Method for converting HV (static) into usable low-voltage power.

[wings]

@ Steven Dufresne
from :

http://books.google.it/books?id=0RmkmrFxHM0C&pg=PA71&lpg=PA71&dq=%22the+super+tube%22+Moray&source=bl&ots=0rOGUKlrkV&sig=6UA4S52QSjdElnDEXD4pXXqlYf0&hl=it&ei=0Y8fS4WqO4St4Qb5n_zfCg&sa=X&oi=book_result&ct=result&resnum=7&ved=0CCoQ6AEwBg#v=onepage&q=%22the%20super%20tube%22%20Moray&f=false

[Steven Dufresne]

@wings,
Thanks. I'm quite familiar with that diagram from Moray King's "Quest for Zero Point Energy". He derives it from Hyde's US patent 4897592 (see attached.) Hyde doesn't show any inductors but for the circuit to serve the function that Moray thinks it serves he adds the inductors.
-Steve
http://rimstar.org   http://wsminfo.org
PS. I highly recommend that book for anyone who's pursuing tapping ZPE and needs some ideas or stimulation.

[the_big_m_in_ok]

@wings,
Thanks. I'm quite familiar with that diagram from Moray King's "Quest for Zero Point Energy". He derives it from Hyde's US patent 4897592 (see attached.) Hyde doesn't show any inductors but for the circuit to serve the function that Moray thinks it serves he adds the inductors.
-Steve
http://rimstar.org   http://wsminfo.org
PS. I highly recommend that book for anyone who's pursuing tapping ZPE and needs some ideas or stimulation.

Here's the patent:
#4,897,592

http://www.google.com/patents/about?id=HX4BAAAAEBAJ&dq=patent:4897592&as_drrb_ap=q&as_minm_ap=0&as_miny_ap=&as_maxm_ap=0&as_maxy_ap=&as_drrb_is=q&as_minm_is=0&as_miny_is=&as_maxm_is=0&as_maxy_is=

Sub-patent references to the main patent above:
#4,622,510
#4,595,852
#4,151,409
#4,127,804
#3,013,201
#2,522,106

--Lee

[Tito L. Oracion]

hi everyone good day 

I just want to share this one here:

http://www.youtube.com/watch?v=sjtwZwBaprc&feature=related
http://www.youtube.com/watch?v=49iwl-GsYaA&feature=related

and this one may help for some body there.
http://www.youtube.com/watch?v=sb4yiW7MFWM&feature=related
http://www.youtube.com/watch?v=7w1nLaOLtTg&feature=related

[Khwartz]

A lot of people have a hard time with evaluating the current in electrostatic systems.
So i felt the need to discuss the curent of an electrostatic discharge.

Now, when we measure electrostatic voltage potential - what we are actually measuring is a charge relative to some theoretical "zero value" being that of the charge held by the object recieving the discharge. We assume that an earth-ground holds a 'true zero' value.

Unlike current from a constant source, such as a generator or battery, the current from an electrostatic discharge is not constant. It takes the form of a symmetrical triangle-wave: increasing from 0 to peak during the first half of the discharge, then decreasing from peak to 0 during the second half. This makes accurately taking physical measurements of the current rather difficult.
Assymetry in this triangle-wave is a function of ionization of the dieletric and will not be included here.

What we will discuss is the method of calculating the peak current rating of a discharge, which occurs at 1/2 of the discharge-time (T)

To do this, we need to know two values:

1) Total Charge ( in Volts)
and
2) Discharge Time ( in seconds)

From this we can determine the rate of discharge, or change in Volts. 
Essentially:  Total Charge / Time of discharge
for example:
a charge of 13,020 V and a discharge time of 21.7 microseconds
gives us a rate of discharge of 600v per microsecond

The Peak Current of the discharge is:
                A(peak) = 1/2 change-in-V * K
where K is the dielectric constant
in the above example:  1/2 (600V/microsecond) * 1 (the dielectric constant of air)
gives us  300 microamps      below i have drawn how this triangle wave would look on a graph.

The 'mean-current' or average current of the discharge is a bit more complex, one would separate this graph into individual measurements, at say 1us intervals, add them together, then divide by the number of intervals to get the average current throughout the discharge.
which im not going to go through right now, but in this example would be something close to ..... 53 microamps give or take?

Hi sm0ky2 ! Just to tell you that you've made a little confusion: you talk about "charge", "discharge" where you should talk about "voltage", "drop of voltage".

Indeed, if you talk about "charge", it means "coulombs" and the ratio of "discharge" versus time is in coulombs/second, thus amps but pretty sure you knew it

Note:

VOLTAGE

is

HOW MUCH A UNIT OF "CHARGE"* IS "ENERGETIC"

in joules/coulomb.

(*a certain number of electrons)

You may not need this note but as I see often basic confusions around, I try to clear up them each time I can


But Very Thanks for sharing these formula, I didn't know

Regards,
Didier