Info
Questa domanda è chiusa. Riaprila per modificarla o per rispondere.
for loop for t
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
t=0.75
a1=2*t*x/(1+t)+t+k;
ad=t/a1;
ap=k/a1;
p(1)=1.01325*10^5;
for i=1:14;
p(i+1)=p(i)+20000;
end
sd=(Ad*p)/(2*rho*g*Veo);
sp=(Ap*p)/(2*rho*g*Vco);
kdd=sd*(ad-q*keb);
kdp=sd*ap;
kpp=sd*o*(q*kcb+ap);
kpd=sd*o*ad;
B=[1];
B(1:15)=B;
C=2*(rw*ep+ed);
C(1:15)=C;
D=(1-kdd)+(rw^2+kpp)+4*rw*ep*ed;
E=2*ed*(rw^2+kpp)+rw*ep*(1-kdd);
F=(rw^2+kpp).*(1-kdd)+kpd.*kdp;
B=B';
C=C';
D=D';
E=E';
F=F';
w=[B C D E F];
[m,n] = size(w) ;
r = zeros(m,n-1) ;
for i = 1:m
r(i,:)= roots(w(i,:))
end
r=real(r)
r=unique(r)
4 Commenti
Image Analyst
il 7 Lug 2020
No question, just an announcement.
I tried to run it but after entering random numbers for lots of missing variables, I finally gave up. It's pretty clear that Ali never read the links Rik sent.
Risposte (0)
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!