Page:EB1911 - Volume 03.djvu/290

v.	m.	log k.	Cr=gp=f(v)=vm/k.
3600	1.55	2·3909520	v 1·55 × log−1  3·6090480
2600	1.7	2·9038022	v 1·7⁠× log−1  3·0961978
1800	2	3·8807404	v 2  × log−1  4·1192596
1370	3	7·0190977	v 3  × log−1  4·9809023
1230	5	13·1981288	v 5  × log−1 14·8018712
970	3	7·2265570	v 3  × log−1  8·7734430
790	2	4·3301086	v 2  × log−1  5·6698914

The numbers have been changed from kilogramme-metre to pound-foot units by Colonel Ingalls, and employed by him in the calculation of an extended ballistic table, which can be compared with the result of the abridged table. The calculation can be carried out in each region of velocity from the formulae:—

(25)  $T(V)-T(v)=k\int _{v}^{V}v^{-m}dv,\ S(V)-S(v)=k\int _{v}^{V}v^{m+1}dv$

$I(V)-I(v)=gk\int _{v}^{V}v^{-m-1}dv,$

and the corresponding integration.

The following exercises will show the application of the ballistic table. A slide rule should be used for the arithmetical operations, as it works to the accuracy obtainable in practice.

Example 1.- Determine the time t sec. and distance s ft. in which the velocity falls from 2150 to 1600 f/s

(a) of a 6-in. shot weighing 100lb, taking n=0.96,

(b) of a rifle bullet, 0.303-in. calibre, weighing half an ounce, taking n=0·8.

V.	v.	T(V).	T(v).	t/C.	S(V).	S(v).	s/C.
2150	1600	28·6891	27·5457	1·1434	20700·53	18587·00	2113·53

	d.	w.	C.	t/C.	t.	s/C.	s.
(a)	6	100	2·894	1·1434	3·307	2113·53	6114 (2038yds.)
(b)	0·303	1/32	0·426	1·1434	0·486	2113·53	900 (300yds.)

Example 2.—Determine the remaining velocity v and time of flight t over a range of 1000 yds. of the same two shot, fired with the same muzzle velocity V=2150f/s.

	S.	s/C.	S(V).	S(v).	v.	T(V).	T(v).	t/C.	t.
(a)	3000	1037	20700·53	19663·53	1861	28·6891	28·1690	0·5201	1·505
(b)	3000	7050	20700·53	13650·53	920*	28·6891	23·0803	5·6088	2·387

* These numbers are taken from a part omitted here of the abridged ballistic table.

In the calculation of range tables for direct fire, defined officially as “fire from guns with full charge at elevation not exceeding 15°,” the vertical component of the resistance of the air may be ignored as insensible, and the actual velocity and its horizontal component, or component parallel to the line of sight, are undistinguishable.

Fig. 1.

The equations of motion are now, the co-ordinates x and y being measured in feet,

(26)  ${\frac {d^{2}x}{dt^{2}}}=-{\mbox{r}}r=-{\frac {gp}{C}},$

(27)  ${\frac {d^{2}x}{dt^{2}}}=-g.$

The first equation leads, as before, to

(28)  $t=C\left\{T(V)-T(v)\right\},$

(29)  $x=C\left\{S(V)-S(v)\right\},$

The integration of (24) gives

(30)  ${\frac {dy}{dt}}={\mbox{constant}}-gt=g\left({\tfrac {1}{2}}T-t\right),$

if T denotes the whole time of flight from O to the point B (fig. 1), where the trajectory cuts the line of sight; so that ½T is the time to the vertex A, where the shot is flying parallel to OB.

Integrating (27) again,

(31)  $y=g\left({\frac {1}{2}}T-{\frac {1}{2}}t^{2}\right)=3/4gt(Tt)={\frac {1}{2}}gt(T-t);$

and denoting T-t by t', and taking g=32f/s²,

(32)  $y=16tt^{\prime },\,$

which is Colonel Sladen's formula, employed in plotting ordinates of a trajectory.

At the vertex A, where y=H, we have t=t'=½T, so that

(33)  $H={\frac {1}{8}}gT^{2},$

which for practical purposes, taking g=32, is replaced by

(34)  $H=4T^{2},{\mbox{ or }}(2T)^{2}.\,$

Thus, if the time of flight of a shell is 5 sec., the height of the vertex of the trajectory is about 100 ft.; and if the fuse is set to burst the shell one-tenth of a second short of its impact at B, the height of the burst is 7.84, say 8 ft.

The line of sight Ox, considered horizontal in range table results, may be inclined slightly to the horizon, as in shooting up or down a moderate slope, without appreciable modification of (28) and (29), and y or PM is still drawn vertically to meet OB in M.

Given the ballistic coefficient C, the initial velocity V, and a range of R yds. or X=3R ft., the final velocity v is first calculated from (29) by

(35)  $S(v)=S(V)-X/C,\,$

and then the time of flight T by

(36)  $T=C\left\{T(V)-T(v)\right\}.$

Denoting the angle of departure and descent, measured in degrees and from the line of sight OB by φ and β, the total deviation in the range OB is (fig. 1)

(37)  $\delta =\phi +\beta =C\left\{D(V)-D(v)\right\}.$

To share the δ between φ and β, the vertex A is taken as the point of half-time (and therefore beyond half-range, because of the continual diminution of the velocity), and the velocity v₀ at A is calculated from the formula

(38)  $T(v_{0})=T(V)-{\frac {{\frac {1}{2}}T}{C}}={\frac {1}{2}}\left\{T(V)+T(v)\right\};$

and now the degree table for D(v) gives

(39)  $\phi =C\left\{D(V)-D(v_{0})\right\},$

(40)  $\beta =C\left\{D(v_{0})-D(v)\right\},$

This value of φ is the tangent elevation (T.E); the quadrant elevation (Q.E.) is φ−S, where S is the angular depression of the line of sight OB; and if O is h ft. vertical above B, the angle S at a range of R yds. is given by

(41)  $\sin S=h/3R,\,$

or, for a small angle, expressed in minutes, taking the radian as 3438′,

(42)  $S=1146h/R.\,$

So also the angle β must be increased by S to obtain the angle at which the shot strikes a horizontal plane—the water, for instance.

A systematic exercise is given here of the compilation of a range table by calculation with the ballistic table; and it is to be compared with the published official range table which follows.

A discrepancy between a calculated and tabulated result will serve to show the influence of a slight change in the coefficient of reduction n, and the muzzle velocity V.

Example 3.- Determine by calculation with the abridged ballistic table the remaining velocity v, the time of flight t, angle of elevation φ, and descent β of this 6-in. gun at ranges 500, 1000, 1500, 2000 yds., taking the muzzle velocity V=2150 f/s, and a coefficient of reduction n=0.96. [For Table see p.274.]

An important problem is to determine the alteration of elevation for firing up and down a slope. It is found that the alteration of the tangent elevation is almost insensible, but the quadrant elevation requires the addition or subtraction of the angle of sight.

Example.- Find the alteration of elevation required at a range of 3000 yds. in the exchange of fire between a ship and a fort 1200 ft. high, a 12-in. gun being employed on each side, firing a shot weighing 850 lb with velocity 2150 f/s. The complete ballistic table, and the method of high angle fire (see below) must be employed.