How do I solve this error: 'Error using alpha. Too many output arguments.' ? It comes up when I try to create a user-defined function for the input: 'de_trim'.

The code is supposed to take the user input 'de_trim' and calculate if creates a stall condition for the plane. None of the output arguments are actually based on 'de_trim', it really just comes into play for with the loop at the end.

As of right now the code as a function works. But when I try to display 2 figures it gives me the error message: "Error using alpha. Too many output arguments." 'alpha' is an vector of 11 entries, and is used as the x-axis in a 'plot' command. It's paired up with 'Cm' which is also 11 units long.

If I hide the top line that defines the code as a function the figures get generated without any issue. 'alpha' and 'Cm' come from a load file which define the variables:

   alpha= [-16, -8, -6, -4, -2, 0, 2, 4, 8, 9, 10], and 
   Cm=[.339, .219, .181, .144, .109, .075, .043, .013, -.048, -.066, -.084]. 
   CL=  [-1.2860, -0.6159,-0.4200,-0.2235,-0.0313,0.1609, 0.3557,0.5533,0.9495,1.0380,1.1170].

function [CL0,CLa,Cm0,Cma]= Dixon_Jordan_M1(de_trim)
load('assignment1.mat')
figure   
  plot(alpha,Cm) %y=-.016*x+.08 (from best fit tool)
   grid on
   title('Cm v AoA')
   xlabel('AoA ')
   ylabel('Cm')
   text(1.25,.350,'y=-.016*x+.08')
figure
   plot(alpha,CL) %y=.094*x+.17 (from best fit tool)       
   grid on
   title('CL v AoA')
   xlabel('AoA ')
   ylabel('CL')
   text(-11.5,.35,'y=.094*x+.17')

   %figure
      %plot(CL,Cm)
      %grid on
      %title('Cm v CL')
      %xlabel('CL ')
      %ylabel('Cm')
   %from a linear best-fit of the above figure (Cm v Cl): y=-.17*x+.11;
    Cmcl=-.17;
    SMs=-Cmcl;
    yint=.11;
    CLtrim=-yint/Cmcl; %CLtrim=.647
    X_NP=(SMs*cbar)+X_CG; %corrected X_NP; returns a value of 2.669
    %Values from CL v AoA and Cm v AoA (linear best fit tool)
     CLa=.094;
     CL0=.17;
     Cma=-.016;
     Cm0=.08;
    %values derived from user input, de_trim
    CLinput= ((de_trim)+(yint/-.03))*(-.03/SMs);
    %trim angle of attack
    Alpha_Trim=(CLinput-CL0)/CLa; %returns value of 5.07 with CLtrim=.647
    %%%%%Stall Condition%%%%%
    CLmax=1.5;
    de_trimM=(yint/.03)+(SMs/-.03)*CLmax;
    fprintf('The slope of CL v AoA, CLa is %.3f , the intercept CL0 is %.3f.\nThe slope of Cm v AoA, Cma %.3f , the intercept Cm0 is %.3f.\n',CLa,CL0,Cma,Cm0)  
   fprintf('\nThe neutral point, X_NP, was found to be %.3f ft.\n',X_NP)
   fprintf('\nWhen considering a non-linear best curve for Cm v AoA, polynomial n=10, Cm0 becomes %.3f.\n',yint)
     if de_trim == de_trimM;
           fprintf('\nFlying at CLmax, at sea level Vstall is 95.17 ft/s')
          elseif de_trim < de_trimM;
        fprintf('\nde_trim does not conflict with stall conditons, the resulting velocity is %.f ft/s\n',Vinput)
          else de_trim > de_trimM;
        fprintf('\nThe value %.2f for de_trim, requires coefficeint of lift greater than max.\nThe max value for de_trim was calculated to be %f\n',de_trim, de_trimM)
    end

Answer 1

[Walter Roberson]

filevals = load('assignment1.mat');
alpha = filevals.alpha;
Cm = filevals.Cm;
CL = filevals.CL;

The JIT changed a couple of releases ago.

Previously if you load() without any outputs and so "poof" a variable into existence and the variable had the same name as a function, then the variable was used -- except in the case where there is an end statement matching the function statement, in which case it was defined that the name-as-a-function would win out.

Now (R2015b? R2016a?) if you poof a variable into existence like that inside a function, and the variable name matches the name of a function, then the name-as-a-function wins out.

The long-term cure is to not poof variables into existence: recode to have load() output a struct and pull the values out of the struct.

The short-term cure would be to assign something to the variable before using the load() to poof the variable into existence:

alpha = 0;
load('assignment1.mat');

would result in alpha being treated as a variable in the code.