Free Energy is Easy - YOU Can Do It!!!

My Friends,

The Magnetic Field is an amazing thing! The Changing Magnetic Field over Time, in other words, a Magnetic Field at Frequency, this means, at an applied Amplitude, Duty Cycle at a regular Cycle, or Frequency.

I have said, for a very long time now: Electromagnetic Induction is Symmetrical, as we understand it in Conventional Science today:

Electromagnetic induction itself is a symmetrical phenomenon, but the resulting effects and applications can sometimes exhibit asymmetry.

 

Symmetry in Electromagnetic Induction

The principle of electromagnetic induction, as described by Faraday's Law of Induction, is fundamentally symmetrical:
\[ \varepsilon = -N \frac{d\Phi}{dt} \]

 

This law applies universally and symmetrically to changing magnetic fields and their interaction with conductors.

 

Asymmetrical Effects

However, the specific applications and implementations can introduce asymmetry:

1. Physical Coil Design: The physical design of coils and transformers can be asymmetrical. For example, a solenoid coil might have different magnetic field strengths along its length due to its geometry.
2. Magnetic Fields in Devices: Magnetic fields in devices like motors and generators may not be evenly distributed, leading to asymmetrical field distributions and forces.
3. Eddy Currents: Induced eddy currents in conductive materials can create asymmetrical heating and electromagnetic effects due to the shape and orientation of the material.
4. Asymmetrical Loads: In electrical circuits, asymmetrical loads can result in uneven distribution of currents and voltages, even though the underlying induction process remains symmetrical.

 

In essence, while the fundamental law governing electromagnetic induction is symmetrical, the practical applications and physical implementations can introduce asymmetrical characteristics.

Calculating the Magnetic Field of a Coil is relatively easy:

The formula for the magnetic field strength \( B \) inside a solenoid:

\[ B = \mu_0 \mu_r \frac{N I}{L} \]

 

Where:

• \(\mu_0\) is the permeability of free space (\(4 \pi \times 10^{-7} \, \text{H/m}\))
• \(\mu_r\) is the relative permeability of the iron core
• \(N\) is the number of turns
• \(I\) is the current
• \(L\) is the length of the solenoid

 

Let's break down the calculations for both the magnetic field strength and magnetic flux density step-by-step.

 

Given Data:

1. Breadth ( Width of Core ): 120 mm = 0.12 m
2. Width ( Height of Core ): 75 mm = 0.075 m
3. Length of Coil: 17 mm = 0.017 m
4. Current ( I ): 200 mA = 0.2 A
5. Number of turns ( N ): 275 turns
6. Core Permeability ( μ ): 200 (relative permeability)

 

Calculations:

 

1. Magnetic Field Strength (H):

The magnetic field strength \(H\) can be calculated using the formula:
\[
H = \frac{N \cdot I}{l}
\]

where:

• \(N\) is the number of turns (275 turns)
• \(I\) is the current (0.2 A)
• \(l\) is the length of the coil (0.017 m)

 

Substituting the values:
\[
H = \frac{275 \cdot 0.2}{0.017} \approx 3235.29 \, \text{A/m}
\]

 

2. Magnetic Flux Density (B):

The magnetic flux density \(B\) can be calculated using the formula:
\[
B = \mu \cdot \mu_0 \cdot H
\]

where:

• \(\mu\) is the relative permeability (200)
• \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{H/m}\))
• \(H\) is the magnetic field strength calculated above (3235.29 A/m)

 

First, calculate the total permeability \(\mu \cdot \mu_0\):
\[
\mu \cdot \mu_0 = 200 \cdot 4\pi \times 10^{-7} \approx 2.513 \times 10^{-4} \, \text{H/m}
\]

 

Then, substituting the values to find \(B\):
\[
B = 2.513 \times 10^{-4} \cdot 3235.29 \approx 0.813 \, \text{Tesla} \, (\text{T})
\]

 

Summary:

• Magnetic Field Strength (H): 3235.29 A/m
• Magnetic Flux Density (B): 0.813 Tesla (T) or 8,130 Gauss (G)

 

This is cool, its very powerful, its clear, and we can use this to go a step further! Lets work out some of the nitty gritty, if we want a Voltage of X Generated:

Let's break this down step by step to ensure the design is as precise as possible.

