Matlab cannot recognize variable

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Jaevon Stewart
Jaevon Stewart 2024-4-2
编辑： Jaevon Stewart 2024-4-2
Ran in:
I have this code here.
%May change entire thing to function
clc;
clear;
%Variables
sigma = .1;
lock = 8;
twist = -8;
a = 5.7*(180/pi);
bloading = .12;
mu = .25;
alpha_tpp = 2.25;
siga = sigma*a; % This variable is seen a lot.
%% solve for inflow
% Change
CT = .0012;
lambdaH = sqrt(CT/2);
muz = mu*tand(alpha_tpp);
aproxerr = 10^5;
esterr = 100;
n = 0;
f = .5;
while aproxerr < esterr
n = n+1;
if n == 1
lambdaold = lambdaH;
else
lambdaold = lambda;
end
%iterating on
lambdain = lambdaH^2/(sqrt(lambdaold^2+mu^2));
lambda = lambdaold - (lambdaold - muz - lambdain)/(1+((lambdain*lambdaold)/(lambdaold^2+mu^2)))*f;
esterr = abs(lambda - lambdaold)/lambda;
end
%% Question 2
delta = 1 - mu^2 + (9/4*mu^4);
%theta_collective =
%(1/delta)*((1+((3/2)*mu^2)))*((6*CT)/(sigma*a)+(.375*mu^2*twist) I
%do not know if this is necessary
Betanaught = ((lock/8)/delta)*(1-((19/18)*mu^2)+(1.5*mu^4))*((6*CT)/siga)+(.005+((29/120)*mu^2)-(.2*mu^4)+(.375*mu^6))*twist + lambda*((1/6)-((7/12)*mu^2)+(.25*mu^4));
Unrecognized function or variable 'lambda'.
Why is Lambda unrecognized here? I ran another code that had pretty much the same iteration up until the end of the while loop, and lambda was defined in it. So what is wrong here? And btw, I appreciate all the help you guys offer me.
Other code for reference
format long
% lambda is x
Rmain=27; %ft
Cmain=1.7; %ft
Nbmain=4;
Tipspeed_main=725; %ft/s
W=16000; %lb
alt=5000;
alpha_tpp=3;
%knots
%%
Amain=pi*Rmain^2;
sigma=(Nbmain*Cmain)/(pi*Rmain);
[~, ~, ~, RHO] = atmosisa(alt*0.3048);
rho=RHO*0.0685218/35.3147;
CT = W/(rho*Amain*Tipspeed_main^2);
xh = sqrt(CT/2);
V = 0*1.68781:10*1.68781:200*1.68791;
Mu = V./(Tipspeed_main);
for k1=1:length(Mu)
mu=Mu(k1);
muz = mu*tand(alpha_tpp);
n = 0;
aperr = 10^-5;
eserr = 100;
while eserr > aperr
n = n + 1;
if n == 1
xold = xh;
else
xold = x ;
end
%iterating on
xin = xh.^2./sqrt(xold.^2+mu.^2);
x = xold - (xold - muz - xin)./(1+((xin.*xold)./(xold.^2+mu.^2))).*.5;
eserr = abs((x - xold)./x);
end
X(k1)=x;
end
figure
plot(Mu,X,'-x')
grid on

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采纳的回答

Dyuman Joshi
Dyuman Joshi 2024-4-2
编辑：Dyuman Joshi 2024-4-2
"Why is Lambda unrecognized here?"
Because the while loop is not initiated, as the condition is not satisfied.
As the while loop does not run, lambda is not defined.
And when you use lambda (undefined parameter) to define Betanaught, it gives you the aforementioned error.
aproxerr = 10^5;
esterr = 100;
while aproxerr < esterr
...
end
Maybe aproxerr is 10^-5 (or something else). Correct the value and your code will run (I tested with couple of smaller values 10^0, 10^-1, etc).
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Jaevon Stewart
Jaevon Stewart 2024-4-2
Thank you! I simply needed to flip the sign in the while loop (facepalm) I completely missed the fact that that was incorrect.
Jaevon Stewart
Jaevon Stewart 2024-4-2
编辑：Jaevon Stewart 2024-4-2
Not flip the sign, change approxerr to 10^-5**

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