# Surface area from a z-matrix

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bobby on 15 Apr 2012
I have x, y axis values and corresponding z values. This allows me to plot the shape using surf function. Example:-
How to calculate surface area of this irregular shape?
Please i'm a novice learning this stuff, i have come along quite well since last week or two. Thank you for all your help. The other answers provided before for irregular shapes have me confused, since i don't know how to triangulate.
Additionally i also want to calculate surface area above a certain height. So i'll cut all the values below a certain height and then possibly calculate surface area above a cut-off height.
Edit:
Thanks to inputs by the community and basically whole programme written by Richard. Here's the code.
function sa(Z, cutoff)
%Credit: Richard Brown, MATLAB central forum (http://www.mathworks.com/matlabcentral/answers/)
dx=0.092; % x-axis calibration
dy=0.095; % y-axis calibration
[m, n] = size(Z);
areas = 0.5*sqrt((dx*dy)^2 + (dx*(Z(1:m-1,2:n) - Z(1:m-1,1:n-1))).^2 + ...
(dy*(Z(2:m,1:n-1) - Z(1:m-1,1:n-1))).^2) + ...
0.5*sqrt((dx*dy)^2 + (dx*(Z(1:m-1,2:n) - Z(2:m,2:n))).^2 + ...
(dy*(Z(2:m,1:n-1) - Z(2:m,2:n))).^2);
zMean = 0.25 * (Z(1:m-1,1:n-1) + Z(1:m-1,2:n) + Z(2:m,1:n-1) + Z(2:m,2:n));
areas(zMean <= cutoff) = 0;
surfaceArea = sum(areas(:));
sprintf('Total surface area is %2.4f\n', surfaceArea)
return
end
Richard Brown on 15 Jul 2013
I've just had it pointed out to me that there is a small mistake in this code. Assuming that x corresponds to columns, and y to rows, then the first two lines (but not the second two lines) of the areas calculation has dx and dy backwards. I've fixed it in my original answer below.

Richard Brown on 16 Apr 2012
Edited: Richard Brown on 15 Jul 2013
OK, well how about splitting each quadrilateral into two triangles, and just summing up the areas? I'm sorry there's no way I can make this look pleasant, but ...
[m, n] = size(Z);
areas = 0.5*sqrt((dx*dy)^2 + (dy*(Z(1:m-1,2:n) - Z(1:m-1,1:n-1))).^2 + ...
(dx*(Z(2:m,1:n-1 - Z(1:m-1,1:n-1)))).^2) + ...
0.5*sqrt((dx*dy)^2 + (dx*(Z(1:m-1,2:n) - Z(2:m,2:n))).^2 + ...
(dy*(Z(2:m,1:n-1) - Z(2:m,2:n))).^2)
surfaceArea = sum(areas(:))
edit: Jul 15, 2013 There was a mistake, and dx and dy were backwards in the first two lines of the code. The code has been corrected now.
Richard Brown on 16 Apr 2012
For each of the areas in the 'areas' array, you could work out the mean Z value
zMean = 0.25 * (Z(1:m-1,1:n-1) + Z(1:m-1,2:n) + Z(2:m,1:n-1) + Z(2:m,2:n))
and then
areas(zMean > cutoff) = 0;
surfaceArea = sum(areas(:));
It's a bit sloppy - you should probably do this for the individual triangles, but it will certainly give you a reasonable approximation

Walter Roberson on 15 Apr 2012
I do not know if you will be able to calculate the surface area of the regular shape interspersed with the irregular tops. When I look at the image with the irregular tops, it looks to me as if the tops are fairly irregular, possibly even fractal. I don't know if a meaningful surface area could be calculated: the surface area of a fractal is infinite.
For the first figure, I can think of a crude way to find the area, but I think there are simpler ways. I would need to think further about good ways to find the area. But to cross-check: are your x and y regularly spaced, or irregularly ?
bobby on 16 Apr 2012
Ah, the good old mandelbrot. Yes, I understand your point now. That is why i edited my post as i did understood that it will be nearly impossible. But the surface area of the surface that i'm after, given in the post, i'm sure a solution exists. Thanks to you i came across a very neat paper by Mandelbrot.

Richard Brown on 16 Apr 2012
If your data is smooth enough (assuming that's what you're after), then there is a really quick way to work out approximately the surface area, using the normals that Matlab computes when plotting the surface. To make it simple I'll assume you have a uniform grid with spacings dx and dy.
dA = dx * dy;
h = surf( ... )
N = get(h, 'VertexNormals');
N_z = squeeze(N(:, :, 3));
% Normalise it
N_z = N_z ./ sqrt(sum(N.^2, 3));
Area = dA * sum(1./N_z(:));
Richard Brown on 16 Apr 2012
Hence my comment about smoothing - if you apply that straight to the mesh you've got there, all bets are off! Anyway, see my next answer, it might be more in line with what you're after.

bobby on 16 Apr 2012
function surfacearea(Z, cutoff)
dx=0.092;
dy=0.095;
[nrow, ncol] = size(Z);
for m=1:nrow
for n=1:ncol
zMean = mean(Z(1:m-1,1:n-1) + Z(1:m-1,2:n) + Z(2:m,1:n-1) + Z(2:m,2:n));
if(zMean >= cutoff)
areas = 0.5*sqrt((dx*dy)^2 + (dx*(Z(1:m-1,2:n) - Z(1:m-1,1:n-1))).^2 + ...
(dy*(Z(2:m,1:n-1) - Z(1:m-1,1:n-1))).^2) + ...
0.5*sqrt((dx*dy)^2 + (dx*(Z(1:m-1,2:n) - Z(2:m,2:n))).^2 + ...
(dy*(Z(2:m,1:n-1) - Z(2:m,2:n))).^2);
end
end
end
surfaceArea = sum(areas(:));
return
end
This is how the final code looks like with a cut-off limit included. Any inputs into speeding up this code? I have 1024*1280 matrix.
bobby on 17 Apr 2012
Yes, the code runs fast, only thing is there should be < sign. Thanks a lot Richard for helping me through this ordeal. I'll definitely shop again. :)