Hi,
I have issues when calculating the values of x and y that make fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2).^(1/2) equal to zero. And I don't know if this is due to the square root. Cause, when taking out the square root there is no problem in the calculation.
This is driving me crazy since if you have a funcition, g(x,y) which is g(x,y) = sqrt(f(x,y)) and you are trying to find the values of x,y that make g(x,y) = 0, those are, at least, equal to those x, y values that make f(x,y) = 0. But according to MATLAB there are not values of x,y (nothing appears in the plot of figure 4 and figure 5).
The code is the following one:
R = 0.35;
L = 0.264;
H = 0.4;
fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2);
figure(1)
hfi = fimplicit(fun, [-pi/2 pi/2 -pi/2 pi/2]);
xlabel('\psi (rad)');
xticks(-pi/2:pi/4:pi/2);
ylabel('\theta (rad)');
figure(2)
fimplicit3(fun, [-pi/2 pi/2 -pi/2 pi/2 0.3 0.8]);
xlabel('\psi (rad)');
ylabel('\theta (rad)');
zlabel('z (m) ');
figure(3)
hfc = fcontour(fun,[-pi/2 pi/2 -pi/2 pi/2]);
LL = hfc.LevelList;
fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2).^(1/2);
figure(4)
fimplicit(fun, [-pi/2 pi/2 -pi/2 pi/2])
figure(5)
fimplicit3(fun, [-pi/2 pi/2 -pi/2 pi/2 0.3 0.8])
figure(6)
fcontour(fun)
xlabel('\psi (rad)');
ylabel('\theta (rad)');
x= linspace(-pi/2,pi/2,100); y = linspace(-pi/2, pi/2, 100);
[X,Y] = meshgrid(x,y);
values = fun(X,Y);
values1 = values;
values1(imag(values1)~= 0) = NaN;
figure(8)
mesh(X,Y,values1);
xlabel('\psi (rad)');
ylabel('\theta (rad)');
zlabel('z (m)');
However for the case fun = @(psi,theta) (sin(psi).*sin(theta)).^(1/2) there is no problem with fimplicit3 but the error with fimplicit() mantains. Any suggestion to obtain the plot or the values without taking out the root?
I know that in this case, to avoid the problem, you can take out the square root, but for other cases or when you don't realize about that...
Thank you and sorry for the long message.
