)20 



HYDROGRAPHICAL SURVEYING [Ari.D. 



Cos 7. Cos (f Vers A 



D. — Til prove Redaction to the Meridian = 



Sin 1" 



Fig. 120. 



Let X be a heavenly body near the Meridian, P the pole, Z the Zenith. 

 Let Hour Angle ZPX = h, Latitude = 90-PZ = ?, Zeuith distance 

 \Z = z. Declination = 90 - P X = c?. 



Cos X Z - Cos P X . Cos P Z 



Then Cos Z P X = 



or Cos h = 



Sin PX. Sm PZ 

 Cos 2 — Sin I . Sin d 



Cos I . Cos d 

 .*. Cos z — Sin I . Sin d = Cos I . Cos d . Cos h 



= Cosl . Cos d .(1 - Vers h) 

 = Cos I . Cos d — Cos I . Cos d . Vers h ; 

 /, Cos z + Cos ? . Cus d . Vers /* = Cos / . Cos d + Sin I . Sm d 



= Cos (? oo (^) 

 = 1 - Vers (Z '^ rf) ; 

 Vers (Z c^ (Z) = 1 - Cos z - Cos Z . Cos d.Yevsh 

 = Vers 2 — Cos Z . Cos d . Vers h. 

 Working with Declination = 90 + P X, we shall get 



= Vers z — Cos I . Cos d . Vers h. 

 But I <^^ d or I + d is the Meridian Zenith Distance = Z. 

 Then Vers Z = Vers z — Cos J . Cos tZ . Vers h. 

 — Cos I . Cos d . Vers h = Vers Z — Vers z 



= 1 — Cos Z — 1 + Cos 2 

 = Cos z — Cos Z 



= - 2 Sin 



2 + Z 



Sin 



2 - Z 



but 2 and Z are nf^arly alike, so 

 and 2-Z is very small .*. 2 Sin 



2 + Z 



2-Z 



may be taken = » 



may be (aken = (z — Z) Sin 1"; 



- Z = 



.-. Cos Z . C js d . Vers A = Sin z . (z - Z) Sin 1", 



Cos 1 . Cos d . Vers li 



tSin 2 Sin 1"' 



but 3 — Z is f'e Reduction to the Meiiilian; 



^ , . , ,, Cos I . Cos d Vers h 



Reduction to Mer. = — ■ • „. — tj, ■ 



bin z hm i 



