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please help me figure out what to do

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Iris Kiebert
Iris Kiebert am 15 Dez. 2019
Bearbeitet: per isakson am 16 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=n
wh=(h+12)*(pi/12); %OPDRAcht 4
sinoaz=-23.45*(pi/180)*cos(((2*pi)/365)*(10+n));
oaz=asin(sinoaz);
sinashadow=cos(oaz).*cosd(wh).*cosd(p)+sinoaz.*sind(p);
ashadow=asin(sinashadow);
sinAshadow=(sind(wh).*cos(oaz))./sinashadow;
Ashadow=asind(sinAshadow);
%Tan(a)=overstaande/aanliggende=height block/lenth shadow
%lenth shadow=heigth block/Tan(a)
lenthofshadow=heigthblock./(tan(ashadow));
x=lenthofshadow.*cos(Ashadow);
y=lenthofshadow.*sin(Ashadow);
figure
[X,Y]=meshgrid(1:11);
figure;
hold on;
plot(X,Y,'k');
plot(Y,X,'k');
x=lenthofshadow.*cos(Ashadow);
y=lenthofshadow.*sin(Ashadow);
axis off
while 15<ashadow||ashadow<90
n_loop=n_loop+1;
if (n_loop >= max_loops)
disp('exceeded max number of loops')
break
end
end
end
i want to calculate how much hours of shade each cell in a grid gets with the azimuth angle (Ashadow) and thelength of the shadow lengthofshadow to aid me in my calculations. But im stuck. please help me
  2 Kommentare
Rik
Rik am 15 Dez. 2019
Have a read here and here. It will greatly improve your chances of getting an answer.
Walter Roberson
Walter Roberson am 16 Dez. 2019
What is the difference between this and your earlier https://www.mathworks.com/matlabcentral/answers/496772-my-figure-doesnt-isnt-right ?

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