coswm should have a 1x365 dimension, but its dimension is 1x1, what am i doing wrong?
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Iris Kiebert
am 15 Dez. 2019
Beantwortet: Walter Roberson
am 15 Dez. 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 Kommentare
Akzeptierte Antwort
Walter Roberson
am 15 Dez. 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 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Visualization finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!