TO THE DETERMINATION OF THE EFFICIENCY OF MACHINERY. 33 
and crank-shaft, acting directly through the centre of the main bearing ; L, 
is the resultant of the weight of the connecting rod, compounded with the 
equilibrant of the force required to give the rod the acceleration (in respect of 
rotation and translation) which it actually has at the given instant. L, is the 
resultant of the weight of the piston compounded with the resistance to acceler- 
ation. The loads shown in the figure have been calculated for an actual small 
direct acting engine, making 1 turn per second. To draw the dynamic frame 
without friction we proceed as follows :—L, is represented by two components /, 
and /’,, acting at the joints ad and bc, L, is wholly borne by the one joint, and 
the actual point of its application is a matter of indifference ; L, is wholly borne 
by the joint cd. The treatment of L, will be explained hereafter. The 
simple dynamic frame without load or friction is shown by lines 1, 2, 
3, 4, 5, 6 in fig. 37a, which also shows the loads referred to the proper 
points. Let it be remarked that /’, is referred not toa joint in the simple frame, 
but to the point through which the actual bearing pressure 28 must pass. 
We may now begin the reciprocal figure 37) by drawing the effort 1a of the 
steam against the piston, the load L, the load L,, and the directions of the 
reaction 5, and the resistance 6a. The line L, is then subdivided into its two 
components /, and /’,, and the line 2 is drawn from the point which subdivides 
L, into its two components /, and /’, parallel to the line / in fig. 37 or 2 in fig. 
37a. Let the point where line 2 intersects 5 be called o ; then the polygon 1a, 
L,, 4, 2, 5 (fig. 376) represents the forces in equilibrium at the joint ab; now, 
returning to fig. 37a, we are able to draw lines 5, 2a and 28: the direction of 2a 
is given by the line of the same name in fig. 37, drawn from o to the inter- 
section of L, with L,; the line 268 is drawn in fig. 37a parallel to the line of 
the same name which in 370 joins o with the intersection of 6a and L,; the 
line 2a and 26 abut against the joints at the end of link 2 (fig. 37a), and meet 
in the line L;. The element ¢ is in equilibrium under the action of the resistance 
6a, the driving force 28 which we have just found, the weight L, and the 
reaction of the main bearing. The load L, is wholly borne by the joint cd, and 
therefore the direction of the remaining component of the reaction at the bear- 
ing is given by the full line 3a, fig. 37a, passing through the centre of the joint 
cd, and the intersection of 6 with 28. From o draw 3a in 378, parallel to 3a in 
37a, and it will cut off a length 6a measuring the resistance which the effort 
is able to overcome; the polygon 28 6a L, and 38 represents the four forces 
under the action of which ¢ is in equilibrium. 3 is the resultant pressure on 
the joint cd. We may complete the reciprocal figure by drawing the force 
18, the load L,, and the forces 68, 4a, and 48. Wemay also complete the 
loaded frame in fig. 37a by drawing 4a and 48 parallel to the lines of the same 
name in 376; a line from o parallel to link 4 of fig. 37a will subdivide L, in 
the ratio in which L; would be subdivided if referred to the two frame joints 
VOL. XXVIII. PART I, I 
