NavList:

Mark Coady you wrote: 

We engineers love our definitions...tend to consider them a life jacket......

I can't help but notice that LHA seems to have two definitions stated in common use. Due to the lovely flowing curves of trigonometry, it all seems to work.....but...for the persnickity types....

A brief search of multiple internet sources last night gave me enough of a statistical split to leave the question open to argument.

The two ways I see LHA defined:  #1 LHA = GHA - Long (simply stated as the "difference in angle", implies to me you would often simply figure by inspection and use the shorter  "included angle")    #2 LHA = GHA +/- Long (+E, -W) (Always measured westward from observer to celestial object)

if I go #1  GHA = 200, Long = 60E   def#1 by inspection I could conclude LHA as the included angle between them: LHA =100 degrees

if I go #2   GHA = 200, Long = 60E  def #2 gives me a strict def as  LHA = 260 degrees circling around from the west.

The cosine value is the same....so...ok....happy enough. (sin & tan are reversed +/-)

Which is the most correct statement:   is it the "difference" only , or "difference measured west". EG as in above the acute angle  (nearest) or obtuse angle measured west the "explementary or conjugate" angle.

Perhaps I am being obtuse more than the angle here....

The first thing to realise is don’t believe everything you read on the internet.  It’s frequently neat, plausible, and wrong!  Very similar explanations then seem to multiply; I wouldn’t dare say are copied.  Secondly, the examples you show aren’t really definitions of LHA; they’re formulae for finding it.  To me, the simplest way of understanding LHA is that celestial bodies go around the earth in as near as damn it 24hours; a bit like the minute hand of a clock.  East is the way the Earth turns, so to an observer on the Earth celestial bodies appear to travel west at 15 degrees of longitude an hour, hence ‘hour angle’.  When a body crosses the observer’s meridian its LHA is zero.  An hour later it will be 15 degrees west of the observer, so its LHA is 015 and so on until 23hours 56 minutes after crossing Greenwich when its LHA will be 359. 

An almanac listing the LHA of a body for every longitude of every observer would be unbearably thick, so they just list its hour angle at Greenwich (GHA) and leave the observer to calculate the rest.  This is easily done by subtracting the observer’s longitude if it’s west or adding their longitude if it’s east. If like me you can never remember whether to add or subtract, just draw a circle as a plan view of the Earth with the pole at the centre and mark on 0E/W, the observer’s meridian, and the body’s meridian as radii, and work it out that way.  If someone only ever operated in the Western Hemisphere, then most of the time they could probably get away with saying LHA=GHA-longitude, and this might be how your first formula came about, but you’d soon be in trouble operating World wide or around the Greenwich Meridian.

It’s probably true to say that LHA is most useful if you’re using navigational tables like AP3270/HO249.  If using pure spherical geometry, to solve the PZX triangle dirctly, then you need the smaller of the two angles between the body’s and the observer’s meridians, which might be LHA or it might be 360-LHA. DaveP