World's first real Free Energy Flashlight - no shaking - no batteries! No Solar

[MileHigh]

Any discussion of a "DC offset" in an electromagnetic wave is "not even wrong."

You are correct, the DC offset of the AM wave on your scope display was because of the DC voltage on the capacitor.  I don't know how the capacitor got charged.  I suppose the two options are somehow the AM wave gets rectified or with more wire in the setup the capacitor can self-charge better.

However, if you suspect that it is from the AM wave being rectified and that's charging the capacitor - talk alone will not cut it.  You would need to show an annotated schematic and scope shots and explain the process with a timing diagram if you want to make a convincing case.

[txt]

I think your a bit out of whack with your power calculations in regards to EM radiation around densely populated area's. Lets look at the video below,and lets assume his antenna is say 20 meters long(around 60 feet),and is say .4mm copper wire. So we would have an antenna with a square area of around 8cm.

Check out the term "effective aperture of an antenna" to correct your calculation. When oversimplified, the capture area of antenna is based on the square root of the wavelength (assuming the antenna can catch the entire ½ wave, like for example the dipole). In other words, make a square around your antenna and you'll be a bit closer. Calculating just the wire thickness is nonsense, of course.

BTW, the 20 m long antenna is not something you can run around with anyway, regardless that it is just 0.4 mm thick, so I would not consider it useful for powering a flashlight.

Do you really think it will take 2 years to charge 3 AA batteries with the energy that is being used to drive the LED and speaker?

Yes, I do. The math is trivial, and I am sure you can succeed the simple multiplying and divisions too. Your system in the video delivers 0.175V at 0.003A. Power = U*I - that gives the continuous charging power of 525μW. For charging 9Wh you hence need 9/525*10^-6 = 17143 hours = 714 days. Which indeed are two years (or pretty close to it). Without counting in the leakage, the charging efficiency, and the possibility that your nearby radio stops broadcasting over the frequency your antenna is tuned to, within that long time.

The video clearly show's that usable light can come directly from the EM radiation-without the need to store that energy.

I do not consider the charging power of 525μW "usable" for any practical flashlight. It is usable for powering low consumption electronics such as remote sensors, or perhaps even the tiny low power LED as a pilot light, but that was never questioned in this thread. EM harvesting for low consumption devices is known and used since decades, and nobody here ever expressed any doubts about it.

The tiny power is simply not useful for high-powered portable devices such as the 3W flashlight. If the ~70 millilumens that you can get out of 525μW, looks usable for you, then I admire your sight, but I can assure you that anyone who would buy a flashlight, finding it emits a few dozens of millilumens, would be highly disappointed.

[skywatcher]

This is not correct.  There is no connection between the transmitter and receiver of regular radio or TV signals.  It doesn't matter how many receivers receive the signal the transmitter doesn't see any difference.   Think of it this way.  You are standing in the middle of a small pond and you rock back and forth to make waves in the water.  Now someone decides to harvest a little of that energy by sitting a model boat in the water and they get to watch it rock back and forth because of the waves you are making.  If we add another hundred people and they each put a boat in the water you will not be able to tell any difference in your efforts to make waves.  Once the waves are made what happens to them after the leave you has no effect at all on your efforts to make waves.

Now if you are talking about sitting up an inductive device near the cross country overhead power lines that is a different thing.  Yes I would agree that taking power from the electric field that surrounds those power lines will cause a need for more energy to maintain the field.  And people have been convicted of theft in the U.S. for doing that.  But an EM field is not the same as a RF field.

Respectfully,
Carroll

At least in the 'near field' (which is many kilometers in case of low-frequency trabsmitters) you have inductive coupling, and this indeed draws additional power from the transmitter. I know that some decades ago some people here in Germany used wires in their garden to get energy for their fluorescent lamps, which worked pretty well nearby big AM transmitter stations, but this has been forbidden because of the fact that it weakened the signal.

[txt]

You have far under estimated the available energy per square CM of EM radiation in populated area's,where we can safely assume there is 1 or more sources of strong EM radiation.

In fact I did not estimate anything at all. I used scientific literature that sums the measurements of multiple teams taking real-life values of ambient EM radiation in several locations, including the inner London and Tokyo. They also specify the distances to the closest high-power transmission towers, and they took care to pick up an average location, not something in a shielded area or a "dead-zone". So no, the values I posted are no estimations, they are hard facts.

[tinman]

 author=txt link=topic=16003.msg475872#msg475872 date=1456727231]

BTW, the 20 m long antenna is not something you can run around with anyway, regardless that it is just 0.4 mm thick, so I would not consider it useful for powering a flashlight.
Yes, I do. The math is trivial, and I am sure you can succeed the simple multiplying and divisions too. Your system in the video delivers 0.175V at 0.003A. Power = U*I - that gives the continuous charging power of 525μW. For charging 9Wh you hence need 9/525*10^-6 = 17143 hours = 714 days. Which indeed are two years (or pretty close to it). Without counting in the leakage, the charging efficiency, and the possibility that your nearby radio stops broadcasting over the frequency your antenna is tuned to, within that long time.

I dont know how you keep coming up with 525uW as the charging power value,but it is wrong.
You keep saying that 525 is our charging power,when in fact,it is the power being sent to the LED. The actual charging power value during the charging cycle is much lower than that.

I do not consider the charging power of 525μW "usable" for any practical flashlight. It is usable for powering low consumption electronics such as remote sensors, or perhaps even the tiny low power LED as a pilot light, but that was never questioned in this thread. EM harvesting for low consumption devices is known and used since decades, and nobody here ever expressed any doubts about it.

It's just a mater of scaling it up,and making the system more efficient. To say it cant be done makes you sound like MHs son,and following in his footsteps. The very same kind of thoughts came from those that some time back,laughed at the thought of turning the suns light energy directly into electrical energy. Like i said,i do not say i believe the ELFE dose what they claim,but i do say it is possible to charge a torch with 3 AA batteries in it over a 24 hour period using the available EM radiation. Who care's if it has to sit on a bench,and be plugged into a fixed harvester due to a long antenna wire,the fact remains it could be done--and this dose not confirm the ELFE is lagig(as i have noted on a number of occasions now).

The tiny power is simply not useful for high-powered portable devices such as the 3W flashlight. If the ~70 millilumens that you can get out of 525μW, looks usable for you, then I admire your sight, but I can assure you that anyone who would buy a flashlight, finding it emits a few dozens of millilumens, would be highly disappointed.

Well it wouldnt be just 70 millilumens if the cap was allowed to charge for a 24 hour period,and we then dissipated that stored energy through an LED in a 3 hour period-now would it.

Did you grasp my experiment at all?-do you know what i was showing?. Can you really say that it is not possible ?.
Below is the circuit from the video. Put it together,and give it a shot. Then you can tell us all how much charging power it actually has,and then convert that into charge time for the AA batteries


Brad