I have this section of code here
az = 0:90:270;
for k = 1:length(az)
rR = 0:.1:1;
L1 = (rR.*rR.*.5)+(mu.*rR.*thetanaught.*sind(az(k)))+(.5.*thetanaught.*mu.^2.*sind(az(k)).^2)+(.5.*twist.*rR.^3)+(mu.*twist.*rR.^2.*sind(az(k)))+(.5.*twist.*mu^2.*rR.*sind(az(k)).^2)+(.5.*rR.^2.*Theta_1c.*cosd(az(k)))+(rR.*mu.*Theta_1c.*sind(az(k)).*cosd(az(k)));
L2 = (.5.*rR.^2.*Theta_1s.*sind(az(k)))+(.5.*mu.^2.*Theta_1s.*sind(az(k)).^3)-(.5.*lambda.*rR)-(.5.*mu.*lambda.*sind(az(k)))-(.5.*rR.*mu.*Beta.*cosd(az(k)))-(.5.*mu.^2*Beta.*cosd(az(k)).*sind(az(k)));
Lc = L1+L2
end
I want to be able to plot Lc for each of the az values. So it looks something like this.
What is an efficient way to get matlab to store the vectors from each loop so I can do this? The way I was doing it before was just saving the values and I ended up with a 1x44 vector.

 Akzeptierte Antwort

Voss
Voss am 4 Apr. 2024
Bearbeitet: Voss am 4 Apr. 2024

0 Stimmen

az = 0:90:270;
rR = 0:0.1:1; % pull rR out of the loop (it never changes)
N = numel(az);
Lc = zeros(numel(rR),N); % make Lc a matrix with N columns
for k = 1:N
L1 = (rR.*rR.*0.5)+(mu.*rR.*thetanaught.*sind(az(k)))+(0.5.*thetanaught.*mu.^2.*sind(az(k)).^2)+(0.5.*twist.*rR.^3)+(mu.*twist.*rR.^2.*sind(az(k)))+(0.5.*twist.*mu^2.*rR.*sind(az(k)).^2)+(0.5.*rR.^2.*Theta_1c.*cosd(az(k)))+(rR.*mu.*Theta_1c.*sind(az(k)).*cosd(az(k)));
L2 = (0.5.*rR.^2.*Theta_1s.*sind(az(k)))+(0.5.*mu.^2.*Theta_1s.*sind(az(k)).^3)-(0.5.*lambda.*rR)-(0.5.*mu.*lambda.*sind(az(k)))-(0.5.*rR.*mu.*Beta.*cosd(az(k)))-(0.5.*mu.^2*Beta.*cosd(az(k)).*sind(az(k)));
Lc(:,k) = L1+L2; % store each result in a column of Lc
end
figure
plot(Lc) % plot N lines (each column of Lc is plotted as a separate line)

2 Kommentare

You always come through for me. I appreciate it!
Voss
Voss am 4 Apr. 2024
You're welcome!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Hilfe-Center und File Exchange

Produkte

Version

R2023a

Tags

Gefragt:

am 4 Apr. 2024

Kommentiert:

am 4 Apr. 2024

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by