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ASSIGNMENT: MATHEMATICALLY MODEL
THE POWER OUTPUT OF THIS: Easier Said than DONE !
However, a reasonable model for PERFECT conditions can be made using known
principals of PHYSICS and MATHEMATICS !
I have been working this out for over two weeks,
12 hour dayze, just to wrap my head around this. Just to let you know,
I have not done it alone, I have had the VERY patient
help of a real Physicist, to be able to present for the FIRST time on
the WEB, a METHOD that everybody can use !! MANY, MANY, THANKS GOES TO OUR DEAR PHYSICIST AT:
Their Email address is:
They have in house Mathematicians and Physicists who will answer questions using the latest
science,known principals, and apply it to your question with logic and humor. Their site is a great read, and fun for everyone, I encourage you to visit !
Their prompt response(s) to me (seen below) allowed me to sleep again at night !
I have had at least four false starts to this final presentation page,
as inevitably I would get bogged down in either a variable or an equation.
My new best friend, the Physicist says:
“Don’t feel self conscious about the math stuff.
A “good” understanding of E and M requires vector calculus, and almost no one gets that far.
I’d guestimate somewhere around 1 in 1,000, and of those very few come away with a decent understanding.”
As for me, if it wasn’t for their help, I’d still be dreaming of calculations literally.
Diving right in, I will lay out the physical facts of our device, lay out the underlying equations, and
then plug in the pertinent variables and show the results, for our mathematical model of this Alternator.
It is a “potential” model, and as such may or may not represent reality,
as the variables can change in real life but it does give a method for all the DIY folks
who build these to “build” them on the blackboard FIRST Then fabricate the device and hopefully come close to the model.
My main reason for undertaking this was that most of the sites are sorta backwards in that
they have years of experience building and testing the devices, and as such they have a feel
for what is going to work.
But they don’t go far enough for my taste in the math and science behind WHY it works.
I wanted to totally understand WHY it works, and
WHAT happens when you change a variable such as:
The load you attach it to also effects it’s output.
It is AMAZING how much is really involved with the Alternator/Generator.
TESLA was brilliant in being able to build on other’s work and envision the generator.
VERY few improvements on his fundamental work have come about since he first introduced the
world to it, that says a LOT about how elemental his work was !
Here we go:
As my DEAR Physicist says:
“Set it up however you like, just keep in mind that magnetic flux is the name of the
game. Picture the magnetic field as flowing water, and the coils as a hoop.
The goal is to get as much water to flow through the hoop as possible
(and then switch the direction over and over).”
Electromagnetic induction:
Electromagnetic induction is the production of voltage across a conductor moving through
a magnetic field.
It underlies the operation of generators, all electric motors, transformers, induction
motors, synchronous motors, solenoids, and most other electrical machines. Michael Faraday is generally credited with the discovery of the induction phenomenon in 1831
though it may have been anticipated by the work of Francesco Zantedeschi in 1829.
Around 1830 to 1832 Joseph Henry made a similar discovery, but did not publish his
findings until later.
Magnetic flux through an open surface:
Faraday’s law of induction
A vector field F ( r, t ) defined throughout space, and a surface Σ bounded by curve ∂Σ moving with velocity v over which the field is integrated. While the magnetic flux through a closed surface is always zero, the magnetic flux through an open surface need not be zero and is an important quantity in electromagnetism. For example, a change in the magnetic flux passing through a loop of conductive wire will cause an electromotive force,
and therefore an electric current, in the loop. The relationship is given by Faraday’s law:

Simplified for our purpose:
The induced voltage depends on the diameter (area) of the coil and the number of turns of the coil. The power is determined as well by the resistance of the coil. Faraday’s law will give you the induced emf in the coil: That is the potential energy per unit charge in the coil.
If the coil is connected to a load, there will be energy consumed.
The current will be I = V/R. The power is OR:

N=the number of turns in the coil
A=area
B is the
magnetic field
dt=time
ΦB is the magnetic flux in Webers or Tesla
( our Physicist suggests sticking with SI units to avoid conversion issues, the SI unit is TESLA)
the phenomenon underlying electrical generators.
converting the mechanical energy of motion to electrical energy.Although Faraday’s law always describes the working of electrical generators,the detailed mechanism can differ in different cases.
Whew !So there we have it, all or most of the tools to adequately determine the potential output of our alternator.What we need now is data about the physical properties of the alternator, reduce that to variable data and begin to plug in the equations !!These are our test magnets:
and
1″ dia. x 3/8″ thick
1″ dia. x 1″ thickGrade N52 – Nickel Plated NdFeB,Grade N52Axially Magnetized
Axial (Poles on Flat Ends)Surface Field: 4440 Gauss
Surface Field: 6619 Gauss 2051.5 @ .5 distance 1/2″
NO noticeable drop off at .5″ (1/2″)3986.8 @ .25 distance 1/4″
3586.8 @.75 3/4″37.65 lbs force
61 lbs Forcehttp://www.kjmagnetics.com/ A GREAT SOURCE !Field’s strength drops off as distance increasesFACTS:14 gauge wire 0.0641 diameter inches (1.628 diameter MM ){ not an important variable, just a fact }
72 turns per coil1″ SQUARE Rotor Plate “winding post”
12 inside coils12 outside coilsOur “design” RPM = 60 RPM { DESIGN is another way of saying
“HIGH HOSEY”, “cuz we said so”… }we need a number to plug in so we say 60 rpm could be close for
an intermittent windturbine,I will also include a set of calculations for the classic powered
GENERATOR of 3600 RPM.
