When theta2 makes a complete cycle, I get error message.

clear all
clc
L2 = 10;
L3 = 20;
theta2 = 0:pi/18:2*pi;
% theta2 = pi/12;
omega2 = 1.6;
theta3 = (2 * pi) + asin((-L2/L3)* sin(theta2))
L1 = L2 * cos(theta2) + L3 * cos(theta3);
C = [L2*sin(theta2)*omega2;-L2*cos(theta2)*omega2]
D = [-L3*sin(theta3) -1 ;L3*cos(theta3) 0]
B = inv(D) * C
plot(theta2,theta3)
grid on
xlabel('theta2(degrees)')
ylabel('theta3(degrees)')
title('crank angle(theta2) vs link 3 angle(theta3)')
figure
plot(theta2,L1)
grid on
xlabel('theta2(degrees)')
ylabel('L1(m)')
title('crank angle(theta2) vs link 1 angle(L1)')

3 件のコメント

Ben Rancici
Ben Rancici 2017 年 2 月 3 日
編集済み: Ben Rancici 2017 年 2 月 3 日
What is the real question here? Are you asking for the community to help you debugging your script? If so, why don't you share with us the error message you get?
KSSV
KSSV 2017 年 2 月 3 日
Your D is not a square matrix, that's why error popped.
Ben Rancici
Ben Rancici 2017 年 2 月 3 日
Indeed, if theta2 is a vector, then D is not square and cannot be inverted.

サインインしてコメントする。

 採用された回答

KSSV
KSSV 2017 年 2 月 3 日

0 投票

clc; clear all ;
clear all
clc
L2 = 10;
L3 = 20;
theta2 = 0:pi/18:2*pi;
% theta2 = pi/12;
omega2 = 1.6;
theta3 = (2 * pi) + asin((-L2/L3)* sin(theta2)) ;
L1 = zeros(1,length(theta3)) ;
for i = 1:length(theta3)
L1(i) = L2 * cos(theta2(i)) + L3 * cos(theta3(i));
C = [L2*sin(theta2(i))*omega2;-L2*cos(theta2(i))*omega2] ;
D = [-L3*sin(theta3(i)) -1 ;L3*cos(theta3(i)) 0] ;
B = D\C ;
end
plot(theta2,theta3) ;
grid on
xlabel('theta2(degrees)')
ylabel('theta3(degrees)')
title('crank angle(theta2) vs link 3 angle(theta3)')
figure
plot(theta2,L1)
grid on
xlabel('theta2(degrees)')
ylabel('L1(m)')
title('crank angle(theta2) vs link 1 angle(L1)')

1 件のコメント

oroskic007
oroskic007 2017 年 2 月 3 日
Thank you so much. Exactly my expectation

サインインしてコメントする。

その他の回答 (1 件)

Walter Roberson
Walter Roberson 2017 年 2 月 3 日

0 投票

When theta2 has a complete cycle from 0 to 2*Pi, then the sin and cos entries go from [0 1] at the beginning, through a bunch of different possibilities and return to [0 1] at the end. Those lead to the entries in D being the same for the first and last element. Duplicate rows in a matrix to be inverted guarantee that the matrix is singular.
You need to end your theta2 just before it completes a cycle to 2*Pi. For example,
theta2 = 0:pi/18:2*pi;
theta2(end) = [];

カテゴリ

タグ

質問済み:

2017 年 2 月 2 日

コメント済み:

2017 年 2 月 3 日

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by