184 
MINUTES OE PKOCEEDINGS OE 
4° to 2° 
The principal steps gave the following results :—■ 
r w 4 = 1351*7, mean vel. for K = 1290, 
(1^0) ? = .95755, » 0 = 1014-56, 
— 9-74,«7 f 0 — 3*9525° gives y = 2| ft., a? = 32*606 ft. 
7 ' b/ 10 = 3*92275° „ y = 6Jft., a? = 75*762ft. 
4/i* 3 = 1653 ft., 4 y 2 = 88*293 ft. 
t = 1136*1, mean vel. for X = 1100, 
= '952205, ti 0 = 1016*46 (true vel. at vertex). 
y = 2*6934, 3^= 1246*56, y = 22*584, 
4 o; 0 = 2899*56, total height = 4 y 0 = 110*877. 
2 ° to 0° \ V 7 
I 
0 ° to 2 ° 
Mean vel. for K= 990, 
7 = 1 * 86 , qXq = 1060*14, 0 y 3 =18*071. 
w 3 = 957'87, mean vel. for K = 940, 
2° to 4° i 
! / 1000 \ 3 _ 
! \ «b / 
972001, .*. w 0 = 1010*29, 
7 = 1*667, 3 ^= 951*02, 0 y 3 = 49*576. 
0 y 9 = 18*071 
2^=49*576 
* 43*23 
4° to end 
L 
f u 4 = 913*34, mean vel. for K = 900, 
(1 ( ' n "J = -983261, « 0 = 1005-64, 
7 = 1-5937, 
Remaining height = 43*23,* 
Tot.110-877 
4^5*21144* 43*2 3 4,^5*21144,— 537*62 
2*25 
4^6*1546 — 40*98 4 ^ 5.1546 —512*99 
3 
- 4^5*07727 == 37*9 8 4^5*07727“ 479*41 
The result of the foregoing is as follows :— 
Projectile reaches a height of— 
ft. 
ft. 
2 J 
at 32*606 from gun, 
0 = 3-9535° 
5J 
• 75*762 „ 
0 = 3-92275° 
110*877 1 
» ,, 2899*56 
O 
o 
II 
-e- 
(vertex) j 
l a 2548*78 from end of range, 
H 
a 58*21 n n 
<p = 5-07727° 
2 J 
n 24*63 n n 
0 = 5-1546° 
0 
„ 5448*34 from gun, 
0 = 5-21144° 
= 5° 12§' 
N.B.—The range tables give 1900 yds. 
