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Main
Date: 28 Jul 2006 10:22:21
From:
Subject: Summer Solstice Calculation Questions
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I was playing around with Starry Night Pro 5, and I thought I'd see if
I could determine the exact time of the summer solstice by watching
the solar coordinates. With SNP, you can select the sun, bring up an
info panel, and then watch the RA and Dec change as you run through
the clock time at the speed of your choice. I saw some things that I
can't explain, and I hope somebody here can help me.
First, in case my problem is with fundamental knowledge, tell me if
I'm wrong about any of these assumptions:
1) The solstice occurs when the sun reaches its maximum declination of
the year, which should be around 23.5 degrees N. The declination
never decreases between the vernal equinox and the summer solstice,
and never increases between the summer solstice and the autumnal
equinox.
2) The sun's RA at the solstice should be very close to 6 hours, but
since the RA is defined by the vernal equinox rather than the
solstice, the solstice may not occur at exactly 6 hours.
3) The sun's RA and Dec do not depend on my location, so if I read in
an almanac that the solstice occurred on June 21 at 8:26 AM EDT, I
need only correct for my time zone, rather than fractions of a time
zone (as I would if I were trying to calculate my local sunrise, for
example).
4) SNP has two sets of RA/Dec coordinates in its info panel, one
labeled J2000, and one labeled JNow. I assume that JNow will be more
accurate for current observations.
OK, assuming all that is correct, here is what I found. All times are
EDT.
1) On June 21, the maximum RA reached was 23 degrees and 26.386
minutes. This was maintained from 2:00 PM to 3:16 PM EDT. The
almanac says the solstice should have been at 8:26 AM EDT.
2) When I ran the time backward from there, the declination slowly
decreased, but it hit a minimum (when it was 23deg 26.314') that
lasted from 2:03 AM to 3:39 AM of June 21, then it began to increase
as I went farther back. It peaked at 23deg 26.332' from about 7:37 PM
to 8:20 PM of June 20, and steadily decreased as I went earlier than
that..
So if SNP is correct, there were three solstices, i.e. a max on both
the 20th and 21st, and a local min between them. I could understand
if round-off errors produced fluctuations right around the true
solstice, but I can't understand
a) the absolute max being nearly six hours off the published time,
and
b) apparently smooth progressions between two maximums nearly a day
apart.
Any explanations, or pointers to URLs, appreciated.
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Date: 28 Jul 2006 21:15:43
From: Sam Wormley
Subject: Re: Summer Solstice Calculation Questions
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no@spam.com wrote:
> I was playing around with Starry Night Pro 5, and I thought I'd see if
> I could determine the exact time of the summer solstice by watching
> the solar coordinates. With SNP, you can select the sun, bring up an
> info panel, and then watch the RA and Dec change as you run through
> the clock time at the speed of your choice. I saw some things that I
> can't explain, and I hope somebody here can help me.
>
> First, in case my problem is with fundamental knowledge, tell me if
> I'm wrong about any of these assumptions:
>
> 1) The solstice occurs when the sun reaches its maximum declination of
> the year, which should be around 23.5 degrees N. The declination
> never decreases between the vernal equinox and the summer solstice,
> and never increases between the summer solstice and the autumnal
> equinox.
>
Good thing I looked this up, as I would have been totally wrong!
Explanatory Supplement - Astronomical Almanac
9.211 Equinoxes and Solstices (pg 477)
The times of the equinoxes and solstices are *defined* when the Sun's
*apparent ecliptic longitude* lambda_s is a multiple of 90°; i.e.,
it is calculated from f(t) = 0, where f(t) = lambda_s -0°, 90°, 180°,
or 270°. Thus in the northern hemisphere, for the spring equinox
f(t) = lambda_s, for the summer solstice, f(t) = lambda_s - 90°, for
the autumn equinox f(t) = lambda_s - 180° and for the winter solstice
f(t) = lambda_s - 270°. At the equinoxes the Sun crosses the equator
when the length of the day exceeds the length of the night due to
refraction, semidiameter, and parallax of the Sun. At that time the
lengths of the day and night are approximately equal everywhere.
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Date: 28 Jul 2006 18:28:08
From: S. Caro
Subject: Re: Summer Solstice Calculation Questions
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Sam Wormley wrote:
>
> Good thing I looked this up, as I would have been totally wrong!
>
> Explanatory Supplement - Astronomical Almanac
>
> 9.211 Equinoxes and Solstices (pg 477)
>
> The times of the equinoxes and solstices are *defined* when the Sun's
> *apparent ecliptic longitude* lambda_s is a multiple of 90°; i.e.,
> it is calculated from f(t) = 0, where f(t) = lambda_s -0°, 90°, 180°,
> or 270°. Thus in the northern hemisphere, for the spring equinox
> f(t) = lambda_s, for the summer solstice, f(t) = lambda_s - 90°, for
> the autumn equinox f(t) = lambda_s - 180° and for the winter solstice
> f(t) = lambda_s - 270°. At the equinoxes the Sun crosses the equator
> when the length of the day exceeds the length of the night due to
> refraction, semidiameter, and parallax of the Sun. At that time the
> lengths of the day and night are approximately equal everywhere.
Does the analemma effect not factor into this ? (Perhaps the math
above explains it :-)
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Date: 28 Jul 2006 23:48:01
From: Sam Wormley
Subject: Re: Summer Solstice Calculation Questions
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S. Caro wrote:
> Sam Wormley wrote:
>
>
>
>> Good thing I looked this up, as I would have been totally wrong!
>>
>> Explanatory Supplement - Astronomical Almanac
>>
>> 9.211 Equinoxes and Solstices (pg 477)
>>
>> The times of the equinoxes and solstices are *defined* when the Sun's
>> *apparent ecliptic longitude* lambda_s is a multiple of 90°; i.e.,
>> it is calculated from f(t) = 0, where f(t) = lambda_s -0°, 90°, 180°,
>> or 270°. Thus in the northern hemisphere, for the spring equinox
>> f(t) = lambda_s, for the summer solstice, f(t) = lambda_s - 90°, for
>> the autumn equinox f(t) = lambda_s - 180° and for the winter solstice
>> f(t) = lambda_s - 270°. At the equinoxes the Sun crosses the equator
>> when the length of the day exceeds the length of the night due to
>> refraction, semidiameter, and parallax of the Sun. At that time the
>> lengths of the day and night are approximately equal everywhere.
>
>
> Does the analemma effect not factor into this ? (Perhaps the math
> above explains it :-)
>
>
The analema is a figure 8-shaped plot of the apparent Sun relative to
the mean Sun. This curve is sometimes seen on globes, maps and the
photography of Dennis di Cicco and most notably, Anthony Ayiomamitis!
And it does change slightly over the millennia.
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Date: 28 Jul 2006 21:24:38
From: Sam Wormley
Subject: Re: Summer Solstice Calculation Questions
|
Sam Wormley wrote:
> no@spam.com wrote:
>
>> I was playing around with Starry Night Pro 5, and I thought I'd see if
>> I could determine the exact time of the summer solstice by watching
>> the solar coordinates. With SNP, you can select the sun, bring up an
>> info panel, and then watch the RA and Dec change as you run through
>> the clock time at the speed of your choice. I saw some things that I
>> can't explain, and I hope somebody here can help me.
