66 
MINUTES OE PKOCEEDINGS OF 
from the glacis, for the purpose of demolishing a casemated redoubt 
situated in the gorge. It appears from available plans, which have 
been verified by reconnaissance and actual observation, that the 
masonry of the redoubt is protected by a traverse placed at a distance 
of 20 metres from it. The crest of the traverse is about one metre 
higher than that of the glacis, and 75 metres from it ; so that the 
total distance of the battery from the crest of the traverse is 925 metres, 
and from the redoubt 945 metres. Finally, the angle of elevation of 
the line joining the gun's muzzle and the glacis crest is ascertained 
to be 6'. 
Since the crest of the glacis is one metre lower than that of the 
traverse, the ground angle of the nearest point of the latter is obtained 
by a simple geometrical calculation. 
tan n x — 
850 tan 6'+ 1 
'925 
2-4838 
925 
0-00268 ; 
whence n x — 10', and y 1 = 2*4838 metres. 
The foot of the redoubt is 6*5 metres below the level of the top 
of the traverse. If we wish to take the former as the lowest point 
of impact we shall have 
tan a = ^ = 0-325, 
and a = 18°. 
A glance at the practice table shows that for this angle the charge 
would have to be taken much too small; we therefore decide to take 
the lowest point of impact half-way up the scarp (which is 5*7 metres 
high), or 6*5 m — 2*85 m = 3*65 metres below the top of the traverse. 
We have, consequently, by formula (2) 
3-65 
tan y 2 = tan a + tan n x — —— + 0'0026 = 0-1852, 
20 
y 2 =10°30'. 
Moreover, we find in the practice tables for the charge of 0*8 k.g. 
that for a range of 945 metres the next larger angle of descent is 
10° 38'. There can be no doubt that 0*8 k.g. is the proper charge for 
the case in question. The corresponding final velocity of 156 metres 
may also be looked upon as sufficient to effect the object in view. 
We now interpolate the angle of elevation for 945 metres between 900 
and 1000 metres, and obtain 0 = 10° 5'. 
This angle would be the one to be employed if the point to be struck 
lay in the same horizontal plane as the muzzle of the gun. . But we 
find that while the mid-height of the scarp is 3"65 metres beloio the 
top of the protecting traverse, the latter is 2*483 metres above the level 
of the gun's muzzle. Consequently the mid-height of the scarp is 
3*65 — 2*483 = 1 *167 metres beloio the horizontal plane in question. 
We must, moreover, unless we would lose the whole effect of the 
projectiles of the lower half of the set of trajectories, fix the point 
