where I am making a mistake while using fsolve to evaluate two unknown with two equation

2 次查看(过去 30 天)
N2 = 6;
N1 = 5;
R2 = (80.02*N1/(N1+N2));
ha = (80.02*N1/(N1+N2))-35.2088;
hd = (80.02*N2/(N1+N2))-23.4725;
u = degtorad(6.291);
R1 = (80.02*N2/(N1+N2));
rho1 = 1.675;
f1 = @(N1, N2, R1, R2, ha, hd, u, rho1, theta2, t)[((R2*theta2-((ha-rho1)./tan(u)-t./sin(u)))./cos(-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))-R1*(1-cos((N2./N1)*(theta2-((ha-rho1)./tan(u)-t./sin(u))./R2)))./sin((-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))+((N2./N1)*(theta2-((ha-rho1)./tan(u)-t./sin(u))./R2)))-hd),
(-R2*theta2+(R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2)-((-((ha-rho1)./tan(u)-t./sin(u))+(R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin((((ha-rho1)./tan(u)-t./sin(u))./R2)))./cos(u)+rho1+t*tan(u))*cos(-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))+((ha-rho1)./tan(u)-t./sin(u)))];
x=fsolve(@(theta2,t)f1(N1, N2, R1, R2, ha, hd, u, rho1, theta2, t),[degtorad(20),1]);
Not enough input arguments.

Error in solution>@(theta2,t)f1(N1,N2,R1,R2,ha,hd,u,rho1,theta2,t) (line 15)
x=fsolve(@(theta2,t)f1(N1, N2, R1, R2, ha, hd, u, rho1, theta2, t),[degtorad(20),1]);

Error in fsolve (line 267)
fuser = feval(funfcn{3},x,varargin{:});

Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.

采纳的回答

Star Strider
Star Strider 2023-7-6
Only one minor problem. The fsolve function needs and returns a vector of paramters.
Changing that call to:
x=fsolve(@(b)f1(N1, N2, R1, R2, ha, hd, u, rho1, b(1), b(2)),[degtorad(20),1])
so that ‘theta2’ is ‘b(1)’ and ‘t’ is ‘b(2)’ gives it what it wants, and it now works —
N2 = 6;
N1 = 5;
R2 = (80.02*N1/(N1+N2));
ha = (80.02*N1/(N1+N2))-35.2088;
hd = (80.02*N2/(N1+N2))-23.4725;
u = degtorad(6.291);
R1 = (80.02*N2/(N1+N2));
rho1 = 1.675;
f1 = @(N1, N2, R1, R2, ha, hd, u, rho1, theta2, t)[((R2*theta2-((ha-rho1)./tan(u)-t./sin(u)))./cos(-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))-R1*(1-cos((N2./N1)*(theta2-((ha-rho1)./tan(u)-t./sin(u))./R2)))./sin((-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))+((N2./N1)*(theta2-((ha-rho1)./tan(u)-t./sin(u))./R2)))-hd),
(-R2*theta2+(R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2)-((-((ha-rho1)./tan(u)-t./sin(u))+(R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin((((ha-rho1)./tan(u)-t./sin(u))./R2)))./cos(u)+rho1+t*tan(u))*cos(-atan(((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*cos(theta2)-R2)./((R2./(cos(((ha-rho1)./tan(u)-t./sin(u))./R2)+sin(((ha-rho1)./tan(u)-t./sin(u))./R2)*tan(u)))*sin(theta2))))+((ha-rho1)./tan(u)-t./sin(u)))];
x=fsolve(@(b)f1(N1, N2, R1, R2, ha, hd, u, rho1, b(1), b(2)),[degtorad(20),1])
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
x = 1×2
1.8001 -3.3181
.

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