DERIVATION OF AZIMUTH, HOUR ANGLE, DECLINATION FROM ASTRONOMICAL TRIANGLE

Having the following parameters below Azimuth, Hour angle, Declination can be obtained from Astronomical solution.

Given: Latitude of the place of Observer- Scaled out from Topographical map

            Altitude obtained from Theodolite

Using Cosine Rule:                         

                                                                 

                                                        

<CAB=angle A, <ABC=angle B,<BCA=angle C    <SZP=angle A, <ZPS=angle t,<PSZ=angle q                                 

Cosa=CosbCosc + SinbSincCosA--------------------------(1)

Cosb=CosaCosc+ SinaSincCosB---------------------------(2)

Cosc=CosaCosb + SinaSinbCosC--------------------------(3)

By making CosA,CosB,CosC the subject of the formula from equation (1),(2)and (3)

CosA= (Cosa-CosbCosc)/ SinbSinc--------------------------(4)

CosB= (Cosb-CosaCosc)/ SinaSinc---------------------------(5)

CosC= (Cosc-CosaCosb)/ SinaSinb---------------------------(6)

To  obtain A,B and C, inverse of cosA,CosB,CosC will be computed for:

A=Cos^-1(Cosa-CosbCosc)/ SinbSinc--------------------------(7)

B=Cos^-1 (Cosb-CosaCosc)/ SinaSinc---------------------------(8)

C=Cos^-1 (Cosc-CosaCosb)/ SinaSinb---------------------------(9)

Note:

Ø= Latitude

∂=Declination

H=Altitude

t=Hour angle

A=Azimuth of a line (if A is on East A=A, if on West A=360°-A)

q=Parallatic angle

(90°-∂)=Co-declination

(90°-Ø)=Co-latitude

90°-H= Co-altitude

Compare to equation (1),(2)and(3)

Cos(90°-∂)=Cos(90°-H)Cos(90°-Ø) + Sin(90°-H)Sin(90°-Ø)CosA------------------(10)

Cos(90°-Ø) =Cos(90°-∂)Cos(90°-H)+ Sin(90°-∂)Sin(90°-H)Cosq-------------------(11)

Cos(90°-H)=Cos(90°-∂)Cos(90°-Ø) + Sin(90°-∂)Sin(90°-Ø)Cost…………..………..(12)

Applying the same approach to equation in (4),(5) and(6) in (10),(11) and (12)

CosA=(Cos(90°-∂)-Cos(90°-H)Cos(90°-Ø))/(Sin(90°-H)Sin(90°-Ø))------------------(13)

Cosq=(Cos(90°-Ø)-Cos(90°-∂)Cos(90°-H))/(Sin(90°-∂)Sin(90°-H))-------------------(14)

Cost=(Cos(90°-H)-Cos(90°-∂)Cos(90°-Ø))/(Sin(90°-∂)Sin(90°-Ø))…………..………..(15)

Using the same approach in equation (10),(11)and(12)

A= Cos^-1 ((Cos(90°-∂)-Cos(90°-H)Cos(90°-Ø))/(Sin(90°-H)Sin(90°-Ø)))------------------(16)

q= Cos^-1 ((Cos(90°-Ø)-Cos(90°-∂)Cos(90°-H))/(Sin(90°-∂)Sin(90°-H)))-------------------(17)

t= Cos^-1 ((Cos(90°-H)-Cos(90°-∂)Cos(90°-Ø))/(Sin(90°-∂)Sin(90°-Ø)))…………..………..(18)

Note that:

Sin(90°-X)=CosX, also Cos(90°-Y)=SinY

Where X and Y  are angles less than less than 90°

A= Cos^-1 ((Sin(∂)-Sin(H)Sin(Ø))/(Cos(H)Cos(Ø)))------------------(*)

q= Cos^-1 ((Sin(Ø)-Sin(∂)Sin(H))/(Cos(∂)Cos(H)))-------------------(**)

t= Cos^-1 ((Sin(H)-Sin(∂)Sin(Ø))/(Cos(∂)Cos(Ø)))…………..………..(***)

Compiled by: Ibitowa O. Victor