Practical Astronomy Focus: Design and Use of Declination Diagrams

 The use of declination diagrams to work out simple positional astronomy problems is incredibly useful and was first noted in Lancelot Hogben's book, Mathematics For The Million, Ch. 7, pp. 302-305, e.g. in his two diagrams shown below:

Hogben used simple algebraic relationships, equations for the key quantities (i.e. zenith distance z.d., latitude L, declination declin. etc,)  to show how unknown quantities could be derived.

For his basic equation for declination, he writes out the full words for quantities (p. 303):

Declination = Observer's latitude + Meridian z. d.

Or:

Declination = Observer's latitude -  Meridian z.d.

Where the top formula holds for all situations if one reckons zenith distances measured south of the zenith as negative and latitudes as south of the terrestrial (or celestial) equator as negative.  The key aspect to note is that the observer's latitude is the same thing as the elevation of the pole (e.g. Polaris) above the horizon.

For the diagram on the left, showing a star which transits north of the zenith, Hogben writes the relationship of the angles as:

ZOP  = AOP  + z.d.

This is since the star's declination is exactly the angle it makes with the celestial equator - which is at right angles to the polar axis.  Then:

declin.  =   90^o  -  AOP

              =    90^o  - ( ZOP   - z.d.)

               =   90^o  - ( 90^o  - lat.)   + z.d.)

            ∴    declin.  = lat. + z.d.

From which all the relevant angular relations can be gleaned.

 In this way providing a useful quantitative handle for astronomy beginners (or "stargazers") to find their way in the night sky.

The basic elements as I applied them to my own teaching of astronomy - based on my book: 

A History of Caribbean Secondary School Astronomy: Stahl, Philip A.: 9781300913399: Amazon.com: Books

are slightly different as shown below:

For example, take this short problem which seeks to find a star’s declination at a given altitude, A:

The altitude of a star as it transits your meridian is found to be 45^o along a vertical circle at azimuth 180^o, the south point.  Find the declination of the star.

 Since this was designed for students at latitude 13 degrees N. the key to the solution rests on recognizing that z, the zenith distance is negative. From the geometry of the diagram one sees that:

90^o = z + A   or z = 90^o  -  A   = 90^o   -  45^o = 45^o

But since we require: |z| =  f  (latitude) and this is also equal to the altitude of Polaris the pole star:  = a =  13^o   

Then z must have a negative value, or: (-45^o), since:

d (decl.) = z +   f = (-45^o) + 13^o    = -32^o

 This makes sense by examining the right side of the declination diagram.  We use the fact that the zenith distance z plus the star’s altitude (A) must equal  90^o .  Further, we know the celestial equator (CE) defines  0^o declination.  Therefore, the a star’s altitude of A = 45^o  shows it to be SOUTH of the CE. 

How much? Well,  90^o  - 45^o = 45^o . 

But this is still 32^o south of CE, hence must be negative in value. (Remember the CE is only 13 degrees from the zenith point). Of course, most zenith diagrams in tests were deliberately drawn not to scale, in order to make sure students grasp the principles and really attend to the geometry.

Suggested Problems:

1) Show the primary angular relationships for Star B in Lancelot Hogben's declination diagram.

2)Draw a declination diagram for London (lat. 51.5 N). What would be the most southerly star visible by declination? Which stars would be circumpolar? What declination parallel would pass through your zenith

3) What is the maximum altitude which would be attained by Alphecca (declination +26 ^o 50’) at Barbados (latitude 13 deg N).

What would the meridian zenith distance of Alphecca be?

See Also:

Selected Questions- Answers From All Experts Astronomy Forum (Design of the Celestial Sphere)