On my YouTube channel page, I have long ago renamed the videos so that the data appear in the titles:

http://www.youtube.com/user/kmarinas86I have made comparisons between several of the motors. Specifically, I looked at six videos meant to achieve the following:

1) To compare the change in performance between one and two coils.

2) To compare the change in performance between two and three coils.

3) To compare the change in performance between three and four coils.

From this I deduced that when minimal other modifications are done:

For each coil added (4000 ft of 28 AWG wire):

For a given voltage, the RPM drops to 90% of the original value.

For a given voltage, The Current / Volt drops to 50% of the original value.

For a given voltage, the power output drops by (90%)^3 or 72.9% of the original value.

From there, I have developed an empirical method to predict what will happen once I get around to adding more coils:

n= newly added coils (past the current 4 coils)

v= voltage increase multiplier (1, or 100%, means the same voltage)

**Note that for conventional technology, (n+4) at this power level is VERY small. It seems, however, that the value of (n+4) is only valid so as long as the wire gauge is the same. It may turn out that there is much more to it than just the number of coils when we change the gauge of the wire. Remember that empirical formulas are only estimates of what is going on. There is not much understanding right now as to the emergence of the relationships.**________________________

Adding a coil

(.9)^3/.5 = change in power out / change in power in

1.458 = change in power out / change in power in

Adding n coils:

[(.9)^3/.5]^n = change in power out / change in power in

1.458^n = change in power out / change in power in

________________________

Cooling effect

n*(backspike current)^2

________________________

Output Power

[(.9)^3]^n*v^3

[(.9)^n]^3*v^3

[v*(.9)^n]^3

If the machine is to operate at the same power output as before:

v^3 = [(10/9)^3]^n

v = [(10/9)^3]^(n/3)

If n=2, v=1.23

________________________

Input power:

(.5)^n*v^2

If the machine is to operate at the same power input as before:

v^2 = 2^n

v = 2^(n/2)

If n=2, v=2

________________________

Efficiency:

[2*(.9)^3]^n*v

[1.458]^n*v

If the machine is to operate at the same efficiency as before:

v = [1/1.458]^n

If n=2, v=0.4704

________________________

Cooling effect / Output power

[n*(backspike current)^2] / [[(.9)^3]^n*v^3]

[n/[(.9)^3]^n]]*[(backspike current)^2/v^3]

[n/[(.9)^3n]]*[(backspike current)^2/v^3]

[n/[(10/9)^(-1)^3n]]*[(backspike current)^2/v^3]

[n/[(10/9)^(-3n)]]*[(backspike current)^2/v^3]

[n*[(10/9)^3n]]*[(backspike current)^2/v^3]

For large n, n*[(10/9)^3n] rises faster than v^3 even if n is proportional to v.

________________________

Input current density (without cooling effect) (wire gauge unchanged):

(.5)^n*v

Restore to original input current density (awg rating unchanged):

2^n = v

If n=2, v=4

________________________

Output power density:

[(.9)^3]^n*v^3 / [n+4]

4 = original coil number

________________________

Output power density / Input current density ::

[[(.9)^3]^n*v^3 / [n+4]]/[(.5)^n*v]

[[2*(.9)^3]^n*v^2] / [n+4]

[1.458]^n*v^2 / [n+4]

If input current density is kept the same, the following would be the projected power density of the device:

[1.458]^n*[2^n]^2 / [n+4]

[1.458]^n*[2^(2n)] / [n+4]

[1.458]^n*[4^n] / [n+4]

[5.832]^n / [n+4]

Here it is graphed:

-3 0.0050413570150648400 0.02

-2 0.0147005970559291000 0.06

-1 0.0571559213534522000 0.23

0 0.2500000000000000000 1

1 1.1664000000000000000 4.67

2 5.6687040000000000000 22.67

1st column = n

2nd column = [5.832]^n / [n+4]

3rd column = 2nd column "normalized" (n=0)

rows 1 to 6: from 1 coil to 6 coils