Fausto:

I will try to respond to AC's points. I may be very technical also.

(An Inductor) will ... raise it's voltage in proportion to the rate of change of the switching mechanism which is a form of resistance.

This is incorrect. AC is modeling a switching-off transistor as a resistance that increases in value with respect to time. That is fine. An inductor will raise it's voltage in proportion to the instantaneous value of the resistance of the switch. At the same time the inductor is discharging its stored energy so the amount of current flowing through the resistance of the switch is also decreasing. Therefore there are two effects happening at the same time. This could be described by a differential equation and you could solve the differential equation to get the plot for the voltage of the discharging inductor with respect to time.

The most important point here is that the inductor outputs a voltage that is proportional to the instantaneous resistance of the load and the instantaneous amount of current flowing through the inductor, and not the rate of change of the resistance of the load with respect to time.

If the resistance is infinite to an inductive discharge then the voltage is infinite ( E=IR) and as the energy cannot change the current (I) must approach zero.

This is incorrect. As the resistance tends towards infinity, the time duration of the current spike tends towards zero. Note that there ALWAYS IS a measurable current flow, but the length of time of the current flow decreases towards zero.

No, the collapsing field induced a potential difference by cutting the coil conductors and if the coil is in a circuit then this potential difference will lead to an electric current thus the voltage came first, it is a voltage source first and the current is a result of the voltage discharging itself into a circuit.

This is incorrect. The magnetic field was built up by the input current flow, and so when the coil is open-circuited, the magnetic field now generates an output current flow. This demonstrates the natural symmetry in the way an inductor stores energy and releases energy. The voltage across the inductor terminals are a result of the current flow. A discharging inductor acts like a current source with an output impedance of infinity.

Well no it cannot thus it would seem obvious that a discharging capacitor has little in common with a discharging inductance.

This is incorrect. Imagine two test setups. The first setup has a 1-Henry inductor connected to a 1-ohm resistor and there is 1 amp of current flowing through the inductor. The second setup has a 1-Farad capacitor connected to a 1-ohm resistor and there is one volt of potential difference across the capacitor. For both cases above when you start the two tests the voltage and current waveforms will be identical. In addition, if you increase the resistance for the inductor test and you decrease the resistance for the capacitor test, then the voltage waveform for the inductor test will be identical to the current waveform for the capacitor test. Also, the current waveform for the inductor test will be identical to the voltage waveform for the capacitor test. In all cases they will be exponentially decaying waveforms. Capacitors and inductors are very similar to each other.

A battery will raise it's voltage in proportion to the voltage of the source charging it and we call this action "charging" where a resistor will not because it is not "charging" it is dissipating energy as heat, they are not the same thing.

This is incorrect. To use the phraseology "a battery will raise its voltage" in this context is a very poor choice of words that would never be spoken by an electrical engineer. I won't go into a detailed technical discussion here. I will say that AC's sentence doesn't even make sense. As a bare minimum, you would have to say that the voltage at a battery's two terminals will increase as a function of the current that you are supplying to the battery. Certainly you can clamp a battery's terminal voltage to an external voltage source that has a low enough output impedance to overcome the battery's output impedance. Then you end up charging the battery because current ends up flowing into the battery. The bottom line is that this is an extremely awkward way of expressing this idea and an extremely poor choice of words that would never be used by someone that truly understands electronics.

Ion posted a model for a battery that I have not looked at yet, but I will assume that the explanation for why a battery terminal voltage will increase when the battery is being charged is explained in the document.

we call this action "charging" where a resistor will not because it is not "charging" it is dissipating energy as heat, they are not the same thing.

Certainly they are not the same thing, but AC is missing the point. In both cases, for a simple load resistor and for a charging battery, the energy source is pumping power into the "load." In one case the "load" is a resistor, and in the other case the "load" is a charging battery. So in both cases it is valid to consider issues related to impedance matching, because impedance matching is all about how well an energy source can transfer power into a load.

One could also ask why all the new battery chargers use boost converters exclusively where 20 years ago they used transformers, one word --efficiency....

AC's comment above about boost converters, and the Tesla comment and other comments in the paragraph I quoted from above have absolutely nothing to do with the discussion.

The fact of the matter is that a battery will raise it's voltage and resistance in response to the action of "charging" and the inductive discharge will raise it's voltage in proportion to the batteries rise in resistance as it must, does this sound like impedance matching?

No, it does not sound like impedance matching at all. Go back and look at my battery example. Impedance matching has to do with maximizing the power transfer between the energy source and the load. In a Bedini motor the load can change impedance and the power transfer remains the same, there is nothing to match.

My battery chargers and the batteries they charge run cool, not hot, not warm

So what, this is purely anecdotal evidence. Chances are that you have never measured the power output of one of your buck-boost chargers so you don't know if the battery should get perceptibly hot or not. Even if the energy transfer is more efficient, the power that gets pumped into the charging battery is still going to be split into power that actually charges the battery and power that becomes waste heat. If that waste heat production is below a certain threshold then you won't be able to perceive it. For all you know after 20 minutes the surface temperature of your battery could increase by 1.5 degrees C and you would not be able to perceive this slow increase in temperature by putting your fingers on the battery.

I have never seen any scientific attempt to back up the argument that "Bedini chargers keep the charging batteries cool." On the other hand, I have read some postings where some Bedini experimenters inadvertently overdo it with their setups and end up boiling their charging batteries.

Milehigh