled directly apart"

I don't know, there are many trajectories, i measured only the one, where the big (standing) magnet was tilted 45 degrees, and there the energy was greater at the side of the pole. And the movement was linear. It may be very different when we approach from some direction, then turn 90 degrees, and leave. Though it may be very difficult to make a mechanical device that will provide such movement.

[Floor]

There is NO TIME ELEMENT in a calculation of mechanical work.

One may derive the work done as in the work done to cause acceleration.

The unit of measurement, Newton of force, is itself derived from / based upon, mass and acceleration.
And acceleration of course has a time component within it.
               because at least
In theory this is a more fundamental and precise method to derive the unit of force (The Newton)  from
mass and acceleration.  F= ma

In actual practice it is almost never done done this way.
...........................
A 101.971 gram mass (roughly 102 grams) exerts as weigh about 1 Newton of force (down) in standard gravity

The lifting of (or the falling of) a 101.971 gram mass, 1 meter in standard gravity is = 1 joule of work done as that lifting or falling.

Lifting 101.971 grams 1 meter against standard gravity in 1 second of time is = 1 Watt of power expended.
..........................
A 1 kilogram of mass exerts 9.80665 Newtons of force (down) in standard gravity (roughly 10 Newtons)

The lifting of (or the falling of) a 9.80665 gram mass, 1 meter in standard gravity is = 9.80665 Joules or about 10 joules of work done as that lifting or falling.
...........................
The speed of the lifting and the speed of the falling do not change the amount of energy or the work done AS THAT LIFTING OR FALLING.
The work done / energy expended, to cause acceleration is not taken into consideration in a calculation of the work done as lifting or of the work done as falling.

If one were accelerating a mass horizontally (no gravity involved) on can arrive at the work done through the formula   Kinetic energy = 1/2 mass x velocity squared.
This is a different matter, and best to leave acceleration out of the. at this time.
...........................

@ EyeEye

Are you going to set and do a presentation of the work done in separating magnet at different angles and so on ?
I think it would be really grate to see some experiments which are confirmations of some of  the other research Ive
seen on the internet.

Lifting 1 kilogram 1 meter against standard gravity in 1 second of time is = 9.80665 Watts (roughly 10 Watts) of power expended.
.............................
1 Joule of energy transferred or work done, in 1 second of time is one watt of  power expended.
10 Joules of energy transferred or work done, in 1 second of time is 10 watts of power expended.
..............................
               best wishes
                       floor


        PS
              Looking at these 2 graphs (attached below) one can see how / why there could be a difference in the two actions.

[ayeaye]

The video  https://archive.org/details/asymmsurf2

The result was not what i expected, it appears that the magnet gets more energy at the left side, but with this linear movement it may be so, and it corresponds to how the magnets were tilted in my other experiment. Something like field lines that go to the other pole, bend away faster, and thus the force at the right side decreases much faster, one can clearly see that in the video.

That confirms my this experiment  https://archive.org/details/Flcm3  in a measured way.

The following was measured from the video. The angles by which the spring scales were tilted, were considered. At both sides the movement was considered so long, that at the end of it there was no measurable force to the magnet.

At the right side the magnet moved 3.57 mm with the force 0.8 N, and 2.29 mm with the force 1.0 N, thus energy at the right side was 5.15 mJ.

At the left side the magnet moved 17.43 mm with the force 0.6 N, and 6.71 mm with the force 0.4 N, thus energy at the left side was 13.14 mJ.

The force when moving was 1.2 N at the right side, and 0.8 N at the left side. Considering that the force when moving was constant, then the forces considering friction were the following.

At the right side, first 3.57 mm -- ( 0.8 + 1.2 ) / 2 = 1 N, friction 0.2 N, 3.57 mJ, next 2.29 mm -- ( 1 + 1.2 ) / 2 = 1.1 N, friction 0.1 N, 2.52 mJ.

At the left side, first 17.43 mm -- ( 0.6 + 0.8 ) / 2 = 0.7 N, friction 0.1 N, 12.20 mJ, next 6.71 mm -- ( 0.4 + 0.8 ) / 2 = 0.6 N, friction 0.2 N, 4.03 mJ.

Thus the energy at the right side was 6.09 mJ, and the energy at the left side was 16.23 mJ, 10.14 mJ more.

By these calculations, only 4 mJ goes to friction during all movement through the field, 6 mJ should be left. Thus the magnet should go through all the field when starting from the left, and gain speed. Yet, when releasing the magnet from the beginning of the field at left, it stops at the neutral position. I have not tried to make it to go through with an initial speed.

What i would like the most, is someone to replicate this experiment, many thanks.

Floor, i don't do fancy presentations. My experiment was described above, plus all the rest of the data in this thread.

The force of separation between the small magnet and a small disc magnet was 2.3 N. The force of separation between two small disc magnets was 1.8 N.

The force of separation between the big magnet and a small disc magnet was 1.6 N. The force of separation between the big magnet and the small magnet was 2.6 N. At that the small disc magnets always attracted to the edge of the big disc magnets.

The force of separation between two big disc magnets was 5.0 N.

The small magnet was two ceramic disc magnets 10 mm in diameter and 5 mm thick, one on another, and the big magnet was 8 ceramic disc magnets 25 mm in diameter and 5 mm thick, one on another.