188 
MINUTES OE PROCEEDINGS OE 
§ 14. Again, suppose tlie velocity to be required after a distance of 1653 
(i.e., when inclination is 2°), 
— s = 3 ' 5 ^ 8 x 1653 = 1294-63. 
w 16 
Add this to tlie number 1574, against vel. 1355, and we get 2868*63, 
against wliicli we find the required velocity, 1139*23 (using propor- 
§ tional parts). Tlie calculated velocity was 1136*1 sec 2°= 1136*8— 
very slightly different. 
These instances have been selected hap-hazard, and not because they 
were found to agree well. 
d 2 
The tables for — t are used in precisely the same manner. 
It must be clearly understood that in all these tables and calculations 
d z . 
the multiplier — is only used to enable one set of tables to hold for all 
calibres. 
Chap. VL 
17. Law op Penetration op Projectiles. 
(Chapter VI.) 
In Art. 83, A. and [x are merely put for ™ and respectively. It 
must not be forgotten that 
Seeeq n (a). 
Art. 83. 
R — Mf and W = Mg ; 
where R is the resistance and f the retarding force or retardation pro¬ 
duced by A in a unit of time. 
/= A _ £ . R = l . (\ + ^ 2 ). 
J M w w > r j 
The forms of integration required for the following are 
*dx 
/?-» 
/ ’ dx 
c 
1 , ! X 
= - tan -1 - • 
Now , 
a* + x* a 
/= 4 y == _^(X + ^); 
ds w 
vdv 
gd 2 
X + /xv 3 
1 d (X -f- jtty”) 
2/x X + /xv” 
w 
ds , 
log — «. . s , for « = V when s = 0. 
X +[xV' z w 
N.B.—2’3026 = loge 10 5 hence follows eq 11 (1) in Bashforth. 
GO 
