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Kieran Kelly wrote:

>To Paul Hirose and others,
>Thank you for your input but I am now completely lost. If I take the USNO
>figures they suggest to me, on an equal altitudes basis, that noon occurs at
>3h 15m 39s. However you appear to demonstrate that this is not correct.
>
>So I went back to first principles and posed the question: If an observer is
>standing at longitude 132d 40'E on July 20, 2002 what time will the sun
>cross his meridian if Meridian

Meridian passage occur when the GP of the Sun is on the observer's meridian
i.e. the Sun GHA is equal to the observer's longitude if West or 360 -
longitude if the observer's longitude is East.

In your case the GHA of the Sun has to be 360 - 132d 40' = 227d 20.0'

 From the almanac for the date July 20, 2002 we find that at 0300 UT the
Sun's GHA is 223d 25.2'

So:
      227d 20.0'
    - 223d 25.2' at 0300 UT
         3d 54.8'

  Searching the Increments and corrections page, we find for this GHA
balance of 3d 54.8' is 15m 39s

So the meridian passage at the observer's location would be 03:15:39 UT
The observer's zone time being -9 the meridian passage time would occur at
12:15:39 (Zone Time) at the observer's longitude

>   Passage quoted in the Nautical Almanac that
>day is 12h 06s? Converting the arc of 132dd 40' to time I derive longitude
>of 8h 50m 40s E.  Subtract his from GMT of Meridian Passage as we are east
>and I derive a mer passage on that day at Long 132dd 40' of 3h 15m 20s GMT.
>This is 19s different than the figure provided by an equal altitudes
>computation from the USNO site.

12h 06m is the meridian passage time at the ZM (Zone Meridian) i.e. at 135d
00.0'E

Difference of longitude between the ZM and the observer's meridian is 2d
20.0' west of the ZM
Converted to time this gives 9m 20s (later because west of ZM)

So  12h06m +  9m20s = 12h15m20s
Bear in mind that the meridian passage times tabulated in the almanac are
rounded to the nearest minute.

Pierre Boucher
Quebec, Canada



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