Thread Subject: Cubed Sphere

Hi All --

Does anyone have an algorithm to generate a cubed sphere? Ideally, I would pass the function a scalar (e.g., # of tiles per face) and would receive the Cartesian coordinates of every vertex on the cubed sphere. Thanks a bunch.

Matt

"Matt" <matt_evans.nospam@ncsu.edu> wrote in message <h61oju$d55$1@fred.mathworks.com>...
> Hi All --
>
> Does anyone have an algorithm to generate a cubed sphere? Ideally, I would pass the function a scalar (e.g., # of tiles per face) and would receive the Cartesian coordinates of every vertex on the cubed sphere. Thanks a bunch.
>
> Matt

  Little is to be gained by presenting a task that has been the subject of many papers. It is not an appropriate problem for this kind of newsgroup.

Roger Stafford

> Little is to be gained by presenting a task that has been the subject of many papers. It is not an appropriate problem for this kind of newsgroup.
>
> Roger Stafford

Roger -- thanks for your reply. I have seen many papers that discuss the USE of the cubed sphere, but none that address the generation of said sphere. Perhaps this problem is trivial for topologists or computer scientists, but I am neither. Could you be so kind as to point me in the direction of some of these references.

Matt

"Matt" <matt_evans.nospam@ncsu.edu> wrote in message <h61s8l$ph0$1@fred.mathworks.com>...
> Roger -- thanks for your reply. I have seen many papers that discuss the USE of the cubed sphere, but none that address the generation of said sphere. Perhaps this problem is trivial for topologists or computer scientists, but I am neither. Could you be so kind as to point me in the direction of some of these references.
>
> Matt

  Perhaps I didn't make myself clear, Matt. What I meant was that this problem undoubtedly involves a very extensive programming effort even if good algorithms have been devised. In other words, it is the opposite of being too trivial; it is not trivial enough. It is for this reason that I deem it inappropriate here in this newsgroup. We prefer answering questions that can be explained in just a few lines of code, and I seriously doubt that that would be true for generating cubed spheres, particularly of the generality of an arbitrary number of tiles per face that you are looking for.

  You might check on the Matlab File Exchange to see if anyone has taken the trouble to deal with it. I do see a www site that seems to be concerned with this area and involves matlab routines:

 http://mitgcm.org/r2_web_testing/latest/online_documents/node227.html

Roger Stafford

"Matt" <matt_evans.nospam@ncsu.edu> wrote in message <h61oju$d55$1@fred.mathworks.com>...
> Hi All --
>
> Does anyone have an algorithm to generate a cubed sphere? Ideally, I would pass the function a scalar (e.g., # of tiles per face) and would receive the Cartesian coordinates of every vertex on the cubed sphere. Thanks a bunch.
>
> Matt

Try this:

n=10; % making (9 x 9 tiles per faces) for one of the six faces

x = 1;
y = linspace(-1,1,n);
z = y;
[X Y Z] = ndgrid(x,y,z);

S = sqrt(1./(X.^2+Y.^2+Z.^2));
X = X.*S;
Y = Y.*S;
Z = Z.*S;

C1 = [X(:) Y(:) Z(:)].';
M = makehgtform('zrotate',pi/2);
C2 = M(1:3,1:3)*C1;
M = makehgtform('zrotate',pi);
C3 = M(1:3,1:3)*C1;
M = makehgtform('zrotate',3*pi/2);
C4 = M(1:3,1:3)*C1;
M = makehgtform('yrotate',pi/2);
C5 = M(1:3,1:3)*C1;
M = makehgtform('yrotate',-pi/2);
C6 = M(1:3,1:3)*C1;

C = [C1 C2 C3 C4 C5 C6].';
tol = 1e-3/(n-1);
Cround = round(C/tol);
[Cround I] = unique(Cround,'rows');
Vertex = C(I,:);

figure(1); clf
plot3(Vertex(:,1),Vertex(:,2),Vertex(:,3),'.');
axis equal

% Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <h621hg$sfs$1@fred.mathworks.com>...
> Try this:
>
Bruno --

Thanks a bunch for the reply.

I think that works -- but it is difficult for me to be certain until I figure out a better way to visualize it. The point cloud "looks right", but I want to see the actual grid to confirm this.

Thanks again.

Matt

% Now you can chek it:

n=17;

x = 1;
y = linspace(-1,1,n);
z = y;
[X Y Z] = ndgrid(x,y,z);

S = sqrt(1./(X.^2+Y.^2+Z.^2));
X = X.*S;
Y = Y.*S;
Z = Z.*S;

C1 = [X(:) Y(:) Z(:)].';
M = makehgtform('zrotate',pi/2);
C2 = M(1:3,1:3)*C1;
M = makehgtform('zrotate',pi);
C3 = M(1:3,1:3)*C1;
M = makehgtform('zrotate',3*pi/2);
C4 = M(1:3,1:3)*C1;
M = makehgtform('yrotate',pi/2);
C5 = M(1:3,1:3)*C1;
M = makehgtform('yrotate',-pi/2);
C6 = M(1:3,1:3)*C1;

[i j] = ndgrid(1:n-1,1:n-1);
i = i(:); j=j(:);
P = [sub2ind([n n], i, j) ...
     sub2ind([n n], i+1, j) ...
     sub2ind([n n], i+1, j+1) ...
     sub2ind([n n], i, j+1)];
offset = n^2*(0:5)';
P = bsxfun(@plus, offset, reshape(P,[1 size(P)]));
P = reshape(P, [], 4);

C = [C1 C2 C3 C4 C5 C6].';
tol = 1e-3/(n-1);
Cround = round(C/tol);
[Cround I J] = unique(Cround,'rows');
Vertices = C(I,:);
P = J(P);

figure(1); clf;
patch('Vertices',Vertices,'Faces',P,'FaceColor','g');
axis equal

% Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <h625m3$7tk$1@fred.mathworks.com>...
> % Now you can chek it:
>

Bruno -- quite nice, indeed. This is perfect. Thanks very much for the help.

Matt

FYI I have updated the code with another grid parametrization (equiangular). It seems more popularly used. It has been submitted to FEX.

While googling, I see a lot of papers around conformal mapping, which has the advantages as well as the inconveniences. I couldn't implement because of lack of a detailed description (selection of the pole, etc...). I can't have access to relevant paper.

Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <h63k5g$oo0$1@fred.mathworks.com>...
> FYI I have updated the code with another grid parametrization (equiangular). It seems more popularly used. It has been submitted to FEX.
>
> While googling, I see a lot of papers around conformal mapping, which has the advantages as well as the inconveniences. I couldn't implement because of lack of a detailed description (selection of the pole, etc...). I can't have access to relevant paper.
>
> Bruno

Bruno -- thanks a ton. I'd like to try the updated code, but I do not see it yet on the File Exchange. How long does it typically take for files to become available? Thanks.

Matt

Matt, normally it'll show up shortly. If in few hours nothing showed up I'll send you the code.

Bruno

I'ts up! Enjoy.

Bruno