>> First, in case my problem is with fundamental knowledge, tell me if
>> I'm wrong about any of these assumptions:
>>
>> 1) The solstice occurs when the sun reaches its maximum declination of
>> the year, which should be around 23.5 degrees N. The declination
>> never decreases between the vernal equinox and the summer solstice,
>> and never increases between the summer solstice and the autumnal
>> equinox.
>>
>
>
> Good thing I looked this up, as I would have been totally wrong!
>
> Explanatory Supplement - Astronomical Almanac
>
> 9.211 Equinoxes and Solstices (pg 477)
>
> The times of the equinoxes and solstices are *defined* when the Sun's
> *apparent ecliptic longitude* lambda_s is a multiple of 90°; i.e.,
> it is calculated from f(t) = 0, where f(t) = lambda_s -0°, 90°, 180°,
> or 270°. Thus in the northern hemisphere, for the spring equinox
> f(t) = lambda_s, for the summer solstice, f(t) = lambda_s - 90°, for
> the autumn equinox f(t) = lambda_s - 180° and for the winter solstice
> f(t) = lambda_s - 270°. At the equinoxes the Sun crosses the equator
> when the length of the day exceeds the length of the night due to
> refraction, semidiameter, and parallax of the Sun. At that time the
> lengths of the day and night are approximately equal everywhere.
>
>
>
>
Yes, as Bill Owen points out, geocentric as opposed to topocentric.
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Date: 28 Jul 2006 13:54:10
From: Bill Owen
Subject: Re: Summer Solstice Calculation Questions
|
no@spam.com wrote:
> I was playing around with Starry Night Pro 5, and I thought I'd see if
> I could determine the exact time of the summer solstice by watching
> the solar coordinates. With SNP, you can select the sun, bring up an
> info panel, and then watch the RA and Dec change as you run through
> the clock time at the speed of your choice. I saw some things that I
> can't explain, and I hope somebody here can help me.
>
> First, in case my problem is with fundamental knowledge, tell me if
> I'm wrong about any of these assumptions:
>
> 1) The solstice occurs when the sun reaches its maximum declination of
> the year, which should be around 23.5 degrees N. The declination
> never decreases between the vernal equinox and the summer solstice,
> and never increases between the summer solstice and the autumnal
> equinox.
The standard definition of the equinoxes and the solstices are actually
in terms of the apparent geocentric *longitude* of the sun, not in terms
of declination. Summer solstice happens when the sun's geocentric
longitude is exactly 90 degrees. (This happens to be equivalent to
RA = 6h.)
> 2) The sun's RA at the solstice should be very close to 6 hours, but
> since the RA is defined by the vernal equinox rather than the
> solstice, the solstice may not occur at exactly 6 hours.
No; see above.
> 3) The sun's RA and Dec do not depend on my location, so if I read in
> an almanac that the solstice occurred on June 21 at 8:26 AM EDT, I
> need only correct for my time zone, rather than fractions of a time
> zone (as I would if I were trying to calculate my local sunrise, for
> example).
Actually the apparent RA and Dec *do* depend on your location, because
of both topocentric parallax and topocentric aberration. The latter
doesn't affect the declination though.
> 4) SNP has two sets of RA/Dec coordinates in its info panel, one
> labeled J2000, and one labeled JNow. I assume that JNow will be more
> accurate for current observations.
JNow is presumably true-of-date coordinates.
> OK, assuming all that is correct, here is what I found. All times are
> EDT.
>
> 1) On June 21, the maximum RA reached was 23 degrees and 26.386
> minutes. This was maintained from 2:00 PM to 3:16 PM EDT. The
> almanac says the solstice should have been at 8:26 AM EDT.
>
> 2) When I ran the time backward from there, the declination slowly
> decreased, but it hit a minimum (when it was 23deg 26.314') that
> lasted from 2:03 AM to 3:39 AM of June 21, then it began to increase
> as I went farther back. It peaked at 23deg 26.332' from about 7:37 PM
> to 8:20 PM of June 20, and steadily decreased as I went earlier than
> that..
>
> So if SNP is correct, there were three solstices, i.e. a max on both
> the 20th and 21st, and a local min between them. I could understand
> if round-off errors produced fluctuations right around the true
> solstice, but I can't understand
> a) the absolute max being nearly six hours off the published time,
> and
> b) apparently smooth progressions between two maximums nearly a day
> apart.
The two local maxima in declination are probably caused by topocentric
effects. The difference between the declination at those maxima is
0.018' by your reckoning, just over 1". The earth subtends an angle of
about 17" as seen from the sun; the difference between geocentric and
topocentric coordinates can amount to over 8".
Can you set your coordinates to the center of the earth and see what
happens?
> Any explanations, or pointers to URLs, appreciated.
>
>
One last thing: As the ecliptic is defined as the mean orbit of the
earth-moon barycenter, and as the moon's orbit is inclined 5 deg to the
ecliptic, the earth bobs up and down each month, and the sun's ecliptic
latitude can reach about 1 arcsec north or south. Add this effect on
top of the earth's orbital motion, and you'll see that the sun's maximum
declination won't necessarily happen at ecliptic longitude 90 deg (or
RA 6h). Likewise, the sun's declination won't be exactly zero at the
equinoxes, when the longitude is 0 or 180.
-- Bill Owen
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Date: 28 Jul 2006 17:21:03
From:
Subject: Re: Summer Solstice Calculation Questions
|
On Fri, 28 Jul 2006 13:54:10 -0700, Bill Owen <wmo@jpl.nasa.gov >
wrote:
>The standard definition of the equinoxes and the solstices are actually
>in terms of the apparent geocentric *longitude* of the sun, not in terms
>of declination. Summer solstice happens when the sun's geocentric
>longitude is exactly 90 degrees. (This happens to be equivalent to
>RA = 6h.)
OK, when I looked at that, it happened at 8:15 for my location (w
Oregon) and 8:17 for the center of the earth. So that's just nine
minutes off the almanac prediction; much better than six hours.
>
>Actually the apparent RA and Dec *do* depend on your location, because
>of both topocentric parallax and topocentric aberration. The latter
>doesn't affect the declination though.
>
>> 4) SNP has two sets of RA/Dec coordinates in its info panel, one
>> labeled J2000, and one labeled JNow. I assume that JNow will be more
>> accurate for current observations.
>
>JNow is presumably true-of-date coordinates.
Yes. The JNow RA was exactly 6 when the Long was exactly 90. The
J2000 lagged a bit.
>
>The two local maxima in declination are probably caused by topocentric
>effects. The difference between the declination at those maxima is
>0.018' by your reckoning, just over 1". The earth subtends an angle of
>about 17" as seen from the sun; the difference between geocentric and
>topocentric coordinates can amount to over 8".
>
>Can you set your coordinates to the center of the earth and see what
>happens?
