### Author Topic: Stanley A Meyer February2021 Q@A  (Read 421 times)

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##### Stanley A Meyer February2021 Q@A
« on: January 29, 2021, 02:20:34 am »
Stanley A Meyer February2021 Q @ A 1 page

The EPG is the topic this month with emphasis on mag-gas production and devices.
Still need more documentation of the 7 tier mag gas EPG especially on output and method
for the electronic extraction circuit See the following post for a detailed explanation.
« Last Edit: April 09, 2022, 14:24:51 pm by jim miller »

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##### Re: Stanley A Meyer February2021 Q@A Page **
« Reply #1 on: February 01, 2021, 10:16:14 am »
A0201-01 Thanks sandia24 . Nice work

A0201-03  I was setting up the variable list for Problem 4 on the power output ad input see if this is reasonable--Q's on
clarification or suggestions to team drop box or pm if your are an ionizationx member
_________________________________________________ _________________________________________________ ________________________
MAIN DISCUSSION FOLLOWS BELOW

VARIABLE  LIST AND VALUE RANGE

VARIABLE                                                                                                                   VALUE             SOURCE/// REFERENCE

V1 = velocity of magnetic field movement per second                                                                      50-90 ips               In 2019 Handout in Bremen Conference
N1 = number of twists per unit length of non-magnetic spiral divider per unit length                        0.3 - 1.2               Estimate for  M4steel considering thickness
and core diameter.
N3 = value of magnetic field strength                                                                                              ?                 TBD by calc. and type of EPG
F1 = value of the frequency pulsing alignment coils for dyne-axis of magnetic field                           60 Hz/sec           Mains frequency in US  50 some parts of EU---  wiki
N4=  number of coils per tier                                                                                                                  1 - 58                 Don Gabel, photogrammetry and SEPG022,
N5 = number of turns in each coil                                                                                               200 - 12000      Estimates using packing fractions, winding depth,
length photogrammetry as secondary verification
P1 = effective cross-section winding factor: random, hexagonal or precision winding            0.78- 0.906      Wiki refs circle packing theory
N6 = number of core sectors enclosed by pick-up coil                                                                3- 4                Don Gabel  images of various  EPG's
N7= number of tiers                                                                                                                      1  - 7                Birth of New Technology   1994 or 1995 ed
A1= cross-sectional area of tubing uses in EPG tier ( in inches0                                               0.218- 0.254     The Copper Handbook

Power Input Variables
)
W1 = watts required for initiation of flow    ( Initial inertial load)                                 Rheological, mass density and volume consideration   TBC
W3 = dyne-axis load                                                                                                                                 see  appendix  TBC

Known values

N1  known
N2  known
N3  calculation to be completed
N4  known
N5 known
N6 known
F1 known
P1  known
V1 known

Stated  design output  was 220 VAC @ 300 amps       ( per seminar notes)

A0202-01    Let's try another attempt  at N3. At one of the conferences in 2019 ( SMC 2019 Bremen Ohio), it was proposed that the Transformer EMF
equation might be used in the mathematical model of the Meyer EPG series regarding the flux density problem.

Through photogrammetry the  maximum number of turns , number of coils, diameter and volume of the core
magnetized slurry/gas can be determined. Since the output power, velocity, and frequency are known with some precision

It may be possible to arrange the transformer EMF equation to obtain a Beta Max for the flux density!!

Another observation was made at the 2019 Bremen Conference that the larger the core volume, the lower value of the magnetic
saturation could be and still maintain the same power output. This is because the total power output for the device is dependent
in part  upon the total amount of flux present in the magnetic core.

A0202-02    Correct, if the other design factors such as the number of coils, number of winds and same velocity of the magnetic
gas or slurry are maintained, the limitations of the maximum level of  magnetic saturation of the EFH series ferrofluids
can be mitigated.  Basically scale up the volume and the magnetic saturation can be lower and still provide the
design power output. While the 400 Hz mil-spec converters are still an option for the magnetic drives I think
you may want to just keep it simple so that operating frequency matches the 50 or 60 Hz standards for output for
residential use.

A0202-03      thnx to thorzpwr

A0203=01      ok, you forgot to hide your location ,metadata  I'll fix it zo the mib's don't get ya....lol

A0205-01       Since output data is only available for the 6Tmaggas EPG and for the velocity of the magnetic medium, 'I think a problem approach might be
to determine the flux in the 6 multi-tier system as if EFH-1 was present and then scale down to the magnetic pump system  and volume of EFH-1
at the stated velocity and use a ,calculated flux density to determine output characteristics of the  magnetic pump  device in terms of output.
The sizing of the bus bars, the parallel  arrangement of the pick-up coils and the  breakdown voltage of the insulation might put some upper
limits to how it was being operated and limits to the possible voltagexs and amps produced

Design output  220 VAC at 300 amps  =  66,000 watts

A020601       So now let's assign values to some of the variables
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Cross sectional area is calculated as follows:

1. determine the diameter of the tubing    0.5" obtained by photogrammetry  0.5 outside diameter
also confirmed by actual measurement by Don Gabel . (see notebook photos)

2. determine the range of. possible internal diameters     Common types of pipe K L and M that have the
same outside diameter but thickness of inner diameter and wall thickness vary.
Stan Meyer may have used  pre-coiled air conditioning or water supply tubing. for ease of construction.

