Hart’s Mechanism:
From Fig,
FC = DE ; CD = EF ;
The point O, A, and B are divide the links FC, CD, and EF in the same ratio.
From similar triangle CFE and OFB,
CE/OF
OB FC
From Similar triangle FCD and OCA
OB = CE X OF x FC
FD x OA
FC x OC x OA = FD X OC x FC
OA X OB = CE X OF X FD X OC FC FC = FD X CE X
OC X OF X FC^2
Since the length OC, OF, and FC are fixed, therefore
OA X OB = FD X CE X constant
Now from point E draw EM parallel to CF and EN perpendicular to FD.
Therefore
FD X CE = FD X FM
= (FN + ND) (FN - MN)
= FN2 – ND2 = (FE2 – NE2) – (ED2 – NE2)
= FE2 – ED2
So , Finally
OA X OB = Constant
