How do I preallocate for speed in the following code?

조회 수: 4 (최근 30일)
Hassan Tawsif
Hassan Tawsif 2023년 3월 7일
답변: Askic V 2023년 3월 7일
clc
clear
N = 9;
S = 18.1;
AR = 7;
lambda = 0.8;
alpha_twist = -1.5;
i_w = 10;
a_2d = 6.3;
a_0 = -3;
a_0_fd = -6;
b = sqrt(AR*S);
bf_b=0.6;
MAC = S/b;
Croot = (1.5*(1+lambda)*MAC)/(1+lambda+lambda^2);
theta = pi/(2*N):pi/(2*N):pi/2;
alpha=i_w+alpha_twist:-alpha_twist/(N-1):i_w;
for i=1:N
if (i/N)>(1-bf_b)
alpha_0(i)=a_0_fd;
else
alpha_0(i)=a_0;
end
end
z = (b/2)*cos(theta);
c = Croot * (1 - (1-lambda)*cos(theta));
mu = c * a_2d / (4 * b);
LHS = mu .* (alpha-alpha_0)/57.3;
for i=1 : N
for j=1 : N
B(i,j) = sin((2*j-1) * theta(i)) * (1 + (mu(i) *(2*j-1)) / sin(theta(i)));
end
end
A=B\transpose(LHS);
for i = 1 : N
sum1(i) = 0;
sum2(i) = 0;
for j = 1 : N
sum1(i) = sum1(i) + (2*j-1) * A(j)*sin((2*j-1)*theta(i));
sum2(i) = sum2(i) + A(j)*sin((2*j-1)*theta(i));
end
end
CL_TO = pi * AR * A(1);

답변 (1개)

Askic V
Askic V 2023년 3월 7일
The most basic approach is given here:
clc
clear
N = 9;
S = 18.1;
AR = 7;
lambda = 0.8;
alpha_twist = -1.5;
i_w = 10;
a_2d = 6.3;
a_0 = -3;
a_0_fd = -6;
b = sqrt(AR*S);
bf_b=0.6;
MAC = S/b;
Croot = (1.5*(1+lambda)*MAC)/(1+lambda+lambda^2);
theta = pi/(2*N):pi/(2*N):pi/2;
alpha=i_w+alpha_twist:-alpha_twist/(N-1):i_w;
alpha_0 = zeros(1,N);
for i=1:N
if (i/N)>(1-bf_b)
alpha_0(i)=a_0_fd;
else
alpha_0(i)=a_0;
end
end
z = (b/2)*cos(theta);
c = Croot * (1 - (1-lambda)*cos(theta));
mu = c * a_2d / (4 * b);
LHS = mu .* (alpha-alpha_0)/57.3;
B = zeros(N,N);
for i=1 : N
for j=1 : N
B(i,j) = sin((2*j-1) * theta(i)) * (1 + (mu(i) *(2*j-1)) / sin(theta(i)));
end
end
A=B\transpose(LHS);
sum1 = zeros(1, N);
sum2 = zeros(1, N);
for i = 1 : N
for j = 1 : N
sum1(i) = sum1(i) + (2*j-1) * A(j)*sin((2*j-1)*theta(i));
sum2(i) = sum2(i) + A(j)*sin((2*j-1)*theta(i));
end
end
CL_TO = pi * AR * A(1);

카테고리

Help CenterFile Exchange에서 MATLAB에 대해 자세히 알아보기

제품


릴리스

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by