64 MR EDWARD SANG ON THE MOTION OF A HEAVY BODY 



N, the point 7 must be approaching to Z, and it must reach Z just when a and (3 

 have met at N. Thus the time in which 7 describes the arc MZ is just equal to 

 that in which it passes from N to M. By the time that a has reached Z again, /3 

 has reached L the extreme point of its motion on the other side, and therefore 

 7 is at P the middle of ZL; lastly, when a has once more descended along 

 ZPLN to N, (3 has descended along LN to the same point, and so 7 also has 

 come to N. 



It thus appears that while the point a makes two complete revolutions, the 

 point 7 makes only one. In the progress of this revolution the velocity of 7 

 varies ; when at N it is half the sum of the initial velocities V A and V B , and 

 when at Z it is reduced to be their difference ; so that the motion of 7 has the 

 general characteristic of one due to the action of gravity upon a heavy body. 



13. If the motion of 7 can be truly represented by the action of gravitation 

 upon a heavy body, we may determine the point from which 7 may be supposed 

 to have descended, and the intensity of the gravitation which must act upon it, 

 by comparing the velocities at the lowest and highest points of its path. Let C 

 be the point from which 7 must have descended in order to acquire at N the 

 velocity \ (V A + V B ), or at Z the velocity \ (V A — V B ), and put G y for the intensity 

 of the gravitation to which it is subjected ; then 



V(G r . NZ . CN) = * (V A + V B ) ; J(G y . NZ . CZ) = J (V A - V B ) . 



Now according to Article 10, 



V A = NE . NZ ; V B = NE . NF 



while if we make ZI a mean proportional between CZ and ZN, and join NI, we 

 have 



NZ . CN = NP, NZ . CZ = IZ 2 , 



so that • 



V(G 7 ) . NI = J NE (NZ + NF) ; J(G y ) . IZ = \ NE (NZ - NF) , 



whence NI : IZ : : NZ + NF : NZ — NF ; a proportion which enables us to deter- 

 mine the position of the point I, and, consequently, that of C. 

 Taking the square of each term of that proportion we have 



NP : IZ 2 : : NZ 2 + 2NZ . NF + NF 2 : NZ 2 - 2NZ . NF + NF 2 



whence 



NI 2 : NZ 2 : : (NZ + NF) 2 : 4NZ . NF 



and consequently 



_ NZ 2 (NZ + NF) 2 , _ NZ + NF _ NZ NE + NZ , 



i>J - ~ 4NZ . NF ».«*-■"* 2V(NZ . NF) 2V(NE . NZ) 



