# Astro Navigation In A Nutshell Part 3

### Part 3 – Calculating Altitude and Azimuth at the DR Position by Spherical Trigonometry.

Astro Navigation In A Nutshell Part Two

Astro Navigation In A Nutshell Part Four

There are several ways of calculating the azimuth and altitude at the DR position; these include the use of sight reduction methods and software solutions. However, the traditional method is by the use of spherical trigonometry which is demonstrated below.

The PZX Triangle

In the diagram above,

X is the position of a celestial body in the Celestial Sphere and U is its Geographical Position on the surface of the Earth.

B is the position of an observer on the surface of the Earth and Z is the zenith of that position.

P is the celestial North Pole and N is the Earth’s North Pole.

PZ is the angular distance from the Celestial North Pole to the zenith of the observer and is equal to 90o – Lat.

PX is the angular distance from the Celestial North Pole to the celestial body and is equal to 90o – Dec.

ZX is the Zenith Distance and is equal to 90o – altitude.

Therefore, altitude is equal to 90o – ZX

The angle ZPX is equal to the Local Hour Angle of the Celestial Body with respect to the observer’s meridian.

The angle PZX is the azimuth of the body with respect to the observer’s meridian.

Summary.

PX = 90o – Dec.

PZ = 90o – Lat.

ZX = 90o – Alt.

Alt = 90o – ZX

<PZX = Azimuth.

<ZPX = Hour angle.

In order to calculate the azimuth and altitude of a celestial body we must solve the triangle PZX in the diagram above.  Specifically, we must calculate the angular distance of side ZX so that we can find the altitude and we must calculate the angle PZX so that we can find the azimuth.

However, because the triangle PZX is on the surface of an imaginary sphere, we cannot solve this triangle by the use of ‘straight line trigonometry’; instead we must resort to the use of ‘spherical trigonometry’ which is explained here.

Example of the use of spherical trigonometry to calculate the azimuth and altitude of a celestial body.

Note.  Traditionally, the ‘half-haversine’ formula was used for this task but this formula does not lend itself well to solution by electronic calculator; therefore, the following solutions which involve the cosine formula have been developed.

Example:  Star Sight.

Scenario:     Greenwich date: 30 June 18hrs 05 mins  33 secs

DR Position:  Lat. 30oN    Long. 45oW

Selected body: Alioth

SHA: 166

Declination: 56oN

GHA Aries:  250

• Calculate LHA

SHA Alioth:   166

Add GHA Aries: 166 + 250 = 416

Subtract Long(W) = 416 – 45 = 371

Subtract 360 = 11

Therefore,  LHA  =  11W

(all results in degrees)

• Calculate PZ/PX/ZPX

PZ = 90o – 30o = 60o      ∴PZ = 60o

PX=  90o – 56o = 34o     ∴PX = 34o

ZPX = LHA = 11west

• Calculate Zenith Distance  (ZX).

As explained here, the formula for calculating side ZX is:

Cos (ZX) =  [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]

∴To calculate zenith distance of Alioth:

Cos (ZX) =  [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]

=  [Cos(60o) . Cos(34o)] + [Sin(60o) . Sin(34o) . Cos(11o)]

=  [0.5 x 0.829} + [0.866 x 0.559 x 0.982]

=  0.415 + 0.475

Cos (ZX) =  0.89

∴ ZX      =  Cos-1 (0.89)  =  27o

• Calculate Altitude.

Altitude  = 90o – ZX   = 90o – 27o  = 63o

• Calculate Azimuth (PZX)

As explained here the formula for calculating angle PZX is:

Cos PZX = Cos(PX) – [Cos(ZX) . Cos(PZ)]  /  [Sin(ZX) . Sin(PZ)]

∴To calculate azimuth of Alioth:

Cos PZX   = Cos(34) – [Cos(27) . Cos(60)]  /  [Sin(27) . Sin(60)]

= 0.829 – [0.89 x 0.5]  /  [0.454 x 0.866)]

=  (0.829 – 0.445) / 0.393

=  0.384   / 0.393   = 0.977

Cos(PZX) = 0.977

∴  PZX   =  Cos-1(0.977)  = 12.31

∴  Azimuth = N12oW (since LHA is west)

In terms of bearing, the azimuth is 348o.

• Summarize results.

LHA =  11west

Declination = 56oN

Azimuth at DR position = N12oW

Altitude at DR position = 63o

Please Note.  This topic is explained in far greater depth in my books ‘Astro Navigation Demystified’ and ‘Celestial Navigation – The Ultimate Course’.