The combined motion of translation and rotation from AB to A’B’ may be assumed to be a motion of entirely rotation about a certain point 0 which is called the instantaneous centre of rotation.

The point 0 can be found out by following procedure.

Draw perpendicular bisectors of chords AA’ and BB’. This is PO and QO. This bisectors meet at 0 which is the instantaneous centre (of zero velocity).

Fig. 4.4 (a) shows different parts of a reciprocating engine. Point A executes linear motion along line AC and point B moves in a circle with ro as angular velocity.

Let w = angular velocity of link AB about 0 .

VA = Linear velocity of point A

= wr A = wAO

V s = Linear velocity of point B

= wr8 = w BO

VA / VB =AO / BO

V_{A} is at right angles to AO at A.

V_{B} is at right angles to BO at B.

Thus, if V_{A} and V_{B} are known, then the instantaneous centre of AB is obtained by drawing perpendiculars to the directions of the velocities at A and B. The point, where these two perpendiculars meet, is the instantaneous centre.