Rapport éventuel avec la sur-unité d'une bobine alimentée par des "pics" de H. T (kicks)
Download 10.18 Kb.
|
| U = 1 / (8 x pi x 10-7) x ( 10-4)2 x pi x 1 / 4 x 1 = 0.003 Joules
If we pulse that energy level at 330 kHz, then we will get 1 kW, and at 660 kHz, 2 kW, etc., thus a higher frequency yields more power. Then the question becomes, how can we pulse the constant magnetic field inside the coil? The answer is simple: by using an external source, we can cancel the Earth's magnetic field inside the coil. There must be power and energy amplification with respect to the external input source. To realise that power amplification, we must do the following: Let the magnetic field variation inside the air coil be given by: B(t) = Bo + Bf x Sin( w x t ) Where Bo is the constant of Earth's magnetic field, Bf is the magnetic field in the coil created by the external power source, and w is the angular frequency of the external source. Replacing B(t) from equation (1) we get the energy variation with time, U(t), and then we can calculate the power as P = dU/dt resulting in: P(t) = Bf x w x V x (Bo + Bf x Sin(w x t) x Cos( w x t ) ) / muo ........ (2) Remember that V is the volume inside the coil. We see here that the output power depends on Bo, the Earth's magnetic field, just as in the case of Bearden's MEG it depends on the magnetic field intensity of the permanent magnet in the circuit. So we can now calculate a COP value with Bo and without Bo, or Bo = 0 Calculating the RMS power for both cases (not reproduced here because it corresponds to a case of basic differential calculus) and using the ratio, the result for the COP is: COP = ( 1 + ( 2 x Bo / Bf )2 )0.5 We see then power amplification, and of course if Bo=0 and not a permanent magnetic field, the maximum COP is 1, input and output powers are equal. In the case of Bearden's MEG, the condition is Bo = Bf for not degaussing the permanent magnet and in that case we have a COP = square root of (5), which is a value between 2 and 3 which corresponds to the practical results for this classic calculation. 7 - 50 But for the Earth's magnetic field, we can get higher values because we can never cause a demagnetisation of the Earth's magnetic field. How many turns on the coil, pulsing frequency, coil diameter, coil length, etc., do we need? The input power to cancel the Earth's magnetic field inside the coil, or the condition Bf = Bo , we calculate using the RMS component of equation (2) above, setting Bo = 0, so we have: Download 10.18 Kb. Share with your friends:
|