Hello everybody, I would like to extract a line from a surface. I already represent quite easily with the commands plot3, mesh and surf my surface, that is divided in three parts because it represents three different domains. My initial input is composed by a matrix (18011*3) of Longitude, Latitude and Moho values. I would like to build a function that in input takes the position of start (A) and end of my profile (B) (xa,ya,xb,yb) and in output return the position of the corresponding z in A and B {e.g. [A,B] = myfunc(xa,ya,xb,yb)}. Finally, I would like to plot the obtained curve after cutting my surface with a vertical plane. I have already achieved this result with another software (Global Mapping Tools, very used in geophysical applications), but now I'd like to obtain the same thing in Matlab environment, of which I am quite new. Someone is familiar with these kind of problems? Do you have any suggestions, please? Thanks in advance,
Leonardo
Hey Leonardo,
I guess your matrix has three columns representin x, y, and moho values. To obtain the z values (which are coming from the Moho column of your matrix right?), you should go through the matrix and extrat the values at the same position of your coordinates. For example, supposing that column 1 is x, 2 is y, and 3 is Moho:
A(1) = matrix(find(matrix(:,1)=xa),3);
A(2) = matrix(find(matrix(:,2)=ya),3);
B(1) = matrix(find(matrix(:,1)=xb),3);
B(2) = matrix(find(matrix(:,2)=yb),3);
In this way you find the moho values for your set of starting and ending point.
For plotting the profile you have two ways of proceeding:
-You interpolate linearly between the two coordinates, with the same gridsize of your mesh such that you obtain the same x and y values you can find in your matrix. Afterwards, you calculate each Moho value in between extracting them from your original matrix
-You take from the beginning each value between xa,xb, and ya,yb, with the corresponding Moho value from the matrix.
Maybe my answer is not completelly clear, if you want more information write me privately!
Cheers,
Igor
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