Hi,
This email will describe the simplest (as far as I know) method of capturing and
storing ambient temperature energy. Hopefully those wanting to reply could
first read the entire email since I'll address various possible questions later
in this email.
I was hoping at least someone would have answered my previously posted question
to nail down their stance if they believe it's possible to capture and store
energy taken from ambient temperature. Since nobody posted his or her stance
I'll just go ahead and post the proof. This could be a fun ride, as debating
experience shows most people won't be nailed, which allows them to weasel out of
any situation, which is probably one reason there are so many formulations of
the 2nd law. There's a well-taken 2nd law quote in the physics community by
physicist P.W. Bridgman, "There are almost as many formulations of the second
law as there have been discussions of it."
Personally it's not my present goal or interest to focus on the 2nd law.
Truthfully, there are too many 2nd law formulations, as one physicist may adhere
to a stricter interpretation than another. My only assertion is that energy can
be captured from ambient temperature, and here is how.
Here is a clear-cut method to demonstrate the assertion. Using a low noise high
gain amp and oscilloscope view a resistors thermal noise. This is an extremely
simple task. I would be more than happy to provide anyone legitimately
interested individual with a simple circuits to view such noise. You will see
the thermal noise voltage fluctuating in a random unpredictable fashion. Guess
what, you are witnessing a direct conversion from ambient temperature energy to
battery storage. A capacitor stores energy in the form of electric potential.
So where's the capacitor you ask. All measuring devices from common amps to
oscilloscopes have input capacitance. If you want more capacitance than simply
place a small capacitor across the resistor. You will still see the thermal
noise voltage, but the average rms voltage amplitude will decrease. There's now
a total of 4 pF if your amp has 2 pF input and you add a 2pF across the
resistor. Lets say at a given moment you see 10 mV across the capacitor. At
that moment you could unplug the capacitor to claim your energy. LOL, indeed
it's a small amount of energy, but it is true that you actually captured energy
from ambient temperature. If you want more energy then simply make more devices.
Please note I am not stating this is your "smoking gun!" This is ***MERELY***
to demonstrate the possibility, to let people know it is indeed possible!! If
you have the money and technology such as IBM then it's possible to make
trillions of such devices in a small area. One device could be a nanometer. One
hundred trillion 2 pF capacitors at 10 mV each contains 10 mJ's of energy. If
memory holds true, the human eye in complete darkness can see a flash of red
focused light of less than 1 nJ. One 780 nm red light photon contains just
2.5E-19 J's!
Ten mJ's may not sound like much, but it merely demonstrates that you can
capture energy from ambient temperature. This is not the best method of
capturing ambient temperature energy, but again it merely proves the assertion.
Again, in the nutshell, a resistor generates thermal voltage noise. All
measuring devices from common amps to oscilloscopes to multimeters always have a
certain amount of capacitance. When you measured that thermal noise voltage that
capacitor in the measuring device is charged to that value. You can also add
your own capacitor across the resistor. Your capacitor would be completely
discharged before you add it, but at any given moment once the capacitor is
connected to the resistor their will be a certain charged voltage on the
capacitor. At any given moment you could unplug the capacitor to retain such
energy. You could perform the same experiment with an inductor since all
measuring devices have inductance.
What you do with such energy is your choice. One hundred 2 pF capacitors charged
to 10 mV is very usable. That's equal to a 200 farad capacitor charged to 10 mV.
You could discharge the cap energy to an inductor followed by a quick field
collapse to generate appreciable amount of voltage across a smaller cap. Or you
could place a percentage of the caps in series to increase the voltage, etc. etc.
Skeptics may wonder just how much energy is required to "unplug" the capacitor.
There is no theoretical limit. How much energy does it require to move a
nanometer filament a fraction of a nanometer? History demonstrates that the
amount of energy required from an electrical switch has drastically decreased.
Consider the FET, which on average has roughly 1E+12 ohms DC resistance. Sure,
the FET has capacitance, but that in itself is stored energy. This is akin to
how much energy is require to stop an object. One might think it requires a lot
pressure to stop the object. Consider a spinning wheel next to a table. On the
table is a hollow metal tube welded to the table. To stop the spinning wheel one
merely needs to slide a metal bar in the hollow tube extending out the other end
of the hollow tube, which jams in the wheels spokes, which abruptly stops the
spinning wheel. The only amount of energy required to stop the wheel merely
depends how much energy was required to slide the metal bar to jam the spokes.
On many occasions I've described a device that has far higher potential for
"free energy" than the aforementioned example. The above is to provide a simple
undeniable clear-cut example. Of course there will always be those who will deny
anything that goes against their beliefs. A more practical device that requires
***NO*** energy such as from a switch would be my resistor and LED device. The
thermal voltage noise from the resistor will generate thermal current in the
LED. All LED's emit photos at any applied voltage. It just turns out the LED is
exponentially more efficient above the forward voltage level. In such a device
the LED would emit more photons when connected to a resistor of high resistance.
Lets consider photovoltaic cells. Even at room temperature in complete darkness
(no solar) there are visible light photons striking the cell. I calculate a 10
cm x 10 cm common solar cell would generate roughly 1E-30 volts. Not much
voltage, lol, but still something nonetheless. The amount of radiated blackbody
energy is small in the visible region. Although the FIR region is another
story. Both sides of a thin sheet of 1m x 1m material radiates roughly 920
watts continuously in complete darkness at room temperature. Technology is
improving, thereby allowing photovoltaic cells to capture lower and lower
frequencies. A Canadian university succeeded in creating a 1355 nm photovoltaic
cell! That's only 1/11th the wavelength away from the peak 15000 nm 920
watts/m^2 blackbody 300 K radiation. BTW, blackbody radiation at 1355 nm is
2E+18 times greater than visible region of 600 nm. To calculate this I compared
the radiation from 16667 to 16677 cm^-1, which is 3.907E-29 watts to 7380 to
7390 cm^-1, which is 7.499e-11 watts.
University of Toronto in Canada achieves 1355 nm photovoltaic cell:
http://nanotechweb.org/articles/news/4/1/7/1
Eventually technology will reach the peak 15000 nm region where a thin double
sided 1m x 1m sheet receives ~920 watts. It's difficult for a person to believe
they are surrounded by a source "free energy" because we don't see such energy
with our eyes.
Regards,
Paul Lowrance