Yes. The max dec was 23deg 26.455', and it happened from 7:39 AM to
8:55 AM EDT, so the 90deg longitude was within that window. I don't
see any deviation from an increase before, and a decrease after, the
max. So I guess the puzzle is solved.
>
>One last thing: As the ecliptic is defined as the mean orbit of the
>earth-moon barycenter, and as the moon's orbit is inclined 5 deg to the
>ecliptic, the earth bobs up and down each month, and the sun's ecliptic
>latitude can reach about 1 arcsec north or south. Add this effect on
>top of the earth's orbital motion, and you'll see that the sun's maximum
>declination won't necessarily happen at ecliptic longitude 90 deg (or
>RA 6h). Likewise, the sun's declination won't be exactly zero at the
>equinoxes, when the longitude is 0 or 180.
>
>-- Bill Owen
Thanks for the help.
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Date: 28 Jul 2006 19:32:22
From: SkySea
Subject: Re: Summer Solstice Calculation Questions
|
>On Fri, 28 Jul 2006 10:22:21 -0700, no@spam.com wrote:
>1) The solstice occurs when the sun reaches its maximum declination of
>the year, which should be around 23.5 degrees N. The declination
>never decreases between the vernal equinox and the summer solstice,
>and never increases between the summer solstice and the autumnal
>equinox.
Correct, for half the year. You can include the whoile year by going
solstice-to-solstice: the Sun increases in dec from winter solstice to
summer solstice, decreases from summer solstice to winter solstice.
The appropriate equinoxes occur in between.
>2) The sun's RA at the solstice should be very close to 6 hours, but
>since the RA is defined by the vernal equinox rather than the
>solstice, the solstice may not occur at exactly 6 hours.
True for times off a chart's given epoch. By definition though, 0h is
the intersection of the equator and the ecliptic, and the
intersections of the ecliptic with the 6, 18, and 12 hour circles mark
the equioxes and solstices.
>3) The sun's RA and Dec do not depend on my location, so if I read in
>an almanac that the solstice occurred on June 21 at 8:26 AM EDT, I
>need only correct for my time zone, rather than fractions of a time
>zone (as I would if I were trying to calculate my local sunrise, for
>example).
The RA and dec do depend on your location. However, given the diameter
of the Earth, it's really, really, miniscule. The solsitces and
equinoxes take the alignments of the centers of the bodies. Your
parallax off of that will vary.
>4) SNP has two sets of RA/Dec coordinates in its info panel, one
>labeled J2000, and one labeled JNow. I assume that JNow will be more
>accurate for current observations.
Those are the epochs. The great circle equator rotates around the sky
as the earth's axis wobbles over 25,600 years. That's enough to cause
errors when using large magnifications in telescopes. Typically,
standard epocjs are given every 50 years. 1950 was a common epoch
until halfway to 2000 (1975), when it became less accurate, but more
available because of already-preinted charts. Software can now display
wahtever epoch you like, down to real-time (now).
>OK, assuming all that is correct, here is what I found. All times are
>EDT.
>
>1) On June 21, the maximum RA reached was 23 degrees and 26.386
>minutes. This was maintained from 2:00 PM to 3:16 PM EDT. The
>almanac says the solstice should have been at 8:26 AM EDT.
>
>2) When I ran the time backward from there, the declination slowly
>decreased, but it hit a minimum (when it was 23deg 26.314') that
>lasted from 2:03 AM to 3:39 AM of June 21, then it began to increase
>as I went farther back. It peaked at 23deg 26.332' from about 7:37 PM
>to 8:20 PM of June 20, and steadily decreased as I went earlier than
>that..
>
>So if SNP is correct, there were three solstices, i.e. a max on both
>the 20th and 21st, and a local min between them. I could understand
>if round-off errors produced fluctuations right around the true
>solstice, but I can't understand
> a) the absolute max being nearly six hours off the published time,
>and
>b) apparently smooth progressions between two maximums nearly a day
>apart.
>
>Any explanations, or pointers to URLs, appreciated.
Dunno offhand. Difference in epochs?
=============
- Dale Gombert (SkySea at aol.com)
122.38W, 47.58N, W. Seattle, WA
http://flavorj.com/~skysea
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Date: 28 Jul 2006 13:36:10
From: Brian Tung
Subject: Re: Summer Solstice Calculation Questions
|
SkySea wrote:
> >2) The sun's RA at the solstice should be very close to 6 hours, but
> >since the RA is defined by the vernal equinox rather than the
> >solstice, the solstice may not occur at exactly 6 hours.
>
> True for times off a chart's given epoch. By definition though, 0h is
> the intersection of the equator and the ecliptic, and the
> intersections of the ecliptic with the 6, 18, and 12 hour circles mark
> the equioxes and solstices.
No, I think the original poster was right: The vernal equinox is at 0h,
but the autumnal equinox and the solstices are all based on properties
of the ecliptic and the celestial equator. So the autumnal equinox is
when the ecliptic crosses the equator going southward, and the solstices
are the two points where the Sun's declination is at one of its two
extremes.
I have no explanation for the bizarre behavior of SNP, though.
--
Brian Tung <brian@isi.edu >
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.html
|
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Date: 28 Jul 2006 14:03:02
From: Bill Owen
Subject: Re: Summer Solstice Calculation Questions
|
Brian Tung wrote:
> SkySea wrote:
>
>>>2) The sun's RA at the solstice should be very close to 6 hours, but
>>>since the RA is defined by the vernal equinox rather than the
>>>solstice, the solstice may not occur at exactly 6 hours.
>>
>>True for times off a chart's given epoch. By definition though, 0h is
>>the intersection of the equator and the ecliptic, and the
>>intersections of the ecliptic with the 6, 18, and 12 hour circles mark
>>the equioxes and solstices.
>
>
> No, I think the original poster was right: The vernal equinox is at 0h,
> but the autumnal equinox and the solstices are all based on properties
> of the ecliptic and the celestial equator. So the autumnal equinox is
> when the ecliptic crosses the equator going southward, and the solstices
> are the two points where the Sun's declination is at one of its two
> extremes.
>
> I have no explanation for the bizarre behavior of SNP, though.
Sorry I didn't see this post before I finished replying to the original
one.
The vernal equinox point on the celestial sphere is indeed exactly at
RA 0h, Dec 0 deg in equatorial coordinates; latitude 0 deg, longitude
0 deg in ecliptic coordinates.
The center of the sun will NOT go through this point in general, because
its geocentric latitude is zero only at the instant when the moon is at
one of the nodes of its orbit. If the sun's latitude happens to be
positive around March 20, it will cross the celestial equator before it
hits longitude 0.
If the moon's "argument of latitude" is between 90 and 270 deg, it is
moving southward in ecliptic coordinates, the earth is moving
northward, and the sun's apparent latitude is decreasing. If this
situation holds around June 21, the sun's declination will max out
before the solstice. Remember, the solstice is defined by ecliptic
longitude, not by RA, not by maximum Dec.
