The book is dedicated to self-propelled mechanical generating devices.

[rakarskiy]

The second rotor is repelled from the first, already rotating rotor with, say, half the speed.
is repelled by electromagnetic forces. We transfer energy for it with conductive rings and brushes.
What increment in the kinetic energy of rotation will the system (second rotor) receive for a motionless, resting observer?

https://youtu.be/QGPwy6eF_UU   - I think this video will be interesting to you.

[rakarskiy]

It doesn't happen like that?! - the reader who is convinced that a generator with a minimum electromagnetic moment (the braking moment of the generator) is impossible will think. In the material, the justification of the possibility is just given, by the method of solving the problem in physics. It took more conditions and data to solve it. But it is solved on the basis of the fundamentals of electrical engineering.


https://rakarskiy-narod-ru.translate.goog/publ/free_energy_systems/motor_generator_cop_gt_1/3-1-0-137?_x_tr_sch=http&_x_tr_sl=ru&_x_tr_tl=en&_x_tr_h l=ru&_x_tr_pto=nui

[rakarskiy]

http://www.energeticforum.com/forum/energetic-forum-discussion/renewable-energy/13619-motor-generators?p=508069#post508069


It is fashionable to read the updated version of the article through Google translator at the link.

It is not difficult to calculate the conversion factor COP = P (generator) / P (engine) = 1 / 0.014 = 71.4 (7140%).
The result looks fantastic, this generator is very real. Using a 100 W drive motor, this would be 0.9 kW of free energy when the electromechanical converter is self-propelled. The main thing that we found out is that turning on is possible, in our case, according to the condition of the problem, we can turn on the engine in the generator circuit with a load, provided that the load power is reduced, for the power consumption of the engine and the frequency of the network and the engine is 563 Hz.

[Floor]

Either the generator or the motor, must be over unity in and
of its self, before this drawing could self run.

Best wishes
floor

[rakarskiy]

Either the generator or the motor, must be over unity in and
of its self, before this drawing could self run.

Best wishes
floor

It's best to decide for yourself. Those who decided to double-check are in slight bewilderment. One even exclaimed: where did the force of Lorentz go!

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Condition and solution to the problem.
It remains to check whether the condition is fulfilled when the electromagnetic moment of the generator (mechanical resistance to rotation) is identical to the output electric power. Let's take a drawing of the simplest electrical circuit with a generator. A generator in the same circuit with incandescent lamps, for example, incandescent lamps with a total power of 1000 W (1 kW), with a mains voltage of 220 volts. It is necessary to calculate the parameters of the generator, and find out the possibility of connecting the engine to rotate the generator of the same circuit.
First, we need the generator parameters. To do this, we will calculate according to the classical scheme of a frame rotating in a magnetic field between two poles.

Solution -1:

1.1 Definition of the initial data of the generator:
P electric power = 1 kW (1000 W);
Mains voltage - U = 220 V;
Determine the load resistance by the formula
Rz = U2 / P = 2202/1000 = 48.4 ohms.
Determine the load current: Iz = P / U = 1000/220 = 4.55 A.
1.2 We choose a wire by cross-section, for current throughput:
Copper wire diameter in varnish = 1 mm, S = 0.79 mm2,
at Аі = 8 A / sq. mm, the current carrying capacity will be:  J = Ai / S (mm²) = 8 / 0.79 = 6.28 A.
Let's define the following missing parameters:
Angular speed of rotation of the shaft / frame: n = 650 rpm;
The diameter of the armature d = 0.35 m (in our case, the distance between the active conductors of the frame relative to the center of rotation);
Magnetic induction: B = 0.4 T, maximum sin (a) = 1.

We need to calculate the EMF [E] and the length of the active wire [l].

