Subject: Kinematics of Machinery

Topic: Special Mechanism

Difficulty: High

This mechanism requires only six links as compared with the eight links required by the peaucellier mechanism.it consist of fixed link OO1 and other straight link O1A, FC, CD, DE and EF are connected by turning pairs at their point of intersection, as shown in fig. 6. The FC and DE are equal in length and the lengths of the links CD and EF are also equal. The points O,A and B divide the links FC,CD and EF in the same ratio. A little consideration will show that BOCE is a trapezium and OA and OB are respectively parallel to FD and CE.

Hence OAB is a straight line. It may be proved now that the product $OA\times OB$ is constant.

From similar triangles CFE and OFB,

$\frac{CE}{FC}=\frac{OB}{OF}\space or\space OB=\frac{CE\times OF}{FC}$...........................(i)

and from similar triangles FCD and OCA

$\frac{FD}{FC}=\frac{OA}{OC}\space or\space OA=\frac{FD\times OC}{FC}$...........................(ii)

Multiplying equation (i) and (ii), we have

$OA\times OB=\frac{FD\times OC}{FC}\times\frac{CE\times OF}{FC}=FD\times CE\times\frac{CE\times OF}{FC^{2}}$

Since the lengths of OC, OF and FC are fixed, therefore

$OA\times OB=FD\times CE\times Constant$....................................(iii) (substituting $\frac{OC\times OF}{FC^{2}}=Constant$)

Now from point E, draw EM parallel to CF and EN perpendicular to FD. Therefore

$FD\times CE=FD\times FM$...............(CE=FM)

$=(FN+ND)(FN-MN)=FN^{2}-ND^{2}$................(MN=ND)

$=(FE^{2}-NE^{2})-(ED^{2}-NE^{2})$.......................(From right angle triangles FEN and EDN)

$=FE^{2}-ED^{2}=Constant$..................................(length FE and ED are fixed)...................(iv)

From equation (iii) and (iv),

$OA\times OB=Constant$

It therefore follows that if the mechanism is pivoted about O as a fixed point and the point A is constrained to move on a circle with centre O1, Then the point B will trace a straight line perpendicular to the diameter OP produced.