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# loop while loop for

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Matthew el 20 de Abr. de 2024
Locked: Rena Berman el 5 de Jun. de 2024
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 comentariosMostrar NingunoOcultar Ninguno
John D'Errico el 21 de Abr. de 2024
Editada: John D'Errico el 21 de Abr. de 2024
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?
Rena Berman el 5 de Jun. de 2024

Sulaymon Eshkabilov el 20 de Abr. de 2024
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]')
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Matthew el 20 de Abr. de 2024
Thank you, I appreciate the help.

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