Mostly Permanent Magnet Motor with minimal Input Power

[gotoluc]

To everyone,

since this topic has been re-opened I may as well share a better performing version of my Mostly Magnet Motor which I compared its pulling force to an off the shelf speaker.

So here is the link to the unlisted video if anyone is interested to see it.

https://www.youtube.com/watch?v=qa1dO8qWPQU

Luc

[gotoluc]

Here is the second part of the demo video with the Signal Generator attached to it.

https://www.youtube.com/watch?v=oDXq8wkhV2c

Luc

[woopy]

Here is the second part of the demo video with the Signal Generator attached to it.

https://www.youtube.com/watch?v=oDXq8wkhV2c

Luc

Hi Luc

thank's for that

i made a replication and it works very well

it is amazing that as the magnetic fields are totally opposing, the coil should behave as per along a single steel non magnetic bar, i mean very weakly, but it seems not to be the case , the coil is very powerfull.
 
Youp i will follow your work as usual

thank's very much for sharing and bravo !!

Laurent

[gotoluc]

Bonjour Laurent,

thank you for posting your replication and results.

What is happening in this design is once the coil is energized it produces a north and south in the core (steel laminations). Now if all 4 magnets are all facing north in the center core the energized coils south side will attract to the PM north and the north side of the coil will repel from the other PM north side. So here we have both poles of the coil performing work and zero cogging from the PM because they don't leave the cores.
However, it doesn't stop there. Some may not know this but half way through the thickness of a coils windings the polarities start to flip. Meaning, if you put core material on the outside of a coil windings and energize the coil the poles in the cores will be opposite from the coils center core. So here we are also taking advantage of the coils opposite outside magnetic field to do as much work as the coils inner magnetic field.
This basically doubles the work output.

Here is an older video that demonstrates the benefits

Link to video: https://www.youtube.com/watch?v=-eTQ49RcFKM


Your test setup would be much more powerful if your coil winding would fill the bobbin so that the windings would be as close to the outside cores as possible. Also, if you add 4 more magnets and 2 more core laminations to cover most of the outside area like in the video I posted in this reply it will do even more work.

The more the permanent magnet force the more the pulling or pushing force is without any increase in power to the coil.
Surface area is important, so again the more surface area the more the force it has.

Glad you enjoyed this old idea

Luc

[Khwartz]

Yes, but that is not a "rate". It is a quantity. The Ampere is a rate, so if you go at a rate for a time, you have what? A quantity. Go 1 km per hour for one hour. How far have you gone? One km. 1 km/hr x 1 hr = 1 km.
  Just as "per" denotes division, "for" denotes multiplication.

Consider what we mean by "three amps per hour".  This is written 3A/H. How I got that number is I started with zero amps and I turned my variac up slowly and at the end of an hour I was up to three amps, and at the end of two hours I was up to six amps, and at the end of three hours I was up to nine amps. I turned the current up at a RATE of 3 A/H, or three amps per hour.  If you want to know my current at, say, 90 minutes, you multiply the rate by the time: 3 a/h x 1.5 h = 4.5 A. This use is in effect the "rate of change of the rate" or the first derivative, the slope, of the graph of the current vs. time.

One Amp-Hour is a _quantity_.  To get there mathematically you _multiply_ the number of amps by the number of hours. So if I have a system that produces one amp-hour at 12 volts and I run it for one hour, I have sent (one amp-hour) x (12 volts) or 12 Watt-hours past my measuring point, for that hour. A quantity, distributed over the hour time. And for that one hour, then, I am dissipating the energy at a _rate_ of 12 Watt-hours per hour.  12 W-h / 1 h = 12 Watts... a _rate_, the power dissipation, the rate of energy dissipation. Power in Watts = Joules of energy per second -- a rate.

Now are we completely confused?

So we agree just Luc was speaking of the quantity of electrical charges provided by the battery then I only correct him on this point; great you went further

Q : quantity of electrical charges in Coulombs (C)

T: duration in second or hour

I: flow of current in Amps (A)

E: voltage in Volts (V)

W: energy in Joules (J)

P: power in Watts (W)

Amps × duration = Q (not//// energy) (Coulombs or Ah, depending of the time unit).

But as the voltage is near steady 12 V, it makes an equivalence for energy consumption as per:

E × Q = W

while A/h makes an indirect equivalence with power by:

E × (I/T) = P

BUT! "A/h" would mean an acceleration of flow of charges, as mathematical dimensions:

I / T = (Q / T) / T = Q / T^2,

which has usually no use for a battery.

Cheers!