please help me figure out what to do

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Iris Kiebert
Iris Kiebert il 15 Dic 2019
Modificato: per isakson il 16 Dic 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 Commenti
Rik
Rik il 15 Dic 2019
Have a read here and here. It will greatly improve your chances of getting an answer.

Accedi per commentare.

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