This code is for a four-bar linkage, but I don't know how to change it to kens linkage.
Point A and D are fixed, AD=Lo, crank is L1, coupler is actually L2=18.75, rocker is L3=9.375
Write a matlab code for a kens linkage that will draw a 15 cm straight line.
clear % clears all variables from the workspace
clc % clears the command window and homes the cursor
close all % closes all the open figure windows
L0 = 7.5;
L1 = 3.75;
L2 = 9.375;
L3 = 9.375;
input_theta = 0:2:360; % input angle vector
% preallocate 1 3-D matrix for the atlas. Plot atlas of points with N
% spans of the coupler. The third index of the matrix is the point.
% Each column is a pair of x-y coordinate of the point
% And each column corresponds to one input angle
atlas = zeros(2,length(input_theta));
% Position of joint A (origin)
xA = 0; yA = 0;
% Position of joint D
xD = L0; yD = 0;
figure;
for ii = 1: length(input_theta)
theta1 = input_theta(ii)*pi/180;% convert angle to radians
% Position of joint B - position of the driver link
xB = xA + L1*cos(theta1);
yB = yA + L1*sin(theta1);
% Distance formula: BC=constant
eqnC2 = '(xB - xCsol )^2 + ( yB - yCsol )^2 = L2^2';
eqnC3 = '(xD - xCsol )^2 + ( yD - yCsol )^2 = L3^2';
% Solve the above equation
solC = solve(eqnC2, eqnC3,'xCsol','yCsol');
% Two solutions for xC and yC.
% Calculate steps of coordinates between B and C.
xC_sol = real(eval(solC.xCsol(1)));
yC_sol = real(eval(solC.yCsol(1)));
x_step = (xC_sol - xB)/2;
y_step = (yC_sol - yB)/2;
atlas(:,ii) = [xB+x_step;yB+y_step];
plot (xA, yA, 'r.',...
[xA, xB],[yA,yB],'g.-',...
[xB, xC_sol],[yB, yC_sol],'r.-',...
[xD, xC_sol],[yD, yC_sol],'b.-',...
'LineWidth', 4, 'MarkerSize', 25);
hold on
plot(atlas(1,1:ii),atlas(2,1:ii));
axis ([-10, 45,-10,35]),axis square;
grid
title(['\theta_1 = ', num2str(input_theta(ii)),' (deg)']);
hold off
pause (0.001)
end
1 answer