How to count the number of iterations in a loop?

Hi everybody,

I'm new in MATLAB. I would like to count the number of iterations in the following loop while distance between particles are lower than a specified minimum (i.e 5NM)

Thanks in advance!

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clc; clear all; close all

%Initial and final points of particles runways

%Particle 1

a1=0; a2=40;

b1=100; b2=40

%Particle 2

c1=45; c2=0;

d1=45; d2=100;

%particles speed (NM/seconds)

va=520/3600;

vb=520/3600;

%Iteration interval

At=1;

%Initial positions

xa(1)=a1; ya(1)=a2;

xb(1)=c1; yb(1)=c2;

%Velocity projection

Vxa= va*cos(0);

Vya=va*0;

Vxb=vb*cos(pi/2);

Vyb=vb*sin(pi/2);

%Loop for particles porsitions

for t=2:1000

%Particle A

xa(t)=xa(t-1)+Vxa*At;

ya(t)=ya(t-1)+Vya*At;

%Particle B

xb(t)=xb(t-1)+Vxb*At;

yb(t)=yb(t-1)+Vyb*At;

Dist = 5;

aux(1,t)=norm([xa(t) ya(t)] - [xb(t) yb(t)]);

% if Dist>aux

% % count the number of iterations (time in seconds) while distance between both particles is less than 5

% end

end

d=sqrt((xa-xb).^2+(ya-yb).^2); %Distance between particles

plot(xa,ya,xb,yb)

axis([0, 100, 0, 100]);

grid on

Answer 1

[Mathieu NOE]

hello

at the very beginnin gof your ode you should initilalize a counter

k = 0;

then in the loop , you do :

     if Dist>aux
    % count the number of iterations (time in seconds) 
    % while distance between both particles is less than 5
        k = k + 1;
     end 

[Andrea Mira]

Thanks @Mathieu NOE for your answer.

if know I have more than one particle in each runway, How I do it?

I'm trying to do it by the same way, but it doesn't work.

Thanks!

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clc; clear all; close all

num_de_iteraciones=700;

At=1;

%Initial and final points of particles runways

%Runway 1

a1=0; a2=40;

b1=100; b2=40;

%Runway 2

c1=45; c2=0;

d1=45; d2=100;

% Number of particles in each runway

n1=6;

n2=10;

cont(n1,n2)=zeros; %to count the times

%projection

rumbo_via1_x=cos(0);

rumbo_via1_y=sin(0);

rumbo_via2_x=cos(pi/2);

rumbo_via2_y=sin(pi/2);

aerovia1_x=zeros(n1,1);

aerovia1_y=zeros(n1,1);

aerovia2_x=zeros(n2,1);

aerovia2_y=zeros(n2,1);

%particles speed (NM/h)

v1=520;

v2=520;

v_via1=520*ones(n1,1);

v_via2=520*ones(n2,1);

%Initial Positions

via1_x(1,1)=a1;

via1_y(1,1)=a2;

via2_x(1,1)=c1;

via2_y(1,1)=c2;

%longitudinal separation between particles in runway1

for i=2:n1

sep_long_via1=5;

dx_ini_via1= sep_long_via1*rumbo_via1_x;

dy_ini_via1= sep_long_via1*rumbo_via1_y;

via1_x(i,1)=via1_x(i-1)-dx_ini_via1;

via1_y(i,1)=via1_y(i-1)-dy_ini_via1;

end

%longitudinal separation between particles in runway2

for j=2:n2

sep_long_via2=5;

dx_ini_via2 = sep_long_via2*rumbo_via2_x;

dy_ini_via2 = sep_long_via2*rumbo_via2_y;

via2_x(j,1)=via2_x(j-1)-dx_ini_via2;

via2_y(j,1)=via2_y(j-1)-dy_ini_via2;

end

dist = zeros(n1,n2);

t_auxmin = zeros(n1,n2);

for t=2:num_de_iteraciones %segundos

via1_x(:,t)=via1_x(:,t-1)+(v_via1(:)./3600).*rumbo_via1_x*At;

via1_y(:,t)=via1_y(:,t-1)+(v_via1(:)./3600).*rumbo_via1_y*At;

via2_x(:,t)=via2_x(:,t-1)+(v_via2(:)./3600).*rumbo_via2_x*At;

via2_y(:,t)=via2_y(:,t-1)+(v_via2(:)./3600).*rumbo_via2_y*At;

for i=1:n1

for j=1:n2

DistColision=5;

aux=norm([via1_x(i,t) via1_y(i,t)] - [via2_x(j,t) via2_y(j,t)]);

if DistColision>aux

%Count the number of iterations (time in seconds) while distance between a pair of particles is less than 5.

(It will be more than one pair of particles which distance between them it will be less than 5. I would like to know, for each pair, the number of seconds)

%ContColisiones(i,j)= ContColisiones(i,j)+1;

t_auxmin(i,j)=t;

end

end

end

[Mathieu NOE]

before answering this question, I was trying to simplify your first code as it was for me not optimal

no need for complicated for loops and if statements

clc; clear all; close all
%Initial and final points of particles runways
%Particle 1
a1=0; a2=40;
b1=100; b2=40
%Particle 2
c1=45; c2=0;
d1=45; d2=100;
%particles speed (NM/seconds)
va=520/3600;
vb=520/3600;
%Iteration interval
At=1;

% time vector (s)
t = 0:1:500;

% %Initial positions
% xa(1)=a1; ya(1)=a2;
% xb(1)=c1; yb(1)=c2;

%Velocity projection
Vxa= va*cos(0);
Vya=va*0;
Vxb=vb*cos(pi/2); 
Vyb=vb*sin(pi/2); 

%Loop for particles porsitions
% for t=2:1000
%     %Particle A
%     xa(t)=xa(t-1)+Vxa*At; 
%     ya(t)=ya(t-1)+Vya*At; 
%     
%    %Particle B
%     xb(t)=xb(t-1)+Vxb*At; 
%     yb(t)=yb(t-1)+Vyb*At; 
%     
%     Dist = 5;
%     aux(1,t)=norm([xa(t) ya(t)] - [xb(t) yb(t)]); 
% %      if Dist>aux
% %     % count the number of iterations (time in seconds) while distance between both particles is less than 5
% %      end 
% end

%Particle A
xa = a1 + cumtrapz(t,Vxa*ones(size(t))); % position (vs time)  = initial position + integral of velocity vs time 
ya = a2 + cumtrapz(t,Vya*ones(size(t))); % position (vs time)  = initial position + integral of velocity vs time 

%Particle B
xb = c1 + cumtrapz(t,Vxb*ones(size(t))); % position (vs time)  = initial position + integral of velocity vs time 
yb = c2 + cumtrapz(t,Vyb*ones(size(t))); % position (vs time)  = initial position + integral of velocity vs time 

d=sqrt((xa-xb).^2+(ya-yb).^2); %Distance between particles

% count the number of iterations (time in seconds) while distance between both particles is less than 5
Dist = 5;
ind = find(d<Dist);
t_dist_below = t(ind);
count = length(ind)

plot(xa,ya,'b',xb,yb,'r',xa(ind),ya(ind),'db',xb(ind),yb(ind),'dr')
axis([0, 100, 0, 100]); 
grid on

maybe you could use it to simplify alos your second code

[Andrea Mira]

Thanks very much @Mathieu NOE

[Mathieu NOE]

you're welcome