-- Bill Owen
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Date: 28 Jul 2006 17:47:03
From: Brian Tung
Subject: Re: Summer Solstice Calculation Questions
|
Bill Owen wrote:
> The center of the sun will NOT go through this point in general, because
> its geocentric latitude is zero only at the instant when the moon is at
> one of the nodes of its orbit. If the sun's latitude happens to be
> positive around March 20, it will cross the celestial equator before it
> hits longitude 0.
>
> If the moon's "argument of latitude" is between 90 and 270 deg, it is
> moving southward in ecliptic coordinates, the earth is moving
> northward, and the sun's apparent latitude is decreasing. If this
> situation holds around June 21, the sun's declination will max out
> before the solstice. Remember, the solstice is defined by ecliptic
> longitude, not by RA, not by maximum Dec.
OK, interesting, I did not know that. What if we observe the Sun from
the Earth-Moon barycenter; does the Sun go through the FPA at the
equinox from that perspective, even if the Moon is not at one of its
orbital nodes?
--
Brian Tung <brian@isi.edu >
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.html
|
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Date: 31 Jul 2006 09:28:43
From: Bill Owen
Subject: Re: Summer Solstice Calculation Questions
|
Brian Tung wrote:
> Bill Owen wrote:
>
>>The center of the sun will NOT go through this point in general, because
>>its geocentric latitude is zero only at the instant when the moon is at
>>one of the nodes of its orbit. If the sun's latitude happens to be
>>positive around March 20, it will cross the celestial equator before it
>>hits longitude 0.
>>
>>If the moon's "argument of latitude" is between 90 and 270 deg, it is
>>moving southward in ecliptic coordinates, the earth is moving
>>northward, and the sun's apparent latitude is decreasing. If this
>>situation holds around June 21, the sun's declination will max out
>>before the solstice. Remember, the solstice is defined by ecliptic
>>longitude, not by RA, not by maximum Dec.
>
>
> OK, interesting, I did not know that. What if we observe the Sun from
> the Earth-Moon barycenter; does the Sun go through the FPA at the
> equinox from that perspective, even if the Moon is not at one of its
> orbital nodes?
No. Here's why:
The ecliptic is defined as the *mean* orbit of the Earth-Moon
barycenter. Short- and long-period perturbations from the other
planets move the EMB off the ecliptic. To put it another way, the
osculating orbit of the EMB will have a small but nonzero inclination,
and the EMB itself will never (well, hardly ever) be at the node.
-- Bill
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Date: 31 Jul 2006 20:14:34
From: Paul Schlyter
Subject: Re: Summer Solstice Calculation Questions
|
In article <44CE2FBB.50809@jpl.nasa.gov >, Bill Owen <wmo@jpl.nasa.gov> wrote:
> Brian Tung wrote:
>> Bill Owen wrote:
>>
>>>The center of the sun will NOT go through this point in general, because
>>>its geocentric latitude is zero only at the instant when the moon is at
>>>one of the nodes of its orbit. If the sun's latitude happens to be
>>>positive around March 20, it will cross the celestial equator before it
>>>hits longitude 0.
>>>
>>>If the moon's "argument of latitude" is between 90 and 270 deg, it is
>>>moving southward in ecliptic coordinates, the earth is moving
>>>northward, and the sun's apparent latitude is decreasing. If this
>>>situation holds around June 21, the sun's declination will max out
>>>before the solstice. Remember, the solstice is defined by ecliptic
>>>longitude, not by RA, not by maximum Dec.
>>
>>
>> OK, interesting, I did not know that. What if we observe the Sun from
>> the Earth-Moon barycenter; does the Sun go through the FPA at the
>> equinox from that perspective, even if the Moon is not at one of its
>> orbital nodes?
>
> No. Here's why:
>
> The ecliptic is defined as the *mean* orbit of the Earth-Moon
> barycenter. Short- and long-period perturbations from the other
> planets move the EMB off the ecliptic. To put it another way, the
> osculating orbit of the EMB will have a small but nonzero inclination,
> and the EMB itself will never (well, hardly ever) be at the node.
>
> -- Bill
The most important perturbations on the EMB comes from Venus and Jupiter.
This will cause the EMB to cross the plane of the ecliptic a few times
each Earth year. I don't think one can say that something "never" or
"hardly ever" happens when it happens THAT often....
The inclination of the EMB's osculating orbit is always very small of
course, but the longitude of the node of the EMB's osculating orbit
will vary a lot, quickly, and irregularly.
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/
|
| | | | | | | |
Date: 31 Jul 2006 14:34:18
From: Bill Owen
Subject: Re: Summer Solstice Calculation Questions
|
Paul Schlyter wrote:
> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov> wrote:
>>The ecliptic is defined as the *mean* orbit of the Earth-Moon
>>barycenter. Short- and long-period perturbations from the other
>>planets move the EMB off the ecliptic. To put it another way, the
>>osculating orbit of the EMB will have a small but nonzero inclination,
>>and the EMB itself will never (well, hardly ever) be at the node.
>>
>>-- Bill
>
>
> The most important perturbations on the EMB comes from Venus and Jupiter.
> This will cause the EMB to cross the plane of the ecliptic a few times
> each Earth year. I don't think one can say that something "never" or
> "hardly ever" happens when it happens THAT often....
>
> The inclination of the EMB's osculating orbit is always very small of
> course, but the longitude of the node of the EMB's osculating orbit
> will vary a lot, quickly, and irregularly.
You're right ... but I didn't write that the EMB will never (well,
hardly ever) *cross* the node, but that it will never (well, hardly
ever) *be at* the node. Yes, crossings happen quite often, but the
set of times when the EMB is *exactly* on the ecliptic is still a set
of measure zero.
Sorry if I confused you.
-- Bill
|
| | | | | | | | |
Date: 01 Aug 2006 11:13:07
From: Paul Schlyter
Subject: Re: Summer Solstice Calculation Questions
|
In article <44CE775A.7090003@jpl.nasa.gov >,
Bill Owen <wmo@jpl.nasa.gov > wrote:
> Paul Schlyter wrote:
>> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov> wrote:
>>>The ecliptic is defined as the *mean* orbit of the Earth-Moon
>>>barycenter. Short- and long-period perturbations from the other
>>>planets move the EMB off the ecliptic. To put it another way, the
>>>osculating orbit of the EMB will have a small but nonzero inclination,
>>>and the EMB itself will never (well, hardly ever) be at the node.
>>>
>>>-- Bill
>>
>>
>> The most important perturbations on the EMB comes from Venus and Jupiter.
>> This will cause the EMB to cross the plane of the ecliptic a few times
>> each Earth year. I don't think one can say that something "never" or
>> "hardly ever" happens when it happens THAT often....
>>
>> The inclination of the EMB's osculating orbit is always very small of
>> course, but the longitude of the node of the EMB's osculating orbit
>> will vary a lot, quickly, and irregularly.
>
> You're right ... but I didn't write that the EMB will never (well,
> hardly ever) *cross* the node, but that it will never (well, hardly
> ever) *be at* the node. Yes, crossings happen quite often, but the
> set of times when the EMB is *exactly* on the ecliptic is still a set
> of measure zero.
>
> Sorry if I confused you.
