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Author Topic: Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?  (Read 5358 times)

Offline BinaryMan

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Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?
« on: November 29, 2005, 09:30:00 AM »
I've been doing some mathematical modeling of faraday disk dynamos using equations found here:http://www.geocities.com/CapeCanaveral/Hangar/5307/Principles.htm. The results are rather interesting. However, a certain assumption is made:

I have read that drawing current in opposite directions on the disk or drawing current from the entire circumference equally will reduce/eliminate the back-torque on the motor. Has anyone been able to verify if this is true or not, and under what condition this works?

There appears to be an optimized value for generator speed that creates the highest net output power. There are several variables to consider in the equations I saw:

B = Field Strength on the disk
R = Resistence of the disk/brushes
r = Radius of the disk
f = Rotational speed of the disk
m = mass of the disk (proportional to (r^2))
T = back-torque ratio on the motor from induction

Through testing different values I discovered the following:

1) The maximum net power gain (maxgain, or output power minus input power) depends on: B, R, and m only (also T)
2) B, R, and m have an exponential relationship to maxgain (R=3rd power, B=6th power, m=2nd power) such that maxgain = (B^6)/[(R^3)*(m^2)]*(1-T)
3) For any given B and r, there is an optimal f, and the system has a COP of 1.5 at this level (ignoring backtorque) regardless of the other variables (although input/output power may change, it maintains the same ratio)
4) If you double r with B constant, then f = f * 4, and maxgain remains the same (no advantage)
5) If you double B with r constant, then f = f * 4, and maxgain is multiplied by 64x (advantageous)

What this really means is that you don't have to increase the radius of the disk to achieve the optimal power output. Even with a small disk, you can get the optimal power output by increasing rotational speed. Of course, there are practical limitations, so you might want to go with a bigger disk to decrease the rotational speed. You can increase net output power by reducing the mass of the disk while maintaining the same radius, or reducing the resistence of the disk/brushes (can also cover a larger % of the disk surface, which should reduce T and R as well). However, the largest gains will be made from increasing the strength of the magnetic field around the disk. If using permanent magnets, then B is relatively constant. What would be better perhaps is to use a solenoid; it won't demagnetize, you can control the magnetic field strength, and you get an exponential (current^6) increase in power (I think).

What are realistic values for these variables? (I am guessing):

B = 0.3 tesla (rare earth magnets sufrace gauss can be 0.4-0.5 tesla, with a small distance to the disk); higher value possible with coils
R = 0.010 ohms (disk surface + brush resistence; good estimate for 1 brush? Multiple or wider brushes in parallel should reduce this)
r = 0.1 meters (any value really, but we will say 0.1 meters (about 4 inch radius))
m = 1.64 kg (if disk is 1/4 inch thick and made of copper; ~2e-4 m3 of copper at 8230 kg/m3. A thinner disk will be lighter of course)
T = 0.5 (full edge coverage is said to reduce this to ~0.25; we will assume only half the surface is covered with brushes)
f = 0.291 rotations/sec (solved from program)

[Program Output]
Rotations/sec: 0.291
Backtorque: 3.76095586728108E-04
Current: 0.2742610386942
Efficiency: 3.00141330911077
NetPowerGain: 1.25483259804341E-04
PowerInput: 2.50612326923767E-04
PowerOutput: 7.52191173456216E-04
Voltage: 0.002742610386942

Extra power produced is negligable. However, if you cover the disk completely with contacts, reduce the thickness to 1/16 inch, and use higher grade materials, then:

B = 0.5
R = 0.001
r = 0.1
m = 0.41
T = 0.2
f = 51.76

Rotations/sec: 51.7579999999701
Backtorque: 132.197950536826
Current: 813.01276292819
Efficiency: 1.87498931555775
NetPowerGain: 176.261925203708
PowerInput: 352.529876943596
PowerOutput: 660.989752684129
Voltage: 0.81301276292819

You would gain 176 watts of extra power despite the backtorque, making the system over-unity. Hopefully there are no errors in my math. Would anyone like to verify it?

Free Energy | searching for free energy and discussing free energy


Offline hartiberlin

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Re: Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?
« Reply #1 on: November 29, 2005, 10:22:12 AM »
With such N-machines the output power is induced inside the
output cables.
There the magnetic field crosses the cables and there is the induction.

