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loop while loop for
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i'm trying to get this while loop to quite based on the rate of change of v being less than .05%.
the while loop quits now just based on t(i)<7; but i'm trying to figure out how i can get this loop to stop based on ((v(2)-v(1))/(t(2)-t(1))<= 0.05.
clc
clear
m = 36/1000;
g=9.81;
T=5;
A=.00044316;
Cd=0.75;
rho=1.203;
a(1)=0;
v(1)=0;
h(1)=0;
t(1)=0;
t_step=0.001;
i=1;
u(1)=0;
while t(i)<7;
a(i+1)= ((T)-(m*g)-(0.5*rho*A*Cd*v(i)^2))/m;
v(i+1)= a(i+1)*t_step+v(i);
h(i+1)= v(i+1)*t_step+h(i);
t(i+1)=t(i)+t_step;
i=i+1;
end
plot(t,v)
2 个评论
John D'Errico
2024-4-21
编辑:John D'Errico
2024-4-21
iF you will remove your question once you get an answer, I'd ask you not to ask a question at all. When you remove the question, you hurt the site, because the answer is no longer meaningful. It does not allow anyone else to learn from your question and this answer. You make it less likely that your next question will find someone willing to spend the time to answer your questions.
If you think you are not supposed to ask a question like this, because your teacher would not want you to do so, then WHY DID YOU ASK THE QUESTION IN THE FIRST PLACE?
采纳的回答
Sulaymon Eshkabilov
2024-4-20
I suppose that what you are trying to get is dv/dt >=0.05. Here is how you can get it done:
clc; clearvars;
m = 36/1000;
g=9.81;
T=5;
A=.00044316;
Cd=0.75;
rho=1.203;
a(1)=0;
v(1)=0;
h(1)=0;
t(1)=0;
t_step=0.001;
i=1;
u(1)=0;
dvdt = 1;
while dvdt>=0.05
a(i+1)= ((T)-(m*g)-(0.5*rho*A*Cd*v(i)^2))/m;
v(i+1)= a(i+1)*t_step+v(i);
h(i+1)= v(i+1)*t_step+h(i);
t(i+1)=t(i)+t_step;
dvdt = (v(i+1)-v(i))/(t(i+1)-t(i));
i=i+1;
end
plot(t,v, 'r-.o')
grid on
text(0.5, 10, ['The simulation is halted after: ' num2str(i) ' iterations and ' num2str(t(end)) ' [seconds]'], 'backgroundcolor', 'w')
xlabel('Time, [s]')
ylabel('Velocity, [m/s]')
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