function [x] = example53(xinitial)

% example53.m: 3-position synthesis for 4-bar linkage

% for the specifications in problem 5-3 in the textbook

% Input:

% xinitial (optional) initial gues, a 3N+10 x 1 matrix

%

% Output:

% x solution, a 3N+10 x 1 matrix

% data from steps 2 and 3 on page 256

% Px,Py,Theta(deg) for each of the N positions

clear global

global specs

specs.PQ = [ ...

0 0 0 ; ... % first position is defined as being at origin and zero angle

-0.244 0.013 -11.34 ; ... % second position

-0.542 0.029 -22.19 ]; % third position

% specified pivots from step 4 on page 257

specs.O2 = [-1.712 0.033]; % specified values for O2x and O2y

specs.O4 = [0.288 0.033]; % specified values for O4x and O4y

% convert column 3 of specs.PQ into radians

specs.PQ(:,3) = specs.PQ(:,3) * pi/180;

% determine how many positions there are (i.e. rows of specs matrix)

N = size(specs.PQ,1);

if (N ~= 3)

error('This code was written for 3-position synthesis only.');

end

% the unknowns are:

% x(1..2) global position of fixed pivot O2

% x(3..4) global position of fixed pivot O4

% x(5..7) link lengths a,b,c

% x(8..9) local coordinates of point P on the coupler

% x(10) angle between PQ and AB

% x(11...) angles theta2, theta3, theta4 in all N positions (3*N unknowns)

Nunknowns = 10 + 3*N;

% if no initial guess was provided, use random numbers

if nargin

xinitial = rand(Nunknowns,1);

end

% we may need many attempts to find a solution, allow 100 attempts

for i = 1:100

% solve the unknowns x (design parameters, and the N positions)

% from a set of simultaneous equations: func(x) = 0

% NRsolve does the same as Matlab's fsolve but the graphics

% is faster if you use NRsolve

[x, info] = NRsolve(@func, xinitial, 40); % allow 40 iterations in each attempt

if (info == 0)

break

end

disp('Hit ENTER to start a new attempt, or CTRL-C to quit');

pause

% make a new random initial guess

xinitial = rand(Nunknowns,1);

end

end

%============================================================================

function draw(x)

% draw the trial solution x

global specs

N = size(specs.PQ,1);

% extract the mechanism design parameters and print them

O2x = x(1);

O2y = x(2);

O4x = x(3);

O4y = x(4);

a = x(5);

b = x(6);

c = x(7);

Pxloc = x(8);

Pyloc = x(9);

PQangle = x(10);

fprintf('Design parameters: ');

fprintf('O2: %8.4f %8.4f ', O2x,O2y);

fprintf('O4: %8.4f %8.4f ', O4x,O4y);

fprintf('Link a: %8.4f ', a);

fprintf('Link b: %8.4f ', b);

fprintf('Link c: %8.4f ', c);

fprintf('Pxloc: %8.4f ', Pxloc);

fprintf('Pyloc: %8.4f ', Pyloc);

fprintf('Angle between PQ and AB: %8.4f deg ', PQangle*180/pi);

% draw the mechanism in the N specified positions

figure(1);

clf;

positioncolors = colormap('lines');

for i = 1:N

% extract the positions of A,B (4 unknowns per position)

theta2 = x(10 + 3*(i-1) + 1);

theta3 = x(10 + 3*(i-1) + 2);

theta4 = x(10 + 3*(i-1) + 3);

% draw the ground link O2-O4 as a solid black line

plot([O2x O4x],[O2y O4y],'k','LineWidth', 3);

hold on

% draw the links O2-A-B-O4 in the color that belongs to position i

Ax = O2x + a*cos(theta2);

Ay = O2y + a*sin(theta2);

Bx = Ax + b*cos(theta3);

By = Ay + b*sin(theta3);

plot([O2x Ax Bx O4x], [O2y Ay By O4y], '-o', 'Color', positioncolors(i,:), 'LineWidth', 1.5);

% do coordinate transformation to calculate where the the coupler point P is

Px = Ax + Pxloc*cos(theta3) - Pyloc*sin(theta3);

Py = Ay + Pxloc*sin(theta3) + Pyloc*cos(theta3);

% draw the coupler APB as a triangle

plot([Ax Px Bx Ax], [Ay Py By Ay], '-o', 'Color', positioncolors(i,:), 'LineWidth', 1.5);

% add some text labels

text(Ax, Ay, ' A', 'FontSize',12);

text(Bx, By, ' B', 'FontSize',12);

text(Px, Py, ' P', 'FontSize',12);

axis('equal');

end

end

%============================================================================

function [f] = func(x,display)

% equations for the N-position synthesis problem, in the form f(x) = 0

% there are

% 3N+10 unknowns x

% 5N equations f(x) = 0 from the linkage and performance specs

% 3N+10-5N equations from free choices

global specs

N = size(specs.PQ,1);

% do some error checking on input x

Nunknowns = 3*N+10;

if (size(x,1) ~= Nunknowns) || (size(x,2) ~= 1)

error('func: x was not a matrix of size (Nunknowns x 1)');

end

% extract the linkage parameters from x

O2x = x(1);

O2y = x(2);

O4x = x(3);

O4y = x(4);

a = x(5);

b = x(6);

c = x(7);

d = sqrt((O4x-O2x)^2 + (O4y-O2y)^2); % length of ground link

Pxloc = x(8);