A very useful free reference is The Copper Tubing Handbook fermi which provides the specifications and measurements
for copper tubing and pipe.

You can google The Copper Tubing Handbook for the pdf  or just click on this link:

https://pbar.fnal.gov/organizationalchart/Leveling/2004%20water%20cage%20work/Cutubehandbook.pdf

One observation concerning the publicly available EPG images, it that there do not seem to be joints on the spiralled sections
themselves although the connecting copper pipes to the pumps or other means of moving the slurry or gas are straight.
I believe Stan Meyerswas practical and tried to keep things simple, so I believe he just used piping that was already coiled when purchased.

So, now let's use the above reference to get a range of possible values for the cross-sections of the copper tubing and pipes
commonly available.. Copper pipe has three basic wall thicknesses: Type K, Type L and Type  M  So even though the outside
diameter may remain the same, a THICKER wall means a SMALLER cross-section inside the tube

So here's the values of cross-sectional area for different copper tubing and pipe in square inches:

Type K    0.218   Type L  0.233   Type M  0.254   So the cross-sectional area for coiled copper pipe  is between 0.218 and 0.254 square inches

Since the 6 tier system is not available for examination at this time, there is a degree of  imprecision for the cross-sectional area value
Because the cross-sectional area is used in volume calculations and in the calculation of total magnetic flux for these systems, the estimates
of system performance depend upon the type of tubing used in the construction
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Length of tubing carrying magnetic slurry/gas

Since the EPGs are of a general circular design, the formula    C = D x Pi  or stated -- Circumference of a circle equals the
diameter of the circle multiplied by Pi    (approximately 3.1416)

Now, if you are trying to find the total length of tubing  used in an EPG which is a spiral, for example(In this case exactly
3 loops, then thinkof this as 3 circles each with a different diameter and circumference
The outer loop is longer than the middle loop which is in turn larger the the innermost ring of loop.

So roughly speaking, let's say you had an EPG like the Magnetic Drive (Red Pump) System and that by examination
or photogrammetryand it was determined that diameter was 17 inches.

If you are using 1/2  inch tubing in the construction, what would be the diameter of the middle loop?

The radius  of the middle loop is moved in by 1/2 inch because of the width of the outer loop or to put it another
way, the diameter of the middle loop would be 16 inches measured across its outside  By a similar reasoning, the innermost  loop
is  or about 15 inches in diameter.

So the length the spiral is approximately ( 15 + 16 + 17) times Pi.   Now Stan Meyer for reasons of type of pump used (B-500 had input and output
connections at right angles)then some portions of the spiral had four loops instead of three so adjustments will have to be
made for this added length.   The total length of is important because this is used in the calculations
for the Volume of gas or ferrofluid being used  and also in the calculations for inductance and the number windings for the
coils as well as the length of wire required for making the  windings
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Coils and length of wire need for project and per coil

End View                "Tube" length

A formula for a single wind around a single circular  core                                                    O                    diameter of wire times 1

1.A formula for multiple winds around a singular tubular core  of length L               O                    diameter of wire  x N  number of windings or wraps

2 A formula for multiple winds around  two adjacent tubular cores of length L        OO                  diameter of wire  x N  number of windings or wraps

.3 A formula for multiple winds around three adjacent tubular cores of length     OOO                    diameter of wire  x N number of windings  or wraps

4 General Formula for multiple winds around  multiple tubes                                         OOOO...               diameter of wire  x N number of windings or wraps

So the length of the tube determines the total number of wraps possible independent of the number of adjacent tubes
(close wraps  no spacing between wraps on tube

Formula  Length of tube (think inductor core) equals the number of wraps times the width or diameter of the wire  L= N times W or    L/ divided by W  = N
ay of determining the number of wraps  that can fit on a given length of  tube or core

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Now for the fun part determining the Length of Wire   needed for one wrap around multiple adjacent cores

Formula for 1 core                                 O           L   = Diameter of core times Pi

Formula for 2 adjacent cores              OO           L = (Diameter of core times Pi)  PLUS  2D  <---    for the wire that bridges the "notch" between the adjacent tubes  (top and bottom)

Formula foe 3 adjacent cores            OOO          L = (Diameter of  core times Pi)  PLUS  4D  <---   to account for the length needed to bridge 2 notches between the adjacent tubes (top and bottom)

Formula for 4 adjacent cores          OOOO          L = Diameter of core times Pi)    PLUS  6D <---    To account for the length  needed to bridge 3 notches between the adjacent tubes (top and bottom)

