필터 지우기
필터 지우기

applying while loop for solving simultaneous equations

조회 수: 1 (최근 30일)
Akshay Pratap Singh
Akshay Pratap Singh 2019년 4월 15일
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

답변 (0개)

카테고리

Help CenterFile Exchange에서 Calculus에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by