Divide Mesh grid by a bisector, MATLAB - matlab

Question

I have a piecewise function, where domain changes for each case. The function is as follows:
For
(x,y)greater than Divider v= f(x,y) (A1)
(x,y)less than Divider v = g(x,y) (A2)
The location of the divider changes with tilt angle of the rectangle given in figures 1 and 2.Figure 1 & 2 The divider will always be a bisector of the rectangle. For example, the divider makes an angle (alpha + 90) with the horizontal.
If the rectangle makes an angle 0, it's easy to implement above functions as I can create meshgrid from
x =B to C & y = A to D for A1
x =A to B & y = A to D for A2
However, when the angles for the rectangle are different, I can't figure out how to create the mesh to calculate the function v using the algorithm A1 and A2 above.
I was thinking of using some inequality and using the equation of the line (as I have the co-ordinates for the center of the rectangle and the angle of tilt). But, I can't seem to think of a way to do it for all angles (for example , slope of pi/2 as in the first figure, yields infinity). Even if I do create some kind of inequality, I can't create a mesh.
1Please help me with this problem. I have wasted a lot of time on this. It seems to be out of my reach
%% Constants
Angle1=0;
Angle1=Angle1.*pi./180;
rect_center=0; % in m
rect_length=5; % in m
rect_width=1; % in m
rect_strength=1.8401e-06;
Angle2=0;
Angle2 =Angle2.*pi./180;
%% This code calculates the outer coordinates of the rectangle by using the central point
% the following code calculates the vertices
vertexA=rect_center+(-rect_width./2.*exp(1i.*1.5708)-rect_length./2).*exp(1i.*Angle2);
vertexA=[vertexA,vertexA+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
vertexB=rect_center+(-rect_width./2.*exp(1i.*1.5708)+rect_length./2).*exp(1i.*Angle2);
vertexB=[vertexB,vertexB+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
za1=vertexA(1:numel(vertexA)/2);
za2=vertexA(1+numel(vertexA)/2:numel(vertexA));
zb1=vertexB(1:numel(vertexB)/2);
zb2=vertexB(1+numel(vertexB)/2:numel(vertexB));
arg1=exp(-1i.*Angle2);
%% This Section makes the two equations necessary for making the graphs
syms var_z
% Equation 1
Eqn1(var_z)=1.5844e-07.*exp(-1i.*Angle1).*var_z./9.8692e-13;
% subparts of the Equation 2
A = 1.0133e+12.*(-1i.*rect_strength.*exp(-1i*Angle2)./(2*pi.*rect_length.*rect_width*0.2));
ZA1 = var_z+za1-2*rect_center;
ZA2 = var_z+za2-2*rect_center;
ZB1 = var_z+zb1-2*rect_center;
ZB2 = var_z+zb2-2*rect_center;
ZAA2 = log(abs(ZA2)) + 1i*mod(angle(ZA2),2*pi);
ZAA1 = log(abs(ZA1)) + 1i*mod(angle(ZA1),2*pi);
ZBB1 = log(abs(ZB1)) + 1i*mod(angle(ZB1),2*pi);
ZBB2 = log(abs(ZB2)) + 1i*mod(angle(ZB2),2*pi);
%Equation 2 ; this is used for the left side of the center
Eqn2= A*(ZA2*(log(ZA2)-1)-(ZA1*(log(ZA1)-1))+(ZB1*(log(ZB1)-1))-(ZB2*(log(ZB2)-1)));
%Equation 3 ; this is used for the right side of the center
Eqn3 = A.*(ZA2*(ZAA2-1)-(ZA1*(ZAA1-1))+(ZB1*(ZBB1-1))-(ZB2*(ZBB2-1)));
%Equation 4 :Add Equation 2 and Equation 1; this is used for the left side of the center
Eqn4 = matlabFunction(Eqn1+Eqn2,'vars',var_z);
%Equation 5: Add Equation 3 and Equation 1; this is used for the right side of the center
Eqn5 = matlabFunction(Eqn1+Eqn3,'vars',var_z);
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
%This vector starts from left corner (minx) to the middle of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec1=minx:(rect_center-minx)/(0.5*nr_x-1):rect_center;
%This vector starts from middle to the right corner (maxx) of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec2=rect_center:(maxx-rect_center)/(0.5*nr_x-1):maxx;
%the y vectors start from miny to maxy
yvec1=miny:(maxy-miny)/(nr_y-1):maxy;
yvec2=miny:(maxy-miny)/(nr_y-1):maxy;
% create mesh from above vectors
[x1,y1]=meshgrid(xvec1,yvec1);
[x2,y2]=meshgrid(xvec2,yvec2);
z1=x1+1i*y1;
z2=x2+1i*y2;
% Calculate the above function using equation 4 and equation 5 using the mesh created above
r1 = -real(Eqn5(z1));
r2 = -real(Eqn4(z2));
%Combine the calculated functions
Result = [r1 r2];
%Combine the grids
x = [x1 x2];
y = [y1 y2];
% plot contours
[c,h]=contourf(x,y,Result(:,:,1),50,'LineWidth',1);
% plot the outerboundary of the rectangle
line_x=real([vertexA;vertexB]);
line_y=imag([vertexA;vertexB]);
line(line_x,line_y,'color','r','linestyle',':','linewidth',5)
The final Figure is supposed to look like this.Final Expected Figure.

Answer 1

I'm not sure which angle defines the dividing line so I assume it's Angle1. It looks like logical indexing is the way to go here. Instead of creating two separate mesh grids we simply create the entire mesh grid then partition it into two sets and operate on each independently.
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
% create full mesh grid
xvec=linspace(minx,maxx,nr_x);
yvec=linspace(miny,maxy,nr_y);
[x,y]=meshgrid(xvec,yvec);
% Partition mesh based on divider line
% Assumes the line passes through (ox,oy) with normal vector defined by Angle1
ox = rect_center;
oy = rect_center;
a = cos(Angle1);
b = sin(Angle1);
c = -(a*ox + b*oy);
% use logical indexing to opperate on the appropriate parts of the mesh
idx1 = a*x + b*y + c < 0;
idx2 = ~idx1;
z = zeros(size(x));
z(idx1) = x(idx1) + 1i*y(idx1);
z(idx2) = x(idx2) + 1i*y(idx2);
% Calculate the above function using equation 4 and equation 5
% using the mesh created above
Result = zeros(size(z));
Result(idx1) = -real(Eqn5(z(idx1)));
Result(idx2) = -real(Eqn4(z(idx2)));
For example with Angle1 = 45 and Angle2 = 45 we get the following indexing
>> contourf(x,y,idx1);
>> line(line_x,line_y,'color','r','linestyle',':','linewidth',5);
where the yellow region uses Eqn5 and the blue region uses Eqn4. This agrees with the example you posted but I don't know what the resulting contour map for other cases is supposed to look like.
Hope this helps.