Saving maximum values from last iteration.
Mostra commenti meno recenti
Hi, I want to save maximum values from last iteration for every "ii". If i run code for a single "ii" it saves up the data properly, but when I run it in range from 1 to 10 it just saves the first one. What am I missing?
*Edited code as asked.
d = 2;
v=700;
om = 500;
dp=5;
dC=20;
m = 1;
S = pi * ((d^2) / 4);
r = 1.500;
g=8;
dt = 1e-5;
i = 1;
t(i) = 0;
t0 = 0;
max_values=zeros(10,3);
v0=zeros(1,10);
numb = [ 0 0 0 0;1 1 1 1; 2 2 2 2;3 3 3 3;4 4 4 4;5 5 5 5;6 6 6 6;7 7 7 7;8 8 8 8;9 9 9 9];
for ii = 1:10
x(i)=numb(ii,1);
y(i)=numb(ii,2);
z(i)=numb(ii,3);
v0(ii) = om * numb(ii,4) * 2 * pi;
data = [0 0 300 240 180 120 60 0 330 300];
if ii==1
vz(i) = v;
vy(i) = 0;
vx(i) = 0;
elseif ii>1 && ii<=7
vz(i) = v;
vy(i) = -v0(ii)*cosd(data(ii));
vx(i) = v0(ii)*cosd(90-data(ii));
elseif ii>7 && ii<=10
vz(i) = v;
vy(i) = -v0(ii)*cosd(data(ii));
vx(i) = v0(ii)*cosd(90-data(ii));
end
while t < 500e-4
dvz(i) = (-1/(2*m))*r*S*v*vz(i);
dz(i) = vz(i);
vz(i+1) = vz(i) + dvz(i) * dt;
z(i+1) = z(i) + dz(i) * dt;
dvy(i) = (-1/(2*m))*r*S*v*vy(i)-g;
dy(i) = vy(i);
vy(i+1) = vy(i) + dvy(i) * dt;
y(i+1) = y(i) + dy(i) * dt;
dvx(i) = (-1/(2*m))*r*S*v*vx(i);
dx(i) = vx(i);
vx(i+1) = vx(i) + dvx(i) * dt;
x(i+1) = x(i) + dx(i) * dt;
t(i+1) = t(i) + dt;
i=i+1;
max_values(ii,1)=max(x(i));
max_values(ii,2)=max(y(i));
max_values(ii,3)=max(z(i));
end
end
4 Commenti
Alexander
il 30 Mar 2024
I don't know what is x, y or z and I don't know what is i and t. Hence, it is impossible to run your code and I don't want to guess what you want. Please supply some numbers to get a script which is showing the problem you have.
Manikanta Aditya
il 30 Mar 2024
@Alexander, I feel here the issue in within the scope of the variables, they are not being reset for each iteration of 'ii', which means the data from the previous iterations are being carried over into the next ones. This should be why only the results from the first iteration are seen.
Can initialise these variables inside the 'ii' loop:
d = 2;
v=700;
om = 500;
dp=5;
dC=20;
m = 1;
S = pi * ((d^2) / 4);
r = 1.500;
g=8;
dt = 1e-5;
max_values=zeros(10,3);
v0=zeros(1,10);
numb = [ 0 0 0 0;1 1 1 1; 2 2 2 2;3 3 3 3;4 4 4 4;5 5 5 5;6 6 6 6;7 7 7 7;8 8 8 8;9 9 9 9];
for ii = 1:10
i = 1;
t = zeros(1, 50000);
t(i) = 0;
x = zeros(1, 50000);
y = zeros(1, 50000);
z = zeros(1, 50000);
vx = zeros(1, 50000);
vy = zeros(1, 50000);
vz = zeros(1, 50000);
x(i)=numb(ii,1);
y(i)=numb(ii,2);
z(i)=numb(ii,3);
v0(ii) = om * numb(ii,4) * 2 * pi;
data = [0 0 300 240 180 120 60 0 330 300];
if ii==1
vz(i) = v;
vy(i) = 0;
vx(i) = 0;
elseif ii>1 && ii<=7
vz(i) = v;
vy(i) = -v0(ii)*cosd(data(ii));
vx(i) = v0(ii)*cosd(90-data(ii));
elseif ii>7 && ii<=10
vz(i) = v;
vy(i) = -v0(ii)*cosd(data(ii));
vx(i) = v0(ii)*cosd(90-data(ii));
end
while t(i) < 500e-4
dvz(i) = (-1/(2*m))*r*S*v*vz(i);
dz(i) = vz(i);
vz(i+1) = vz(i) + dvz(i) * dt;
z(i+1) = z(i) + dz(i) * dt;
dvy(i) = (-1/(2*m))*r*S*v*vy(i)-g;
dy(i) = vy(i);
vy(i+1) = vy(i) + dvy(i) * dt;
y(i+1) = y(i) + dy(i) * dt;
dvx(i) = (-1/(2*m))*r*S*v*vx(i);
dx(i) = vx(i);
vx(i+1) = vx(i) + dvx(i) * dt;
x(i+1) = x(i) + dx(i) * dt;
t(i+1) = t(i) + dt;
i=i+1;
max_values(ii,1)=max(x);
max_values(ii,2)=max(y);
max_values(ii,3)=max(z);
end
end
This way, for iternation of 'ii' it starts with fresh variables, and the results from the previous iterations won't affect the current one.
Alexander
il 31 Mar 2024
So I think you are pleased now.
Daniel J
il 31 Mar 2024
Risposte (0)
Categorie
Scopri di più su MATLAB in Centro assistenza e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!