MATLAB: Inverse tangent inaccuracy

Question

I wrote a program to calculate the open and crossed angle values of a four bar mechanism and everything seems to work fine until I get to the final calculation in which I try to take the inverse tangent of the quadratic equation and the resulting answers are incorrect. Here is my code.

L1=6;

L2=2;

L3=7;

L4=9;

theta2=30*(pi/180);

a=L2;

b=L3;

c=L4;

d=L1;

K1=d/a;

K2=d/c;

K3=(a^2-b^2+c^2+d^2)/(2*a*c);

K4=d/b;

K5=(c^2-d^2-a^2-b^2)/(2*a*b);

A=cos(theta2)-K1-(K2*cos(theta2))+K3;

B=-2*sin(theta2);

C=K1-((K2+1)*cos(theta2))+K3;

D=cos(theta2)-K1+(K4*cos(theta2))+K5;

E=B;

F=K1+((K4-1)*cos(theta2))+K5;

theta4a=2*(atand(-B+sqrt((B^2)-(4*A*C))/(2*A)))

theta4b=2*(atand(-B-sqrt((B^2)-(4*A*C))/(2*A)))

theta3a=2*(atand(-E+sqrt((E^2)-(4*D*F))/(2*D)))

theta3b=2*(atand(-E-sqrt((E^2)-(4*D*F))/(2*D)))

The values I get for answers are:

theta4a = -106.6958

theta4b = 146.7016

theta3a = -31.0711

theta3b = 132.5987

The resulting correct answers should be:

theta4a = -143.67

theta4b = 117.32

theta3a = -115.21

theta3b = 88.84

I can't seem to find the problem any help would be appreciated. Thanx!

Answer 1

Hi Robert,

All you need do is replace the expression for theta4a by

theta4a = 2*atand((-B +sqrt((B^2)-(4*A*C)))/(2*A))

and similarly for the other three. You forgot that 2A divides -B as well as the sqrt term.