THE ROYAL ARTILLERY INSTITUTION. 
68 
be taken into consideration that the foot of the exterior slope of the 
parapet is about 1 metre from the cordon, and that., owing to the berm 
thus formed, the mass of earth above the escarp will not be thrown 
down in sufficient quantity. It therefore appears more judicious to 
fix the mean point of impact at half the height of the scarp and the 
lowest point of impact half the 50 per cent, vertical spread below it.* 
As we have already estimated the charge from the magnitude of the 
angle a, half the 50 per cent, vertical spread for 0*8 kilogrammes at 
between 800 and 900 metres can be taken as 1*25 metres. 
The calculation can now be made by formula (2) as follows •-— 
for the difference in height, 
u — I m + S*75 m + T25 m = 6 metres; 
and for the horizontal distance, 
30 
30 
cos 30° 0-886 
= 34*7 metres; 
therefore 
tan y 2 = tan a + tan n x = ——' + tan 5' = 0-17293 + 0-00145 = 0-17438, 
34*7 
T2 =9° 58*'. 
The distance x 2 is 909 m + 34*7 m = 943*7 m , or in whole numbers 944 m . 
The angle of descent for this distance is found by interpolation; thus 
we have 
0*8 k.g. 0*9 k.g. 
10° 37', 9° 15', 
of which the arithmetical means are 
0-85 k.g. 
9° 56'. 
As this angle is only slightly larger than the required angle of 
descent (9° 50'), the charge of 0*85 k.g., although somewhat too small, 
may be considered as the correct one. 
If the value of y 2 lay between 10° 37' and 9° 56' the charge of 
0*8 k.g. would have to be used: on the other hand, for all values from 
9° 56' to 9° 15' we should use 0”85 k.g. 
To calculate the angle of elevation for 944 metres, we have, by inter¬ 
polation, for 
0*8 k.g. 0-9 k.g. 
0 = 10° 4', <t> = 8° 45', 
therefore, taking the arithmetic means, 
for O’85 k.g. 
0 = 9° 25'. 
* “Half the 50 pqr cent, vertical spread ” is equivalent to the 'probable vertical deviation (either 
above or below the point aimed at) of any particular projectile.—T k. 
9 
