Asked by Kevin Krause
on 1 Nov 2015

One of the main problems is that there will never be a sign change on the interval so it'll always execute the first part of the if statement. I commented out the err>errtol part for testing. I don't know what i'm doing wrong for calculating using the bisection method. I have to use while loops.

% A scale is made with two springs, and when an object is attached to the

% two springs, the springs stretch and the ring at the end of the two

% springs is displaced downwards a distance x.

%

% The relationship between the weight of the object and the distance x is:

% W=(2k/l)(l-lo)(b+x)

% The initial spring length

% lo = sqrt(a^2+b^2)

% The stretched spring length

% l = sqrt(a^2+(b+x)^2)

% for the given scale a=8.0 in. b=5.0 m. and the spring constant k=16.0 lb/in.

% b=5.0 m.=196.85 in.

% Trying to find the deflection x.

%

% Have the user input lower and upper values of the starting

% interval (xl, xu), as well as the weight of the object

% Initialize the error tolerance (errtol)

% and the maximum number of iterations (maxiter)

%

% Check the interval to see if there's a sign change

%

% while (iterations < maxiter) and (err > errtol)

% Calculate midpoint of the interval xr=(xl+xu)/2

% Calculate error estimate (err)

% Evaluate the function at the midpoint f(xr)

% If the sign of f(xr) is equal to the sign of f(xl)

% if f(xr)*f(xu)>0

% then xr becomes the new xl.

% else xr becomes the new xu.

% end while loop

%

% If err <= errtol

% then solution found, output result

% else display ‘root not found within tolerance

W = input('\nWhat is the weight of the object attached to the ring in pounds (lb)? ');

xl = input('\nWhat is the lower initial bracket estimate in inches? '); %always 0

xu = input('\nWhat is the upper initial bracket estimate in inches? '); %always 10

a = 8.0;

b = 5.0 ;

lo = sqrt(a^2+b^2);

k = 16.0;

errtol = 0.005;

xr = (xl+xu)/2;

maxiter = 10;

iterations = 0;

f = @(x) (2*k/(sqrt(a^2+(b+x)^2)))*((sqrt(a^2+(b+x)^2))-lo)*(b+x);

err = abs((xu-xl)/xu);

xrold = abs((xu-xl)/xu);

while (iterations<maxiter) %|| (err>errtol)

iterations=iterations+1;

xr=(xl+xu)/2;

err = abs((xr-xrold)/xr)*100;

if f(xr)*f(xl)>0

xu = xr;

else

xl = xr;

end

xrold = xr;

end

Answer by Mischa Kim
on 6 Nov 2017

Edited by Mischa Kim
on 6 Nov 2017

Kevin, I see two problems with your algorithm/function f(x):

if f(xr)*f(xl)>0

needs to be

if f(xr)*f(xl)<0

and the function f(x) does not (I believe) have any roots for positive x and for the set of parameters you chose. In other words, the starting xl needs to be negative for the algorithm to converge to a root. According to your comments, xl is always 0.

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