1.3 EMF [E] is calculated from the current formula, for a complete circuit: Ii = E - U / R + Rz, (where: R is the resistance of the active wire of the generator frame). To calculate the EMF, the formula will take the form: E = (U + I) * (R + Rz) = 220 + 4.55 * 48.4 + Rz (?), To solve this equation, we do not know the resistance of the active wire, the frame.
1.4 There is another option, to calculate the EMF [E] according to the formula: E = U * k, where: k is the ratio of the network voltage to the EMF of the phase. It cannot be less than 2. When k = 2, it is an ideal state when the phase resistance is zero. Let's check: E = U + (I * R + Rz) = 220 + 4.55 * 48.4 = 440 V; E = U * k = 220 * 2 = 440 V.
1.5 Now we need to determine the length of the wire and its resistance. We find it from the formula EMF [E] for a linear wire: E = B * l * v. From the formula, we derive the calculation for the active wire: l = E / v / B.
1.6 For further calculation, we need the parameter of the rate of change of the magnetic flux - v (m / s), we find it by the formula: v = πnD / 60 = 3.14 * 650 * 0.35 / 60 = 11.9 m / s.
1.7 Find the length of the wire l = E / v / B = 440 / 11.9 / 0.4 = 92.44 meters.
1.8 Determine the resistance of the active phase wire: R = Ri * l [where Ri = ρ * (1 / mm²) = 0.017 * (1 / 0.79) = 0.0276 Ohm] = 92.44 m * 0.0276 Ohm = 2.55 ohm.
1.9 Next, we can calculate the current in the loop:
Mains current Ii (at E = 440 V): Ii = E - U / Rz + R = 440 - 220 / 48.4 + 2.55 = 4.32 A.
1.10 Calculate the difference between the load current and the current of the circuit: 4, 55 - 4.32 = 0.23A. We can adjust our coefficient:
k = 2 + (1- Ii / Iz) = 2 + (1 - 4.32 / 4.55) = 2.05054945055.
1.11 We correct our EMF of the generator frame: E = U * k = 220 * 2.05054945055 = 451 V.
1.12 Correct the length of the active wire of the generator frame: l = E / v / B = 451 / 11.9 / 0.4 = 94.75 meters.
1.13 We check the EMF value according to the original formula: E = B * l * v = 0.4 * 94.75 * 11.9 = 451 V.
1.14 We check the value of the current in the circuit: I = E - U / R + Rz = 451 - 220 / 48.4 + 2.55 = 4.55 A.

As a result: the load power of 1000 W with a resistance of 48.4 Ohm, 220 V of the mains voltage, provides the phase of the generator, the length of the active wire of which is 94.75 meters, the resistance of the active wire is 2.55 Ohm, the current in the loop is 4.55 A.

Solution - 2:
Determination of the possibility of connecting the engine to a circuit with a generator and load:
2.1 We can use the classical formula to calculate the electromagnetic moment of the generator (assuming that Re = Pk): T motor = Re * 9.55 / n = 1000 * 9.55 / 650 = 14.69 N * m.
2.2 It remains to clarify the coefficient of the idle torque of the generator (what resistance to rotation has the mechanical part of the generator when rotating at a given speed without connecting the frame to the circuit with a load), traditionally they are 20% 100% of the electromagnetic torque under load, as a result, the engine torque: T = 100% + 20% = 120% of the electromagnetic moment of the generator created by the current induction: k = 1.2. The electromagnetic torque of the drive motor will be = 1.2 * M = 17.63 N * m. Determine the power of the drive motor Pm = 17.63 * 650/9550 = 1.20 kW.
2.3. Determine the conversion efficiency ratio: COP = Pgenator / Rmotor = 1 / 1.2 = 0.83.

As you can see, everything coincides with the traditional statements. Conversion coefficient: COP = 0.83 <1. We cannot include to the load, additionally, a drive motor for rotating the generator, proceeding from the condition of equilibrium of the kinetic and electrical powers.

But that's not all, let's go ahead and calculate the electromagnetic moment of the generator through the ampere force and the radius vector

Solution - 3.
Determination of the electromagnetic moment of the generator through the electromagnetic force and the radius vector:
To calculate the electromagnetic moment, through the Ampere force, we apply the basic formula: M = Fa * (d / 2): where Fa is the Ampere force, in Newtons.
Ampere force is calculated by the Formula: Fа = B * In * La, where: magnetic induction B = μ₀ * (In / 2π * r), in T; current strength in ampere turns: In = n * I, in amperes; and La is the sum of the lengths of the two active sides of the frame, divided by the number of turns: La = 2 * l / n, in meters.
Let's imagine that our frame is elongated to the full length 1/2 of the generator phase length:
la = 94.75 meters / 2 = 47.36 meters, we get the length of one edge of the frame.
In fact, we received not the real length of the generator armature, for an electromechanical device. Let's say that we have it with a minimum or very large - Idling torque.
Let's calculate the electromagnetic moment of one frame of an unreal generator, with a gap between the magnetic pole and the wire = 1 mm (0.001 m):
M = ((μ₀ * I * n / 2π * r) * (I * n) * (La * n) * (d / 2) = (0.00000126 * (4.55 * 1/2 * 3.14 * 0.001)) * (4.55 * 1) * (2 * 47.36 * 1) * (0.5 * 0.35) = 0.136 N * m
We got a completely unrealistic small dimensionality of the electromagnetic moment of 0.136 N * m, against EMM = 14.69 N * m, subject to the equilibrium of powers!
That's the result! I double-checked for an error that I did not find. I made a few more calculations: putting the length of the wire in turns and reducing the length of the frame (armature / rotor). Results in the table: (see article)