>
> -- Bill
From a strictly mathematical point you're of course right. Otoh this
does not apply just to zero latitude, it applies to *any* given
ecliptic latitude: the EMB is never (well hardly ever) there....
So one might then wonder: if the EMB never (well hardly ever) is
anywhere, where *is* it ???? And if the EMB is on one side of the
ecliptic, and later on the other side of the ecliptic, without ever
having been on the ecliptic in between -- how did it get through the
ecliptic? Some tunnel effect? :-)
Since we're dealing with measureable quantities which all have some
measurement error, the purely mathematical point of view isn't
particularly relevant. A more relevant point of view would be to e.g.
consider when the EMB is close enough to the plane of the ecliptic to
be, with our current measurement capabilities (or with some tolerance
level), indistinguishable from being exactly on the ecliptic. And
that will be a time interval longer than zero, and it will happen a
few times each year.
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/
|
| | | | | | |
Date: 04 Aug 2006 07:29:58
From: Odysseus
Subject: Re: Summer Solstice Calculation Questions
|
In article <44CE2FBB.50809@jpl.nasa.gov >, Bill Owen <wmo@jpl.nasa.gov>
wrote:
<snip >
> The ecliptic is defined as the *mean* orbit of the Earth-Moon
> barycenter. Short- and long-period perturbations from the other
> planets move the EMB off the ecliptic. To put it another way, the
> osculating orbit of the EMB will have a small but nonzero inclination,
> and the EMB itself will never (well, hardly ever) be at the node.
What is the greatest (most extreme) ecliptic latitude of the Sun?
--
Odysseus
|
| | | | | | | |
Date: 04 Aug 2006 08:30:24
From: Bill Owen
Subject: Re: Summer Solstice Calculation Questions
|
Odysseus wrote:
> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov>
> wrote:
>
> <snip>
>
>> The ecliptic is defined as the *mean* orbit of the Earth-Moon
>> barycenter. Short- and long-period perturbations from the other
>> planets move the EMB off the ecliptic. To put it another way, the
>> osculating orbit of the EMB will have a small but nonzero inclination,
>> and the EMB itself will never (well, hardly ever) be at the node.
>
> What is the greatest (most extreme) ecliptic latitude of the Sun?
I looked through 9 years of the _Astronomical Almanac_, and the greatest
excursion I could find was 1.06 arcsec, in early November 1999.
The moon contributes about 0.6 arcsec, and it's easy to see that monthly
signature in the tables. If you take that out, what's left changes on
a time scale of months, due (as Paul pointed out earlier in this thread)
to perturbations mostly by Venus and Jupiter.
Venus was in inferior conjunction on August 20, 1999, when it was near
its maximum southern heliocentric latitude (i.e. about as far below the
ecliptic as it can get). This circumstance would produce a downward
perturbation on the earth lasting several months, and the maximum
solar latitude indeed occurred that November.
Meanwhile, Jupiter was at the southernmost point of its orbit in March
of that year, and at opposition in ober. Pretty much the same story.
So the two planets which perturb the earth's orbit the most were both
well south of the ecliptic when they were closest to us, moving the
earth-moon barycenter southward as well. Then in early November, the
moon was at its *northernmost* ecliptic latitude, so the earth (being
on the other side of the barycenter) was at the southern point of its
monthly motion. All three -- Venus, Jupiter, Moon -- moved the earth
southward. The result was that the sun appeared at a latitude of
+1.06 arcsec.
-- Bill
|
| | | | | | | | |
Date: 05 Aug 2006 08:41:32
From: Odysseus
Subject: Re: Summer Solstice Calculation Questions
|
In article <44D36810.3050503@jpl.nasa.gov >,
Bill Owen <wmo@jpl.nasa.gov > wrote:
<snip >
> So the two planets which perturb the earth's orbit the most were both
> well south of the ecliptic when they were closest to us, moving the
> earth-moon barycenter southward as well. Then in early November, the
> moon was at its *northernmost* ecliptic latitude, so the earth (being
> on the other side of the barycenter) was at the southern point of its
> monthly motion. All three -- Venus, Jupiter, Moon -- moved the earth
> southward. The result was that the sun appeared at a latitude of
> +1.06 arcsec.
Thanks for the detailed analysis! It certainly appears that there must
be few occasions on which those factors would combine to greater effect
than they did in 1999.
--
Odysseus
|
| |
Date: 28 Jul 2006 10:52:03
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
The oldest European monument standing is an astronomical clock which
registers the mid-point between the Sept/Mar Equinoxes.
http://www.iol.ie/~geniet/eng/newgrang.htm
The builders constructed the roofbox to register the midpoint
(solstice) at exactly the same time each year at roughly 9:11 AM each
and every year.
Now,the Ra/Dec system is based on 3 years of 365 days and 1 year of 366
days hence it is impossible to construct a monument like the solstice
marker of Newgrange based on the calendar system.
Astronomy and timekeeping,being a Universal heritage of humanity,has
descended so far from the careful reasoning of those people of
Newgrange,Stonehenge,the pyramid builders,the Ptolemaic astronomers and
the heliocentric astronomers.The great intutive insights which balance
the intutive with the observational faculties are now almost lost .
In short,if you believe a computer program (an observational
convenience) can substitute for astronomy and its insights,it ain't
astronomy you are doing.
no@spam.com wrote:
> I was playing around with Starry Night Pro 5, and I thought I'd see if
> I could determine the exact time of the summer solstice by watching
> the solar coordinates. With SNP, you can select the sun, bring up an
> info panel, and then watch the RA and Dec change as you run through
> the clock time at the speed of your choice. I saw some things that I
> can't explain, and I hope somebody here can help me.
>
> First, in case my problem is with fundamental knowledge, tell me if
> I'm wrong about any of these assumptions:
>
> 1) The solstice occurs when the sun reaches its maximum declination of
> the year, which should be around 23.5 degrees N. The declination
> never decreases between the vernal equinox and the summer solstice,
> and never increases between the summer solstice and the autumnal
> equinox.
>
> 2) The sun's RA at the solstice should be very close to 6 hours, but
> since the RA is defined by the vernal equinox rather than the
> solstice, the solstice may not occur at exactly 6 hours.
>
> 3) The sun's RA and Dec do not depend on my location, so if I read in
> an almanac that the solstice occurred on June 21 at 8:26 AM EDT, I
> need only correct for my time zone, rather than fractions of a time
> zone (as I would if I were trying to calculate my local sunrise, for
> example).
>
> 4) SNP has two sets of RA/Dec coordinates in its info panel, one
> labeled J2000, and one labeled JNow. I assume that JNow will be more
> accurate for current observations.
>
> OK, assuming all that is correct, here is what I found. All times are
> EDT.
>
> 1) On June 21, the maximum RA reached was 23 degrees and 26.386
> minutes. This was maintained from 2:00 PM to 3:16 PM EDT. The
> almanac says the solstice should have been at 8:26 AM EDT.
