Finding the closest value in an large array

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

Answer 1

[Ameer Hamza]

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);

[AeroEng]

Sweet! That worked for the minimum! That helps so much! Thank you very very much!!!

Any clue on the maximum of lambda?

[Ameer Hamza]

You can use max() in a similar way for lambda

[max_val, index] = max(lambda, [], 'all', 'linear');
[row, col] = ind2sub(size(F), index); 

[AeroEng]

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!

[Ameer Hamza]

I am glad to be of help.