Magnetic Electron Pump MEP open source overunity device !

[hartiberlin]

Hi,
just in from Sterling D.Allan:

The MEP:

http://peswiki.com/index.php/OS:Magnetic_Electron_Pump
and
http://tech.groups.yahoo.com/group/pes_MEP/

Looks good so far !

Many thanks to Sterling and the anonymous inventor !

Regards, Stefan.

[linda933]

Interesting, but looking at figure 1,  we can see a large error in his calculation of input power.  He shows an input pulse measured on a 0.01 ohm shunt and peaking at 0.3V.  This is 30A.

Following the rest of the math, and giving him an advantage by calculating the area under the curve as if it were a linear ramp (50%), we have a pulsed power of 30A x 24V x 1/2 or 360W.  Now he says the pulse duty factor is 12.4% giving a total average input power of 44.64W not 8W as he states!

In reality the figure would probably be more like 55W based on the exponentially-rising (>50% probably more like 63%) curve of the current waveform, which is clearly larger area under the curve than a linear ramp.

This looks like a simple case of bad math.  I think he used a figure of 3A peak (off by 10X) then failed to reduce this based on less than full area under curve and then multiplied by 24V, getting 72W and then multiplied by .124 (12.4% duty cycle) to get his approximate 8W figure.

50+W input for 32.8W output is not so impressive when you add up the numbers correctly!  I'm surprised no one (especially the excited inventor) and Mr. Allen et al at Peswiki never noticed this gaping error.  I'm only a second year engineering student and it leapt right off the page at me!

Linda

[markdansie]

Hi Linda,
you should ask Sterling about joining the NEC, you would be most welcome there.
Mark

[linda933]

Turns out I was wrong on my speculation as to how the error was made.  As it turns out, the inventor has more fully described his math now over on that Yahoo forum.

He made a little mistake and a big one:  First, he calculated the average current within the pulse based on using half the square root of two or 0.707 to get the average from the peak.  Those who have some training in this know that constant only applies to finding the average value of a sine.  The rising exponential shape of the current pulse he drew is not a sine shape but could be fairly close to 0.707 of the full 30A peak value.  I would guess between .6 and .7 from his drawing.  Small error.

The big error came from calculating the duty cycle factor in wrong.  The right way is to find the power in one pulse (V x I) and then multiply that by the duty cycle to find the average power.  What this inventor did wrong was he mistakenly multiplied the voltage by the duty cycle and then again multiplied the current by the duty cycle and then multiplied his two results. That's definitely cheatiing!

It's like having a 100W (10A x 10V) pulse lasting one second and happening at 10% duty cycle (once every ten seconds).  The total average power is simply 100 x 1/10 or 10W.  He was saying it was 10V x 1/10 x 10A x 1/10 = 1W.  No fair squaring the duty cycle!

Science and physics don't have all the answers and there are things we don't yet understand.  Multiplying and dividing properly in terms of calculating power and energy do not happen to fall into that category of mysterious phenomena, however.  Too many "inventions" get quite far along in terms of work and emotional attachment before some huge error is discovered in the basic math.

Forums like this are an excellent place to put forth ideas for inspection and test.  That's what research is all about, after all.  I was glad i was able to spot this error quickly and, while the inventor might be saddened to realize his error, just think of all the time and grief he will save by moving ahead quicker to more productive plans and designs!  That's the whole idea...keep it heading forward!

Linda

[linda933]

Hi Linda,
you should ask Sterling about joining the NEC, you would be most welcome there.
Mark

Thank you, Mark.  I was pleasantly surprised on discovering the answer to this puzzle and posting it over there as the first joining member (after Sterling of course) to find myself in the good company of several well-known experts who all concurred and two even congratulated me for being the first to ferret out the math errors.  It was just a little bit thrilling to be so acknowledged by Dr. Peter Lindemann and Ken Rauen. 

I will think about what you suggested and look into the requirements and responsibilities of joining such a widely respected and credible group.  Of course they'd have to promise never to assign me to any of Panacea's projects...

Linda