| 1 |
% Test of the function intbet2outcrops |
| 2 |
clear |
| 3 |
|
| 4 |
% Theoritical fields: |
| 5 |
eg = 1; |
| 6 |
|
| 7 |
switch eg |
| 8 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 9 |
case 1 % The more simple: |
| 10 |
% Axis: |
| 11 |
lon = [200:1/8:300]; nlon = length(lon); |
| 12 |
lat = [0:1/8:20]; nlat = length(lat); |
| 13 |
|
| 14 |
% chp goes linearly from 20 at 0N to 0 at 20N |
| 15 |
[a chp] = meshgrid(lon,-lat+lat(nlat)); clear a c |
| 16 |
[a chp] = meshgrid(lon,-lat+2); clear a c |
| 17 |
chp(14:16,:)=1; % Make the integral proportional to the surface |
| 18 |
|
| 19 |
% Define limits: |
| 20 |
LIMITS(1) = -1 ; |
| 21 |
LIMITS(2) = -1 ; |
| 22 |
LIMITS(3:4) = lat([15 15]) ; |
| 23 |
LIMITS(5:6) = lon([1 nlon]) ; |
| 24 |
|
| 25 |
% Expected integral: |
| 26 |
dx = m_lldist([200 300],[1 1]*1.75)./1000; |
| 27 |
dy = m_lldist([1 1],[1.625 1.875])./1000; |
| 28 |
Iexp = dx*dy/2; % Unit is km^2 |
| 29 |
|
| 30 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 31 |
|
| 32 |
end %switch |
| 33 |
|
| 34 |
|
| 35 |
% Get integral: |
| 36 |
[I Imat dI] = intbet2outcrops(chp,LIMITS,lat,lon); |
| 37 |
|
| 38 |
disp('Computed:') |
| 39 |
disp(num2str(I/1000^2)) |
| 40 |
disp('Approximatly expected:') |
| 41 |
disp(num2str(Iexp)) |
| 42 |
|
| 43 |
break |
| 44 |
figure;iw=1;jw=2; |
| 45 |
subplot(iw,jw,1);hold on |
| 46 |
pcolor(chp);shading flat;canom;colorbar;axis tight |
| 47 |
title('Tracer to integrate'); |
| 48 |
|
| 49 |
subplot(iw,jw,2);hold on |
| 50 |
pcolor(double(Imat));shading flat;canom;colorbar;axis tight |
| 51 |
title('Points selected for the integration'); |