VOL. XXXI.] , PHILOSOPHICAL TRANSACTIOXS. 513 



Scholium. — Because the point e may be taken on both sides of the point d, 

 there are two points, k, k, where gh meets the trajectory. 



Prop. vii. Given the Projectile velocity ; to Jind the Direction to cause the 

 trajectory to pass through a Given Point. — Let the body be projected from a, 

 fig. 4, and let k be the point through which the trajectory must pass. Draw 

 CA in the direction of gravity, and equal to the altitude due to the given pro- 

 jectile velocity. Draw ce perp. to ac, and ke perp. to ce. With the centres 

 A and K, and radii ac and ke, describe two circles meeting in f. Draw ap, 

 and bisect the angle cap by the line ab. Then will ab be the direction 

 sought. Ql. e. I. 



For CA is the 4th part of the parameter at the point a, by prop. 1 ; and, by 

 construction, af = ac, and kf = ke ; therefore p is the focus of a para- 

 bola passing through a and k. And the same will touch ab in a, because of 

 the equal angles pab, cab. Therefore the body projected from the point a, in 

 the direction ab, with the velocity due to the altitude ac, will pass through the 

 point K. a. e. d. 



N. B. With the centres a, k, and radii ac, ke, of the described circles, 

 there will be two concourses, f, f ; and bisecting the angles pac, fAC, there 

 will be two directions, which will cause the trajectory to pass through the given 

 point K. 



Prop. viii. Given the Projectile Directions to Jind the Velocity, to cause the 

 trajectory to pass through a Given Point. — Let the body be projected from a, 

 fig. 6, in the direction ab, and k the given point to be passed through. Draw 

 AK, which bisect in c, and draw bc in the direction of gravity ; also join bk. 

 Draw AD, KE parallel to cb ; and draw ap, kf, making the angle fab = dab, 

 and the angle fkb = ekb. Then will ap be eqjual to the altitude due to the 

 velocity sought, q. e. p. 



For because bc is in the direction of gravity, it is a diameter of the parabola; 

 and because ac is = ck, bc is the diameter to the ordinate ak. Hence, since 

 ab is a tangent to the parabola at a, kb will be a tangent also at k. Hence, 

 because ad is in the direction of the diameters, and the angle fab is =: dab, 

 therefore ap passes through the focus of the parabola. In like manner kp 

 passes through the focus. Therefore p is the focus of the parabola, and pa the 

 4th part of the parameter at the point a, which is thence equal to the altitude 

 due to the velocity sought. 



Prop. ix. To ^nd the Least Velocity, and the Direction belonging to that 

 velocity, to cause the projectile to pass through a Given Point. — Let a be the 

 point of projection, fig. 7, and k the given point to be passed through. Draw 

 CA, ek in the direction of gravity; and joining ak, bisect the angles cak, eka, 



VOL. VI. 3 U 



