---+-----------------
(a) | 6 | 100 | 2.894 | 1.1434 | 3.307 | 2113.53 | 6114 (2038 yds.)
(b) | 0.303 | 1/32 | 0.426 | 1.1434 | 0.486 | 2113.53 | 900 (300 yds.)
----+-------+------+-------+--------+-------+---------+-----------------

_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 = 2150 f/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 15deg," 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.

[Illustration: FIG. 1.]

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

(26) d^2x/dt^2 = -r = -gp/C,
(27) d^2y/dt^2 = -g.

The first equation leads, as before, to

(28) t = C{T(V) - T(v)},
(29) x = C{S(V) - S(v)}.

The integration of (24) gives

(30) dy/dt = constant - gt = g(1/2T - t),

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 1/2T is the time to the
vertex A, where the shot is flying parallel to OB.

Integrating (27) again,

(31) y = g(1/2Tt - 1/2t^2) = 1/2gt(T - t);

and denoting T - t by t', and taking g = 32f/s^2,

(32) y = 16tt',

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

At the vertex A, where y = H, we have t = t' = 1/2T, so that

(33) H = 1/8gT^2,

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

(34) H = 4T^2, 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-ten