Solutions To Practical Astronomy Focus Problems: Use Of Declination Diagrams

 The 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?

Solutions:

1) for Star B which transits south of the zenith:

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

∴    declin.  = lat. -  z.d.

2) The declination diagram constructed for London, latitude 51. ^o 5 N, is shown below:

The most southerly star by declination is marked on the southern horizon. Since CE (the celestial equator) and NCP (the north celestial pole) must be 90 degrees apart, and CE defines the circle for 0 degrees declination, then the most southerly declination is: - 38. ^o5  or 38.5 degrees south of the CE.

The circumpolar stars would be those observed from London which over time describe a circle around the NCP. Thus the condition for circumpolarity would be:  51^o.5  < d < 0

where d denotes declination. The declination of all the stars that would pass through the Londoner's zenith is given by the angle delta extending from the CE (at 0 degrees) to z.

This is:  (90 ^o - 38. ^o 5)  = + 51. ^o 5

Hence, all objects with a declination of +51. ^o5

3) This solution requires a declination diagram for Barbados (lat. 13 deg N) which is shown below:.

 

The star of interest is Alphecca: declination +26 ^o 50’. The maximum altitude a, attained by Alphecca, is easily computed from:

a = (90 ^o - z)

where z denotes the zenith distance (or distance from the observer's zenith).

We know that all objects with declination d = 13 ^o pass through the zenith, therefore the zenith distance z for an object of declination +26 ^o 50’ must be:

z = (26 ^o 50’ - 13 ^o) = 13 ^o 50’.

Note this is also the 'meridian zenith distance'.

Therefore, the maximum altitude, a:

a = (90^o - 13.^o 50’) = 76 ^o 10'