coswm should have a 1x365 dimension, but its dimension is 1x1, what am i doing wrong?
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Iris Kiebert
le 15 Déc 2019
Réponse apportée : Walter Roberson
le 15 Déc 2019
close all
clear all
h=[1:24];%hours in day
m=[1:14400];%minutes in day
n_param=5;
n=[1:365];%days in year
az=0.0001; % degrees %a=0,001 , because a is near zero during sunrise and sunset.
ndec=355; %21 december is de 335e dag van het jaar
njul=202; %21 juli is de 202e dag van het jaar
p=52.3667; %altitude amsterdam
heigthblock=25; %heigth of the block in mETER
sino=zeros(1,365);
max_loops = 100;
for n_days=n %azimuth angle
sinoaz=-23.45*(pi/180)*cos(((2*pi)/365)*(10+n));
oaz=asin(sinoaz);
%wm=(m+7200)*pi/7200;
%sin(a)=cos(o)*cos(wh)*cos(p)+sino*sin(p)
%sin(a)-sino*sin(p)=cos(o)*cos(wh)*cos(p)
%cos(wh)=(sin(a)-sino*sin(p))/((cos(o)*cos(p)))
coswm=(sind(az)-(sinoaz*sind(p)))/((cos(oaz)*cosd(p)));
wmaz=acos(coswm);
sinaaz=sind(az);
sinAz=(sind(wmaz)*cosd(oaz))/sinaaz;
Az=asind(sinAz);
end
0 commentaires
Réponse acceptée
Walter Roberson
le 15 Déc 2019
The / operator is Matrix Right Divide, which is least squares fitting effectively. If you did not intend to do fitting at that point then you probably want the ./ operator which is element by element division.
0 commentaires
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Visualization dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!