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Finding the closest value in an large array

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AeroEng
AeroEng 2020-5-9
评论： Ameer Hamza 2020-5-9
I have an array of 1001x1001. I would like to pull out a single value that is closest to 0 out of my "F" array. I do not know how to do this. Everytime I do it, it returns an entire column but I only want a single value.
The same thing happens for when I try to do the max of my "lambda". Would you please help me find the max and value closest to zero of these two variables. Thank you very much!
H_max = 20000*0.3048 ; % Meters
SM = 2;
amax = 10;
ML = 1 ; % kg
rhos = 2700 ; % kg/Meter^3
rhop = 1772 ; % kg/Meter^3
sigma = 60*10^6 ; % Pascal
N = 3 ; % Number of Fins
R = 287 ; % J/kg
T = 298 ; % Kelvin
g = 9.81 ; % Meters/Second^2
gamma = 1.4 ; % ND
a = (gamma*R*T)^(1/2) ; % Meters/Second
Pa = 101.325*10^3 ; % Pascal
% Solving for R_max, W_eq, t_b with linear burn rate
Rmax = 1 + amax;
Weq = ((H_max*g)/((log(Rmax)/2)*(log(Rmax)-2)+((Rmax -1)/(Rmax))))^(1/2);
Meq = Weq/a;
tb=((Rmax -1)*Weq)/(g*Rmax);
P0_Pa = (1+Meq^2*((gamma-1)/2))^(gamma/(gamma-1)) ; % ND - Pressure Ratio
P0 = P0_Pa*Pa ; % Pa - Pressure
% Starting with an Initial Lambda Max
lambdamax = 0;
D = 0:0.001:1; % i
L = 0:0.001:1; % j
% Preallocating each variable
delta = zeros(size(D));
r = zeros(size(D));
Mfb = zeros(size(D));
Mn = zeros(size(D));
Mcone = zeros(size(D));
Mcyl = zeros(size(D));
Mfin = zeros(size(D));
Ms = zeros(size(D));
M0 = zeros(size(D));
Mp = zeros(size(D));
Lp = zeros(size(D));
Xcp = zeros(size(D));
Xcg = zeros(size(D));
xp = zeros(size(D));
lambda = zeros(size(D));
F = zeros(size(D));
for i = 1:length(D)
for j = 1:length(L)
delta(i) = D(i)*P0/(2*sigma);
r(i) = D(i)/2;
Mfb(i) = pi*D(i)*rhos*D(i)*delta(i) ;
Mn(i) = delta(i)*rhos*pi*(D(i)/2)*((D(i)/2)+(D(i)^2+ (D(i)^2)/4)^(1/2));
Mcone(i) = rhos*delta(i)*(pi*r(i)*(r(i)+r(i)^2)^(1/2));
Mcyl(i,j) = rhos*pi*D(i)*delta(i)*L(j);
Mfin(i) = rhos*D(i)^2*delta(i)*pi +(3/2)*rhos*delta(i)*D(i)^2;
Ms(i,j) = (Mcyl(i,j) + Mfin(i) + Mcone(i));
Mp(i,j) = (Rmax - 1)*(Ms(i,j)+ML);
Lp(i,j) = Mp(i,j)/(pi*D(i)^2*rhop/4);
Ckn = (4*N*(4/3))/(1+sqrt(6)) ; % ND
Xcp(i,j) = (1.33*D(i) + Ckn*(D(i)+L(j)+(D(i)/3)))/(2+Ckn) ; % Meters
xp(i,j) = 2*D(i)+L(j)-Lp(i,j)/2;
Xcg(i,j) = ((2/3)*D(i)*Mn(i) + (2/3)*D(i)*ML + (D(i)+L(j)/2)*Mcyl(i,j)...
+ Mp(i,j)*xp(i) +((3*D(i)+2*L(j))/2)*Mfin(i))/(Mfin(i) + Mp(i,j)...
+ ML+Mcyl(i,j)+Mn(i));
lambda(i,j) = ML/(Ms(i,j)+Mp(i,j));
F(i,j) = Xcp(i,j)-Xcg(i,j)-D(i)*SM;
end
end

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Ameer Hamza
Ameer Hamza 2020-5-9
You need to specify 'all' to min() and max() to find the minimum over the entire matrix. For example, this returns the minimum value and its location
[min_val, index] = min(abs(F-0), [], 'all', 'linear');
[row, col] = ind2sub(size(F), index);
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AeroEng
AeroEng 2020-5-9
Oh shoot! I didnt know if that was specific to the minimum but I understand now! Thank you very much! I appreciate all your help!

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