 

Magnetic Flux ( Phi )

First, calculate the magnetic flux:
\[ \Phi = B \cdot A \]

 

Where \( B \) is the magnetic field strength and \( A \) is the cross-sectional area of the core.

 

Magnetic Field Strength ( B )

For a coil:
\[ B = \mu \cdot H \]

 

Where:

• \( \mu \) is the permeability of the core material.
• \( H \) is the magnetic field intensity, defined by the current passing through the coil and the number of turns (which we need to design).

 

Induced Voltage ( V )

The voltage induced in a coil is given by Faraday's Law of Induction:
\[ V = -N \cdot \frac{d\Phi}{dt} \]

 

Where:

• \( N \) is the number of turns in the coil.
• \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux.

 

Since we need to generate 48 Volts, let's move toward calculating the necessary parameters for this:

 

1. Calculate the cross-sectional area of the core:
\[ A = 120 \times 75 = 9000 \text{ mm}^2 = 9 \times 10^{-3} \text{ m}^2 \]

 

2. Permeability of the core ( μ ):
\[ \mu = 200 \times 4\pi \times 10^{-7} \approx 2.51 \times 10^{-4} \text{ H/m} \]

 

3. Assuming the magnetic field intensity ( H ):
\[ H = \frac{I \cdot N}{l} \]

 

Where:

• \( I \) is the current through the coil.
• \( l \) is the length of the magnetic path.

 

Let’s assume \( l \) (length of magnetic path) as 0.17 m (17 mm).

Using \( V = 48 \text{ Volts} \) and the fact that \( V = I \cdot R \), let’s assume \( I = \frac{V}{R} = \frac{48}{40} = 1.2 \text{ Amps} \).

 

4. Voltage generation and coupling:
\[ V_2 = k \cdot N_2 \cdot \frac{d\Phi}{dt} \]

 

Given coupling \( k = 0.87 \), we need the total number of turns \( N \) for the coils to ensure the voltage generation is precise.

Given your load resistance and power requirement:
\[ P = V^2 / R \]
\[ P = 48^2 / 40 = 57.6 \text{ Watts} \]

 

Note that the actual power utilized depends on the effective voltage in use, not exceeding the design limits.

 

5. Turns Calculation:

Using Faraday’s Law, considering a sinusoidal change in flux,
\[ V_2 = k \cdot N \cdot 2 \pi f \cdot \Phi \]

 

Where:

• \( f \) is the frequency of the alternating current.
• \( \Phi = B \cdot A \).

 

Assuming frequency \( f \) to be standard 50Hz.

 

From:

\[ V = 2 \pi f \cdot N \cdot \Phi \]

 

We rearrange to find \( N \).

 

Given that \( \Phi = B \cdot A \):

\[ \Phi = \mu \cdot H \cdot A = 2.51 \times 10^{-4} \cdot H \cdot 9 \times 10^{-3} \]

 

Ultimately,

\[ V = 48 = N \cdot 2 \pi \cdot 50 \cdot 2.51 \times 10^{-4} \cdot H \cdot 9 \times 10^{-3} \]

 

Rearrange to solve for \( N \).

