AUTO-SIGHTS. 
435 
the tangent elevation due to R, (h being a constant) the line of sight is 
altered through 6 the gun will thereby have imparted to it the tangent 
elevation due to GC. 
Let the gun be again correctly laid on T, and suppose the whole 
mounting to be tilted either to the front (Fig. I.) or to the rear (Fig. 2) 
along the line of fire through the angle 0. The line of sight will then 
become GB, but the tangent elevation will remain that due to R; and 
to bring the line of sight back to T, the gun-layer must move it through 
0 . In so doing he must inevitably impart to the gun the tangent 
elevation due to GC. 
Make GD equal to GC; then, theoretically, when the line of sight 
is again on T, the trajectory will pass through D. Let the curve 
which (produced in Fig. 2.) cuts SL in O represents this trajectory. 
Then, when the range is R and the want of level is 9 , TO is the 
resultant error measured along the sea level—as we invariably calcu¬ 
late errors in range. 
Now, I cannot find the value of TO directly; but I can arrive at 
a very close approximation to it, in this way: from D with the angle 
of descent due to the range GC (or the angle /3 ) lay off DE to cut SL 
in E, and through T erect a perpendicular to cut the trajectory in P. 
Join PD, cutting SL in F. 
It is obvious that O must lie between E and F, and EF is very 
small. Calculating it out for the ranges and for the conditions named, 
I found that its maximum value was 14*7 yards; and that this was 
when the range was 4,000. If, then, we assume that TO is the mean 
of TE and TF, we must up to 4,000 yards be within considerably less 
than 8 yards of its actual value. It only remains to find TE and TF. 
The angle GTS, or the range finding angle, is known since its sine 
= —; the angle /3 is given in the range tables, and 0 is assumed 
3 1 * 
to be 4'. Let us call the range finding angle a, and assign to R any 
value. 
Then GD = GC = 
R sin a 
sin (a-0) 
(Fig- 1 ) 
or 
R sin a 
sin (a + 0) 
(Fig. 2.)= 
h 
3 sin (a ±0) 
and TD = GD - R (Fig .1.), 
OP 
R-GD (Fig. 2.) and TE 
TD sin 3 
sin (a +3) 
Next, TP = l / 2 gt (T— t) where T and t are the times of flight due 
to the ranges GC and R, T being always the greater. Thus, TP 
being known, the angle TDP (which we will designate y ) can be 
