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Astronav. question:

My position is:
               Lat: 00 D
              Long: 000 D

Today, the lenght of the true solar day equals the lenght of the mean solar day.
At 1200h ship's time (zone time 0), the azimuth of the sun was 195 (T).

What date is it today?

Jacquelin Hardy

jhardy@zone.ca


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From mail Mon Oct  7 21:06 EDT 1996
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On Mon, 7 Oct 1996 11:54:11 -0400, M. Jacquelin Hardy wrote:

>Astronav. question:
>
>My position is:
>               Lat: 00 D
>              Long: 000 D
>
>Today, the lenght of the true solar day equals the lenght of the mean solar day.
>At 1200h ship's time (zone time 0), the azimuth of the sun was 195 (T).
>
>What date is it today?
>

Easy one,  02 / 11 / 96

Cheers,
John
=====================================
He who has never been lost is no navigator
=====================================


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>
>My position is:
>               Lat: 00 D
>              Long: 000 D
>
>Today, the lenght of the true solar day equals the lenght of the mean solar
day.
>At 1200h ship's time (zone time 0), the azimuth of the sun was 195 (T).
>
>What date is it today?
>
        The answer is November 2.

        The days of the year when the length of the true(or apparent) solar
day equals the length of the mean solar day can be found by examining a
graph of the Equation of Time (see Bowditch-84, page 491).  Since the
Equation of Time equals the accumulated difference between the apparent and
mean solar days, when the Equation of Time reaches a minimum or maximum that
difference is not changing, that is the mean and apparent solar days are of
equal length.  This occurs four times each year, minima on aproximately May
13 and Nov 2, maxima  on Feb 12 and July 25.

        Next from the Nautical Almanac find the GHA of the sun at noon GMT
(ship's time in the problem) for each of the four possible days.

        Date            GHA
        Feb 12          356d 26.2
        May 13          0d 55.3
        July 25         358d 22.6
        Nov 2           4d 6.5

By inspection it looks like Nov 2 is the right answer as the sun will be
well to the west of the prime meridian (same as the ship's meridian in this
problem).  Actual calculation gives an azimuth of 195d.


        Where did you get these problems?  Send more.


Wes Murfin                 Email: cn1907@coastalnet.com
1409 Cando Place           Voice: 910-455-8746
Jacksonville, NC 28540



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Subject: Astronav. quiz...
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>Astronav. question:
>
>My position is:
>               Lat: 00 D
>              Long: 000 D
>
>Today, the lenght of the true solar day equals the lenght of the mean solar
day.
>At 1200h ship's time (zone time 0), the azimuth of the sun was 195 (T).
>
>What date is it today?
>
>Jacquelin Hardy
>
>jhardy@zone.ca


I get Nov. 2nd. Could be any year, but I'll go for 1996 even though it's not
November yet.

Resoning simular to Wes.=20

Noticed something from our English speaking cousins. Nov 2, 1996 is 2/11/96.
In American it's 11/2/96. Maybe we should all use 2/Nov/96. What do you=
 think?

Gordon

+------------------------------------------------------------+
|  Gordon Talge WB6YKK            e-mail: gtalge@pe.net      |
|  Department of Mathematics      QTH: Loma Linda, CA        |=20
|  Mt. San Jacinto College        Lat.  N  34=B0 03' 03.6720"  |=20
|  San Jacinto, CA                Long. W 117=B0 15' 09.5760"  |
+------------------------------------------------------------+


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To: jhardy@zone.ca (M. Jacquelin Hardy)
Cc: navigation@gomoku.ronin.com
Subject: Re: Astronav. quiz...
Date: Tue, 08 Oct 1996 11:05:21 GMT
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On Mon, 7 Oct 1996 11:54:11 -0400, you wrote:

>Astronav. question:
>My position is:
>               Lat: 00 D
>              Long: 000 D
>
>Today, the lenght of the true solar day equals the lenght of the mean =
solar day.
>At 1200h ship's time (zone time 0), the azimuth of the sun was 195 (T).
>What date is it today?
>
>Jacquelin Hardy
>jhardy@zone.ca
>
Great problem !

We think it is 2.11.1996, same reasoning as in the other
answers we received while puzzling.
Checking with sight reduction formulea gave

LAH sun (12:00:00 UTC) =3D 004d 06.5'
DEC sun =3D 14d 55.3' S
Azimuth =3D 195.05 d
hc =3D 74d 32,1'

Is there more stuff like this out there ??

Best regards from
Tom & Monika


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