I am trying to create a surface plot of a matrix I created, but I need to make lattitude and longitue in to 180x360 matrices as well in order for the function to work. My code is below:
%Plot Slopematrix
addpath('C:\Users\Falon Treis\Documents\MATLAB\CE1_2C_slopematrix.mat');
%addpath('C:\Users\Falon Treis\Documents\MATLAB\CE2_2C_slopematrix.mat');
load('CE1_2C_slopematrix.mat')
k = 1:180; %lat make have 180x 360 matrix
m = 1:360; %long
z = slopematrix; %180x360
surf(k, m, z)
%surf(k, m, z)
save CE1_2C_slope_plot.mat
%save CE2_2C_slope_plot.mat

Antworten (1)

Walter Roberson
Walter Roberson am 28 Apr. 2021

0 Stimmen

surf(k, m, z.')
Remember that the rows correspond to y not to x

5 Kommentare

Falon Treis
Falon Treis am 28 Apr. 2021
I am not sure how that helps me change the fact that the lattitude matrix has 180 and the longitude matrix has 360
A demonstration that transposing z does work:
k = 1:180; %lat make have 180x 360 matrix
m = 1:360; %long
slopematrix = sort(rand(length(k), length(m)));
size(slopematrix)
ans = 1×2
180 360
z = slopematrix;
try
surf(k, m, z)
disp("(k,m,z) succeeded!");
catch ME
disp("(k,m,z) failed!");
end
(k,m,z) failed!
try
surf(k, m, z.')
disp("(k,m,z.') succeeded!");
catch ME
disp("(k,m,z.') failed!");
end
(k,m,z.') succeeded!
Falon Treis
Falon Treis am 28 Apr. 2021
I don't understand why you are saying to transpose z when I am trying to transpose k and m to fit z. What exactly does transposing z do and won't I be messing with the placement of the values in the slope matrix?
lat is vertical, not horizontal, so you should not be using it as your x.
k = 1:180; %lat make have 180x 360 matrix
m = 1:360; %long
slopematrix = sort(rand(length(k), length(m)));
size(slopematrix)
ans = 1×2
180 360
z = slopematrix;
surf(m, k, z)
xlabel('long')
ylabel('lat')
zlabel('slope')
Falon Treis
Falon Treis am 28 Apr. 2021
Ok. I see that now. Thank you very much for your time and patience.

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