514 PROFESSOR R. H. SMITH ON A NEW GRAPHIC 



The solution of the next example in fig. 7 is not quite so direct, because 

 here the velocity assumed as known, namely, pa that of A, is that of a joint in 

 the pentagon PxABCP;,. First, pa is drawn of the known magnitude and 

 perpendicular to P : A ; and then aft of indefinite length perpendicular to AB. 

 Then, pS and pc are drawn of indefinite length perpendicular to P 3 D and P 2 C, 

 that is, in the directions of the velocities of D and C. The points b and d now 

 sought for are known to lie on the lines a/3 and pB, and also it is known that 

 the line joining b and d is perpendicular to BD. Any point /3 on a/3 is chosen, 

 and from it /3S drawn perpendicular to BD ; and then the triangle /3Sy is con- 

 structed similar to BDC, corresponding sides being perpendicular. The triangle 

 bdc that is sought for must evidently be similarly placed to /3Sy between the 

 the lines pB and a/3. Therefore, y is joined with the intersection of pS and a/3, 

 and this line is produced to intersect pc, drawn from p perpendicular to P 2 C. 

 This gives the true position of c, and the triangle deb is then completed by 

 drawing cd and cb perpendicular to CD and CB to meet p8 and a/3. The 

 point e is obtained by drawing pe and de perpendicular to P 3 E and DE. 



The acceleration diagram has, in this case, to be constructed according to a 

 similar indirect method. The acceleration of A being supposed known can be 

 plotted at once. Then the radial components of the accelerations of B round 

 A, of C round P 2 , and of D round P 3 , are calculated and plotted off in their 



proper directions from a and p'; their magnitudes being W-, ^L, ^j- . 



r ° ° BA P 2 C P 3 D 



From the three points so obtained, three lines, which we may call /3, y, and 8, 



are drawn of indefinite length perpendicular to BA, P 2 C, and P 3 D. The 



tangential components of the above three accelerations lie along these lines, 



which, therefore, contain the three points b' ', c , and d' sought for. On the 



line 8, any two points, B 1 and 8 2 , are chosen, and from each the centripetal 



acceleration \& of C round D is plotted parallel to CD ; and from the two 



points thus obtained are drawn two lines perpendicular to CD, to meet the 

 line y in two points, say y x andy 2 . On the two bases, B x y x and S 2 y 2 , are con- 

 structed two triangles similar to DCB, whose two vertices may be called /3, and 

 /3 2 . Neither of these points, /3,, /3 2 , will be found to lie on the line /3, and their 

 distances from this line may be taken as measures of the errors involved in the 

 two guesses, 8 l5 S 2 , at the position of d. The error thus found in the resulting 

 position of /3, is a linear function of the error in the guessed position of 8 ; and, 

 therefore, the interpolation between these two errors in order to reduce them 

 to zero is to be performed by simple proportion. This linear interpolation is at 

 once effected graphically by drawing a line through /3 X and /3 2 , and producing it 

 until it meets the line /3. The intersection thus found will be the true position 

 of* b\ Or, otherwise, the two error-distances of ft and /3 2 from the line /3 may 

 be plotted off from the points B { and 8 2 perpendicularly to the line 8 (or both in 



