Internal Capacity Squarings for 1-32 winds.
>also pairs with two past movements, and not merely one.
> And the return starting with 29;
> Previously 28 was paired with 5, so 29 must be paired with 4, but it
> can only do this horizontally, instead of vertically, and now it
> The vertical code of 33 is broken.starts on a horizontal basis of addition to seven square, or 49. Thus
> However on move 33, after the midpoint of windings; a new code
on seventh quadrant move 33 must land horizontally next to the former
move of 16 formerly made in the fourth quadrant journey.
Thus during the former last two quadrants of travel the journey has
followed itself backwards from the path established by the first 16
moves, but this method breaks down in the 7th quadrant; which is why I
became confused. However the patterns are identical in that they
consist of a clockwise chess knight pattern along the outside squares
that circles the inside four squares, which is repeated above.
Now if this were structured as a coil using insulated square
windings where 64 volts appears across 64 winds, we can calculate the
internal capacity on the basis of the voltage difference between
windings. For a laterally wound coil of 64 square shaped windings in a
64 array, the first layer of 8 has 7 differences of 1 volt between
them. The electric energy stored between these square surface areas is
1/2*C*V^2. The portion representing the capacity can be measured as
the C value between just one adjacent surface area between two
windings,and this is then repeatedly multiplied by the squarings of
the voltage difference between windings. So here it can be seen that
of these voltage differences between windings, there are seven sets of
eight, or 56 on both a horizontal and vertical basis, for a total of
110 internal capacities, of which for a laterally wound coil half of
these consists of one volt differences in one dimension.
On the other hand for the knight model the first time the winding
is laid adjacent to another winding occurs on winding 17
as the fourth quadrant placement shows, making for the first voltage
measurement difference of 3^2. On the next wind 18, this is held in
isolation from the rest of the return layer. However the next winding
19 has both 3^2 and 5^2 voltage differences with former windings. And
to finish for this quadrant another 5^2 difference is found between 15
This gives 2(3^2 + 5^2) for the fourth quadrant.
For the fifth quadrant of travel;
Here we initially See that a voltage difference BETWEEN the quadrants
exists between 9 and 20, or 11^2. 9^2 exists on the first movement of
21, then 11^2, 13^2,15^2 summing to
Returning again to the seventh quadrant to finish the second layering;
25-4=21, between quadrants, this seems to be duplicated as a voltage
difference in the previous quadrant as 27-6=21
For the first movement into the quadrant;
and finally 31 is held in isolation for reference to future wind layers.
this ends with 32-3= 31.
I will later add these squarings to find the internal capacity for the
first folded layer in this array.