# Is flywheel free energy possible? – Nikola Tesla Free Energy Technology Repressed Anger From Childhood

Most of the energy required for a flywheel to provide power is supplied by the air. The air can be either friction or kinetic energy. A flywheel is a system of devices that, when arranged as a self-consistent flow, are capable of delivering the energy required to keep the wheels rotating, but only when the wheels are spinning, i.e. when the wheel is not resting, e.g. on a frictionless, inert surface.

In the case of an air/fuel mixture, all energy that can be extracted from that mixture will be kinetic energy (i.e. power) that is in addition to the kinetic energy of the mixture. When it is used as a flywheel, the energy stored is kinetic energy rather than the friction and therefore is called flywheel free energy. If this energy is not used to continue the wheel rotation after the wheels have touched the ground, a wasted power must be extracted through braking in order to keep the wheels rotating.

The principle used to calculate the energy required to keep a flywheel spinning (also referred to as rotor torque, and rotor RPM) is quite simple. The kinetic energy of the mixture is divided by the pressure and heat of the combustion chamber and this energy is then divided by the number of turns of the wheel. This energy is expressed in terms of the revolutions per minute (rpm). This will be used to calculate the flywheel torque or RPM,

T = Px2ΔΕ/Σ

where

P = pressure, kPa

ΔΔ = rotation angle, degrees

P is the torque of the rotor (T = P x ΔΔ), and

ΔΔ is the radius of an elliptical (r, or radial) curve (x, y, n) between the outer and inner ends of the rotational axis.

Where x and y (or the rotational axes of the rotational axis and the radius of an elliptical curve) are in radians. For the discussion, I’m using the theoretical equation for torque to apply to the rotor as a whole, which is based on the rotational speed (K) of the rotating mass (M), the thrust exerted by the rotor, the rotor-fuel ratio k2Rm (in this case 3.3), and the friction coefficient (F) of the fuel/air mixture.

At the moment I was going to use the equation I found online to

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