This is always forgotten.
In my opinion N-machines are a dead end to go.

Many have proven, that the N-machine concept is flawed and Tewari and dePalma
were wrong...
I myself visited dePalma in 1987 and he could not show anything...

Regards, Stefan.

Offline BinaryMan

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Re: Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?
« Reply #2 on: November 29, 2005, 11:17:58 PM »
Does it make any difference if the magnets are not attached to the rotor (they are stationary actually, it's not an N-machine)? My actual inspiration is from a page that addresses the N-machine but goes in another direction http://www.stardrivedevice.com/over-unity.html. I have not been able to find Mark Tomion (the site owner) on any of the "fraud-watch" sites, so it might be legitimate. The real question is if one can reduce the motor drag from the induced current. I'm almost done with building a small model to test the Faraday Induction idea, but it seems to small to really cover it with brushes as suggested on the site. I have two sources which state that back-torque can be reduced:

"Taken together, the preceding material clearly shows that ? with ideal stator shielding ? back-torque will only be expressed in an induction dynamo (with a fixed stator) to the extent that eddy currents are allowed to circulate in "unused" radial sectors of the disk*, and this principle can now be used to figure what percentage of the classical eddy current back-torque will be produced: it is simply (but not strictly) proportional to the ratio of that portion of the disk's circumference which is not 'covered' by its collector brushes to its entire circumference!"

http://www.stardrivedevice.com/over-unity.html.

"Note that the current will be extremely high and according to the Biot-Savart law a force will result from this current and magnetic field. The force results according a charge at velocity 'v' crossed by the Magnetic Induction - F=qv x B. Knowing this direction of force (out of the disk perpendicular to the axis) we can cancel it out by using symmetrical brushes - one on each side of the disk and extract current in parallel. This will then cancel the effects of torque and minimize any frictional drag that just one brush may have caused."

http://www.geocities.com/CapeCanaveral/Hangar/5307/Principles.htm

Is this effect real, or mathematical BS? Unfortunately after a month of researching I still don't really understand the interactions of magnetic fields of magnets with those produced by current in a wire. I even created a program to model fields in 3D but what puzzles me is where the "north and south" are in a single straight section of wire; in coils it's more clear. Any suggestions where to to go from here?

Free Energy | searching for free energy and discussing free energy

Re: Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?
« Reply #2 on: November 29, 2005, 11:17:58 PM »
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Offline BinaryMan

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Re: Faraday Disk Dynamo Analysis - Overunity Possible Perhaps?
« Reply #3 on: November 30, 2005, 01:49:06 AM »
If the magnets and brush/circuits are stationary and only the disk is rotating, how does the disk cut the flux? The external circuit has no relative motion to the magnets...

-----

I did some more thinking to see if the power output truly relates to the magnetic field to the sixth power (seemed too high an exponent). I discovered the following:

[base equations (from website)]
Output power: (pi^2) (f^2) (B^2) (r^4) / (R)
Input power: (2) (pi^3) (m) (r^4) (f^3)
Efficiency = (B^2) / [ (R) (2) (pi) (f) (m) ]
Voltage = (pi) (f) (B) (r^2).

I also know the max efficiency (COP) for any given setup is 1.50 from testing. Therefore the optimal rotational speed (f) is given by:
f = (B^2) / [(3) (pi) (R) (m)]

and also:
m = (r^2) (d)
where (d) is the density of the disk per unit surface area (kg/m^2).

Substituting for (f) and (m) into the other equations:

Output power: [ (3/27) (B^6) ] / [ (R^3) (d^2) ]
Input power:  [ (2/27) (B^6) ] / [ (R^3) (d^2) ]
Voltage:      [  (1/3) (B^3) ] / [ (R) (d) ]

Which shows that the magnetic field strength is the most significant variable. Since the strength of the magnetic field of a coil is proportional to the current running through it, we can get a significant boost in output power from the disk (at a rate faster than we add current to the coil). This means the generator would be cheaper (magnets are expensive vs. coils) and last longer (no degaussing of magnets to worry about), and you have more control over the magnetic field strength to prevent overheating etc.

 

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