John:
This has been a favorite "problem" of mine also since developing an interest
in the topic. In different texts, the sign of the EOT differs; this according
to the authors point of reference.
Most I have read however use the convention that a "positive" value means "sun
fast" ; and a negative value means "sun slow". Given this convention one can
write:
WT = LMT -/+ L(e/w) - EOT
where WT = watch time
LMT = local mean time; that is watch time if you were on the
meridian of the zone
- Longitude east of the zonal meridian OR + Longitude west of
the zonal meridian
- the EOT ( where November is positive; ie, sun fast; ie,
sun on the zonal meridian at say, 1145 hours watch time )
Believe me, I've stood staring into space, imagining all sorts of spacial
contortions to work it out in my brain. I try to avoid use of the term LMT
since it can mean "truly local" or "zonal meridian" I prefer "watch time"
or Zonal Meridianal Time to avoid cerebral confusion.
One could of course reverse my convention of the sign on the EOT. Call
November negative; then Longitudinal correction to watch time is +East and
-West.
It adds a bit to some confusion if you use the Nautical Almanac which gives
GHA (hour angle) . My origins and perspective is navigation so the almanac and
conversions are second nature. I've found that in dialling ( new to me) the
thinking is often the reverse (?inverse) of navigational thought.
Good Luck
DAVE
Lat 33º 39' N Long 118º 04'W [ by my own sights and subject to
correction ! :) ]
John Harding wrote:
> I was wondering if there was a convention about the sign of the equation
> of time. Waugh describes 'dial fast' and 'dial slow' but doesn't have a
> sign. His graph looks negative in February. I want to make a graph that
> is positive in February and negative in November and I want to label it as
> such so that users can simple look up the value to apply to LAT to get
> LMT.
>
> So I have two questions... Is it OK to call the equation of time a
> negative number and 2. Am I confused? Actually I already know the answer
> to that one but I mean do I have it right about the sign as described
> above.
>
> Thank you,
> John