 

1. Given constants:

• Frequency, \(f = 50 \text{Hz}\)   
• Desired voltage, \(V = 48 \text{V}\)
• Coupling factor, \(k = 0.87\)
• Permeability, \(\mu = 2.51 \times 10^{-4} \text{H/m}\)
• Cross-sectional area, \(A = 9 \times 10^{-3} \text{m}^2\)

 

2. Magnetic field intensity ( H ):
\[ H = \frac{I \cdot N}{l} \]

 

Given:
\[ I = 1.2 \text{A} \]
\[ l = 0.17 \text{m} \]

Let's solve for the flux (\(\Phi\)):
\[ \Phi = \mu \cdot H \cdot A \]

 

3. Combining all variables:
\[ V = N \cdot 2 \pi \cdot f \cdot \Phi \]

 

Plug in the values:
\[ 48 = N \cdot 2 \pi \cdot 50 \cdot \mu \cdot H \cdot A \]
\[ 48 = N \cdot 2 \pi \cdot 50 \cdot 2.51 \times 10^{-4} \cdot \left( \frac{1.2 \cdot N}{0.17} \right) \cdot 9 \times 10^{-3} \]

 

4. Solving for \(N\):
\[ 48 = N \cdot 2 \pi \cdot 50 \cdot 2.51 \times 10^{-4} \cdot \left( \frac{1.2 \cdot N}{0.17} \right) \cdot 9 \times 10^{-3} \]

\[ 48 = N^2 \cdot 2 \pi \cdot 50 \cdot 2.51 \times 10^{-4} \cdot \left( \frac{1.2}{0.17} \right) \cdot 9 \times 10^{-3} \]

\[ 48 = N^2 \cdot 2 \pi \cdot 50 \cdot 2.51 \times 10^{-4} \cdot 7.0588 \times 10^{-2} \]

\[ 48 = N^2 \cdot 5.577 \times 10^{-3} \]

\[ N^2 = \frac{48}{5.577 \times 10^{-3}} \]

\[ N^2 \approx 8606 \]

\[ N \approx \sqrt{8606} \]

\[ N \approx 92.75 \]

 

Therefore, the number of turns \(N\) required is approximately 93 turns for each coil to generate the desired 48 Volts.

Here, you can see, using Standard, Conventional Science, finding the Holes, and making use of where the holes lay, and filling these holes, we can very very easily change the world! You have a reasonably solid Mathematical Theory, that shows you, that we can do what we have shown many thousands of people for a very long time! It works, and no one else has ever figured it out, and shared it with the world, we have, AboveUnity.com's Members and I have given you Truth and a solid, FREE, Future to look forward to!

Want more?

The Energy Density of these machines is very very high, if you convert to the same units, compared to anything else ever open to the public:

Power Density:

Power density is the power per unit volume. Assuming the coil volume \( V \) is given by:
\[
V = \text{Cross-Sectional Area} \times \text{Length}
\]
\[
V = 6.669 \times 10^{-3} \text{ m}^2 \times 17 \times 10^{-3} \text{ m} = 1.13373 \times 10^{-4} \text{ m}^3
\]

Given the load power:
\[
P = \frac{V_s^2}{R} = \frac{48^2}{40} \approx 57.6 \text{ W}
\]

So, power density:
\[
\text{Power Density} = \frac{P}{V} = \frac{57.6 \text{ W}}{1.13373 \times 10^{-4} \text{ m}^3} \approx 508,256 \text{ W/m}^3
\]

 

To give you some comparison, here is an example: Approximately 250-300 Wh/kg

Lithium-Ion Batteries

Example: Tesla Model S Battery Pack

• Energy Source: Lithium-ion cells

• Energy Density of Battery Pack: Approximately 250-300 Wh/kg

• Use Case: Electric vehicles

• Why It's Impressive: Lithium-ion batteries store a large amount of energy relative to their weight and volume, making them perfect for applications where both efficiency and portability are crucial, such as in electric vehicles and portable electronics.

This post should absolutely astound you! This post IS the answer to the worlds Energy Crisis that never was! satan's children have always been the Crisis on the world, nothing else have ever been a Crisis except them!

If you want to go and check through the Math, and work out what I have shared, please feel free and please point any mistakes if you find any!

Best Wishes,

   Chris