Turning on and off a Pernament Magnet Field

[JackH]

Hello z_p_e,

""2) You originally stated that the means of turning a magnet OFF was by way of applying a ~180kHz square wave across the magnet (perpendicular to the field).""

The above quote was not any of my idea.  I did not talk about using ~180kHz square wave across the magnet, that must have been someone else.  I have only two ways to turn on/off a permanent, one is by dc power(8 watts), the other is by means of a mechanical lever.  Both have been patent applied for. The dv voltage is not applied across the magnet, this I think would totally destroy the permanent magnet.

Later,,,,JackH

[z_p_e]

Hi Jack, thanks for your response.

I see that I have made a mistake in thinking that this topic, as well as the "Permanent Magnet: ON/OFF Mechanism..." topic were speaking of the same inventor. Being that you had posted in that topic, I assumed it was a more or less duplicate topic. I wonder if others made the same erroneous assumption.

Anyway, that definately cleared some things up...thank you.

You mentioned that 8 watts is required to do your switching, and again I'm wondering, what is the percentage reduction in the field?

You also mentioned that your method increases the strength of the magnet by 4X. In this "ON" state, are you still using 8 Watts of power? In other words, when using the DC voltage means of switchng the magnetic field, is 8 Watts of power required for both the "ON" AND "OFF" states?

Regards,
z_p_e

[JackH]

Hello z_p_e,

Well I'm sorry but I just do not know what you mean when you say, what is the percentage reduction in the field?

8 Watts of power is only needed for the ON cycle. No power is used during the off state, in fact it gives power back from back EMF during the off state. Currently I am just feeding this back EMF to ground through a capaciter, but it will be used in the future to help run the motor.  During the off state the magnetic field is at about 99% gone.

Later,,,,JackH

[z_p_e]

Hi Jack. Sorry for the confusion.

What I mean by the ON state, is the default state of every magnet, with no influence to try and increase or decrease it's field by some external effect. In other words, take any normal magnet in your hand...this magnet its in it's ON state.

Now take that same magnet, influence it by some external effect (such as your invention), and the magnet's field is either reduced significantly, or eliminated all together....the magnet is now in what I have been calling an "OFF" state.

I am referring to the state the magnet's field is in, not when you are using the electric power (8 Watts) to achieve your effect.

What I meant by percentage reduction of the field, could be stated by a ratio such as: "100% - Boff/Bon", the magnetic field ratio of OFF to ON, taken from 100%. You have already answered that by saying 99%, thanks.

8 Watts of power is only needed for the ON cycle. No power is used during the off state

Having said what I meant by ON and OFF, and reference to the magnet vs. applied power, does your quote still stand, or is it now reversed?

Taking a stab:

[400% magnetic field strength, magnet ON state] = no electrical power required or applied
[1% magnetic field strength, magnet OFF state] = 8 Watts electrical power applied to achieve effect

Is this correct?

Regards,
Darren

[hartiberlin]

No power is used during the off state, in fact it gives power back from back EMF during the off state. ....
During the off state the magnetic field is at about 99% gone.

Later,,,,JackH

Ahh Jack, now I have understood your design.
You norrmally have shorted out the flux of the statormagnet
via a shortout core at the OFF state, no power required.

And then use an 8 Watts energy pulse via a coil around this
shortout core to redirect the flux
from the shortout core to a different core, where your rotor
magnet is driven , right ?

Looks like an "inverted" Flynn design then...
Regards, Stefan.