Thanks Bob for taking the time to follow my discussion. If we are lucky others will help to inform the skeptics that Dr. McKubre is doing his experimentation in a manner that we can agree is appropriate. Don't beat yourself up too badly since I have a hunch that very few others would have believed what I was attempting to explain. At first it just did not pass the smell test. I had the same reservations that you had until I used the same basic line of reasoning that you have just written below.
Now if we can only settle the temperature and radiated power questions from the latest testing! Dave -----Original Message----- From: Bob Higgins <rj.bob.higg...@gmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Mon, Oct 27, 2014 10:58 pm Subject: Re: [Vo]:questions on McKubre cells and AC component Well Dave, you have made a good and convincing argument. My hat is off to you and I need to eat it with a big public helping of crow. It seems like if we go back to basics, the average power is integral((I dot V)dt)/integral(dt). If I is a constant, then you can pull I outside the integral and you get the average power as I x integral(Vdt)/integral(dt), which means the average power is I times the average voltage. Thank you for taking the challenge, making me rethink, and putting me straight! Regards, Bob On Mon, Oct 27, 2014 at 5:53 PM, David Roberson <dlrober...@aol.com> wrote: Bob, I take that as a challenge. I am not offended my friend, but find this a great opportunity to prove what I am saying is correct. I predict that you will agree with me once you have an opportunity to dig deeper into the subject. It is not clear to me what you are showing in your example, perhaps due to a problem with my display. Let me choose an example for you to consider. Again, we can assume that the current being delivered into the load is exactly 1 amp. If we further assume that the load resistance is 1 ohm, then under DC conditions we will measure precisely 1 volt across the load resistor. I and I assume you would calculate the power as being 1 watt delivered to the load resistor under this static condition. Now, suppose that the resistance changes to .5 ohms. In that case the voltage becomes exactly .5 volts. This results in a power being delivered to the resistor of .5 watts. For the other half of the AC square waveform the resistor becomes 1.5 ohms. In that case the power delivered becomes 1.5 watts since 1 amp x 1.5 volts = 1.5 watts. Since we are assuming a symmetrical AC waveform, this is a pretty good example of that with numerous harmonics that also get into the act. The assumed waveform is therefore a 1 volt peak to peak square wave that is riding upon a 1 volt DC average. So the total power average becomes (.5 watts + 1.5 watts) / 2 = 1 watt. Each half of the waveform makes its contribution and they balance each other out about the normal DC average of 1.0 watt. This is true for all AC waveforms, regardless of the harmonic content provided that the current retains a constant DC value. I have stated this on numerous occasions and it is a general concept. Power can only be extracted from a source current that flows at the same frequency as the source voltage. In this case the current is at a DC frequency, so no power can be extracted from the source except into a DC(0 Hertz) voltage related load. Dr. McKubre essentially made the same statement with respect to his experimental setup. Another feature of a constant current environment is that the power delivered into the load varies directly with the load voltage and not proportional to the square of the voltage as is normally encountered. That is what allows the average to be used in this case instead of having to deal with the messy RMS waveform additions. If you have reservations about what I have stated I strongly suggest that you put together a Spice model. That will prove that what I am saying is right on target. Dave -----Original Message----- From: Bob Higgins <rj.bob.higg...@gmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Mon, Oct 27, 2014 3:11 pm Subject: Re: [Vo]:questions on McKubre cells and AC component I hate to say this, but what you say is absolutely wrong. You could only do as you describe if the "voltage" being averaged is the RMS voltage. You cannot take the average voltage and multiply it by the average current to get average power. For example, suppose that the voltage was V=1_0.5sin(wt). The average of this voltage is 1. Lets say we have a constant current of 1A. By your method, the power would be 1 watt. However, the actual power is: P = (1A) sqrt(mean(1+0.5sin(wt))^2)) = (1A) sqrt(mean(1 + sin(wt) + 0.25sin(wt)^2)) = (1A) sqrt(1+.25 (mean(.5 - .5cos(2wt)))) P = (1A) sqrt(1.125) = 1.0607 Watts On Mon, Oct 27, 2014 at 11:58 AM, David Roberson <dlrober...@aol.com> wrote: The instantaneous power being delivered by the source is equal to the product of the current and voltage. When the current is constant, only DC voltage loads can accept power and thus energy from the source. All of the AC voltages that appear across the source terminals integrate to zero during a full cycle and do not enter into the input power equation. This understanding seems to escape most people until they review the theories carefully. I had to prove roughly the same issue to several skeptics that thought that DC due to load rectification of the AC power source could be used to sneak extra power into the earlier ECAT. They thought this was possible since the input power meter did not monitor DC directly.
