712 PROFESSOR FLEEMING JENKIN’S APPLICATION OF GRAPHIC METHODS 
ale 3 ; a’ : 
Differentiating again, and remembering that - = 0, we obtain finally 
LO _ _ rsina ?—r’) (da\? 
d@ (=r sin%a)s @) F 
Again 
AB= AN + BN 
x=rcosa+lcos@ 
ae —rsin a —/sin 6% 
Clam bu Pay ox 60 : 
and qo? cosa (G)’ Zcos 6 (i) Zsin 8 = 
Sere dé ae é : d 3 
Substituting for 7, and Gz their values as determined above, and putting 7 sin a 
for / sin 0, and ,//? — 7? sin 2a for J cos 6, we obtain finally 
Ce da\ ° 7? cos 2a + 7° sin ta 
dt ha) | esas (2? — 7* sin a)? \ 
The forces required to accelerate the piston and the connecting-rod may 
now be calculated as follows :— 
For the piston.—If M be the mass of the piston and piston-rod in Ibs., the 
force in lbs. is 

M dz 
g We 
When this quantity is negative, the force acts towards the centre of the crank 
shaft. 
For the connecting-rod.—The motion may be looked at as a translation of ? 
the whole rod in the direction of motion of the piston, combined with a rotation” 
of the rod about the crosshead. Hence the force producing acceleration is the 
resultant of three components :—F,, the force required for the linear accelera- 
tion in the direction of motion of the piston ; F,, the force required for rotation — 
about the crosshead at the angular velocity which the rod has at the instant 
under consideration. This acts towards the centre of rotation, and is equal and 
opposite to the so-called “centrifugal force ;” and, lastly, F;, the force required 
to give the rod the angular acceleration which it has at the given instant. 

Fig. 56. 
Let M’ be the mass of the connecting-rod in lbs., and G its centre of mass, 
(Note. —In figs. 55 and 56 the engine has been represented as seen from behind, if the engine in 
the previous figures be considered as viewed from the front.) P 
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