Plotting end points from different arrays
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This program has been used to plot graphs of various variables against time to see how they change. The variable Qe is changed by repeatedly decreasing by 0.1 and then the plots are put together on one graph. I was wanting to plot the end points of these variables (a, e, de, da) against Qe however I have been unable to do so. Any help is appreciated.
format long
g = 36;
M = 10000002;
a = zeros(M,1);
e = zeros(M,1);
da = zeros(M-1,1);
de = zeros(M-1,1);
T = [0:10000000];
e(1) = 0.0549;
a(1) = 0.3844e9;
for m = 0:10000000
de(m+1) = -(4.04562756e41 + ((2.75561858e40)/g)) * ((a(m+1))^(-13/2)) * e(m+1);
a(m+2) = a(m+1) + (da(m+1) * 3.154e7);
end
e = e(1:end-1);
a = a(1:end-1);
plot(T,e)
hold on
采纳的回答
Clay Swackhamer
2022-5-10
Hi AJ,
I'm not entirely sure about your question, since you ask to "plot the end points of these variables (a, e, de, da) against Qe" however Qe is a scalar (just a single number) in your code. So maybe you want to plot the variables (a, e, de, da) against Qe, while varying Qe from 6 to 15? That is what I did below. Hope it helps.
%% plotting_endpoints_from_arrays
%% Setup
close all
clear
clc
tic
format long
%% Loop over Qe
Qe = 6:1:15
% Run the calculation one time for each value of Qe
for i = 1:1:length(Qe)
M = 10000002;
a = zeros(M,1);
e = zeros(M,1);
da = zeros(M-1,1);
de = zeros(M-1,1);
T = [0:10000000];
e(1) = 0.0549;
a(1) = 0.3844e9;
for m = 0:10000000
de(m+1) = -(4.04562756e41 + ((2.75561858e40)/Qe(i))) * ((a(m+1))^(-13/2)) * e(m+1);
da(m+1) = (((8.09125512e41)*(e(m+1)^2)) + ((1.16026046e40)/Qe(i))) * (a(m+1))^(-11/2);
e(m+2) = e(m+1) + (de(m+1) * 3.154e7);
a(m+2) = a(m+1) + (da(m+1) * 3.154e7);
end
e = e(1:end-1);
a = a(1:end-1);
% Save endpoints
aEnd(i) = a(end);
eEnd(i) = e(end);
deEnd(i) = de(end);
daEnd(i) = da(end);
end
%% Plot endpoints of each variable vs Qe
subplot(2,2,1)
plot(Qe,aEnd, 'bo')
title('a vs Qe')
subplot(2,2,2)
plot(Qe,eEnd,'bo')
title('e vs Qe')
subplot(2,2,3)
plot(Qe,deEnd,'bo')
title('de vs Qe')
subplot(2,2,4)
plot(Qe,daEnd,'bo')
title('da vs Qe')
toc
2 个评论
AJ99
2022-5-11
Clay Swackhamer
2022-5-11
Glad it helped. If this answers your question can you go ahead and accept the answer above? Thanks
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