Elbow problem matlab

Elbow problem matlab - clc clear h = 6.3048 w =...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
clc; clear; h = 6 * .3048; % Person's height in meters w = 165*4.448221628; % Person's weight in Newtons Lf = .19 * h; % Length of forearm in meters y_bic = .22*h; % Origin length of biceps brachii in meters y_bra = .17*h; % Origin length of Brachialis in meters y_brd = .02*h; % Origin length of Brachioradialis in meters x_bic = .05*h; % Insertion length of biceps brachii in meters x_bra = .05*h; % Insertion length of Brachialis in meters x_brd = .11*h; % Insertion length of Brachioradialis in meters A_bic = 4.6; % Given Area of biceps brachii in meters^2 A_bra = 7.0; % Given Area of Brachialis in meters^2 A_brd = 1.5; % Given Area of Brachioradialis in meters^2 angle_min = 10*pi/180; % Minimum and Maximum angles between humerous and ulna in radians angle_max = 170*pi/180; angle = angle_min:.01:angle_max; % Vector assignment of angles between minimum and maximum in radians and degrees degrees = angle .* 180 ./ pi; F_forearm = .03*w; % Force from weight of forearm in Newtons F_weight = 20*9.81; % Force from the weight in the hand in Newtons l_bic = sqrt(y_bic^2+x_bic^2-2*y_bic*x_bic*cos(angle)); % Finding the length of biceps brachii using Law of Cosines l_bra = sqrt(y_bra^2+x_bra^2-2*y_bra*x_bra*cos(angle)); % Finding the length of Brachialis using Law of Cosines l_brd = sqrt(y_brd^2+x_brd^2-2*y_brd*x_brd*cos(angle)); % Finding the length of Brachioradialis using Law of Cosines angle_bic = acos((x_bic.^2-y_bic^2-l_bic.^2)./(2.*y_bic.*l_bic)); % Finding the angle between the humerous and the biceps brachii angle_bra = acos((x_bra.^2-y_bra^2-l_bra.^2)./(2.*y_bra.*l_bra)); % Finding the angle between the humerous and the Brachialis angle_brd = acos((x_brd.^2-y_brd^2-l_brd.^2)./(2.*y_brd.*l_brd)); % Finding the angle between the humerous and the Brachioradialis d_bic = y_bic.*sin(angle_bic); % Finding the perpendicular disance from the joint to the biceps brachii in Meters d_bra = y_bra.*sin(angle_bra); % Finding the perpendicular disance from the joint to the Brachialis in Meters
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
d_brd = y_brd.*sin(angle_brd); % Finding the perpendicular disance from the joint to the Brachioradialis in Meters Fm_bic = [Lf*sin(angle).*(.5*F_forearm+F_weight)]./[d_bic]; % Magnitude of the Force in the biceps brachii in Newtons Fm_bra = [Lf*sin(angle).*(.5*F_forearm+F_weight)]./[d_bra]; % Magnitude of the Force in the Brachialis in Newtons Fm_brd = [Lf*sin(angle).*(.5*F_forearm+F_weight)]./[d_brd]; % Magnitude of the Force in the Brachioradialis in Newtons Fx_bic = Fm_bic .* -(x_bic.*sin(angle_bic))./l_bic; % Force in the x- direction of the biceps brachii in Newtons Fx_bra = Fm_bra .* -(x_bra.*sin(angle_bra))./l_bra; % Force in the x- direction of the Brachialis in Newtons Fx_brd = Fm_brd .* -(x_brd.*sin(angle_brd))./l_brd; % Force in the x- direction of the Brachioradialis in Newtons Fy_bic = Fm_bic .* (y_bic-x_bic.*cos(angle_bic)./l_bic);
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 15

Elbow problem matlab - clc clear h = 6.3048 w =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online