Re: Electo London Gravity ?
- From: "Significant Zero" <paulpsremove@xxxxxxxxxx>
- Date: Mon, 11 Jul 2005 14:37:55 +0100
"sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
news:42d25d14$0$18639$14726298@xxxxxxxxxxxxxxxxxx
|
| "Significant Zero" <paulpsremove@xxxxxxxxxx> wrote in message
news:1121078108.29029.2@xxxxxxxxxxxxxxxxxxxxxxx
| >
| > "sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
| > news:42d198f4$0$18647$14726298@xxxxxxxxxxxxxxxxxx
| > |
| > | "Significant Zero" <paulpsremove@xxxxxxxxxx> wrote in message
| > news:1121027931.10233.1@xxxxxxxxxxxxxxxxxxxx
| > | >
| > | > "sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
| > | > news:42d162be$0$18639$14726298@xxxxxxxxxxxxxxxxxx
| > | > |
| > | > | "Significant Zero" <paulpsremove@xxxxxxxxxx> wrote in message
| > | > news:1121013883.18182.2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | > | >
| > | > | > "sue jahn" <susysewnshow@xxxxxxxxxxxx> wrote in message
| > | > | > news:42d0fd5a$0$18637$14726298@xxxxxxxxxxxxxxxxxx
| > | > | snip
| > | > | > | > | > >
| > | > | > | > | > > 1/r^1 = 1 so anything with 1/r^1 is an inertial
ballistic
| > | > object ?
| > | > | > so
| > | > | > | > I
| > | > | > | > | > > don't know why that guy was quoting Coloumbe at 1/r
below.
| > ?
| > | > | > | > |
| > | > | > | > | It is the same as gravity. If you were a proton in the
center
| > of
| > | > an
| > | > | > | > | electron cloud half the e- minus would be on you right,
half
| > on
| > | > your
| > | > | > left.
| > | > | > | > | If you were on the surface of the e- cloud, 100% the
| > electrons
| > | > would
| > | > | > | > | be on one side of you. Same as gravity, 1/r^1 inside the
| > cloud.
| > | > | > | > | 1/r^2 as you move away from the cloud.
| > | > | > | >
| > | > | > | > Its not 1/r^1 see the internal gravity slope graph I mention
| > above
| > | > its
| > | > | > more
| > | > | > | > complex than that, inside a planet anyway. I will thing
about
| > your
| > | > | > example
| > | > | > | > as I don't think that equates to the field law slope inside
a
| > | > planet. I
| > | > | > | > think your example at particle level to equate to a planet
would
| > | > have to
| > | > | > be
| > | > | > | > stating the coulomb force experienced by an electron as it
| > burrowed
| > | > into
| > | > | > the
| > | > | > | > centre of a proton and to make the gravity and coulomb
examples
| > | > parallel
| > | > | > you
| > | > | > | > would have to be talking about the forces experienced by a
| > proton as
| > | > it
| > | > | > | > burrowed its way to a cloud of electrons which would not be
1/r
| > but
| > | > a
| > | > | > | > complex slope as per gravity so the bloke who quoted 1/r
below
| > has
| > | > not
| > | > | > | > thought about it enough ?
| > | > | > |
| > | > | > | You won't like to hear this, but I am not sure protons have
| > gravity
| > | > | > either.
| > | > | > | Perhap they do if we can detect FQHE.
| > | > | >
| > | > | > They do but it would just be a component of the charge and would
| > | > probably be
| > | > | > immeasurable small between two hydrogen atom ? Clue me FQHE I'm
| > terrible
| > | > | > with abbreviations sometimes.
| > | > | http://www.google.com/search?hl=en&q=FQHE&btnG=Google+Search
| > | > |
Interesting
| > | > | >
| > | > | > |
| > | > | > | Regarding R and r
| > | > | > | The mathymatishun in you, wants to see a smooth curve but see
| > here:
| > | > | > | http://www.physicsclassroom.com/Class/circles/u6l3e5.gif
| > | > | > |
| > | > | > | Missing on that graph is a nearly vertical red line
representing
| > | > | > | force below the surface.
| > | > | >
| > | > | > The force below the surface is equal to the acceleration and
that is
| > | > what
| > | > | > that upside down tick is displaying.?
| > | > | No. You need to draw a line from 0,0 up toward the
| > | > | one o'clock position to meet with the highest point on the
| > | > | red line.
