moonrise definition

For an observer at sea level (or I guess maybe with eyes 5 ft above
sea level) and the point on the horizon at the azimuth where the
moon is rising at sea level, is moonrise defined as the time the
moon first begins to peek over the horizon or the time the moon
is fully over the horizon?

I have been using the site http://aa.usno.navy.mil/data/
(but that site is currently down, at least to me, are there
other good ones?) to get moonrise by generating a table of
moon altitude and azimuth versus time (to do that if your
location is not in their database of place names you need
to put in latitude and longitude) and then interpolating
to get the time of upward zero altitude crossing. Do
you know if their table is defined such that that would
give the same as moonrise probably given directly on
other sites (i.e. if moonrise is defined as when the moon
is fully over sea level horizon that table's zero altitude
is the same and not when the moon is e.g. halfway over
the horizon such that the midpoint is in line with sea level)?

OK, that web page is back up so in a few minutes I will
check the moonrise time for tomorrow evening here in
St. John's, Newfoundland Canada (exact time of full moon
is 6:03 p.m. Newfoundland Daylight Time [UTC-2.5h] here)
in a few minutes if it stays up. Hmmm, maybe I should
do Argentia given it is a US naval web site (my parents
worked at the Argentia base so maybe some American
naval musicians jammed with my father).

David
http://www.nfld.com/~dalton (with section on Newfoundland and Labrador
Travel and Music; George Bush Sr. prefers Labrador but I
haven't been there yet myself but hope to reasonably soon)

I would assume it is the same for the moon as it is the sun.

Sun rise is defined as the instant the very first ray of light makes it over
the horizon.

Same for sunset, when the last ray of light disappears over the horizon.

These values are Longitude/Latitude and elevation dependent.

I'm just betting it's the same for the moon.

F Marion

In article p7xlc.22331$Ik.1620004@attbi_s53,
Francis Marion wrote:

I would assume it is the same for the moon as it is the sun.

Sun rise is defined as the instant the very first ray of light
makes it over the horizon.

Same for sunset, when the last ray of light disappears over the
horizon.

Actually the most common definition is "when the upper limb touches a
mathematical horizon". For the Sun this is the same as "when the
very first/last ray makes it over the horizon". But in the case of
the Moon, the upper limb may be dark, i.e. not illuminated by the
Sun. But in spite of that, the "upper limb" definition is normally
used also for the Moon; I've seen no-one bothering to compute when
the first part of the _sunlit_ portion of the lunar disk makes it
over the horizon. If someone actually did that, they'd encounter
another problem at new moon when _no_ part of the lunar disk seen
from the Earth is sunlit....

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://www.stjarnhimlen.se/
http://home.tiscali.se/pausch/

In article ,
David Dalton wrote:

For an observer at sea level (or I guess maybe with eyes 5 ft above
sea level)

Assuming a perfect sea horizon and no refraction in the air, rusing
from sea level to 5 ft above sea level actually makes your local
horizon drop 2 arc minutes.....


and the point on the horizon at the azimuth where the
moon is rising at sea level, is moonrise defined as the time the
moon first begins to peek over the horizon or the time the moon
is fully over the horizon?

Most countries defines sun/moonrise/set as the moment when the upper
limb of the sun/moon touches a mathematical horizon, i.e. an
imaginary line exactly 90 degrees from your local zenith. A few
countries (my own country Sweden as well as our neighbour Finland)
has a different definition: sun/moonrise/set is defined as when the
center of the solar or lunar disk touches the mathematical horizon.

Using the "upper limb" definition for the moon has one disadvantage:
the upper limb of the moon may be dark, i.e. not illuminated by the
sun.


I know of no country which defines rise/set as when the lower limb
touches the horizon.

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://www.stjarnhimlen.se/
http://home.tiscali.se/pausch/

Paul Schlyter wrote:

In article ,
David Dalton wrote:

For an observer at sea level (or I guess maybe with eyes 5 ft above
sea level)

Assuming a perfect sea horizon and no refraction in the air, rusing
from sea level to 5 ft above sea level actually makes your local
horizon drop 2 arc minutes.....

Assuming a linear approximation, just for fun, at my elevation of 9000ft
that would displace the local horizon by 60 degrees! Cool! :-)

I think you mean 2 arc seconds.

--
Greg Crinklaw
Astronomical Software Developer
Cloudcroft, New Mexico, USA (33N, 106W, 2700m)

SkyTools Software for the Observer:
http://www.skyhound.com/cs.html

Skyhound Observing Pages:
http://www.skyhound.com/sh/skyhound.html

To reply have a physician remove your spleen

Greg Crinklaw wrote:
Assuming a linear approximation, just for fun, at my elevation of 9000ft
that would displace the local horizon by 60 degrees! Cool! :-)

I think you mean 2 arc seconds.

Actually...he doesn't.

The thing is that it doesn't drop anywhere *near* linearly. A rule of
thumb is that the distance of the horizon (assuming a spherical Earth,
no refraction, etc) is equal to 90 miles times the square root of the
altitude in miles. This is fairly accurate so long as the altitude
isn't very high; any mountaintop should be just fine.

