How to make the for loop the length as the array inside the for loop?
6 ビュー (過去 30 日間)
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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
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回答 (1 件)
Star Strider
2023 年 10 月 10 日
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 件のコメント
Star Strider
2023 年 10 月 10 日
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.
.
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