>
> 2) When I ran the time backward from there, the declination slowly
> decreased, but it hit a minimum (when it was 23deg 26.314') that
> lasted from 2:03 AM to 3:39 AM of June 21, then it began to increase
> as I went farther back. It peaked at 23deg 26.332' from about 7:37 PM
> to 8:20 PM of June 20, and steadily decreased as I went earlier than
> that..
>
> So if SNP is correct, there were three solstices, i.e. a max on both
> the 20th and 21st, and a local min between them. I could understand
> if round-off errors produced fluctuations right around the true
> solstice, but I can't understand
> a) the absolute max being nearly six hours off the published time,
> and
> b) apparently smooth progressions between two maximums nearly a day
> apart.
>
> Any explanations, or pointers to URLs, appreciated.
|
| | |
Date: 28 Jul 2006 19:53:56
From: SkySea
Subject: Re: Summer Solstice Calculation Questions
|
Wrong.
To say that the solstice is reached at a particlar time (I assume you
also mean on a particualr day, but regardless) each year is a
statement that the Earth's rotation period is an exact fraction of the
orbital period. It isn't. It also implies that the Earth's orbital
period and rotational periods are constant. They aren't. Or it assumes
that the variations in the the orbital period and rotation are somehow
locked, so they remain in synchronization. They aren't.
Look through some almanacs and find out at what times the solstices
occured. It's not always at 9:11, and the liklihood that it ever will
be again is 1:1,440.
>"oriel36" <geraldkelleher@yahoo.com> piddled:
>The builders constructed the roofbox to register the midpoint
>(solstice) at exactly the same time each year at roughly 9:11 AM each
>and every year.
=============
- Dale Gombert (SkySea at aol.com)
122.38W, 47.58N, W. Seattle, WA
http://flavorj.com/~skysea
|
| |
Date: 29 Jul 2006 05:15:06
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Sam Wormley wrote:
> S. Caro wrote:
> > Sam Wormley wrote:
> >
> >
> >
> >> Good thing I looked this up, as I would have been totally wrong!
> >>
> >> Explanatory Supplement - Astronomical Almanac
> >>
> >> 9.211 Equinoxes and Solstices (pg 477)
> >>
> >> The times of the equinoxes and solstices are *defined* when the Sun's
> >> *apparent ecliptic longitude* lambda_s is a multiple of 90=B0; i.e.,
> >> it is calculated from f(t) =3D 0, where f(t) =3D lambda_s -0=B0, 90=
=B0, 180=B0,
> >> or 270=B0. Thus in the northern hemisphere, for the spring equinox
> >> f(t) =3D lambda_s, for the summer solstice, f(t) =3D lambda_s - 90=
=B0, for
> >> the autumn equinox f(t) =3D lambda_s - 180=B0 and for the winter sol=
stice
> >> f(t) =3D lambda_s - 270=B0. At the equinoxes the Sun crosses the equ=
ator
> >> when the length of the day exceeds the length of the night due to
> >> refraction, semidiameter, and parallax of the Sun. At that time the
> >> lengths of the day and night are approximately equal everywhere.
> >
> >
> > Does the analemma effect not factor into this ? (Perhaps the math
> > above explains it :-)
> >
> >
>
> The analema is a figure 8-shaped plot of the apparent Sun relative to
> the mean Sun. This curve is sometimes seen on globes, maps and the
> photography of Dennis di Cicco and most notably, Anthony Ayiomamitis!
> And it does change slightly over the millennia.
The analemma is a consequence of using a human devised principle that
is the equable 24 hour day to register the Sun against the
Equator/axis,it is so childish that it can only mean the noble
discipline of astronomy has reached a dire state.
Natural noon sunlight strikes the sundial on the exact same line
throughout the year,the shadow may be long or short depending on the
orbital orientation of the Earth at specific points in its annual
orbit but there is no figure 8..
The reason that the determination of natural noon is vital using the
line on a sundial is that the Equation of Time correction is applied to
the natural noon event to reduce the total length of the natural
unequal day to the equable 24 hour day.
Those horrible creeps in the 17th century,in attempting to use the
celestial sphere to gauge terrestial longitudes forced an axial tilt
component to the Sun and kept the Earth orientated to a celestial
sphere.A 21st century person would marvel at the audacious vandalism
wrought on the Equation of Time correction and move on but the utter
incapacity of men to be clinical is removing the hideous corruption
only highlights the dominance of the destructive empirical cult.
How I enjoy turning a scrap of astronomical information into something
entirely different while you plod along with dull and dreary comments
based on a celestial sphere.
|
| |
Date: 29 Jul 2006 03:44:06
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
People would be satisfied to know what happens the quarter of a day
each year that goes into the calendrical correction every fourth year
as a leap day on Feb 29th.
Instead of recognising that the calendrical system cannot be justified
astronomically and exists as a complimentary extension of the Equation
of Time system which generates the equable 24 hour day from the
relationship of axial and orbital motions/orientations over the course
of an annual orbit,you bluff and bluster to those who do not care and
no know better.
You are a damgerous and silly group of people at this stage in human
existence,when humanity deserves to be treated better just as most life
sciences try to present the wonders of nature,incompetence still
dominate the astronomical landscape with an awful version of the
Earth's motions,even the basic one.
Not even the Ptolemaic astronomers held on to the stellar background in
drawing conclusions on observed planetary motions yet you do it for
the Sun as well.
Bill Owen wrote:
> Brian Tung wrote:
> > SkySea wrote:
> >
> >>>2) The sun's RA at the solstice should be very close to 6 hours, but
> >>>since the RA is defined by the vernal equinox rather than the
> >>>solstice, the solstice may not occur at exactly 6 hours.
> >>
> >>True for times off a chart's given epoch. By definition though, 0h is
> >>the intersection of the equator and the ecliptic, and the
> >>intersections of the ecliptic with the 6, 18, and 12 hour circles mark
> >>the equioxes and solstices.
> >
> >
> > No, I think the original poster was right: The vernal equinox is at 0h,
> > but the autumnal equinox and the solstices are all based on properties
> > of the ecliptic and the celestial equator. So the autumnal equinox is
> > when the ecliptic crosses the equator going southward, and the solstices
> > are the two points where the Sun's declination is at one of its two
> > extremes.
> >
> > I have no explanation for the bizarre behavior of SNP, though.
>
> Sorry I didn't see this post before I finished replying to the original
> one.
>
> The vernal equinox point on the celestial sphere is indeed exactly at
> RA 0h, Dec 0 deg in equatorial coordinates; latitude 0 deg, longitude
> 0 deg in ecliptic coordinates.
>
> The center of the sun will NOT go through this point in general, because
> its geocentric latitude is zero only at the instant when the moon is at
> one of the nodes of its orbit. If the sun's latitude happens to be
> positive around March 20, it will cross the celestial equator before it
> hits longitude 0.
>
> If the moon's "argument of latitude" is between 90 and 270 deg, it is
> moving southward in ecliptic coordinates, the earth is moving
> northward, and the sun's apparent latitude is decreasing. If this
> situation holds around June 21, the sun's declination will max out
> before the solstice. Remember, the solstice is defined by ecliptic
> longitude, not by RA, not by maximum Dec.
