320 APPLIED SCIENCE 



small direct-acting engine, making 1 turn per second. To draw 

 the dynamic frame without friction we proceed as follows : L 6 

 is represented by two components l b and l' b , acting at the joints ab 

 and be. L a is wholly borne by the one joint, and the actual 

 point of its application is a matter of indifference ; L c is wholly 

 drawn by the joint cd. The treatment of L d 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 l' b is referred not to a joint in the simple frame, but to the 

 point through which the actual bearing pressure 2/3 must pass. 

 We may now begin the reciprocal figure 37b by drawing the effort 

 la of the steam against the piston, the load L a the load L 6 and the 

 directions of the reaction 5, and the resistance 6a. The load L 6 

 is then subdivided into its two components l b and I' , and the line 

 2 is drawn from the point which subdivides L 6 into its two com- 

 ponents l b and l\ parallel to the line b in Fig. 37 or 2 in Fig. 37 a. 

 Let the point where line 2 intersects 5 be called o; then the polygon 

 la, L a , l b , 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 2/3 : the direction of 2a is given by the line of 

 the same name in Fig. 376, drawn from o to the intersection of 

 L rt with L 6 ; the line 2/3 is drawn in Fig. 37a parallel to the line 

 of the same name which in 37b joins o with the intersection of 

 6a and L 6 ; the lines 2a and 2/3 abut against the joints at the 

 end of link 2 (Fig. 37o), and meet in the line L 6 . The element c 

 is in equilibrium under the action of the resistance 6a, the 

 driving force 2/3 which we have just found, the weight L c and 

 the reaction of the main bearing. The load L c is wholly borne 

 by the joint cd, and therefore the direction of the remaining 

 component of the reaction at the bearing is given by the full 

 line 3a, Fig. 37a, passing through the centre of the joint cd, and 

 the intersection of 6 with 2)8. From o draw 3a in Fig. 37 b, parallel 

 to 3a in Fig. 37a, and it will cut off a length 6a measuring the 

 resistance which the effort is able to overcome ; the polygon 2/8 

 6a L c and 3/3 represents the four forces under the action of which 

 c is in equilibrium. 3/9 is the resultant pressure on the joint 

 cd. We may here complete the reciprocal figure by drawing the 

 force 1/8, the load L d , and the forces 6/3, 4a, and 4/3. We m;iy 



