how to display on increment 0.5 for time

how to only show printf only increment of 0.5? right now it's shown in increment of 0.001. its gonna long from 0 to 10.

i still need increment 0.001 for my calculation if i use 0.5. it will be different.

i just want to show a smaller set of data. rightnow, the earlier data can't be scroll.

%le1=1;
%le2=1.5;
%d=10;
%kr=25000;
%kf=20000;
%j=800;
%m=1000;
clc;
h=0.001; z=0;
v=0; x=0; x2=0;
theta=0;theta2=0;
fprintf('time\t\tx\t\ttheta\n--------------------------------------------------------------------------------\n')
maxbounce=0;
maxpitch=0;
tbounce = 0;
tpitch=0;
for t=0:0.001:10
%figure(1);
%plot(t,x,'*');
%hold all;
%figure(2);
%plot(t,theta,'*');
hold all;
j1=h*z;
k1=h*(1/1000)*(1000*sin(3*pi*t)+1250*sin(3*pi*t-0.2*pi*(2.5))-(45000)*x-(17500)*theta);
j2=h*(z+0.5*k1);
k2=h*(1/1000)*(1000*sin(3*pi*(t+0.5*h))+1250*sin(3*pi*(t+0.5*h)-0.2*pi*(2.5))-(45000)*(x+0.5*j1)-(17500)*theta);  
j3=h*(z+0.5*k2);
k3=h*(1/1000)*(1000*sin(3*pi*(t+0.5*h))+1250*sin(3*pi*(t+0.5*h)-0.2*pi*(2.5))-(45000)*(x+0.5*j2)-(17500)*theta);  
j4=h*(z+k3);
k4=h*(1/1000)*(1000*sin(3*pi*(t+h))+1250*sin(3*pi*(t+h)-0.2*pi*(2.5))-(45000)*(x+j3)-(17500)*theta);   
x1=x+(j1+2*(j2+j3)+j4)/6;
z=z+(k1+2*(k2+k3)+k4)/6 ; 

n1=h*v;
m1=h*(1/800)*(1875*sin(3*pi*t-0.5*pi)-1000*sin(3*pi*t)-17500*x-76250*theta);
n2=h*(v+0.5*m1);
m2=h*(1/800)*(1875*sin(3*pi*(t+0.5*h)-0.5*pi)-1000*sin(3*pi*(t+0.5*h))-17500*x-76250*(theta+0.5*n1));
n3=h*(v+0.5*m1);
m3=h*(1/800)*(1875*sin(3*pi*(t+0.5*h)-0.5*pi)-1000*sin(3*pi*(t+0.5*h))-17500*x-76250*(theta+0.5*n2));
n4=h*(v+m1);
m4=h*(1/800)*(1875*sin(3*pi*(t+h)-0.5*pi)-1000*sin(3*pi*(t+h))-17500*x-76250*(theta+n3));
theta1=theta+(n1+2*(n2+n3)+n4)/6;
v=v+(m1+2*(m2+m3)+m4)/6 ; 
theta=theta1;
x=x1;

if(x>maxbounce)
 maxbounce =x;
 tbounce = t;
endif
if(theta>maxpitch)
 maxpitch=theta;
 tpitch=t;
endif


printf('%.4f\t\t%.6f\t\t%.4f\n',t,x1,theta1);
end
printf('Maximum value for x = %.4f at t= %d\n',maxbounce,tbounce);
printf('Maximum value for theta = %.8f at t = %d',maxpitch,tpitch);

[Rik]

Edit restored from Google cache. @ihtifazuddin Nashruddin Don't edit away important parts of your question. That is extremely rude to the people helping you.

Answer 1

[Star Strider]

Change this:

for t=0:0.001:10

to:

for t=0:0.5:10

That is the only line that defines ‘t’ that I can see, so that one change should do what you want. .

[ihtifazuddin Nashruddin]

nope. the output of on the middleand right will be different. can't do that

[Star Strider]

One option is to print only if the result of a test is true:

t=0:0.001:10;                                           % Original Vector

tprint = t.*(rem(t,0.5)<0.0001)>0;                      % Print If ‘tprint’ Is ‘true’

CheckResult = t(tprint);                                % Check (Delete Later)

In this example, test if ‘tprint’ is true at each iteration, then if so print only that result. The loop continues to iterate otherwise.

Here:

CheckResult = [0.5000    1.0000    1.5000    2.0000    2.5000    3.0000    3.5000    4.0000    4.5000    5.0000    5.5000    6.0000    6.5000    7.0000    7.5000    8.0000    8.5000    9.0000    9.5000   10.0000]

so it would only print at those values for ‘t’.

In partial context:

tprint = t.*(rem(t,0.5)<0.0001)>0;                      % Print If ‘tprint’ Is ‘true’
if tprint
    fprintf('%.4f\t\t%.6f\t\t%.4f\n',t,x1,theta1);
end

or equivalently:

if (rem(t,0.5)<0.0001)>0
    fprintf('%.4f\t\t%.6f\t\t%.4f\n',t,x1,theta1);
end

.

[ihtifazuddin Nashruddin]

tq. it working

[Star Strider]

As always, my pleasure!

Answer 2

[Steven Lord]

Either create a counter that counts how many iterations you've performed and only fprintf if the counter indicates the current value of the loop variable is (close enough to) a multiple of 0.5 or use ismembertol or rem to check if the loop variable is "close enough to" a multiple of 0.5.

You do not want to simply use == or ismember as 0.01 is not exactly representable in floating point.