How to make the for loop the length as the array inside the for loop?
Show older comments
I am making a for loop and I want to subtract each element by each other. I was able to do this but my array is now 1x200 and my original array was 1x201. How do I make it so the for loop array is 1x201?
x1 = [1e6:20000:5e6];
L = 4e6; % in m
D_c = 5000;% in m
D1 = [0:25:5000];
x_o = 1e6; % in m
x_2 = [x_o:20000:5e6];
Tw_o = 300; % in K
LR_w = 0.004; % in K/m
LR_c = -0.007; % in K/m
pz_0 = 1000; % in mb
D_1 = D_c.*(sin((pi/L).*(x1-x_o)));
T_c2 = Tw_o-(((LR_w-LR_c).*D_1)-((LR_c.*D1)));
g = 9.81;
Rd = 287;
p1 = pz_0*(((((T_c2)-(LR_c.*D1))./(T_c2))).^((g)/(Rd*LR_c)));
for i = 1:length(p1)-1
deltap(i) = p1(i+1)-p1(i);
end
Answers (1)
x1 = [1e6:20000:5e6];
L = 4e6; % in m
D_c = 5000;% in m
D1 = [0:25:5000];
x_o = 1e6; % in m
x_2 = [x_o:20000:5e6];
Tw_o = 300; % in K
LR_w = 0.004; % in K/m
LR_c = -0.007; % in K/m
pz_0 = 1000; % in mb
D_1 = D_c.*(sin((pi/L).*(x1-x_o)));
T_c2 = Tw_o-(((LR_w-LR_c).*D_1)-((LR_c.*D1)));
g = 9.81;
Rd = 287;
p1 = pz_0*(((((T_c2)-(LR_c.*D1))./(T_c2))).^((g)/(Rd*LR_c)));
deltap = diff(p1)
for i = 1:length(p1)-1
deltap(i) = p1(i+1)-p1(i);
end
deltap
If you want to calculate the numerical derivative at eash point instead, use the gradient function —
deltap = gradient(p1)
.
2 Comments
Matthew
on 10 Oct 2023
Star Strider
on 10 Oct 2023
One way would be to take the absolute value (the abs function), the other, since they are uniformly negative, would be to multiply them by -1 or just use a unary negative:
deltap = -gradient(p1)
Use whatever works best in your application.
.
Categories
Find more on Loops and Conditional Statements in Help Center and File Exchange
on 10 Oct 2023
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
