GUNNERY. 



N s, the sine of A C s, is the sine of twice 

 the angle s A F ; half the impetus being 

 radius. 



Whence, at the directions of 15 or 75, 

 the amplitude is equal to the impetus ; 

 for, from what has been said, half the im- 

 petus being radius, a fourth part of the 

 amplitude is the sine of twice the angle 

 of elevation ; but the sine of twice 15, 

 that is, the sine of 30, is always equal to 

 half the radius , or in this case a fourth 

 part of the impetus is equal to a fourth 

 part of the amplitude. From this and the 

 preceding proposition, there are two 

 easy practical methods for finding the im- 

 petus of any piece of ordnance. The 

 fourth part of the amplitude is a mean pro- 

 portional between the impetus at the 

 curve's principal vertex and its altitude. 

 For MN:Ns::NA=sF==T?p 



The altitudes are as the versed sines of 

 double the angles of elevation, the impe- 

 tus remaining the same. For making half 

 the impetus radius, A N the altitude is the 

 versed sine of the angle A C s =s twice 

 angle * A F. And also, radius : tangent 

 angle elevation :: one-fourth amplitude : 

 altitude: that is, R : tangent angle s A. f 



Projections on Ascents and Descents ,fig. 8, 9. 



If the mark can be hit only with one di- 

 rection A G, the impetus in ascents will 

 be equal to the sum of half the inclined 

 plane and half the perpendicular height, 

 and in descents it will be equal to their 

 difference ; but if the mark can be reach- 

 ed with two directions, the impetus will 

 bq greater than that sum or difference. 

 For when A G is the line of direction, the 

 angle g G A being =MAG = GA < -; 

 G g = A g, and g z added to or substract- 

 ed from both, makes G z half the impetus 

 equal to the sum or difference of A g> a 

 fourth part of the inclined plane, and g z 

 a fourth part of the perpendicular height. 

 In any other direction F P is greater than 

 F o = A F ; and F/, added to or subtract- 

 ed from both, makes /P half the impetus 

 greater than- the sum or difference of 

 A F, a fourth part of the inclined plane, 

 and F/a fourth part of the perpendicular 

 height. Whence, if in ascents the impe- 

 tus be equal to the sum of half the inclin- 

 ed plane and half the perpendicular 

 height, or if in descents it be equal to 

 their difference, the mark can be reached 

 only with one direction ; if the impetus is 

 greater than that sum or difference, it may 

 be hit with two directions; and if the im- 

 petus is less, the mark can be hit with 

 none at all. 



Prob. II. The angles of elevation, tin 

 horizontal distance, and perpendicular 

 height, being given, to find the impetus. 

 Fig. 8, 9. 



From these data you have the angle of 

 obliquity, and length of the inclined plane ; 

 then as 



A * : A M : : S. angle A M s : S. angle 

 A s M : : S. angle s A F : S. angle M A F, 

 and A F : A s : : S. angle M A S : S. angle 

 M A F ; whence, by the ratio of equality, 

 A F : A M : : S. angle s A F x S- a "gle 

 M A s : S. angle M A F x S. angle M A F, 

 which gives this rule. 



Add the logarithm of A F to twice the 

 logarithmic sine of the an.^le MAP; from 

 their sum subtract the logarithmic sines 

 of the angles s A F and M A s, and the 

 remainder will give the logarithm of A M 

 the impetus. 



When the impetus and angles of eleva- 

 tion are given, and the length of the in- 

 clined plane is required, this is the rule. 

 Add the logarithm of A M to the logarith- 

 mic sines of the angles s A F and M A s ; 

 from their sum subtract twice the loga- 

 rithmic sine of angle M A F, and the re- 

 mainder will give the logarithm of A F, 

 the fourth part of the length of the inclin- 

 ed plane. 



If the angle of elevation t A H, and its 

 amplitude A B (fig. 11,]) and any other 

 angle of elevation t A H is given ; to find 

 the amplitude A b for that other angle, 

 the impetus A M and angle of obliquity 

 D A H remaining the same. 



Describe the circle A G M, take A F a 

 fourth part of A B, and A /a fourth part 

 of A b ; from the points F, /, draw the 

 lines F s, and / p parallel to A M, and 

 cutting the circle in the points g, // ; then 

 A F : A M : : S. angle s A F x S. angle 

 M A * : S. angle M A F X S. angle MAP; 

 and A M : A/ : : S. angle MAP X S. angle 

 M A F : S. angle p A/ x S. angle p A M ; 

 whence by the ratio of equality, 



A F : A/: : S. angle s A FX S. angle 

 M A s : S. angle p A/x. angle/* A M, 

 which gives this rule. 



Add the logarithm of A F to the loga- 

 rithmic sines of the angles/; Af, p A M ; 

 from their sum subtract the logarithmic 

 sines of the angles s A F, s A M, and the 

 remainder will give the logarithm of A/, 

 a fourth part of the amplitude required. 



Prob. III. To find the force or velocity 

 of a ball or projectile at any point of the 

 curve, having the perpendicular height of 

 that point, and the impetus at the point of 

 projection given. From these two data 

 find out the impetus at that point; then 



