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Lets Talk AC(did i figure it out)?

Last response: in Wireless Networking
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October 18, 2010 3:35:53 AM

)+-+-+-+-+- )-+-) -+)
-+-+-+-+-+-+) -+-) +-+)
o-+-+-+- (Cu) )(Fe)) (Cu) + + + + +
-+-+-+-+-+-+) -+-) +-+)
)+-+-+-+-+- )-+-) -+)


- go up
+ go down

o = core (-||+)

- is perpetual to + as + is perpetual to -

thier fore if you have a core that is 99.9% circular sphere you could direct a "signal" of ether "-" or "+" depends on where the .01% stands if you look at a "Globe" you would see that it has a "N" and a "S"( N == End == +)while (S == Start == -) but in order to have a circle you would need to start from ponit 1 and go tho point 9 and land on point 0 but when their is a 0 their is a 1. but the only reason that thier must be a 0 after or before a 1 is cause even if you turn off a "Current" "Connection" you must turn it back on or the signal will disappear... but if you have it unable to turn it back off than the signal will grow and grow until it find a "equal" "signal" of the opposite response.

so if you had a (-||+) in the middle of a "Electromagnetic Field" that creates another (-||+) but instead it goes (Cu(|Fe|)Cu) as Cu = - and Fe turns + but (Fe +) must have a (-) (which it does (Cu -)) but (Cu -) is only "transmitting ((-) 66.6% ) to (Fe +) their fore the 33.3% of the Fe that is left uncovered creates it own (-+) but the signal from (Fe (-||+)) goes tho a direction that does not interfere with the (66.6% (-)) but if you "fuse" to make a (--) or a (++) you would eventually have the (--||++) find another (--||++) but than soon (-^inf||+^inf) will eventually find a "Equivalent Opposite"

so 99.9% creates a unstable amount of energy.
therefore you must make the .1% trans-splitted to equal the amount of the unstable energy produce to equal a stable flow of energy.....
soo this was just top of my head.... may you correct me and add to it ^^ ? :pt1cable: 

More about : lets talk figure

October 18, 2010 12:36:50 PM

you are forgetting the angular momentum in electrodynamics.

Remember L=r*p is not guage invariant; so you have to use:

p-(eA)/c where: p = kinetic momentum, e=electric charge, A=vector potential, and c=speed of light, of course.

from this we use the hamilitonian of electric charge to calculate the field being:

H=(1/2m)(p-[eA/c])^2+eφ where m=mass, and φ = scalar potential.

From that we reach the Lorentz law, which is of course:
K=r*(p-[eA/c])

Hope this clears things up.

[/wiki was used as a source to clarify the hamiltonian as it has been awhile since i have used it.]
October 19, 2010 1:28:18 AM

:o  :o 

ahhh... ok... thanks for my homework... >.<

i going to have to decypt L=r* Kinetic Momentum... what is L and what is r?

but thank you for some answers. no i just have to decypt this equation to my understanding... hmmm thank you :love: 
October 19, 2010 1:52:26 AM

)+-+-+-+-+- )-+-) -+)
-+-+-+-+-+-+) -+-) +-+)
o-+-+-+- (Cu) )(Fe)) (Cu) + + + + +
-+-+-+-+-+-+) -+-) +-+)
)+-+-+-+-+- )-+-) -+)


Look more closely.. this not now going ----> "Straight" this is going 360(z) x 360(y) x 360(x)

- != -
+ != +

- = +

look forget the .



O = Compressed Feild
....|>+>+>+>+>pos>>>>>>>>>>>>>>>>|
....^__________________........................|
<--()+-+-+-+-+-+-+(spin ^)<><><><><>()
....^-----------------------------........................|
....|<-<-<-<-<-<neg<<<<<<<<<<<<<<<<|

add conductor to O and the Conductor will create it own energy which will turn to (o)><><o><><(o)

Conduct.. Transmit... Receive... Transmit... Conduct...so on so on so on.......can you achieve this?


(PI x PI) != 10

-+- +-+ != +
but what if I = -+- to have a electron signal you must have another - after the +

draw 4 dots to form a square ... now slept the dots into 4 so you would have 1/4 x 1/4 circle. each 1/4 will transmit either a - or a + but + cant touch + nor - can touch - now create a 1/2 out of the 2 of the 1/4 pieces you would not have a 1/2 that has a 180(z) x 180(x) x180(y). this is a puzzling subject for me...... i have a hard time explaining what i think... >< :fou:  :pfff: 


i think i have found my answer... now on to the next problem/equation http://en.wikipedia.org/wiki/Superparamagnetism
!