3d Hemisphere transformation Problem from A to 90 degrees in Z axis to B direction

%code for A

N=500;

r = 6;

center = [2,1,3];

plot3(center(1),center(2),center(3),'ro')

hold on

P = zeros(N,3) ;

for i=1:N

x1 = rand;

x2 = rand;

Px = r*sqrt(x1*(2-x1))* cos(2*pi*x2);

Py = r*sqrt(x1*(2-x1))* sin(2*pi*x2);

Pz = r*(1-x1);

P(i,:) = [Px, Py, Pz]+center;

end

hold on

plot3(P(:,1), P(:,2), P(:,3),'.r')

to

Answer 1

[David Goodmanson]

Hi Musa,

In the figure you don't say which axis is which, but I will assume +x axis out of the page toward the viewer, +y axis right, +z axis up. With the figure rotated as shown, the old z coordinates are along the +y axis and the old y coordinates are along the -z axis. It's possible to use rotation matrices for this, but in the 90 degree case it's easier to just change the coordinate labels as shown in plot 2 below.

N=1000;
r = 6;
center = [2,1,3];
figure(1)
plot3(center(1),center(2),center(3),'b*')
hold on
P = zeros(N,3) ;
for i=1:N
    x1 = rand;
    x2 = rand; 
    Px = r*sqrt(x1*(2-x1))* cos(2*pi*x2);
    Py = -r*sqrt(x1*(2-x1))* sin(2*pi*x2);
    Pz = r*(1-x1);
    P(i,:) = [Px, Py, Pz]+center;
end
hold on
plot3(P(:,1), P(:,2), P(:,3),'.r')
axis equal
xlabel('x'); ylabel('y'); zlabel('z')
hold off

figure(2)
plot3(center(1),center(2),center(3),'b*')
hold on
P = zeros(N,3) ;
for i=1:N
    x1 = rand;
    x2 = rand;
    Px = r*sqrt(x1*(2-x1))* cos(2*pi*x2);
    Pz = -r*sqrt(x1*(2-x1))* sin(2*pi*x2);
    Py = r*(1-x1);
    P(i,:) = [Px, Py, Pz]+center;
end
hold on
plot3(P(:,1), P(:,2), P(:,3),'.r')
view([4 -1.2 .8])
xlabel('x'); ylabel('y'); zlabel('z')
axis equal
hold off

.