How to build a Permanent Magnetic Shielding device applied to a workable motor

[Dr.Wlazlak]

A light coming out of an LED from a joule Thief - Well: interesting because
just yesterday: I was holding a 4 watt Led light that runs on 120 volts in my hand standing on a rug on the floor 
doing nothing but holding the light in my hand by the glass bulb - and the bulb ( glowed ) 
The big question - What experiment done does not make a LED light glow - 

Nothing moreover to say about that - Just only the topic has nothing to do with magnetic switches - 
( the topic has to do with ) Magnetic Shielding - just a little different than a switch -

But again - people should not Quote other people's words - or items that may conflict in the future with their own 
confusions that may be sorted out as being owned by the confused ones presenting someone else's ideas that may
not be as wished to be confused as theirs of ideas, in the first place ( just a thought )

Always in the most confused state of mind, Din ma ( moss ) retired engineer and aka Dr What his name?

There are so many to remember - somewhat confusing ) Well: at least at this time of the day: 

( Finally -- I forgot where I parked my Car )  The Finally Theory 

[Classic]

@Dr.Wlazlak if you don't mind, I would like to correct your calculus regarding capacitor charging.

First of all when we charge a capacitor we measure energy transfer and not energy dissipated. This guy have assumed there an ideal transfer and not accounted for any loss, just for the sake of demonstration.

Now, power consumed or dissipated if you prefer is calculated as 1Joule = 1Watt in 1 second; and we can assume is power dissipated as it is energy taken out from source.
We can now translate 1000 Volts at 0.12 Amperes the output of transformer without any loss and observe that your calculus is correct regarding energy PER PULSE and indeed, power drawn is very small per pulse ! This is why we need to add time.

Power= watt
Energy=watt x time

It really doesn't matter how many times in 1 second we have pulsed the power out, the result will be the same in 1 second (without accounting for loss). What really matters is the fact the capacitor will get this power and when we account for energy stored (without accounting for loss) we calculate power delivered in unit of time. This is when you store a quantity of electric charges by each pulse.

In the demonstration the author have assumed the capacitor will reach the 1000 V by adding a quantity of electric charges per each pulse to match much slower rate of discharge of the same quantity of electric charges at 1000 V.
Now, it is just a matter of calculus to find out how much time we need to fill a capacitor at its rated voltage (time constant) and capacity of capacitor can be determined based on this calculus.

It is now obvious that if we try to drain the capacitor at the same rate as charging rate we can only obtain a loss in real world where loss is accounted for.

Of course in real life we need to account for loss if we aim to have an efficient system and make required adjustments in a timely manner and most of the effort will be placed on switching mechanism.
As in your magnetic shielding motor the effort to switch at right angle is much less than trying to power up the whole system and futile attempt to obtain more energy output then energy output.

Energy conserved in permanent magnets versus energy conserved in capacitors. So, if I keep the capacitor open to be drained continuous I only harvest the loss and the same in permanent magnet if I keep the shield static and your aim is to use stored magnetic energy with a required raport of transmission.

[Classic]

A very simple demonstration can be performed very easy: we take 4 aa 1.5 V batteries and measure the current and voltage to make sure the batteries are almost equal in terms of power.

First set of 2 battery in series is connected direct to an LED and when set a time counter the moment they are connected.
Second set of 2 batteries in series is connected to an electrolitic supercapacitor at least 10 Farads, then this capacitor connected to an identical LED whit a time counter set separately. We measure voltage of capacitor to match the voltage of first set of batteries without capacitor.

Measurements: first we run capacitor set until capacitor is discharged and LED can't be lit. Note the time it takes.
Second we run the LED powered direct by batteries for the same time as capacitor set was run.

Or, much easier: take 2 aa 1.5 V batteries in series and direct charge 1 F supercapacitor. Measure voltage of batteries and current before charging capacitor and after. When voltage has reach maximum voltage in capacitor, measure current. Normally you should get at least double amps in capacitor then what was initial in your batteries at almost same voltage.

Use short and very thick wires, eventually multistrand.

We can run the test until both sets of batteries are depleted, or a single run and measure battery sets after usage. Draw your own conclusions.

[citfta]

Classic. 
 I am going to wait for Dr. Wlazlak to respond since this is his thread.  After his response, unless he says otherwise your very off topic posts will be deleted.  Keep your discussion on topic or in your own thread.

Respectfully,
Carroll

[Classic]

Classic.
 I am going to wait for Dr. Wlazlak to respond since this is his thread.  After his response, unless he says otherwise your very off topic posts will be deleted.  Keep your discussion on topic or in your own thread.

Respectfully,
Carroll

I guess I will be put on mute again or maybe banned from forum.