# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Moon - Antares
From: NavList
Date: 2009 Jan 26, 07:38 -0800

Mr. Reed, I believe that civil discourse requires a bit of formality,
particularly due to some of the invective found in the archives.  Topics that
involve intellectual discourse can become testy, so for now, I will stick to
the nee-plus-ultra formality.

----------

I have been considering the statement by Mr. Reed that the changes in
elevation at the moon's limb may result in "seconds" of variation with
respect to the time of an event

The minimum distance from the earth's center to the moon's center is given by
the Astronomical Almanac (USNO, 2009) as 0.00238 astronomical units.
Further, the Astronomical Almanac gives an astronomical unit as 149597871
kilometers.  Therefore, the distance betwixt centers is

356,043 km = 14959871 km * 0.00238

The radius of the moon is given as 1737.4 km, so we can calculate the semi-diameter as

16.775 arc-minutes = DEGREES(arctangent (1737.4 km / 356043 km))* 60

This agrees, in general, with the Nautical Almanac (USNO, 2009)

The radius of the earth is given as 6378.14 km, so we can calculate the horizontal parallax as

61.5775 arc-minutes = DEGREES(arctangent (6378.14 km / 356043 km)) *60

Again this agrees, in general, with the Nautical Almanac.

The angular rate of the moon is given in the Nautical Almanac,  in the
increments section, as 14 degrees 19 minutes per hour (examine the increment
for 59 minutes and 60 seconds).  Because mountains on the moon will be small
compared to the radius, we must convert to arc-seconds of arc per seconds of
time

14.31666 arc-seconds/second = 14.31666 degrees/hour * (3600 arc-seconds/degree) / 3600 seconds/hour

We can therefore calculate the height of the mountain required to obtain 1
full second, which is the tangent of the angular rate multiplied by the
distance from the earth to the moon.

24.712 km = tangent(RADIANS(14.31666 arc seconds/second)) * 356,043 km

Next we should find the highest mountains on the moon.  While this is subject
to some debate, as there is no sea level on the moon, we can find tall
mountains quite readily by internet search.  We will ignore the placement of
these mountains on the moon, for the purposes of this argument, we will
assume that they are precisely on the moon's limb.  The tallest mountains
are:

Mons Huygens        5.5 km

These mountains fall short of the 24.712 km threshold for 1 second of time.
We can calculate the angle subtended by these mountains, and by application
of the angular rate of the moon, the time.

Angle Subtended, arcseconds = DEGREES(arctangent(mountain height / earth moon distance))*3600

Time, seconds = Angle Subtended * Moons Angular rate

Mons Huygens        0.222 seconds

In short, the tallest mountains on the moon, even if on the limb, fall far
short of "seconds" of time.  One can make arguments that the numbers can be
juggled to affect the result.  For example, we can subtract the earth's
radius from the distance, because the observer may be examining at the
zenith, yet these only affect the result in a marginal way.

From a different standpoint, lunar calculation for longitude could not have
been successful had the limb provided such variation in precision as seconds
of time.  Observers, of necessity, bring the object to the moon at different
limb locations as a function of the close proximity of the moon to the earth.
If seconds of variation existed, then the longitude could not have been
determined with the degree of accuracy necessary.

I look forward to a spirited, if polite, discussion of this topic.

Best Regards

-------------------------------------------------
[Sent from archive by: bmorris-AT-tactronics.com]

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