514 PHILOSOPHICAL TRANSACTIONS. [aNNO 1721. 



by the lines ab, kb, intersecting in b. Draw bc perp. to ac ; then ac will be 

 the altitude due to the velocity sought, and ab the corresponding direction. 



For draw bp perp. to afk, and cb meeting ke in e. Because of the equal 

 angles bac, bap, also bke, bkf, and the right angles c, e, f, it will be ac = 

 AP, and KE = KP. Hence the points a and k are in a parabola touching ab 

 in A, and whose parameter is equal to 4ac, the focus being p. Therefore a 

 body projected from a, in the direction ab, with the velocity due to the altitude 

 CA, will describe the said parabola, by prop. 2. I say also that is the least 

 velocity, or ca the 4th part of the least of all the parameters, by which the 

 parabola can be described, to pass through the points a, k. 



For, if possible, in ca take the less altitude ca, which may be the 4th part 

 of the parameter at a. Draw ce perp. to ca, meeting ke in e, and with the 

 centre a and radius ac, describe a circle meeting ak in f. Because ca is said to 

 be the 4th part of the parameter at A, the focus of the parabola will be some 

 point p, in the circumference of the circle cpf, described with the centre a and 

 radius ac. if then k be a point in that parabola, pK will be = ck. But pk 

 is = EK. Hence, since ck is less than ek, pK will also be less than fk. But 

 pK is greater than fK, and fK greater than fk, because fA is less than fa by 

 hypothesis ; hence pK is greater than pk. But it was said that pK is less than 

 PK ; which is absurd. Therefore no parabola can be described through a and 

 K, that can have a less parameter than in the solution, a. e. d. 



Prop. x. Given the Projectile Velocity, to find the Direction which shall cause 

 the body to range to the Greatest Distance on a Given Plane ; and to determine 

 that distance. — Let the given plane be ak, fig. 8, and ak the greatest distance 

 sought. Draw ca in the direction of gravity, and equal to the 4th part of the 

 parameter at the point a. Then bisect the angle cak with the line ab, which will 

 be the direction of the projectile sought. Draw cb perp. to ca, making be = cb. 

 Then draw ek parallel to ca, so will ak be the greatest distance required. 



For with the centre a and radius ac describe a circle, meeting ak in f ; and 

 draw BP, bk. Because the angles bac, bap are equal, by construction, and 

 AP = AC, then will bp = bc = be by constr. also the angles at f are right : 

 hence kp = ke. Therefore the points a, k are in a parabola whose focus is p, 

 and which touches ab in a, because of the equal angles bac, bap, and having 

 the 4th part of the parameter at a equal to ac. Therefore the body projected 

 from A, in the direction ab, with the velocity due to the altitude ca, will pass 

 through the point k, by prop. 2. a. e. d. 



I say also, that ak is the greatest distance, to which the body can be pro- 

 jected from A, with the same velocity. — For, if possible, let a parabola be de- 

 scribed with the same parameter, passing at a greater distance k, from the point 



