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

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.

Answer 1

[Star Strider]

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

.