how to plot a function on a 3D sphere?
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N0=100;
r_med=[0.445 0.889 1.445 2];
sigma_g=7;
N_ang=91;
theta = (0:1:360)/180*pi;
phi = (0:1:360)/180*pi;
[x,y]=meshgrid(theta,phi);
Num_r = 50e3;
r = linspace(1,50,Num_r)./2;
I0=1;Q0=0;U0=0;V0=0;
for i = 1:length(r_med)
% [P11(i,:),P12(i,:),P33(i,:),P34(i,:),Qsca_c(i,:),~,~] = ZK_W_Cloud_PhaseFunc(N0,r_med(i),sigma_g,N_ang);
P11_t(i,:) = [P11(i,:) fliplr(P11(i,2:end))];
P12_t(i,:) = [P12(i,:) fliplr(P11(i,2:end))];
P33_t(i,:) = [P33(i,:) fliplr(P11(i,2:end))];
P34_t(i,:) = [P34(i,:) fliplr(P11(i,2:end))];
P1=repmat(P11_t(i,:),length(phi),1);
P2=repmat(P12_t(i,:),length(phi),1);
P3=repmat(P33_t(i,:),length(phi),1);
P4=repmat(P34_t(i,:),length(phi),1);
z = P1.*I0+((P2.*Q0.*cos(2*y))+(P2.*U0.*sin(2*y)));
end
[v,u,w]=sph2cart(x,y,z);
v=z.*cos(y).*cos(x);
u=z.*sin(y).*cos(x);
w=z.*sin(y);
figure,
surf(v,u,w),xlabel('x'),ylabel('y'),zlabel('z')
%set(gca,'zscale','log');
I want to obtain the figure as shown

However, the output of my code is just a circle

Please correct me.
2 Kommentare
Walter Roberson
am 3 Jan. 2022
How does this differ from https://www.mathworks.com/matlabcentral/answers/1620195-how-to-plot-a-3d-function?s_tid=srchtitle ?
Wiqas Ahmad
am 3 Jan. 2022
Antworten (1)
My recommendation would be to build these surfaces using the cylinder function, e.g.,
R=1; %radius of circular cross-section
d=1.5; %radius of torus
fn=@(z) 2*R*(z-0.5);
x=fn( linspace(0,1) );
r1=d-sqrt(R^2-x.^2); %inner half surface
r2=d+sqrt(R^2-x.^2); %outer half surface
[X1,Y1,Z1]=cylinder(r1,40);
[X2,Y2,Z2]=cylinder(r2,40);
h(1)=surf(X1, Y1, fn(Z1)); hold on
h(2)=surf(X2, Y2, fn(Z2)); hold off
rotate(h,[1,0,0],90)
axis equal
xlabel X; ylabel Y; zlabel Z;
view(35,30)
5 Kommentare
Wiqas Ahmad
am 3 Jan. 2022
Matt J
am 3 Jan. 2022
If your scenario is different, you will need to edit your question, so that people can understand how. The cylinder command works for both th surfaces you say you want ot plot.
Wiqas Ahmad
am 4 Jan. 2022
Matt J
am 4 Jan. 2022
Yes, but that doesn't explain why my posted solution is insufficient.
Wiqas Ahmad
am 4 Jan. 2022
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