The first two columns are actually what we counted. With a generator no-load ratio of 1.25, the power of the drive motor will be: 0.012 kW (12 W) !!! To reach parity with the traditional formula (conditions that, REM = Pk) is achieved with an idle speed = 130. Obviously, an anchor with two support points, of such length, cannot be made. A rotor with intermediate supports is possible. How much the moment of force will be, for the rotation of this rotor at idle, we will not determine.
We have two more calculations. The wire was laid in turns (we will take an inactive wire for zero resistance, and we will not take it into account). To achieve parity with the condition: REM = Pk, we got 107 turns. The length of the edge of the frame was reduced from 47 meters to 0.445 meters (which is quite acceptable for the size of the generator). It turns out that the current and magnetic induction have grown 107 times: Current - [486.85 / 4.55 = 107 times]; The magnetic induction is [0.1941 / 0.0018 = 107 times], but the length of the wire or bundle has decreased 107 times: The length of the frame edges is [94.8 / 0.89 = 107 times].
Logically, the length can be attributed to the section, the overall result should not change, but it changes. If we compare it with a nozzle through which the jet flows, the smaller the cross section, the higher the speed and the actual pressure. If you add more turns and shorten the length, this pressure will increase even more. For a motor, this is not bad, but for a generator, it is the cost of current induction from the side of the drive motor. It is necessary to look for a solution for the optimal size of the generator with a minimum number of turns, for a given wire length.

The solution is in the design of the generator!
What to do? There is always a solution, especially since physics indicates where to look for this solution.
What you need to do: recalculate the result for 104 meters of active wire. We will have 104/2 = 52 pairs of poles The length of the rotor will be 0.5 m, the diameter is 0.35 m. Lay the entire length of the wire in a wave on these poles. Does it matter if the wire has a straight line or a zigzag, the main thing is that the phase is laid in one wire. We will have a frequency parameter since it will already be an alternator.
Let's calculate the frequency by the formula: f = p * n / 60, where: p is the number of pole pairs; n - angular speed, in rpm, f = 52 * 650/60 = 563 Hz.
In this variant of laying, one more parameter will be added, this is the electromagnetic traction force arising from the focused magnetic flux of the coil, which is formed between the wires with a multidirectional current vector, which will not exceed 0.157 N, when laid in one wire. Wave wire laying, patent reference. In our case, this is laying on a frame that is neutral to magnetic flux, for example, plastic, a 3D printer. Calculation by formulas:  http://rakarskiy.narod.ru/_fr/0/8775586.jpg

The electromagnetic moment with the sum of the Ampere forces and the electromagnetic traction force will preliminarily have a value: M = (0.78 + 0.157 = 0.937) * (0.35 / 2) = 0.163 N * m, which will be, in the need for the drive power of the generator motor , taking into account the coefficient 1.2 at idle, P = 1.2 M * n / 9500 = 0.014 kW (14 W).
It is not difficult to calculate the conversion factor:
COP = Rgenator / Rmotor = 1 / 0.014 = 71.4 (7140%).

The figure looks fantastic, this generator is quite real. If a 100 W drive motor comes out, it will be 0.9 kW of free energy during the self-propelled operation of the electromechanical converter. The main thing we found out is that switching on is possible, in our case, according to the condition of the problem, we can turn on the motor in the generator circuit with a load, subject to a decrease in the load power, for the power consumption of the motor and the frequency of the network and the motor of 563 Hz.

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http://rakarskiy.narod.ru/publ/free_energy_systems/motor_generator_cop_gt_1/3-1-0-137