Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: High

The sliding between a pair of teeth in contact at Q occurs along the common tangent T T to the tooth curve as shown in fig. The velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact.

The velocity of point Q , considered as a point on wheel 1, along the common tangent T T is respectively by EC. From similar triangles QEC and $O_{1}$MQ,

EC/MQ =ν/$O_{1}$Q =$\omega_{1}$ OR EC =$\omega_{1}$ .MQ

Similarly the velocity of point Q , considered as a point on wheel 2, along the common tangent T T is respectively by ED. From similar triangles QCD and $O_{2}$NQ,

ED/QN =$ν_{2}$/$O_{2}$Q =$\omega_{2}$ OR ED =$\omega_{2}$.QN

Let $v_{S}$ = velocity of sliding at Q.

$v_{S}$ = ED-EC = $\omega_{2}$.QN - $\omega_{1}$.MQ

= $\omega_{2}$ (QP+PN) - $\omega_{1}$(MP-QP)

($\omega_{1}+ \omega_{2}$)QP+ $\omega_{2}$.PN - $\omega_{1}$MP …(i)

Since $\omega_{1}$/ $\omega_{2}$ =$O_{2}$P/$O_{1}$P =PN/MP OR $\omega_{1}$.MP = $\omega_{2}$.PN therefore equation (i) becomes

$v_{S}$ =($\omega_{1}+ \omega_{2}$)QP …(ii)

we see from equation (ii), that the velocity of sliding is proportional to the distance of the point of contact from the pitch point.