We have already discussed that in a cross belt drive, both the pulleys rotate in ** opposite** directions as shown in Fig.

Let *r*1 and *r*2 = Radii of the larger and smaller pulleys,

*x *= Distance between the centres of two pulleys (*i.e. O*1* O*2), and* L *= Total length of the belt.

Let the belt leaves the larger pulley at *E* and *G* and the smaller pulley at *F* and *H*, as shown in Fig. Through *O*2, draw *O*2*M* parallel to *FE*.

From the geometry of the figure, we find that *O*2*M* will be perpendicular to *O*1*E*. Let the angle *MO*2 *O*1 = α radians

We know that the length of the belt,

*L *= Arc* GJE *+* EF *+ Arc* FKH *+* HG*

= 2 (Arc *JE* + *EF* + Arc *FK*)

It may be noted that the above expression is a function of (*r*1 + *r*2). It is thus obvious that if sum of the radii of the two pulleys be constant, then length of the belt required will also remain constant, provided the distance between centres of the pulleys remain unchanged.

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