On Jul 16, 2009, at 9:33 AM, Taylor J. Smith wrote:
Hi All, 7-16-09
Is there anything to this?
Jack Smith
http://veritasshow.blogspot.com/2009/06/german-scientist-posts-
complete-free.html
The PDF is here:
http://www.wbabin.net/physics/turtur1e.pdf
This makes no sense to me. Even if the theory behind the rotor is
valid, the test device is way too large.
The energy density rho(w) is characterized by H. E. Puthoff [H. E.
Puthoff, "The Energetic Vacuum: Implications for Energy Research",
Speculations in Science and Technology, vol. 13, no. 4, pp. 247-257,
1990.] by:
rho(w) dw = [w^2/pi^2*c^3]/[hw/2] dw = (hw^3) / (2*pi^2*c^3) dw
(resulting in units of joules/m^3)
Rearranging we have:
rho(w) dw = (h/(2*pi^2*c^3)) w^3 dw
rho(w) dw = K w^3 dw, where K = (h/(2*pi^2*c^3))
Integrating over w=0 to w=B to get cumulative energy density f(B) to
cutoff frequency B:
f(B) = K/4 B^4
This indicates that the total energy density of the vacuum (though
not constant if tapped) is proportional to the fourth power of the
cutoff frequency being tapped. A device built on the scale of cm or
even mm is going to limit the cutoff frequency B for the device to a
very small number, and thus the available energy density. If lambda
is the wavelength, then c/lambda gives the cutoff frequency B, and
f(B) = ((h/(8*pi^2*c^3)) * (c/lambda)^4
and plugging in lambda = 1 cm I get an energy density f(B) of
2.515865x10^-19 J/m^3. Using lambda = 1 mm, I get 2.515865x10^-15 J/
m^3. Even if that energy is utilized 100 percent, and replaced at
the speed of light, 3x10^8 m/s, the maximum replacement rate for a 1
mm cube device is (3x10^8 m/s)/(10^-3 m) = 3x10^11 Hz, so the maximum
output is ((3x10^11 Hz)*(1 mm)^3)*(2.515865x10^-15 J/m^3) =
7.54x10^-13 W.
I don't think 10^-12 watts is going to make the turbine move, even if
there is a net force from the design, which I doubt for unrelated
reasons.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
