56 
MINUTES OF PROCEEDINGS OF 
Let the length uf its side, as hereafter determined, 
= a. 
The mean errors in deflection and range, having been determined (quad- 
ratically, so as to give the quantity Captain Noble defines as e 2 ) from the 
shots on the ground, must then be applied to the normal target, the former 
unchanged, giving our quantity h ; the latter multiplied by the sine of the 
descending angle to give k. The problem is then reduced to finding the 
probability of hitting a square target whose centre coincides with the centre 
of impact, and whose side = a. 
To find this we must revert to the reasoning in Art. 21. 
30. Let the following figure represent the group of shot referred to 
the normal position, and let the small square in the centre formed by the 
intersection of the four lines of a, b } c } d 9 represent the size of the test 
target we are now considering. 
c 
O 
© 
9 
0 
£> 
0 
o 
k * 
© 
0 ci. 
© 
® © 
o 
V 
o 
6 
* 
o 
b 
• 
0 
9 
* 
« 
c 
1 
Taking first the errors in a horizontal direction, we have seen (Equation 10) 
that the probability of any shot falling between the distances x and x-\-dx to 
the right of the centre, will be 
= e~P* x * dsc, 
v 7 r 
whence the probability of any shot falling between the centre and any 
distance = ^ will be 
= - 4 = f 2 e-r^dx .( 17 ) 
v 7 rJ o 
And, by the symmetry of position, to the right and left of the centre, the 
probability of any shot falling within the vertical band formed by the lines c 
and d (being a distance apart = a) will be 
r"- e-t^dx 
VirJ o 
(15) 