In summary, we now can calculate the length of a single wrap of wire around multiple adjacent cores and if we multiply that by the number of wraps  or turns that can be wrapped on a given  linear length of core
it is helpfun in deciding amount of wire needed

General Formula for Single Layer 1 wrap or turn around multiple adjacent tubes

L length equals ( Diameter of core or tube) plus ( ( N or number of cores minus 1) times 2)

So now is possible to calculate the number of winds or wraps (single layer0 around an EPG if we know the diameter of the outermost core of a spiralled EPG, the number of "loops" in the spiral, the outside
diameter of the core tubing and the gauge, diameter or width of the wire used to  wrap the core
t
So lets give a quick try for the multitier 6TmaggasEPG

1 tier is about 17 inches in diameter.   Since the line drawing of the 7 tier system  and photographs show the drain/connecting tubes are 180 degrees apart so its possible to keep the number of loops for a tier to
be 2.5 3.5 or 4.5 loops or if the connecting tubes are all  exact  integers of loops the connecting tube could be all on one side.  Or the direction of the flow could be counterclockwise  one one tier and clockwise in the other tier. So based on the line drawing lets say that that each tier has 3.5 loops

Length of core for 1 tier       [ ( 15+16+17)]times Pi ]  plus( 1/2 times 14 times Pi) = 150.78 + 29.99 =  172.77 inches   6 tiers 1036 inches
172.77 inches  divided by .025 inches per turn  (22 gauge wire by photogrammetry  = maximum 6910 turns per tier
6 times 6910  = 41,460 turns  or if you use exactly 3 loops per tier    150.78 times 6 = 904  inches  904 divided by  0.025 = about 36,191 turns  6 tiers 906 inches

Image a n inductor with between 36 and 41 thousand turns of wire and  between  75 and 86 feet long !! depending on method of construction

Design parameters                                                                                                                                   Metric

The design output is 220 volts at 300 amp draw  66,000 watts  (Watts)
(W) 220 times 300 amp draw   = 66,000 watts

The cross-sectional area of the core is between 0.218 and  0.254 square in
or (A)  =  1.406 to 1.634  sq cm or 0.0001406 0.0001634  square meters

F (frequency) is  60 cycle/ second AC

V (voltage) is 220 volts AC output

K Constant  = 4.44

Solving of Bm =BetaMax

Basic equation

V   =  voltage
F  = supply frequency
N = number of turns
A = cross sectional area in square meters
B = peak  magnetic flux density in Weber / meter squared or T tesla
K = 4.44

V = 4.44  x F x N x A x B or rearranging this

B =   divided by( 4.44    x  f x  N    x   a )

so let's try plugging in a few figures for a six tier device

V = voltage       220 VAC
F=  60 hertz per second in the US
N= 11,873
A =  0. 000468  sq m area         3 channels of pipe  x 0.242 sq inches divided by conversion factor 1550  =  000468 square metres
4.44 = constant
Bmax  =   220/ 1480  or      0.1486  Wb/M squared  or Tesla     for the 5/8" six tier system4.44 times F*N * BetaMax * A

Rearranging: BetaMax  =  V  divided by ( 4.44 x F x  N x  A)

220 divided by(  4.44 times  60 Hz/sec  frequency times 36191 x .218 A sq inches   =  .0001046

now to work on units. with a different diameter..
V = 220 VAC...
F = supply frequency
N = number of turns
A = cross sectional area in square meters
B = peak  magnetic flux density in Wb / meter squared or T tesla
K = constant

V = 4.44  x F  x N  x A x Bmax,   or rearranging this

Bmax  =        V divided by( 4.44 * F *  N * A )

so let's try plugging in a few figures for a six tier device with a  5/8"  OD copper spiral

V = voltage       220 VAC
F= 60 hertz per second in the U
N = 11,87
A =   0. 000468  sq m area                                       3 channels of pipe  x 0.242 sq inches divided by conversion factor 1550  =  000468 square metres
B = BetaMax
K = 4.44   ( constant )

Thus Bmax  =   220/ 1480  or      0.1486  Wb/M squared  or Tesla     for the 5/8" six tier system

[    b]Next Topic Multiple layer coils

In terms of construction if the cross sectional area is changed because of using a larger diameter tubing but keeping N number of turns the same and the length of the
spiralled coils is the same and other factors the same (same desired output) t because the output is related to the amount of flux of the core, the larger the core in terms
of cross section (and volume) means that a lower Beta value in the core of  the upsized EPG can  still result in the desired power output.  Basically if more power is
needed the large core can allow for a lower amount of flux to be used if there is a limit to magnetic saturation for the slurry or mag-gas matrix.

This is more useful to calculate wire requirements for the Mechanical Pump EPG .
Since it's possible to estimate the thickness of the coils, the length of the original coils,
the gauge of the wire and velocity of the ferrofluid 50 ips  and using a flux value estimate
a power output for the Mechanical Pump EPG.
« Last Edit: January 12, 2023, 03:34:31 am by jim miller »