On finding the Maximum Range of Projectiles in vacuo. 539 



therefore 



0T2 = 0B2 + B II' . T:\I + 20B . ON - 20B« 



= 0B2 + BH'(TM + 0N-0B) 



= 0B2+BH'.BP'. 



From this equation it is evident that OT is given in length, 

 and therefore that for a given velocity of projection the locus of 

 T is a circle. The centre and radhis of this circle may be found 

 by the following construction : — 



Draw through B the 

 lines PBH vei-tical, and 

 P'BH' perpendicular to 

 the plane; cut off BH 

 = 4A, and draw HH' ho- 

 rizontal ; bisect BH' in 

 0; O is the centre of the 

 circle. To find the ra- 

 dius, draw BQ parallel 

 to the plane PR; on 

 P'H' describe a semi- 

 circle, cutting this line 

 in Q ; the distance OQ 

 is the radius of the circle. 



If the range be given =Pr suppose, draw through r a vertical 

 line cutting the circle in the points / and l' ; Bt and B/' will be 

 the two directions of the gun answering to this range. If the 

 maximum range be sought, it is evident that the vertical through 

 the extremity of the range must touch the circle ; therefore, if 

 the horizontal line T'T be drawn through 0, cutting the circle in 

 T' and T, the lines BT and BT' will be the directions of the gun 

 for the best range, the one down, the other up the plane. 



The isoscelism of the triangle BRT may be proved as follows : — 

 Imagine the line OR dra\vn in the figure, then 



OR2-RB2=OP'^-BP'2 = (OP'-fBF)(OP'-BP') 

 = (0B + 2BP').0B 

 = 0B2 + 20B.BP' 

 = 0B2-i-BQ2. 

 But 



0B2 + BQ2=0Q2 = 0T2 = 0R2-Rr. 

 Therefore 



0R2-RB2=0R2-RT2. 

 Therefore 



RB = RT. 

 Similarly 



R'B = R'T'. 

 2 N 2 



