by observations of their Depression. 185 



Extracting the square root, and there results 

 2sin 2 iH-l = ±secD / {sin(D + D / )sin(D-D / )}* . (12.) 



2 sin 2 I H = 1 ± sec D, {sin (D + D y ) sin (D - D,) }*, and 



sin£H = {|(1 ±secD / [sin(D + D / )sin(D-D / )]*)} i . (13.) 



From formula (16.) it appears that H — 90° = 2 sin 2 | H — 1, 

 the radius, or sin 90° being unity, therefore calling H — 90° 

 = d, we have 



sin d m sec D, {sin (D + D,) sin (D - D,)}* . . (14.) 

 the most useful form, since d in this case is the arc required. 

 Remarks. — 1. If the negative sign in formula (13.) be em- 

 ployed, the resulting value of H will be the supplement of that 

 by the positive. 



2. Since D is the depression of the given point, and D, that 

 of the horizon, D — D / will then be the difference between the 

 depression of the given point D and that of the horizon D^, 

 which difference may be measured with a sextant or reflect- 

 ing circle, while the depression of the horizon may be com- 

 puted by the usual formulae ; and then D and D — J), become 

 known without an altitude circle on shore, as required in the 

 first method. 



3. From the same figure the method of determining the 

 height above the level of the sea may be very readily derived. 



Let H' be the visible horizon, the angle EAH' = D the 

 depression, A H' a tangent to the horizon at H' passing 

 through the point of observation A, and A B = h, the re- 

 quired height. 



From the property of the circle E A H' = H' C A = 2 H'FD, 

 and H' F B m \ D. The triangles ABH',H'BF give 



shUDicosD:: AB:BH' = AB x -^L 

 a sin I D 



and 



sinD:cos±D::BH':H'C = AB x COS ]? COS f^ 

 2 sin D sin £ D 



= A B cot D cot J D. 



Whence h = H' C tan D tan \ D = £ g tan 2 D nearly . (15.) 

 Introducing the effects of refraction, n, 



h = %g{l + w) 2 tan 2 D (16.) 



or h = l§{l +n) sin 2 l" D" = 0'5832 g sin 2 1" D" (17.) 

 when the dip is expressed in seconds and n — 0'08, 



const, log or log of 0-5832 g sin 2 1" = 6'457582. 



