152.] PLANE MOTION. 



v OR sin (d + 6) 

 ' = = - 



If, as is usually the case, the connecting rod is much longer 

 than the crank arm, </> will be a small angle, and we may substi- 

 tute sin (f> for tan 0. But from the triangle OPQ we have 



Hence v = u( sin0 + cos0- sin0) = u( sin + sin 20V 

 \ m J \ 2m J 



150. The motion of the piston head being rectilinear, we find 

 its acceleration j by differentiating the expression for v found 

 in the preceding article with respect to t : 



. dv f . n I *\du f . i A 



; = -T = sm0H sm20)- 7 - + | cos0H cos 20 



* dt \ 2m ) dt \ m J 



J0 

 dt 



or, since = &> = 



/=( sin 0H sin 2 ) - + ( cos 0H cos 2 ] > 



\ 2m J dt \ m ) a 



where =o if the crank motion can be regarded as uniform. 



151. If the connecting rod were of infinite length so as to 

 make PQ (in Fig. 35) parallel to A^A^ the position of the 

 piston corresponding to the position Q of the crank pin would 

 l>e represented by the projection M of Q on A^A^ ; that is, NM 

 would be = o. This length NM is therefore called the devia- 

 tion due to the obliquity of the connecting rod. 



With NM=o the expression for the acceleration (Art. 150) 

 would reduce to dv/dt(t^la) cos0, representing a simple har- 

 monic motion (see Art. 179). 



152. The slide valve of a steam engine is generally worked 

 by an eccentric whose radius is set on the shaft at such an 