>
> -- Bill Owen
|
| |
Date: 29 Jul 2006 02:57:36
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Brian Tung wrote:
> SkySea wrote:
> > >2) The sun's RA at the solstice should be very close to 6 hours, but
> > >since the RA is defined by the vernal equinox rather than the
> > >solstice, the solstice may not occur at exactly 6 hours.
> >
> > True for times off a chart's given epoch. By definition though, 0h is
> > the intersection of the equator and the ecliptic, and the
> > intersections of the ecliptic with the 6, 18, and 12 hour circles mark
> > the equioxes and solstices.
>
> No, I think the original poster was right: The vernal equinox is at 0h,
> but the autumnal equinox and the solstices are all based on properties
> of the ecliptic and the celestial equator. So the autumnal equinox is
> when the ecliptic crosses the equator going southward, and the solstices
> are the two points where the Sun's declination is at one of its two
> extremes.
>
> I have no explanation for the bizarre behavior of SNP, though.
>
Why your silliness shades off into an unstable and dangerous condition
is that in matters of investigating global climate temperature norms
from astronomical astronomical causes,your cartoon sidereal
justification makes no room for the Keplerian asymmetry in the
relationship between axial and orbital motions/orientations from
Mar/Sept as distinct from Sept/Mar.This asymmetry is quite apart from
cyclical hemispherical weather patterm (seasons) insofar as it reflects
the change speed of the Earth and the rate of change of orbital
orientation as constant axial rotation is passing through the orbital
change.
The Equation of Time system reflects the rate of change of orbital
orientation and affirms Keplerian orbital motion and geometry and has
no axial tilt component.The only thing that reflects the return of a
meridian to natural noon is how far the orbital orientation has changed
as axial rotation returns to face the Sun/Earth line at 90 degrees to
orbital orientation -
http://www.mhhe.com/physsci/astronomy/fix/student/images/04f15.jpg
While you are looking at that graphic,enjoy the hideous spectacle of
your constant .986 deg sidereal justification.Yoiu cannot make it fit
into an elliptical framework for the Earth would cover a greater
distance along its orbital circumference the further from the Sun it is
under such a Newtonian scheme.
http://astrosun2.astro.cornell.edu/academics/courses//astro201/images/sidereal_day.gif
Perhaps you are better suited to showing pretty pictures to each other
and calling yourselves astrophotographers.
> --
> Brian Tung <brian@isi.edu>
> 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.html
|
| |
Date: 29 Jul 2006 02:43:37
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
SkySea wrote:
> Wrong.
>
Vacuous
> To say that the solstice is reached at a particlar time (I assume you
> also mean on a particualr day, but regardless) each year is a
> statement that the Earth's rotation period is an exact fraction of the
> orbital period.
What you are doing is forcing the annual orbital motion and axial
rotation to suit the calendar system whereas the correct way to
approach this topic is the same way the people who constructed the 5
200 year old Newgrange monument did.
It isn't. It also implies that the Earth's orbital
> period and rotational periods are constant. They aren't. Or it assumes
> that the variations in the the orbital period and rotation are somehow
> locked, so they remain in synchronization. They aren't.
>
You unfortunately have copnstructed a system where the orbital motion
is neutralised and all motion is referenced from axial rotation.The
17th century vandalism is based on borrowing a .986 degree/3 min 56 sec
from axial rotation and terrestial longitudes and forcing it into an
orbital displacement -
http://www.pfm.howard.edu/astronomy/Chaisson/AT401/IMAGES/AACHCIR0.JPG
This dumb justification which requires a cycle based on 3 years of 365
days and 1 year of 366 days is little more than a convenience for
Ra/Dec observing based on all motions moving with a celestial sphere.
In short,you could not build the gorgeous Newgrange monument which
marks the midpoint between the Equinoxes.
> Look through some almanacs and find out at what times the solstices
> occured. It's not always at 9:11, and the liklihood that it ever will
> be again is 1:1,440.
>
My astronomical ancestors,5200 years ago mind you,created a structure
based on the return of the Sun to a specific
position,heliocentrically,it means that axial and orbital motions of
the Earth were and can be gauged without referencing it to the calendar
system based on the stellar background.
The Equation of Time system is based strictly on the motions of the
Earth using the Sun as a reference but the dumb 17th century
maneuvering fixed axial orientation to the celestial sphere and worked
on the principle that the Earth has a variable axial tilt to the Sun.
It is worthwhile to reproduce the system and structure which is the raw
material for the creation of the equable 24 hour day day and from that
point of departure on to the complimentary addition of the calendar
system.Unfortunately you try to justify the calendrical system
astronmomically and subsequently the return of a star to a meridian in
23 hours 56 min 04 sec.
>
> >"oriel36" <geraldkelleher@yahoo.com> piddled:
> >The builders constructed the roofbox to register the midpoint
> >(solstice) at exactly the same time each year at roughly 9:11 AM each
> >and every year.
>
> =============
> - Dale Gombert (SkySea at aol.com)
> 122.38W, 47.58N, W. Seattle, WA
> http://flavorj.com/~skysea
|
| |
Date: 31 Jul 2006 13:34:48
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Climate changes reflect a changing relationship between axial rotation
and orbital motion depending on whether the orbit tends to be more
circular or more elliptical.
http://ircamera.as.arizona.edu/NatSci102/lectures/kepler.htm
The natural asymmetry that exists between axial and orbital motion
between Mar to Sept as distinguished from Sept to Mar,due to variations
in the Earth's orbital speed becomes more pronounced as the Earth's
orbital geometry becomes motre elliptical.
Newtonians,with their perturbations,do not recognise Keplerian orbital
geometry insofar as their celestial sphere justification for axial and
orbital motion is based on a constant .986 degree orbital displacement
which makes the Earth travel faster in its orbital circumference in an
elliptical framework at the aphelion -
http://astrosun2.astro.cornell.edu/academics/courses//astro201/images/sidereal_day.gif
Enjoy yourselves with your complicated linguistic perturbative
fireworks based on celestial sphere geometry,the real astronomical
insights are there for whoever wishes to take them up and use them.
Paul Schlyter wrote:
> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov> wrote:
> > Brian Tung wrote:
> >> Bill Owen wrote:
> >>
> >>>The center of the sun will NOT go through this point in general, because
> >>>its geocentric latitude is zero only at the instant when the moon is at
> >>>one of the nodes of its orbit. If the sun's latitude happens to be
> >>>positive around March 20, it will cross the celestial equator before it
> >>>hits longitude 0.
> >>>
> >>>If the moon's "argument of latitude" is between 90 and 270 deg, it is
> >>>moving southward in ecliptic coordinates, the earth is moving
> >>>northward, and the sun's apparent latitude is decreasing. If this
> >>>situation holds around June 21, the sun's declination will max out
> >>>before the solstice. Remember, the solstice is defined by ecliptic
> >>>longitude, not by RA, not by maximum Dec.
> >>
> >>
> >> OK, interesting, I did not know that. What if we observe the Sun from
> >> the Earth-Moon barycenter; does the Sun go through the FPA at the
> >> equinox from that perspective, even if the Moon is not at one of its
> >> orbital nodes?
