can i get rid of rectangular coordinates and get something like a blade shape image in matlab as shown in figure. If yes then please refer me to some example

Answer 1

[Bjorn Gustavsson]

yes. Use:

surf(X,Y,Z,C),shading flat.view(0,90)

Where you might set the Z-matrix to all zeros (or something else if you'd rather like that)

HTH

[Rizwana]

I need to plot contours of z ordinates. I cant set them to zero. I have attached a file which reads ca and cptot. radius is given in range r = [24.94 25.20 25.73 26.25 26.77 27.30]; I need to interpolate these radius values and finally pot contours of cptot varying with radius along y-axis and ca along x-axis. But not in rectangular box. It should look like a blade shape. Something like cfd figures

[Bjorn Gustavsson]

OK, make an array with the radius of each [ca, cptot]-point so that you can make an array XYZ:

XYZ = [R.*cos(ca),R.*sin(ca),cptot];

from there you simply use the delaunay function to get a triangulation:

tri = delaunay(XYZ(:,1),XYZ(:,2));

After that you'll have to go to the file exchange and donwload the tricont, tricontf, or tricontour functions and use those to make your contour plot. Alternatively you could use griddata or TriScatteredInterp to resample your function to a suitable regular cartesian grid and then use contour on that, but resampling might be undesirable.

HTH

[Rizwana]

Sir: It is giving error saying dimensions of the matrix are not matching right after the 1st line i tried. XYZ = [R.*cos(ca),R.*sin(ca),cptot];

[Bjorn Gustavsson]

You have to check that their sizes are the same, first check:

whos R ca cptot

then hopefully you'll only have to transpose R:

XYZ = [R.'.*cos(ca),R.'.*sin(ca),cptot];

HTH

Answer 2

[Walter Roberson]

Have a look at "Obtaining Angular Directions in a Projection Space" in the document http://www.mathworks.com/help/map/accessing-computing-and-inverting-map-projection-data.html#f11-12738 . The graph shown there in 6. -- isn't that the same kind of graph that you need, except with a greater mapping angle? If so then you can use the setup shown in 1. but with different lat and lon limits.

Answer 3

[Image Analyst]

If it's just a rubber sheet stretch, then imtransform() can do that.