Bedini Electrostatic Self Running

11 months ago
51

In this video, we explore the potential of a Bedini motor to power an electrostatic capacitor charging circuit and create a self-looped energy system., it is known for its high efficiency and the ability to increase efficiency with small mechanical loads.

I propose using the Bedini motor to drive the spin of electrostatic plates, which can charge a set of Leyden jars to over 150kV within seconds. Instead By using an SCR and neon trigger to dump the charge of a capacitor into a 12V charging/run battery, we create a self-looped system that takes advantage of the extra electrostatic energy input.

I done the math to calculate the input and output power of this system, based on assumptions of a 100mA input current and a capacitance of 100uF, with a discharge frequency of 0.5Hz. Based on these values, we calculate an input power of 1.2 watts and an output power of 2.5 watts, indicating that the system could potentially produce more output power than input power.

I incorrectly calculated the energy stored in the capacitor as 5 joules, which was an order of magnitude too high. Me bad!!!! So let's do this again :)

If the energy stored in the capacitor is 0.5 J and it is released over a 2 second period, the average power output of the system would be:

P = E / t
P = 0.5 J / 2 s
P = 0.25 watts

So the power output would be 0.25 watts. Less then our input. So in order to get more in output we need to modify the system as follows:

To achieve an output power of 1.3 watts, assuming an input power of 1.2 watts, the power gain needs to be 1.08, which means the output power needs to be 1.08 times the input power.

Using the same formula as before:

P = 0.5 x C x V^2

We can rearrange it to solve for capacitance C:

C = (2P) / V^2

Where P is the desired output power (1.3 watts) and V is the voltage at which the capacitor will discharge (100 volts).

Plugging in these values:

C = (2 x 1.3) / (100)^2 = 0.0000026 Farads, or 2.6 microfarads

So a capacitance of 2.6 microfarads would be sufficient to achieve an output power of 1.3 watts with a voltage of 100 volts. With a cap dump of one second.

Same idea! Sorry about that. I might just delete this video in a little while and update with the corrected version. :)

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