Our “design” R, for LOAD/resistance is
a 100 watt lightbulb: 144 Ω (ohms) (ohm=SI unit as well)Our “design” GAP between the rotor and stator is 1/2 inch = .5″this gives us the greatest Gauss/TESLA, magnetic fluxIt increases the magnetic flux the closer (smaller) the gap is, so bear that in mind, the opposite is true too, it drops off with distance.Now, in my previous attempt in plugging the information in there were several variables that were offI had read the Gauss ratings for the magnets from a “general” table and not specifically for the magnet itself, I found the specific table for them this time.There is some difficulty with the site’s calculator in determining drop off Gauss results for the larger magnets IE: the calculator says you are inside the magnet for a .25 distance, either that or the Gauss is equal to it’s surface (which I will use as a plug in)To Convert Gauss to T(esla):{One tesla is equal to 10^{4 }gauss}OR{1 Gauss = .0001 Tesla, or 1 Tesla = 10,000 Gauss}Original Article at: http://www.watchman2012.com
For our 3/8″ thick magnet and gap of .5″the Gauss =2051.5 T= 2051.5 X .0001 = 0.20515TFor our 3/8″ thick magnet and gap of .5″the TESLA Unit = 0.20515TFor our 1″ thick magnet and gap of .5″
the Gauss = 6619 T=6619 X .0001 = 0.6619TFor our 1″ thick magnet and gap of .5″the TESLA Unit = 0.6619TFinally, I was missing the cross sectional
area of A, of each coilFrom our Dear
Physicist:A = (pi*D^{2})/4 where D is the diameter of that pipe.
If you were to wrap it around a pipe, thenA= (pi*D^{2})/4 where D is the diameter of that pipe.You mentioned that the coils were square.
The area for a square is just A=D^{2}.
Also, make sure to always use SI units.
So, A = (1″)^{2} = (0.0254m)^{2} = 0.00064516 m^{2}.
THE EQUATIONS:So we have a simple yet eloquent equation of:V = (dΦ / dt) = N x A x ( dB / dt)Where:V = VoltagedΦ = the magnetic fluxdt = timeequals:N = the turns in the coilA = the cross sectional area of the coildB = the magnetic fluxdt = timeOur physicist transposes this for us:The magnetic field in the coils should be roughly sinusoidal.Definitely not exactly, but close enough.There will be higher harmonics, but they shouldn’t make a huge difference.
B = aSin([2 pi]fT)
Where:
“a” is the amplitude of the field,“f” is the frequency in Hz,“T” is time.
The time derivative of this is:dB/dt = 2*pi*f*a Cos([2pi] fT)
The “T” in the Cos is just time, not a particular amount of time.It’s saying that, as time goes on, Cosine is going to go up and
down.There’s nothing for you to plug into it.For example:The voltage coming out of a wall socket would beV=170 Sin(2*pi*60 T).Knowing the T is only important if you want to know what the
voltage is “right now”.But it’s not useful, because it’ll be completely different in a
couple hundredths of a second.AC power goes up and down a lot (how much is exactly is just the
frequency).So, you don’t concern yourself with the momenttomoment power
output, and focus instead on the average power output.Happily, the average of Sin^{2} and Cos^{2} is
1/2, (.5) which makes things easier.I grabbed this from wiki, it shows the
relationship nicely:
Here begins my second and third tangents LOLI happily went assigning values for variables, and for the first time
ever, switched my microsoft desktop calculator to “scientific” !I guessed correctly, I might add, what the various buttons do LOL and
went about going thru a great number of calculations to
square variables, multiplication and sum things up for BOTH the 3/8″
magnets, and the 1″ magnets.What I calculated took me over 12 hours to complete !It looked VERY coolI was precisely wrong LOL
I had quickly phoned my friend to warn him to be extremely careful when running experimental tests because if my math and understanding was correct, he was generating HIGH voltage and Amperage.
It wasn’t correct, but caution is advised anyway, since onetenth of an Amp, for onetenth of a second can stop your heart !
