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Solution is 202° = 0.
It is 50.5 null degrees each zero.
Get 0.25° each degree in a circle quarter range: 1° - 90°.
Get 0.25° each degree for 4 times: 4° - 360°.
The 50.5 business is the business of the zero.
The zero business four times is a circle solution in equilibrium.
The zero business is the circle: 4 * 90° = 360°.
The quadruple circle is the circle solution in equilibrium: 16 * 90° = 1440°
Statement: 1 = 4 --> 0 = 0000
The 5 zero business is the circle with an equilibrium solution: 1800°.
The inverter 10 times is 5 circles like equilibrium circle.
The program find 4 circle equilibrium points.
Program output is one or more this type solutions:
M1 = M(:,:,1) --> 1° zero = 1° degree
M2 = M(:,:,2) --> 2° zero = 2° degree
M3 = M(:,:,3) --> 3° zero = 3° degree
M4 = M(:,:,4) --> 4° zero = 4° degree
1° equilibrium point = 1° degree
2° equilibrium point = 2° degree + 90
3° equilibrium point = 3° degree + 180
4° equilibrium point = 4° degree + 270
zero of the four equilibrium points = 135
The zero of the four equilibrium points is always three circles in eight sectors.
The sum of the four circle equilibrium points is always 742°.
The circle equilibrium solution more one is always 7 times the zero of the four equilibrium points less the sum of the four circle equilibrium points.
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| MATLAB release |
MATLAB 6.1 (R12.1)
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| Zip File Content |
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FZS.m, license.txt
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