error: too many output arguments

Hi! I'm new to MATLAB, so I'd appreciate it if someone could help me. Here is my code for the position analysis of a slider-crank mechanism. Whenever I try to run the code, the command window says: "Error using UnitVector: Too many output arguments. Error in SliderCrank2 (line 29): [e1,n1] = UnitVector(0);"

% SliderCrank_Position_Analysis.m 
% performs a position analysis on the slider-crank linkage and 
% plots the piston position as a function of crank angle 
% by Karolina Kolodziej, February 20, 2019
 
clear, clc;

a = 0.10; % crank length (m)
b = 0.26; % connecting rod length (m)
c = 0.0; % vertical slider offset

% Ground pins

x0 = [0;0]; % ground pin at A (origin)

% solving for conn. rod angle (theta 3) and piston pos. (d)

N = 361; % # of times to perform calc.
[xB,xC] = deal(zeros(2,N)); % allocate space for pins B and C
[theta2, theta3, d] = deal(zeros(1,N)); %".." for link angles

% Main loop
for i = 1:N
    theta2(i) = (i-1)*(2*pi)/(N-1);
    theta3(i) = asin((c-a*sin(theta2(i)))/b);
    d(i) = a*cos(theta2(i)) + b*cos(theta3(i));
    
    % calculate unit vectors
[e1,n1] = UnitVector(0);  
[e2,n2] = UnitVector(theta2(i));  
[e3,n3] = UnitVector(theta3(i));
    
    % solve for position of point B
    xB(:,i) = FindPos(x0,a,e2);
    xC(:,i) = FindPos(xB(:,i),b,e3);
end

plot(theta2*180/pi,d*100,'o','Color',[153/255 153/255 153/255]) 
hold on 
plot(theta2*180/pi,xC(1,:)*100,'Color',[0 110/255 199/255]) 
title('Piston Position versus Crank Angle for Slider-Crank') 
xlabel('Crank angle (degrees)') 
ylabel('Position (cm)') 
legend('d','xC') 
grid on

[Stephen23]

And what exactly is UnitVector ? How is it defined?

[Karolina Kolodziej]

I think the foundation of it should look something like this.. I am not sure if I'm supposed to write 3 separate files for each variable e1, e2, e3, or if its all the same one.

function [e,n] = UnitVector(theta)
 
e = [ cos(theta); sin(theta)]; n = [-sin(theta); cos(theta)];

Answer 1

[Karolina Kolodziej]

I figured it out. There was a missing function to find position:

Main Code:

% SliderCrank_Position_Analysis.m 
% performs a position analysis on the slider-crank linkage and 
% plots the piston position as a function of crank angle 
% by Karolina Kolodziej, February 20, 2019
 
clear, clc;

a = 0.10; % crank length (m)
b = 0.26; % connecting rod length (m)
c = 0.0; % vertical slider offset

% Ground pins

x0 = [0;0]; % ground pin at A (origin)

% solving for conn. rod angle (theta 3) and piston pos. (d)

N = 361; % # of times to perform calc.
[xB,xC] = deal(zeros(2,N)); % allocate space for pins B and C
[theta2, theta3, d] = deal(zeros(1,N)); %".." for link angles

% Main loop
for i = 1:N
    theta2(i) = (i-1)*(2*pi)/(N-1);
    theta3(i) = asin((c-a*sin(theta2(i)))/b);
    d(i) = a*cos(theta2(i)) + b*cos(theta3(i));
        % calculate unit vectors
[e1,n1] = UnitVector(0);  
[e2,n2] = UnitVector(theta2(i));  
[e3,n3] = UnitVector(theta3(i));
     
    % solve for position of point B
    xB(:,i) = FindPos(x0,a,e2);
    xC(:,i) = FindPos(xB(:,i),b,e3);
end

plot(theta2*180/pi,d*100,'Color',[153/255 153/255 153/255]) 
hold on 
plot(theta2*180/pi,xC(1,:)*100,'linewidth',2,'Color',[0 110/255 199/255]) 
title('Piston Position versus Crank Angle for Slider-Crank') 
xlabel('Crank angle (degrees)') 
ylabel('Position (cm)')  
grid on

UnitVector:

function [e,n] = UnitVector(theta)
e=[cos(theta);sin(theta)];
n=[-sin(theta);sin(theta)];
end 

FindPos:

function x= FindPos(x0,L,e)
x=x0+L*e;