applying while loop for solving simultaneous equations

1 vue (au cours des 30 derniers jours)
Akshay Pratap Singh
Akshay Pratap Singh le 15 Avr 2019
I am trying to include a condition that 'tan(alphac)<(1/lam)' in the following code. Here, ''alphac'' is dependent of ''x''. and the value of x is determined at last. I want to write a code such that the value of x is determined by applying the condition.
code:
dbstop if error
clear all
clc
format longEng
syms x y lam
a=[4;0.55];
% The Newton-Raphson iterations starts here
LAM=linspace(0,10,11);
h=4;
q=20;
gma=18.4; nq=2*q/(gma*(h+x));
delta=26;
phi=39;
A=lam*nq/(1+nq);
kv=0;
kh=0;
da1=delta*(pi/180); da2=-delta*(pi/180); pha1=phi*(pi/180); pha2=phi*(pi/180);
dp1=delta*(pi/180); dp2=delta*(pi/180); php1=phi*(pi/180); php2=phi*(pi/180);
psi=atan(kh/(1-kv));
m=pha1+da1;
b=pha1-psi;
c=psi+da1;
alphac=atan((sin(m)*sin(b)+(sin(m)^2*sin(b)^2+sin(m)*cos(m)*sin(b)*cos(b)+A*cos(c)*cos(m)*sin(b))^0.5)/(A*cos(c)+sin(m)*cos(b)))
kg=(tan(alphac-pha1)+(kh/(1-kv)))/(tan(alphac)*(cos(da1)+sin(da1)*tan(alphac-pha1)));
r=1-lam*tan(alphac);
kq=r*kg;
A2=0;
alphac2=atan((sin(m)*sin(b)+(sin(m)^2*sin(b)^2+sin(m)*cos(m)*sin(b)*cos(b)+A2*cos(c)*cos(m)*sin(b))^0.5)/(A2*cos(c)+sin(m)*cos(b)))
kg2=(tan(alphac2-pha1)+(kh/(1-kv)))/(tan(alphac2)*(cos(da1)+sin(da1)*tan(alphac2-pha1)));
pg=0.5*gma*(1-kv)*kg2*(h+x)^2;
A=1;
alphac1=atan((sin(m)*sin(b)+(sin(m)^2*sin(b)^2+sin(m)*cos(m)*sin(b)*cos(b)+A*cos(c)*cos(m)*sin(b))^0.5)/(A*cos(c)+sin(m)*cos(b)))
kg1=(tan(alphac1-pha1)+(kh/(1-kv)))/(tan(alphac1)*(cos(da1)+sin(da1)*tan(alphac1-pha1)));
r1=1-lam*tan(alphac1);
kq1=r1*kg1;
pq=(1-kv)*(q*kq*(h+x)+0.5*q*(kq1-kq)*(h+x));
va2=asin(sin(da2)/sin(pha2))-asin(sin(psi)/sin(pha2))-da2-psi;
ka2=(1/cos(psi))*(cos(da2)*((cos(da2)-sqrt(sin(pha2)^2-sin(da2)^2)))/(cos(psi)+sqrt(sin(pha2)^2-sin(psi)^2)))*exp(-va2*tan(pha2));
vp1=asin(sin(dp1)/sin(php1))+asin(-sin(psi)/sin(php1))+dp1+psi;
kp1=(1/cos(psi))*(cos(dp1)*((cos(dp1)+sqrt(sin(php1)^2-sin(dp1)^2)))/(cos(psi)-sqrt(sin(php1)^2-sin(psi)^2)))*exp(vp1*tan(php1));
vp2=asin(sin(dp2)/sin(php2))+asin(-sin(psi)/sin(php2))+dp2+psi;
kp2=(1/cos(psi))*(cos(dp2)*((cos(dp2)+sqrt(sin(php2)^2-sin(dp2)^2)))/(cos(psi)-sqrt(sin(php2)^2-sin(psi)^2)))*exp(vp2*tan(php2));
sinda1=sin(da1); sindp1=sin(dp1); sinda2=-sin(da2); sindp2=sin(dp2);
cosda1=cos(da1); cosdp1=cos(dp1); cosda2=cos(da2); cosdp2=cos(dp2);
pp1=kp1*gma*0.5*(x^2);
pa2=ka2*gma*(x*y+0.5*(y^2)); pp2=kp2*gma*(y*(h+x)+(0.5*(y^2)));
zp1=x/3;
zp2=((0.5*(h+x)*(y^2))+((y^3)/3))/(((h+x)*y)+(0.5*(y^2)));
za2=((0.5*x*(y^2))+((y^3)/3))/((x*y)+(0.5*(y^2)));
e2=(pp1*cosdp1)+(pa2*cosda2)-(pg*cosda1)-(pp2*cosdp2)-pq;
e3=(pp1*cosdp1*zp1)+(pp2*cosdp2*zp2)-(pg*cosda1*((h+x)/3))-(pa2*cosda2*za2)-pq*(1/3)*(h+x)*((kq+2*kq1)/(kq+kq1));
g=[e2; e3];
J=jacobian([e2, e3], [x, y]);
A=zeros(2,numel(LAM));
for i=1:numel(LAM)
del=1;
indx=0;
lam=0;
while del>1e-6 && tan(alphac)<(1/lam)
gnum = vpa(subs(g,[x,y,lam],[a(1),a(2),LAM(i)]));
Jnum = vpa(subs(J,[x,y,lam],[a(1),a(2),LAM(i)]));
delx = -Jnum\gnum;
a = a + delx;
del = max(abs(gnum));
indx = indx + 1;
end
Z(:,i)=double(a)
end

Réponses (0)

Catégories

En savoir plus sur Symbolic Math Toolbox dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by