THE VALUE OF A HIGH SITE FOR COAST ARTILLERY. 
229 
With the same notation as before, let OXT be the true and OBT 
the exaggerated trajectory, its ordinates being p times as great. 
Through X draw the ordinate FXB, so that 
BE=pXE, 
Join BP and produce it to meet OH produced in F. 
Let HF=kOH, and the angle HPF or BPX=y. 
Now, in any particular case, where R and h and consequently f3 are 
given, a knowledge of 7 will enable us to find the position of X, and to 
evaluate the difference between PT and PX , from a plotting of the 
trajectories, such as is given in fig. 12 . 
Now 
and 
Again 
and as 
tan 0=tan XPE= J ^ ; , 
BE 
tan (j8+y) = tan BPE=j^, 
.*. tan (ft + y) —p tan ft. 
tan y = 
k.OH 
HP 
= k tan ft, 
tan/3+tan y 
1 — tan/3 tan y 
= tan (ft+y), 
p tan/3 = 
tan ft-\-kta,nft 
1 - k tan 2 ft 9 
.-. k=-. 
y - 1 
1 +p tan 2 /3 ’ 
tan y — 
p 1 
1 +p tan 2 (3 
tan /3, 
p - 1 
cot j3 +p tan ft ’ 
-y = cot _1 ^ L ^ (cot ft+pta,n ft], 
an expression from which its value can be easily found. 
Suppose now P is the position of any point of aim, and it were 
required to determine where the shot would strike the water, if the 
range were wrongly estimated by any number of yards PT; let PB be 
drawn at the angle (/3 + 7 ) due to the height and the range to meet 
29 
