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verdigris
Posted: Mon Mar 12, 2007 9:51 am
Guest
On 11 Mar, 20:19, "scerir" <sce...@libero.it> wrote:
Quote:
'The accordance between revealed 4He and produced energy
seems to be a clear signature of a nuclear process
occurring in condensed matter.'

The bottom line is how can we get a temperature high enough - at least
10^7K to cause fusion.The paper at

http://lenr-canr.org/acrobat/DeNinnoAexperiment.pdf

suggests that the temperature can be lower because a plasma of d
electrons from palladium lowers the coulomb barrier ( presumably the
plasma is caused by the
high temperature generated in a small volume by the saturation of
palladium with deuterium atoms ).An electric current passes through
the palladium.Now this could raise the temperature a bit more by
frictional heating,but if there are cavities in palladium

http://lenr-canr.org/acrobat/DeNinnoAexperiment.pdf

perhaps hydrogen is turned into an ultraviolet laser

( but where would the relativistic electrons this document mentions
come from in the experimental setup we are dealing with).
Dirk Bruere at NeoPax
Posted: Mon Mar 12, 2007 6:27 pm
Guest
verdigris wrote:
Quote:
On 11 Mar, 20:19, "scerir" <sce...@libero.it> wrote:
'The accordance between revealed 4He and produced energy
seems to be a clear signature of a nuclear process
occurring in condensed matter.'

The bottom line is how can we get a temperature high enough - at least
10^7K to cause fusion.The paper at

http://lenr-canr.org/acrobat/DeNinnoAexperiment.pdf

suggests that the temperature can be lower because a plasma of d
electrons from palladium lowers the coulomb barrier ( presumably the
plasma is caused by the
high temperature generated in a small volume by the saturation of
palladium with deuterium atoms ).An electric current passes through
the palladium.Now this could raise the temperature a bit more by
frictional heating,but if there are cavities in palladium

http://lenr-canr.org/acrobat/DeNinnoAexperiment.pdf

perhaps hydrogen is turned into an ultraviolet laser

( but where would the relativistic electrons this document mentions
come from in the experimental setup we are dealing with).

Laser acceleration of electrons in plasma to relativistic velocities can
be done over remarkably short distances. Whether this is relevant is moot.

--
Dirk

http://www.onetribe.me.uk - The UK's only occult talk show
Presented by Dirk Bruere and Marc Power on ResonanceFM 104.4
http://www.resonancefm.com
Autymn D. C.
Posted: Sat Mar 17, 2007 2:32 am
Guest
Autymn D. C.
Posted: Sun Mar 18, 2007 6:53 pm
Guest
On Mar 6, 8:55 pm, Richard Saam <rds...@att.net> wrote:
Quote:
of their candidacy examinations. I recall one question that makes
me
shudder to this day. It simply asked to calculate the height of the
ocean
tides. There wasn't a single candidate who took that exam who had
any idea
how to do this on the back of an envelope.

G = gravitational constant
Me = earth mass 5.97E+27 g
Mn = moon mass 7.35E+25 g
Re = earth radius 6.371315E+08 cm
Rem = distance between moon and earth 3.844E+10 cm
T = tide height
Wm = water test mass

Gravity works on a water test mass
with force from earth and moon:

G*Me*Wm/Re^2 = G*Me*Wm/(Re+T)^2 + G*Wm*Mm/(Rem)^2

Why not 0 = the two weihts? What has the Me--Wm attraction to do with
them?
Richard Saam
Posted: Mon Mar 19, 2007 6:14 am
Guest
Autymn D. C. wrote:
Quote:
On Mar 6, 8:55 pm, Richard Saam <rds...@att.net> wrote:

of their candidacy examinations. I recall one question that makes

me

shudder to this day. It simply asked to calculate the height of the

ocean

tides. There wasn't a single candidate who took that exam who had

any idea

how to do this on the back of an envelope.

G = gravitational constant
Me = earth mass 5.97E+27 g
Mn = moon mass 7.35E+25 g
Re = earth radius 6.371315E+08 cm
Rem = distance between moon and earth 3.844E+10 cm
T = tide height
Wm = water test mass

Gravity works on a water test mass
with force from earth and moon:

G*Me*Wm/Re^2 = G*Me*Wm/(Re+T)^2 + G*Wm*Mm/(Rem)^2


Why not 0 = the two weihts? What has the Me--Wm attraction to do with
them?


Look at it this way:

G*Me*Wm/Re^2 - G*Me*Wm/(Re+T)^2 = G*Wm*Mm/(Rem)^2

As an added note:
This is a potential condition.
The ~30 meter bulge as BOTE previously calculated
will not be achieved on the rotating earth (24 hr).

Actual tide heights continually chase this potential
attempting to overcome water kinetic and viscous effects.

Richard
Richard Saam
Posted: Wed Mar 21, 2007 5:10 am
Guest
Autymn D. C. wrote:

Quote:
On Mar 6, 8:55 pm, Richard Saam <rds...@att.net> wrote:

of their candidacy examinations. I recall one question that makes

me

shudder to this day. It simply asked to calculate the height of the

ocean

tides. There wasn't a single candidate who took that exam who had

any idea

how to do this on the back of an envelope.

G = gravitational constant
Me = earth mass 5.97E+27 g
Mn = moon mass 7.35E+25 g
Re = earth radius 6.371315E+08 cm
Rem = distance between moon and earth 3.844E+10 cm
T = tide height
Wm = water test mass

Gravity works on a water test mass
with force from earth and moon:

G*Me*Wm/Re^2 = G*Me*Wm/(Re+T)^2 + G*Wm*Mm/(Rem)^2


Why not 0 = the two weihts? What has the Me--Wm attraction to do with
them?

Look at it this way:


G*Me*Wm/Re2 - G*Me*Wm/(Re+T)2 = G*Wm*Mm/(Rem)2

As an added note:
This is a potential condition.
The ~10 meter bulge as BOTE previously calculated
will not be achieved on the rotating earth (24 hr).

Actual tide heights continually chase this potential
attempting to overcome water kinetic and viscous effects.

Richard

(note correct 10 meter rather than 30 meter on prior post)
 
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