%Blahblah Uncle Don's T-Line using Cylindrical Coordinates
%First build A Matrix for 3 circles divided into 8 pies

romax=3; %romax is the total ro values we take
simax=8; %simax is the divisions of the coordinate system

siz = romax*simax; %size of the A matrix

A = eye(siz)*0; %this sucka is just the matrix with no values in yet. It's reset to zero
B = zeros(siz,1);

%And now let's build the functions to calculate the constants for our 
%cylindrical coordinate finite difference method calculations
delro = 0.005; %5 cm
delsi = pi/4; %psi increment in radians
ronot = 0.01; %the circle layer distance from origin

%inner node ~closer to origin
Ao = 1-delro/2/ronot;

%outer node ~towards edge
Bo = 1+delro/2/ronot;

%counterclockwise node and clockwise node
Co = (delro/ronot/delsi)^2;

%original node ~the center node
Do = -(2+2*(delro/ronot/delsi)^2);

%Symmetry Considerations: by examining the transmission line, it can be
%seen that several nodes are equal by symmetry
%nodes 1,5
%nodes 7,3
%nodes 8,2,4,6
%nodes 9,13
%nodes 15,11
%nodes 16,10,12,14
%nodes 17,21
%nodes 23,19
%nodes 24,20

%nitty gritty ~crunchin numbers by each ronot
ronot = 0.01;
Ao = 1-delro/2/ronot;
Bo = 1+delro/2/ronot;
Co = (delro/ronot/delsi)^2;
Do = -(2+2*(delro/ronot/delsi)^2);

%node 1
B(1) = -65*Ao; %inner node known constant will be moved to B matrix
A(1,9) = Bo*0; %outer node
A(1,2) = Co;   %counterclockwise node
A(1,8) = Co;   %clockwise node
A(1,1) = Do;   %center node

%node 8

%node 3
B(3) = -65*Ao;
A(3,11) = Bo;
A(3,4) = Co;
A(3,2) = Co;
A(3,3) = Do;

%node 4
B(4) = -65*Ao;
A(4,12) = Bo;
A(4,5) = Co;
A(4,3) = Co;
A(4,4) = Do;

%node 5
B(5) = -65*Ao;
A(5,13) = Bo;
A(5,6) = Co;
A(5,4) = Co;
A(5,5) = Do;

%node 6
B(6) = -65*Ao;
A(6,14) = Bo;
A(6,7) = Co;
A(6,5) = Co;
A(6,6) = Do;

%node 7