76 
MINUTES OF PROCEEDINGS OF 
to demolish the upper half of a scarp,, as good a breach was produced 
by means of 190 21 c.m. shells (80 k.g. in weight) as by 484 shells from 
the last-named gun. The total weight of the ammunition (shells and 
powder together) was 15,900 k.g. in the case of the 21 c.m., and 
13,990 k.g. in that of the 15 c.m. The quality of the masonry was in 
both cases much the same. The 21 c.m. had therefore the advantage 
as regards the smaller number of projectiles fired. This may possibly 
be in its favour, owing to time being thus gained, notwithstanding the 
greater weight of the ammunition expended. This result, however, 
affords no information as to the suitability of this gun for breaching 
purposes in the case where, with high angles of descent, the final 
velocity has to be reduced considerably below the 160 metres which is 
considered the lowest allowable when using the short 15 c.m. against 
tolerably strong masonry. 
This gun is very suitable for destroying small enclosed works ( Kohl - 
bauten), detached loopholed walls, &c., at ranges of from 750 to 
1500 metres. In the case of larger objects the long 15 c.m. gun of 
1861 can be used up to a maximum range of 1800 metres, and with a 
minimum final velocity of 130 metres. 
11. To DETERMINE THE PROPER CHARGE AND ELEVATION WHEN THE 
COVERING MASS LIES NEARER TO THE GUN THAN TO THE POINT TO BE 
STRUCK. 
This case occurs, as was mentioned in the introduction, in curved 
defensive fire and in curved dismounting fire. 
If the relative heights of the cover and of the object with reference 
to the level of the guffs muzzle are given, and also-.their horizontal 
distances from it, the required angle of elevation can be calculated 
exactly by the formula 
tan (j) 2 = tan n 1 + tan a.* 
The tangent of % will, under the given profile conditions, be always 
positive. The angle ^ 2 is reckoned from the line joining the centre 
of the muzzle and the lowest point of impact (Fig. 6). In order, 
therefore, to obtain the angle of elevation above the horizontal plane 
through the muzzle, the angle </> 2 , found by the above formula, must be 
diminished, first by the angle of jump (A), and further by the ground 
angle (%) of the point of impact, if this point lies below the level of the 
muzzle. If the contrary is the case the angle </> 2 — A must be increased 
by the ground angle (?? 2 ). 
In order to calculate the values of tan n x and tan a by the formulas 
tan n x — —, and tan a = 
aq x 2 — x t 
it must be remembered that the ordinate y x must be taken larger than 
* The derivation is given in the Handbook for the Austrian Artillery, Part I., Section 4, page 28. 