Thus, 5 feet, which is roughly an altitude of 0.001 miles, gives a local
horizon at a distance of 90 miles times sqrt(0.001), or about 3 miles.
The angle by which the horizon drops is equal to the angle which 3 miles
subtends from the center of the Earth. This is 1 part in about 1,300,
and therefore, in fact, somewhat *more* than 2 arcminutes.

At an elevation of 9,000 feet (about 1.7 miles), the local horizon is
90 miles times sqrt(1.7) or about 117 miles. This is only about 40
times greater than at an altitude of 5 feet, and subtends an angle of
just about 1.7 degrees.

(If you prefer things in meters, the rule of thumb is 113 kilometers
times the square root of your altitude in kilometers.)

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

Brian Tung wrote:
Greg Crinklaw wrote:

Assuming a linear approximation, just for fun, at my elevation of 9000ft
that would displace the local horizon by 60 degrees! Cool! :-)

I think you mean 2 arc seconds.


Actually...he doesn't.

The thing is that it doesn't drop anywhere *near* linearly.

Geeze Brian, I was kidding.

A rule of
thumb is that the distance of the horizon (assuming a spherical Earth,
no refraction, etc) is equal to 90 miles times the square root of the
altitude in miles. This is fairly accurate so long as the altitude
isn't very high; any mountaintop should be just fine.

Which is zero for sea level... I see: it's a trick.

Well, that's all well and good, but of little practical use because it's
not likely anyone is going to observe from a hole in the ground at
exactly sea level with a horizon distance of zero. :-)

--
Greg Crinklaw
Astronomical Software Developer
Cloudcroft, New Mexico, USA (33N, 106W, 2700m)

SkyTools Software for the Observer:
http://www.skyhound.com/cs.html

Skyhound Observing Pages:
http://www.skyhound.com/sh/skyhound.html

To reply have a physician remove your spleen

Greg Crinklaw wrote:
Actually...he doesn't.

The thing is that it doesn't drop anywhere *near* linearly.

Geeze Brian, I was kidding.

Oh. Well, since you seemed to seriously correct him, you must admit
it wasn't obvious. It sure seemed like your 2 arcsec suggestion was
based on linear dropping--as though what was wrong was Paul's figure
for the drop at 5 feet, not the linearity.

If not, never mind.

Well, that's all well and good, but of little practical use because it's
not likely anyone is going to observe from a hole in the ground at
exactly sea level with a horizon distance of zero. :-)

Ah, but you see:

1. Paul was responding to someone who implied that there wasn't
any difference in sunrise/sunset between 0 feet of altitude
and 5 feet; and

2. It's Paul.

(Note to Paul: Don't shoot. g)

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

In article , Brian Tung wrote:

Greg Crinklaw wrote:
Assuming a linear approximation, just for fun, at my elevation of 9000ft
that would displace the local horizon by 60 degrees! Cool! :-)

I think you mean 2 arc seconds.

Actually...he doesn't.

The thing is that it doesn't drop anywhere *near* linearly. A rule of
thumb is that the distance of the horizon (assuming a spherical Earth,
no refraction, etc) is equal to 90 miles times the square root of the
altitude in miles. This is fairly accurate so long as the altitude
isn't very high; any mountaintop should be just fine.

Thus, 5 feet, which is roughly an altitude of 0.001 miles, gives a local
horizon at a distance of 90 miles times sqrt(0.001), or about 3 miles.
The angle by which the horizon drops is equal to the angle which 3 miles
subtends from the center of the Earth. This is 1 part in about 1,300,
and therefore, in fact, somewhat *more* than 2 arcminutes.

At an elevation of 9,000 feet (about 1.7 miles), the local horizon is
90 miles times sqrt(1.7) or about 117 miles. This is only about 40
times greater than at an altitude of 5 feet, and subtends an angle of
just about 1.7 degrees.

(If you prefer things in meters, the rule of thumb is 113 kilometers
times the square root of your altitude in kilometers.)

There's a similar rule of thumb for the horizon depression when you
rise above sea level: assuming negligible atmospheric refraction, the
horizon depression in degrees is very nearly the square root of your
altitude above sea level in kilometers. Thus at 1 km above sea, the
sea horizon is 91 degrees from the zenith. And at 100 km it's 100
degrees from the zenith. Of course this rule remains valid only when
fairly close to the Earth.

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://www.stjarnhimlen.se/
http://home.tiscali.se/pausch/

In sci.astro.amateur Brian Tung wrote:
Greg Crinklaw wrote:
Assuming a linear approximation, just for fun, at my elevation of 9000ft
that would displace the local horizon by 60 degrees! Cool! :-)

I think you mean 2 arc seconds.

Actually...he doesn't.

The thing is that it doesn't drop anywhere *near* linearly. A rule of
thumb is that the distance of the horizon (assuming a spherical Earth,
no refraction, etc) is equal to 90 miles times the square root of the
altitude in miles. This is fairly accurate so long as the altitude
isn't very high; any mountaintop should be just fine.

Thus, 5 feet, which is roughly an altitude of 0.001 miles, gives a local
horizon at a distance of 90 miles times sqrt(0.001), or about 3 miles.

A fact which has sometimes been said to relate to the traditional
"3 mile limit" of territorial waters. On the other hand, that limit
has also been attributed to the range of typical shore cannon during
the time when these international conventions first became common...

Bill Keel