sphere with center and radius

how to plot a sphere with center(6500000,6500000,6500000) and radius 650000

 Accepted Answer

How about this?
[x,y,z] = sphere;
x = x*650000;
y = y*650000;
z = z*650000;
figure
surf(x+650000,y+650000,z+650000)

More Answers (2)

Try this:
% Make unit sphere
[x,y,z] = sphere;
% Scale to desire radius.
radius = 650000;
x = x * radius;
y = y * radius;
z = z * radius;
% Translate sphere to new location.
offset = 6500000;
% Plot as surface.
surf(x+offset,y+offset,z+offset)
% Label axes.
xlabel('X', 'FontSize', 20);
ylabel('Y', 'FontSize', 20);
zlabel('Z', 'FontSize', 20);
axis equal;

8 Comments

thank you very much! it works perfect
Hii Image Analyst,
i have drawn the sphere from the given coordinates and radius , now i want to check the pixel value of color map so that i can verify if some pixels are present inside the sphere or not .
Just check the distance from the offset or center
distance = sqrt((x-xCenter).^2+(y-yCenter).^2+(z-zCenter).^2);
if distance is less than the radius, it's inside.
Hi Image Analyst, could you please explain, what is xCenter?
@M.S. Khan xCenter is the x location of the center of the sphere. For the example I gave it's 6500000.
@Jeson Lonappan, you can't get them from this. It's not the earth. And there is mo Merdian line defined so there is no absolute longitude. Any point on the sphere could therefore be any longitude.
Alberto Acri
Alberto Acri on 24 Jan 2024
Edited: Alberto Acri on 24 Jan 2024
Hi @Image Analyst! How can I extract the coordinates of the sphere nodes? The x,y,z matrices are 21x21 doubles. I would like to use plot3!
I don't know from that function, but you can of course easily generate your own coordinates on the shell of the sphere. For any given z value, it's simply a circle. See the FAQ:
See also:
That is one way to use plot3() to draw a spherical shell. Once you have the x,y,z data, you can offset it to the desired center.
If it's not actually necessary to use plot3() specifically, or if the sphere needs to be drawn in some other manner, then you'll have to say.

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Ankita De
Ankita De on 11 Jul 2022
can you please tell nearest neighbour and contact distance distribution using binomial point process

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