Loop Closure Equations for Four-Bar Linkages ref : pp. 175-184 • We will represent each vector by a length ri and an angle θi • All angles are measured counterclockwise from a line that remains parallel to the fixed x axis attached to the reference frame
r p = r2 + r3 = r1 + r4 or
r2 + r3 − r4 − r1 = 0 r2 cos θ 2 + r3 cos θ 3 − r4 cos θ 4 − r1 cos θ1 = 0 r2 sin θ 2 + r3 sin θ 3 − r4 sin θ 4 − r1 sin θ1 = 0
2005/4/13
機構學 (C. F. Chang)
National Kaohsiung University of Applied Sciences, Department of Mechanical Engineering
2
1
r2 cos θ 2 + r3 cos θ 3 − r4 cos θ 4 − r1 cos θ1 = 0 r2 sin θ 2 + r3 sin θ 3 − r4 sin θ 4 − r1 sin θ1 = 0
Position Analysis
r3 cos θ 3 = − r2 cos θ 2 + r4 cos θ 4 + r1 cos θ1
(a)
r3 sin θ 3 = − r2 sin θ 2 + r4 sin θ 4 + r1 sin θ1
(b)
(a)2+(b)2!
r32 = r12 + r22 + r42 + 2r1r4 (cos θ1 cos θ 4 + sin θ1 sin θ 4 ) − 2r1r2 (cos θ1 cos θ 2 + sin θ1 sin θ 2 ) + 2r2 r4 (cos θ 2 cos θ 4 + sin θ 2 sin θ 4 )
(c)
Combining the coefficient of cosθ4 and sin θ4 yields A cos θ 4 + B sin θ 4 + C = 0
(d)
where
function of θ2 (ri and θ1 are constants)
A = 2r1r4 cos θ1 − 2r2 r4 cos θ 2 B = 2r1r4 sin θ1 − 2r2 r4 sin θ 2 C = r12 + r22 + r42 − r32 − 2r1r2 (cos θ1 cos θ 2 + sin θ1 sin θ 2 ) 2005/4/13
機構學 (C. F. Chang)
3
Position Analysis (cont) A cos θ 4 + B sin θ 4 + C = 0
θ θ 2 tan 4 1 − tan 2 4 2 2 , cos θ = sin θ 4 = 4 θ θ 1 + tan 2 4 1 + tan 2 4 2 2
θ θ (C − A) tan 2 4 + 2 B tan 4 + ( A + C ) = 0 2 2
2 θ − 2 B ± 4 B − 4(C − A)(C + A) tan 4 = 2(C − A) 2
=
− B ± B 2 − C 2 + A2 (C − A)
“± ±” corresponding to the two possible positions of the P for a given value of θ2 2005/4/13
機構學 (C. F. Chang)
National Kaohsiung University of Applied Sciences, Department of Mechanical Engineering
4
2
Position Analysis (cont)
2 2 2 θ − B ± B − C + A tan 4 = (C − A) 2