| > | >
| > | > I have lost you here as we are debating what happens inside a planet
or
| > body
| > | > and this according to the link you gave earlier is an inverted tick
and
| > does
| > | > not match 1/r^2 or 1/r
| > |
| > | What happen inside is induced dipoles in the planet coupled to induced
| > | dipoles in the miner's body. All you have to do is ratio the dipoles
above
| > him
| > | to the dipoles below him to calculate his weight. At earth center it
is
| > | 50% above 50% below for a weight of zero.
| > | >
| > | >
| > | > ~ acceleration inside earth
| > | >
| > | > /\ surface
| > | > /
| > | > /
| > | > center
| > |
| > | That *%#!0# patent examiner has you hooking rockets on our
| > | poor little miner...don't he? You trying to smash his helmet on
| > | the cave ceiling or ya gonna let him stand in the shaft before ya
| > | light the fuse? :o)
| > |
| > | There ain't no *acceleration* in those VDW and London
| > | calculations... just FORCE. :o)
| >
| > Acceleration against an energy state change (surface) equals force?
| > The above graph is correct as you have been looking at uniformed bodies
and
| > the acceleration is down not up.
|
| OK... If you say so. I'd much rather call it force unless we can
| can quantify time and displacement.
|
How do you quantify force except by time and displacment ?
| >
| > snip
| >
| > | A point ? Ouch again!
| >
| > Just a mathematical point as no real physical points exist except
| > subjectively {:-).
| >
| > snip
| >
| > | Decode the holy grail first.
| > |
| > | (1-(1/(1/3))^2)/(1 + (1/(1/3))^2) = - 0.08 = FTL ? -p<+p or
| > | > (m*-v)<(m*+v) or (m*-c^2)<(m*+c^2) =g?
| >
| > The first bit is the product of the reaction of an accelerating mass and
by
| > recursion equalling FTL?
| Big deal! I am moving FTL right now wrt one of the particles at fnal.
| If I was a bit closer to one of those particles, Coulombly speaking,
| it would be a big big deal.
Great it was just a remaining point from a SR gung ho were the point was no
FTL so I arithmaticed it out of their math {:-)
Just seems to give loads of possibilities if you don't put your mind in that
rat trap.
| Don't cry. I understand. That mean old Maxwell made you do that
| didn't he. He thinks simply accelerating particles makes 'em radiate.
| I guess it never occured to him what kind of gloves he was gonna
| wear before he took a firm grasp on a particle to accelerate it. ;-)
No I was fighting with an SRist at the time, FTL had never been a hang up of
mine {:-) Who's Maxwell {:-)
|
| and the second is the asymmetric response to a
| > symmetric effect ?.
| Sounds like a chaotic butterfly makin' cyclones.
Its one of the operating bases of gravity as well so get out your butterfly
net {:-)
|
| >
| > snip
| >
| > | > | Who would have tho't someone would be a national leader of the
| > | > | English speaking people that doen't even speak the languge?
| > | > |
| > | >
| > | > Who are you referring to ?
| > | http://www.google.com/search?hl=en&q=dubya+english&btnG=Google+Search
| > |
| >
| > Some of his twit is mind blowing" It's clearly a budget - it's got lots
of
| > numbers in it." G.W. Bush that's the sort of guy we need running another
| > country not the S.H types who never make jokes and just tear your
toenails
| > off, give me GWB every time over SH, GWB a hoot with a boot and a man
after
| > my own bum. I expect if you are a Cubanio then you might have some
| > historical issue with the US P. but then every body has their cross to
bear
| > and I hope for your sake you can get over it, codolences to go with the
| > fishing.
|
| Before 9/11 WTC Dubya had big plans to go to Mars. I am just sorry for
| his sake that he had to settle for a quick junt out to a carrier instead.
| Hopefully he'll get his travel plans back on schedule soon. ;o)
| http://www.wepsite.de/Dubya_USS_Abraham_Lincoln_Chip_Bok_May_2_2003.gif
| http://www.wepsite.de/Dubya_economyBorgman_May_5_2003.jpg
|
Your just bitching about the number of air miles he's got saved\pinched {:-)
just think of the laughs he's generated, he's the Tony Han*** of the US.
| Sue...
|
| >
| > snip
| > | > --
| > | > Significant Zero E-field = Electric field, M-field =Magnetic
field,
| > two
| > | > unbound field effects
| > | > http://home.freeuk.com/paulps/
| > | > Maybe updates. (1-(1/(1/3))^2)/(1 + (1/(1/3))^2) = - 0.08 = FTL
? -p<+p
| > or
| > | > (m*-v)<(m*+v) or (m*-c^2)<(m*+c^2) =g?
| > | >
| > | >
| > | >
| > |
| > |
| >
| >
|
|
.
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