Pad a matrix with additional rows and concatenate another column
    5 vues (au cours des 30 derniers jours)
  
       Afficher commentaires plus anciens
    
    Bill Symolon
 le 3 Mai 2018
  
    
    
    
    
    Commenté : Walter Roberson
      
      
 le 3 Mai 2018
            Hello, all.
I asked this question a couple of days ago and you provided some excellent help. I've taken your code and scaled it up to my full data set, but it's still not working quite right. The goal is to end up with a matrix that is 7776x7.
The code runs through without errors, but the resulting matrix is 7776x6. The 7th column (time) isn't being concatenated properly. I hoping you can tell me what I'm missing. Here is my full code.
if true
% Define initial conditions
mu = 398600.4415;
Alt = 400:400:1200;
Ree = 6378;
a = Alt + Ree;
e = 1.7E-16;
i = 20:20:60;
RAAN = 0:60:300;
omega = 0:60:180;
nu = 0:60:300;
% Calculate the orbital period in minutes
C = unique(a);
for z = 1:length(C)
  if C(z) == 6778.14
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  elseif C(z) == 7178.14
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  else
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  end
end
% Divide the orbital period into 6 equal time steps
T_step = zeros(length(T), 6);
T_step(1,:) = [0:T(1)/5:T(1)];
T_step(2,:) = [0:T(2)/5:T(2)];
T_step(3,:) = [0:T(3)/5:T(3)];
T_step = T_step';
% Nested for-loops
index = 1;
for j = 1:length(a)
  for k = 1:length(i)
      for l = 1:length(RAAN)
          for o = 1:length(omega)
              for p = 1:length(nu)
                  parameters(index,:) = [a(j) e i(k) RAAN(l) omega(o) nu(p)];
                  index = index + 1;
              end
          end
      end
  end
end
% Add the time steps to the parameters matrix
R=length(parameters);
for index=1:R
  temp = padarray(parameters(index,:), [m-1 0], 'replicate','pre');
    if parameters(index,1) == 6778.14
        temp = [temp, T_step(:,1)];
    elseif parameters(index,1) == 7178.14
        temp = [temp, T_step(:,2)];
    elseif parameters(index,1) == 7578.17
        temp = [temp, T_step(:,3)];
    end
    if(index == 1)
        M = temp;
    else
        M = [M; temp];
    end
end
parameters = M;
  end
Thanks again for any help you can provide!
0 commentaires
Réponse acceptée
  Walter Roberson
      
      
 le 3 Mai 2018
        Those comparisons on the parameter values will probably never be true.
2 commentaires
  Walter Roberson
      
      
 le 3 Mai 2018
				I suggest you debug with
vals = [6778.14, 7178.14, 7575.14];
for K = 1 : length(vals)
  [~, idx] = min( abs(parameters(:,1)-vals(K)) );
  fprintf('%closest to %g is at index %d and is %g which is a difference of %g\n', vals(K), idx, parameters(idx,1), parameters(idx,1)-vals(K));
end
Plus de réponses (1)
  Yuvaraj Venkataswamy
      
 le 3 Mai 2018
        Check this line,
temp = padarray(parameters(index,:), [m-1 1], 'replicate','pre');
Voir également
Catégories
				En savoir plus sur Axis Labels dans Help Center et File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


