The lengths of the links comply with the conditions: E͞C=G͞F, E͞F=C͞G, D͞E=D͞C=a, C͞B=B͞G=E͞H=H͞F=b and A͞B=A͞D. The mechanism always satisfies the inversion condition DB×DH=b²-a²=k², where k is the inversion constant. When crank 1 turns about fixed axis A, point H of link 2 travels along straight line q-q which is perpendicular to base AD and is at the distance h from point D: h=k²/(2A͞B). Figure EFGC forms a crossed-crank linkage.
$706$LW,GI$