Kapanadze Cousin - DALLY FREE ENERGY

[Dog-One]

Hello Dog One    ..... Yess     2 times 19M is  38M

Okay good.  Dielectric lines of force are balanced.

So in terms of wire length, the grenade is neutral.  Same meters of wire
clockwise as counter-clockwise.


Now for magnetic lines of force, my follow-up questions are...

Total turns of layers 1 + 2 =    what?

Total turns of layers 3 + 4 + 5 + 6 =    what?

[verpies]

I'm surprised no one noticed what happened when the metal shunt on the end of the stick, was shorted across the terminals next to one of the light bulbs. One would expect a shower of sparks, killing the whole machine and welding the shunt to the two terminals

One would expect that if the device acted as a voltage source.
One would not expect that if the device acted as a current source.

^{Back to Basics: Charged capacitors act as voltage sources and charged inductors act as current sources.}

[T-1000]

A simple answer, yes or no...

Does the wire length of layers 1 + 2 = 3 + 4 + 5 + 6?

The simpler answer - L1+2 inductance = -L3+4+5+6 .. This is what we aiming for with winding/unwinding L5+L6 for reaching total inductane close to 0
And the standing wave have to be under inductor of grenade which makes scalar field there.

Cheers!

[AlienGrey]

Okay good.  Dielectric lines of force are balanced.

So in terms of wire length, the grenade is neutral.  Same meters of wire
clockwise as counter-clockwise.


Now for magnetic lines of force, my follow-up questions are...

Total turns of layers 1 + 2 =    what?

Total turns of layers 3 + 4 + 5 + 6 =    what?

Don't quote me on this but I will now try and explain the idea or totally confuse you

lets suppose,  (i will now use use L = layer ok), L6 &5 = length 1 (1/4) so it's 2 1/4 = 2  of our 14 sections, With it ?

L3 & L4 are twice length of L5 L6 = 1/2 X2 = 4 still with it ?

L1 &L2 are twice  L3 & L4  = 1wave x 2 = 8 still with it  so 8 + 4 + 2 = total of 14 equal sections of = length cable of our 38.5 meters of cable.(all lengths of cable 14 in all have to be the same length x the amount 1/4 waves used in that section of cable your wingdings

lets assume our wave is 38 meters ok so if 38.5m / 14 = 2.5 meters approx per 1/4 wave

Note L3 4 5 6 are used to try to zero the inductance of the first two wingdings
but all the 37 meter should be used in the wind with the rest used as a connector length = 37.5 total.
all info is from information collected from original source.

L6  1/4 wave  total wire length 2.75 m                      *per layer **     1/4 wave
L5  1/4 wave total wire length  2.75 m                             ''                 1/4 not wave

L4  1/4 + 1/4  total wire length 5.5 m                              ''                 1/2 wave
L3  1/4 + 1/4 total wire length 5.5 m                               ''                 1/2 not  wave (inverted)

Note the lower L1&2 have to 'spot on'
L2  1/4 + 1/4 +1/4 +1/4 total  wire length 11  m            ''                   1.0 wave
L1  1/4 + 1/4 +1/4 +1/4  total wire length 11 m             ''                   1.0 not wave (inverted)
     
Golden ratio ? wave guide
confused ( wait till you find out how un linier in inductance it is as you wind it !

[wattsup]

@verpies

http://overunity.com/12736/kapanadze-cousin-dally-free-energy/msg501064/#msg501064

Just wanted to thank you for your post above. I will look into this to see how they derived this data.

wattsup