182 
MINUTES OF PROCEEDINGS OF 
Take the example in Art. 71. 
Here F = 1335, 
angle of projection = 5°; 
.\ u 5 = 1335 cos 5° = 1329*9, 
^ == 6 i?i^ = ; *56935. 
w 70*6 
Let the sections be of 1° interval. 
(1) Estimate mean velocity for K at 1300, .*. K = 107*9. 
m O is known = ( 
1000\ 3 K W 
•P e 
5 / £ W 
= (S) S+ 3 - 352 x ' 56935 x ^ 
= *92736. 
A 
The logarithms of — and of P 5 are given in Tables I. & III. 
.*. w n = 1025*5. 
p. 81, 
Table VI. 
H y = f X W36 = 3 - 058 - 
Now, 
Similarly, 
( 1 ) 5^4 — l (°^- 2 - 058 4 )' 
LYo 4 = *10992 — *08294 = — *02698, 
= *11167 - *08383 = - *02784, 
, 5 X 2 . 058 4 = — *02698 + *58 x *00086* = *02748, 
/. 5& 4 = 897*7 ft. 
5^4 = 70 * 8 , 
f - 7" 
5 f 4 — ‘ * 
/K v /1000V /1000X 3 
(5) 
. p„ 
ft? 
*52635; 
^ 4 = 1238*5. 
(6) Estimate afresh for K, and proceed as before. 
For the descending branch, 0 is negative ; 
.*. p = tan 0 is negative, 
P = (3 J rp z ) changes its sign* 
* By proportional parts: • 00086 is the difference between '02698 and *02748. 
