THE ROYAL ARTILLERY INSTITUTION. 
45 
With regard to the first assumption, since the points at which the 
angle a is subtended by the base BC all lie in the arc BHAKC } therefore if 
A 
perpendiculars be drawn from B and C to meet the circumference again 
in H and K, all points directly opposite the base BC, at which that base 
will subtend an angle a } lie in the arc HAK. 
Join BK, IIC; then their intersection will be G, the centre of the circle* 
Join AG and produce it to L. Then BL is equal to LC, and the angles at L 
are right angles* 
By our method of working, our actual range taken is AB , and our shortest 
possible range is IIB, our longest KB; for it may be easily shewn that K 
and K are the points in the arc HAK nearest to and furthest from B. 
Hence the greatest error in excess we can have will be 
Et&AB^HB 
= BC 1 
^ 2 * . a 
sin — 
-BC 
COS Ct 
sin a 
BC /a, \ 
- ( COS -T- — cos a J 
sm a \ 2 / 
, , a n 9 a 
BC 1+cos--2 cos*- 
1 — cos 2 ^ 
2 a / 
C0S 2V 
_BC ( 1 + a cos |) (l — C QS |) 
2 a / n 
COS-.y 1_ C0S 2_ 
= X '( 2 + ^|)^i 
+ cos| 