Pyloc = x(9);

PQangle = x(10);

% generate the 5 equations from each of the N positions

f = zeros(Nunknowns,1);

for i = 1:N

% extract the positions of points A,B in position i, from x

theta2 = x(10 + 3*(i-1) + 1);

theta3 = x(10 + 3*(i-1) + 2);

theta4 = x(10 + 3*(i-1) + 3);

% do coordinate transformation to calculate where the the coupler point P is

Ax = O2x + a*cos(theta2);

Ay = O2y + a*sin(theta2);

Px = Ax + Pxloc*cos(theta3) - Pyloc*sin(theta3);

Py = Ay + Pxloc*sin(theta3) + Pyloc*cos(theta3);

% 2 equations from the vector loop

f(5*(i-1) + 1) = O2x + a*cos(theta2) + b*cos(theta3) - c*cos(theta4) - O4x;

f(5*(i-1) + 2) = O2y + a*sin(theta2) + b*sin(theta3) - c*sin(theta4) - O4y;

% 2 equations that require point P to be at specified locations in position i

f(5*(i-1) + 3) = Px - specs.PQ(i,1); % specified Px is in row i, column 1 of specs.PQ matrix

f(5*(i-1) + 4) = Py - specs.PQ(i,2); % Py is in column 2

% 1 equation that requires line PQ to be at specified angle in position i

% (orientation of line PQ is orientation of line AB plus PQangle)

f(5*(i-1) + 5) = theta3 + PQangle - specs.PQ(i,3);

end

% now generate extra equations that represent the free choices

if (N == 2)

error('Code for 2-position synthesis is not available in this version.');

elseif (N == 3)

% for 3-position synthesis, there are 4 free choices

% here we require specified locations for the fixed pivots O2 and O4

f(5*N+1) = O2x - specs.O2(1);

f(5*N+2) = O2y - specs.O2(2);

f(5*N+3) = O4x - specs.O4(1);

f(5*N+4) = O4y - specs.O4(2);

else

error('Free choices were not programmed yet for the %d-position synthesis', N);

end

% at each iteration, print x and f side by side

if (nargin > 1)

[x f]

% also print the norm of f (which should go to zero during the solution process

fprintf('Norm of f: %f ', norm(f));

draw(x);

pause(0.001);

end

end

%=======================================================================

function [x, info] = NRsolve(fun, x0, maxiterations)

% solves f(x) = 0 using Newton-Raphson method

% input:

% fun...........pointer to function that computes f(x)

% x0............initial guess for solution

%

% output:

% x.............solution of f(x) = 0

if (nargin

maxiterations = 20; % maximum number of iterations

end

tolerance = 1e-8; % iterations will stop if norm(f)

x = x0; % start with x at initial guess

for i = 1:maxiterations

[f,A] = evaluate(fun, x); % evaluate function and Jacobian matrix

if norm(f)

disp('solved.');

info = 0; % info=0 indicates success

return; % return to main program

end

dx = A\f; % solve A*dx = f

x = x - dx; % move x towards the solution of f(x) = 0

% if norm f did not decrease, undo half of the step dx, and so on

nbacktrack = 0;

while ( norm(fun(x)) > norm(f) ) && nbacktrack

dx = dx/2;

x = x + dx;

nbacktrack = nbacktrack + 1;

end

end

% if we arrive here, we have done too many iterations

disp('a solution could not be found');

% info = 1 indicates that the solution was not found

info = 1;

end

%========================================================================

function [f,A] = evaluate(fun, x)

% evaluates a function fun(x) and estimates its Jacobian matrix at x

Nx = numel(x);

f = fun(x,1); % the "1" causes something to be printed within fun

h = 1e-5; % perturbation step size for derivative estimation

for i = 1:numel(x)

x(i) = x(i) + h; % perturb x(i)

newf = fun(x);

x(i) = x(i) - h; % restore original x(i)

A(:,i) = (newf - f) / h; % one column of Jacobian matrix

end

end

Homework 7.pdf 12 MICE 260 Spring 2018 PRESENT CLEARLY HOW YOU DEVELOPED THE SOLUTION TO THE PROBLEMS Each problem is worth up to S points. Points are given as follows: points: points: Work was complete, but not deariy presented or some errors in cakcuation work w complete and presented deernthe answer is correct 3 points: Some errors oromissians in methods or presentation 2 points: Majorerrors or omissions in methods or presentation 1 point Problm was understoed but incorrect approach was usod 1. Modify the Matlab code example53m to design a fourbar inkage that moves the box through the three positions shown. Use the fixed pivots that are shown in the disgram. snore the attachment points A and B lbut don't be surprsed your design process produkes similar attachment points). 2. Modify the Matlab code examples3.m to design a fourbar Inkage that moves the box through the three posltions shown. All maving links must have lengths that are mult ples o 10mm. 3. Modify the Matlab code exarmple53opt m to desnaourbar lirkage that moves the box through the three positions shown, Make an objective function x) to get the fixed piwots as close as passible to the IMPORTANT: For problems 1-3, your sclution should include: lal the design parameters of the linkage, (b) the Figure produced by Matlab, (c) the lines of Matlab coce that you changed (and no other codel), and (d) do you think there are ary cireuit, branch, ar ordor (CRO) detects (based on visual inspection only)? contiued on next page) Page 1 of 2