break out of for loop help

I have this for loop which works perfectly fine except that I want it to stop the for loop when y=0. The loop is calculating trajectory so I want it to stop when the ball hits the ground (or y=0). I have tried everything I can think of but nothing works. It either doesn't stop the loop or it returns the completely wrong y values. I've tried if y==0, y<=0, y(n)==0, y(n)<=0, y(n+1)==0, y(n+1)<=0, and putting it before and after the calculations. nothing worked Please help.
for n=1:N
ax(n+1) = -D*vx(n)*v(n);
ay(n+1) = -D*vy(n)*v(n)-g;
v(n+1) = sqrt(vx(n)^2+vy(n)^2);
x(n+1) = x(n)+T*vx(n)+0.5*T^2*ax(n);
vx(n+1) = vx(n)+T*ax(n);
y(n+1) = y(n)+T*vy(n)+0.5*T^2*ay(n);
vy(n+1) = vy(n)+T*ay(n);
if y(n+1)==0, break; end
end

2 comentarios

Jan
Jan el 20 de Oct. de 2011
Please post your initial values such that we can run your code.
Sean Smith
Sean Smith el 20 de Oct. de 2011
h=1;
C=0.5;
R=0.0366;
p=1.2;
m=0.145;
g=9.81;
D=C*p*pi*R^2/(2*m);
vo=90*0.44704;
th0=30*(pi/180);
Tmax=vo*sin(th0)/g+sqrt(2*h/g+vo^2*sin(th0)^2/g^2);
T=1/100;
t=0:T:Tmax;
N=floor(Tmax/T);
v(1)=vo;
x(1)=0;
vx(1)=vo*cos(th0);
y(1)=h;
vy(1)=vo*sin(th0);
ax(1)=-D*vx(1)*v(1);
ay(1)=-D*vy(1)*v(1)-g;

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 Respuesta aceptada

Jan
Jan el 21 de Oct. de 2011

1 voto

The loop does stop. You can check this by a disp statement before the break.
I assume you have written the code into a script and y is defiend from an earlier run. Then a "clear" statement on top would be helpful - not "clear all". Or you could convert it to a function by inserting this as first line:
function myIntegrator
or what ever your file is called.
Note: Using several commands in one line impedes debugging. You cannot set a breakpoint on the break command in:
if y(n+1)==0, break; end
But this would have revealed the problem.

1 comentario

Sean Smith
Sean Smith el 21 de Oct. de 2011
i didn't mean to hit answer accepted, oops.
how does it fill it with only 1 iteration? shouldn't it do it from 1 to N? what would you suggest to get it to stop when y=0?

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Más respuestas (1)

Image Analyst
Image Analyst el 20 de Oct. de 2011

0 votos

It probably won't hit 0.0000000000000000000 exactly. It might get to 0.000000000000002, so check if it's close:
if abs(y(n+1)) < 0.0001 % Some tolerance.
break;
end

1 comentario

Sean Smith
Sean Smith el 20 de Oct. de 2011
still not stopping it for some reason

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