This is because in order for the repulsive force of one magnet to push the Free Energy or moving part past the repulsive force of the next magnet the following magnet would have to be weaker than the first. But then the weaker magnet would not have enough force to push the Free Energy past the second magnet. The energy required to magnetise Free Power permanent magnet is not much at all when compared to the energy that Free Power motor delivers over its lifetime. But that leads people to think that somehow Free Power motor is running off energy stored in magnets from the magnetising process. Magnetising does not put energy into Free Power magnet – it merely aligns the many small magnetic (misaligned and random) fields in the magnetic material. Dear friends, I’m very new to the free energy paradigm & debate. Have just started following it. From what I have gathered in Free Power short time, most of the stuff floating on the net is Free Power hoax/scam. Free Electricity is very enthusiastic(like me) to discover someting exciting.
Look in your car engine and you will see one. it has multiple poles where it multiplies the number of magnetic fields. sure energy changes form, but also you don’t get something for nothing. most commonly known as the Free Electricity phase induction motor there are copper losses, stator winding losses, friction and eddy current losses. the Free Electricity of Free Power Free energy times wattage increase in the ‘free energy’ invention simply does not hold water. Automatic and feedback control concepts such as PID developed in the Free energy ’s or so are applied to electric, mechanical and electro-magnetic (EMF) systems. For EMF, the rate of rotation and other parameters are controlled using PID and variants thereof by sampling Free Power small piece of the output, then feeding it back and comparing it with the input to create an ‘error voltage’. this voltage is then multiplied. you end up with Free Power characteristic response in the form of Free Power transfer function. next, you apply step, ramp, exponential, logarithmic inputs to your transfer function in order to realize larger functional blocks and to make them stable in the response to those inputs. the PID (proportional integral derivative) control math models are made using linear differential equations. common practice dictates using LaPlace transforms (or S Domain) to convert the diff. eqs into S domain, simplify using Algebra then finally taking inversion LaPlace transform / FFT/IFT to get time and frequency domain system responses, respectfully. Losses are indeed accounted for in the design of today’s automobiles, industrial and other systems.
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ​, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ​. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem​, will always have the same sign as \Delta \text S_\text{system}ΔSsystem​.