G?6 



' " Mi>-.-x^vN- ■ [ 538 ],^;^^i^^V,,>f . ^^ .,: .^s-,^ i>0 



LXIX, A general Construction for finding the vnaximum Range of 

 Projectiles in vacuo. Bij the Rev. Joseph A. Galbraith, 

 M.A., Erasmus Smith's Professor of Natural and Experi- 

 mental Philosophi/ in the University of Dublin^. 



IN the last Number of the Philosophical Magazine I read with 

 much pleasure a communication from Professor Sylvester, 

 in which he investigates the maximum range of a projectile on 

 an inclined plane, the point of projection being situated above 

 the plane. At the end of his article he gives a geometrical con- 

 struction for finding the elevation for the best range on a hori- 

 zontal plane. 



The following investigation, which gives a general construction 

 for planes not horizontal, may prove interesting to those readers 

 of the Philosophical Magazine who feel pleasure in following a 

 train of reasoning strictly geometrical in all its steps. The con- 

 struction I propose is very easily effected, so that if drawn to 

 scale it is capable of furnishing very good practical results. 

 Moreover, it may be said from its generality to command every 

 case which may be proposed with regard to projectiles in vacuo. 



Let B be the battery from which 

 the shot is fired, at a given vertical 

 height BP above the descending 

 plane PR ; let PR be the range of 

 the ball, BT the direction of the 

 gun, and RT a vertical line drawn 

 through R to meet this line in Tj 

 then the locus of the point T for a 

 given velocity of projection shall be 

 a circle. For cut off BH equal to 

 4/i, or four times the height due to 

 V, the velocity of projection, and 

 draw HH' horizontal to meet in H' 

 the perpendicular BP', let fall on 

 the plane, bisect BH' in and join 



OT ; from T draw TM and TN perpendicular and parallel to 

 the plane. 



Let t be the time of flight; then Wll = vt, and TR = ^^T^ 

 E liminating t, and substituting for v^ its equal 2gh, we obtain 



Br^=4/ixTR=BHxTR. 

 Since the triangles HBH' and MTR are similar, 



BHxTR=BH'xTM; 

 therefore BT2 = BH' x TM. In the triangle 0TB, 



OT2 + OB2-20B.ON = BT2=BH'.TM; 

 * Communicated by the Author. 



