Effacer les filtres
Effacer les filtres

Cross-section (Need help to correct the code or modify it )

2 vues (au cours des 30 derniers jours)
Moustafa Abedel Fattah
Moustafa Abedel Fattah le 12 Mai 2024
Dear all
I need to correct or modify the following code to give the attached figure: clc
close all;
clear;
% Define the given arrays
x1 =[0, 1, 2, 3, 3.7];
z1 =[6, 6, 6, 6, 6];
x2 =[4.1, 5, 6, 7, 8, 9];
z2 =[4, 4, 4, 4, 4, 4];
x3 =[9, 10, 10, 11,12, 13, 14, 15, 16, 16.4];
z3 =[6, 6, 6, 6, 6, 6, 6, 6, 6, 6];
x4 =[15.3, 16, 17, 18, 19];
z4 =[3, 3, 3, 3, 3];
x =[x1,x2,x3,x4];
z =[z1,z2,z3,z4];
%x = [0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21];
%z = [2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 1 1 1 1 1 1];
% Find sudden changes in depth
sudden_changes = diff(z);
% Find indices where sudden changes occur
fault_indices = find(sudden_changes ~= 0) + 1;
% Initialize variables to store fault information
fault_locations = [];
fault_angles = [];
fault_types = {};
% Plot the top of the rock formation
plot(x, z, '-o');
hold on;
grid on;
% Earth's surface
earth_surface = zeros(size(x));
% Plot fault lines and gather fault information
for i = 1:length(fault_indices)
idx = fault_indices(i);
plot([x(idx-1), x(idx)], [z(idx-1), z(idx)], 'r--');
% Calculate intersection point with Earth's surface
intersection_x = x(idx-1) - z(idx-1) * (x(idx) - x(idx-1)) / (z(idx) - z(idx-1));
plot(intersection_x, 0, 'ro'); % Plot intersection point
set(gca, 'YDir', 'reverse');
% Calculate dip angle using the tan-rule
dip_angle = atand((z(idx) - z(idx-1)) / (x(idx) - x(idx-1)));
% Determine fault type
if dip_angle <= 90
fault_type = 'Normal';
plot(intersection_x, 0, 'go'); % Plot normal fault intersection point
elseif dip_angle > 90
fault_type = 'Reverse';
plot(intersection_x, 0, 'mo'); % Plot reverse fault intersection point
else
fault_type = 'Strike-Slip';
end
% Store fault information
fault_locations = [fault_locations; intersection_x];
fault_angles = [fault_angles; dip_angle];
fault_types{end+1} = fault_type;
end
% Display fault information in a table
if isempty(fault_indices)
% Display a message if there are no faults
disp('No fault lines detected.');
else
% Display fault information in a table
T = table(fault_locations, fault_angles, fault_types', ...
'VariableNames', {'Location', 'Angle', 'Type'});
disp(T);
end
Location Angle Type ________ ______ __________ 4.9 -78.69 {'Normal'} 9 90 {'Normal'} 14.2 69.864 {'Normal'}

Réponse acceptée

Walter Roberson
Walter Roberson le 12 Mai 2024
atand() with one parameter returns values in the range -90 to +90, not in the range 0 to 90 to 180. No single-parameter atand() values will be greater than 90
You should consider whether you want to use the two-parameter form of atand().
And you should take into account negative results from atand()
  4 commentaires
Walter Roberson
Walter Roberson le 15 Mai 2024
should be "Reverse" instead:"Normal"
Not according to your code, it should not be. The atand is 69.864 which is <= 90 and you have defined <= 90 as being Normal.
From the documentation: atand
For real values of X, atand(X) returns values in the interval [-90, 90].
-90 to +90 is <= 90 so for real angles, 'Normal' will always be satisfied.
Moustafa Abedel Fattah
Moustafa Abedel Fattah le 15 Mai 2024
You are right Sir

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Coordinate Systems dans Help Center et File Exchange

Produits


Version

R2014b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by