446 
MINUTES OE PROCEEDINGS OE 
Tig. 2. 
N 
O 
To apply this formula, let" it be supposed that an escarp wall covered 
by a counterguard (Fig. 2) is to be destroyed. Let the revetment be 
at a distance of 3640 ft., and the crest of the covering breastwork at a 
distance of 3521 ft. from the gun; let the vertical height of the crest 
of the breastwork be 7 ft., and that of the revetment 16 ft.; the height 
of the cordon being 5 ft. above the horizontal plane. 
The most favourable height to make a good breach is about J of 
the total height of the wall from its base; the point of aim will then be 
5*67 ft. below the horizontal plane and 12*67 ft. below the crest of the 
counterguard. The horizontal distance between the crest of the 
counterguard and the wall to be breached will be 119 ft. The ratio 
12*67 
— - - = 0*1065 will give a first approximation to the tangent of the 
angle of incidence. On examining the practice table above given (A), 
we find that the tangent of the angle of incidence 0*1075 at a range of 
1225 yds. = 3675 ft,, with a charge of 2*08 lbs. is the nearest. But as 
2*08 lbs. is almost midway between 2*48 lbs. and 1*8 lbs., we may expect 
3*3 _}_ 3*9 
to have a mean vertical deviation of-^-= 3*6 ft. (see Table C), 
midway between the mean vertical deviations with a 1*8 lb. and a 
2*48 lb. charge at a mean range of 3640 ft. But as the height of the 
wall to be breached is more than double the above deviation, and as 
| of the wall is to be fired at, we may take § of 3*6 = 2*4 ft. and add 
it to 5*67 ft.—the height of the point to be hit below the horizontal 
plane—to obtain the height H = 5*67 x 2*4 = 8*07 ft. below the hori¬ 
zontal line through the piece. To obtain the tangent of the angle of 
incidence we have 
A = 3640 ft.; « = 3521 ft.; L/=8*07ft.; h = 7 ft. 
^(7 + 8*07) — 8*07 
tan 0 = —-- = 0*1287. 
3521 
On referring to the table (B) we find that at a range of 1225 yds., 
with charges of 1*9 and 1*7 lb., the tangents of the angle of incidence 
are 0*1213 and 0*1388 respectively; that is, that with a mean charge 
of 1*8 lb. we may expect to have a tangent of 0*1300, approximating to 
0*1287. The mean height on the tangent scale (Table B) will be 