I could see that something was off, and made several more hours of
calculations to see if I could change the results.I must say I had great fun doing it.It’s why blackboards come with erasers LOLSo I wrote my Dear Physicist again, and asked more detailed
questions about the equations as I was not seeing thelogic in some of them, and misapplying Cosine, not knowing that
the pretty subequation was there for VERY minute moments in time.Skipping over the incorrect equations and right into the CORRECT
equations for each magnet size with each coilI now have THE DEFINITIVE methodto calculate the POTENTIAL power of a magnet/coil pair:The boiled down facts/variables:Magnets are:B = 1″ dia. x 3/8″ thick =2051.5 Gauss @ .5 ” =0.20515T squared = 0.0420865225and:B = 1″ dia. X 1″ thick =6619 Gauss @ .5″ = 0.6619T squared = 0.43811161N = 72 turns per coilA = 0.00064516^{2 }of each coil =0.0000004162314256R = 144 Ω (ohms)f = frequency in Hertz (Hz)The FINAL EQUATION:Here is our Dear Physicist’s example of the full blown, final
equation @60 RPM (1 Hz) :P = (72^{2}0.00064516^{2}/144) (2*pi*1*0.6619 Cos(2*pi*1*T) )^{2} =
(72^{2}*0.00064516^{2}*2^{2}*pi^{2}*0.6619^{2}/144) [Cos(2*pi*1*T)]^{2} =0.000259 (Cos(2*pi*1*T))^{2}So to make a blank, fill in the variables, equation, we start with:P = ( N^{2} x A^{2} / R) ( 2 x pi x f x B Cos(2 x pi x f x T) )^{2 }=(N^{2} x A^{2} x 2^{2 }x pi^{2} x B^{2} / R) [ Cos( 2 x pi x f x T ) ]^{2}Further:since we know the Happy average of Sin^{2} and
Cos^{2} = 1/2 = .5We simplify it to:P = ( N^{2} x A^{2} / R) ( 2 x pi x f x B x .5) ^{ }=( N^{2} x A^{2} x 2^{2 }x pi^{2} x B^{2} / R) x .5( N^{2} x A^{2} x 2^{2 }x pi^{2} x B^{2} / R) x .5 <—– THIS is INTERNET GOLD for ALL DIY WINDTURBINE DESIGNERS !!!!!!WHERE:N = Number of turns in the coil squaredA = The cross sectional Area of the coil squared in: (m^{2 }) an SI Unitpi = pi^{2 }= YUMMY !B = Magnetic flux of the magnet in T(esla) Units – SquaredR = The Resistance of the connected load in Ω (ohms) an SI unit.5 = our Happy average of Sin^{2} and Cos^{2}AGAIN:MANY, MANY, THANKS GOES TO
OUR DEAR PHYSICIST AT:Their Email address is:They have in house Mathematicians and Physicists who will answer questions using the latest science, known principals, and apply it to your question with logic and humor.Their site is a great read, and fun for everyone, I encourage you to visit !Their prompt response(s) to me allowed me to sleep again at night !Original Article at: http://www.watchman2012.com
The Calculations:For the 1″ x
3/8″ magnet coil pair @ 60 RPM:P = ( 5184 x 0.0000004162314256 x 4 x 9.865881 x
0.0420865225 / 144) = (0.000024887213527721514387891264 x .5) =
0.00001244 Watts per coilP = 0.00001244 Watts per coilx 12 coils =0.00014932 Watts @ 60 RPMx 120 Volts = 0.0000012443 AMPERES===================================================For the 1″ x 1″ magnet coil pair @ 60
RPM:P = ( 5184 x 0.0000004162314256 x 4 x 9.865881 x
0.43811161 / 144) = (0.000259 x 0.5) =
0.0001295 Watts per coil !!!!!!!!!!!!P = 0.0001295 Watts per coilx 12 coils = 0.001554 Watts @ 60 RPMx120 Volts = 0.00001295 AMPERES===================================================For the typical Generator Speed of
3600 RPM (60 Hz) :We need to adjust the
equation for the increased Hz:( N^{2} x A^{2} x 2^{2 }x pi^{2} x 60^{2} x B^{2} / R) x .5( 60 RPM = 1 Hz, so you take a Hz reading of YOUR alternator, Multiply by 60, then Square it to plug into the equation )For the 1″ x 3/8″ magnet coil pair @ 3600 RPM:P = ( 5184 x 0.0000004162314256 x 4 x
9.865881 x 3600 x 0.0420865225 / 144) =
(0.0895939686997974517964085504 x .5) =
0.0447969843498P = 0.0447969843498987258982042752 Watts per
coil
====================================================
0.43811161 / 144) = (0.9326538649839237318887998464 x .5) =
0.4663269324919618659443999232
design 120 Volts to calculate the Amperes of the coils, just to show
what COULD potentially be produced.
perimeter rotor:
on TOP of the 3/8″ magnets to increase the magnetic flux.
concerned that the field may not reach the inner coils
properly.
3/8″ magnets looks like:
on paper over JUST the 3/8″ magnets and the field looks like
this:
inside of the ring where the magnets are placed.
field looks like.
would reach the inner coils:
inner coils, but we wanted it stronger
3/8″ magnets to increase the field strength
trailer axel and spun that puppy up !
following:
have a good comparison of calculated vs actual
Dear Physicist !
for comparison.