> >
> > No. Here's why:
> >
> > The ecliptic is defined as the *mean* orbit of the Earth-Moon
> > barycenter. Short- and long-period perturbations from the other
> > planets move the EMB off the ecliptic. To put it another way, the
> > osculating orbit of the EMB will have a small but nonzero inclination,
> > and the EMB itself will never (well, hardly ever) be at the node.
> >
> > -- Bill
>
> The most important perturbations on the EMB comes from Venus and Jupiter.
> This will cause the EMB to cross the plane of the ecliptic a few times
> each Earth year. I don't think one can say that something "never" or
> "hardly ever" happens when it happens THAT often....
>
> The inclination of the EMB's osculating orbit is always very small of
> course, but the longitude of the node of the EMB's osculating orbit
> will vary a lot, quickly, and irregularly.
>
> --
> ----------------------------------------------------------------
> Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
> e-mail: pausch at stockholm dot bostream dot se
> WWW: http://stjarnhimlen.se/
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Date: 31 Jul 2006 10:42:20
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Bill Owen wrote:
> Brian Tung wrote:
> > Bill Owen wrote:
> >
> >>The center of the sun will NOT go through this point in general, because
> >>its geocentric latitude is zero only at the instant when the moon is at
> >>one of the nodes of its orbit. If the sun's latitude happens to be
> >>positive around March 20, it will cross the celestial equator before it
> >>hits longitude 0.
> >>
> >>If the moon's "argument of latitude" is between 90 and 270 deg, it is
> >>moving southward in ecliptic coordinates, the earth is moving
> >>northward, and the sun's apparent latitude is decreasing. If this
> >>situation holds around June 21, the sun's declination will max out
> >>before the solstice. Remember, the solstice is defined by ecliptic
> >>longitude, not by RA, not by maximum Dec.
> >
> >
> > OK, interesting, I did not know that. What if we observe the Sun from
> > the Earth-Moon barycenter; does the Sun go through the FPA at the
> > equinox from that perspective, even if the Moon is not at one of its
> > orbital nodes?
>
> No. Here's why:
>
> The ecliptic is defined as the *mean* orbit of the Earth-Moon
> barycenter. Short- and long-period perturbations from the other
> planets move the EMB off the ecliptic. To put it another way, the
> osculating orbit of the EMB will have a small but nonzero inclination,
> and the EMB itself will never (well, hardly ever) be at the node.
>
> -- Bill
Perturbations are a destructive Newtonian invention and it is not that
difficult to gauge why that is so.
The anomalous motion of Jupiter's satellite Io was accounted for by Ole
Roemer's insight that the speed of light is finite.the apparent
variations in speed of Io was wonderfully resolved by taking account of
the variation in orbital distances between Earth and Jupiter as seen
from an orbitally moving Earth.No perturbations or any attempt to
explain the anomalous motion of Io directly,just a clear and exquisite
refinement of Copernican heliocentricity and the Keplerian refinement.
So,what had Isaac to say on finite light speed and Io -
" Some inequalities of time may arise from the Excentricities of the
Orbs of the Satellites; [etc.]... But this inequality has no respect to
the position of the Earth, and in the three interior Satellites is
insensible, as I find by computation from the Theory of
their Gravity. " NEWTON Opticks 1704
The Mora Luminis or Equation of Light is an astronomical correction
based on finite light speed,it does not beg a physical solution and
before anyone bothers to genuflect when they hear the 'Theory Of
Gravity',Io is Jupiter's innermost satellite hence Newton is talking
through his arse.
The stunning pictures of Io and its shadow make the Roemerian insight
lovely in a contemporary setting,should astronomers care to drop the
ballistic agenda of Newton and try to grasp the impact of Roemer's
insight then we can get back to doing astronomy without 17th century
perturbation theories.
http://www.deskpicture.com/DPs/Astronomy/ioshadow.jpg
To make this a productive thread,the orientation of the shadow will
depend on whether the Earth is approaching or receding from the orbit
of Jupiter in our common heliocentric orbit and the anomaly can still
be seen regardless of how clear the images of Io now are.
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Date: 04 Aug 2006 05:39:34
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Odysseus wrote:
> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov>
> wrote:
>
> <snip>
>
> > The ecliptic is defined as the *mean* orbit of the Earth-Moon
> > barycenter. Short- and long-period perturbations from the other
> > planets move the EMB off the ecliptic. To put it another way, the
> > osculating orbit of the EMB will have a small but nonzero inclination,
> > and the EMB itself will never (well, hardly ever) be at the node.
>
> What is the greatest (most extreme) ecliptic latitude of the Sun?
>
> --
> Odysseus
There is an natural asymmetry between axial and orbital motions that
are asymmetrical between Mar/Sept and Sept/Mar due to variations in the
Earth's orbital speed.The solstices are inclined to obscure this
asymmetry because the orbital geometry and its relationship with axial
motion indicates a symmetry between June/Dec and Dec/Jun
Breaking the annual orbit into 4 seperate and identical sections based
on axial tilt to the Sun/orbital plane is counter-productive,insofar as
the loss of the natural orbital/axial asymmetries either side of the
Equinoxes prevents a proper investigation of the terrestial
consequences of that asymmetry.
There is so much material left unattended so theorists can sound
impressed with each other,even out of unfamiliarity,there is no problem
making as many mistakes as you wish in order to raise the standard and
become comfortable with the accurate working principles for the Earth's
motions.I try to shut off these pretensious and useless empirical
linguistics which prevent people from approaching climatology through
astronmomy in a meaningful and exciting way.It just needs a few
participants to fiight for that higher standard without fear of the
silly ostracism or intimidation from optical astrophotographers and
theorists.Then again,anyone who is afraid to push out into the stormy
seas of astronomy was never really an astronomer to begin with.
EMB indeed !!.
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Date: 04 Aug 2006 03:28:40
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
You poor buggers are still working off celestial sphere coordinates
based on axial orientation whereas the real substance is bypassing the
axial orientation and concentrating on orbital motion and subsequently
the change in orbital orientation in accordance with Keplerian orbital
geometry.
To make this enjoyable for people who wish to work with global climate
and the oscillation of temperature bands signifying the change in
orbital orientation against fixed axial orientation,I provide the
necessary graphics.
http://www.mhhe.com/physsci/astronomy/fix/student/images/04f15.jpg
http://www.climateprediction.net/images/sci_images/annual.gif
The pair of you are impressed with your own linguistic fireworks while
I detest my own inability to explain this wonderful new way to mesh
astronomy with climatology by meshing the oscillation of the global
temperature bands with the underlying causes based on constant
radiation received from the Sun and the accurate relationship between
axial and orbital motions and orientations.
This is important,this is very important,even for people who are
optical astrophotographers who worry about clouds and streetlights.They
should at least recognise that the Ra/Dec system is a calendrically
based convenience which should not be justified in order to permit the
correct astronomical working methods tto be applied to climatological
investigations.
Then and only then will you do humanity a big favor.
Paul Schlyter wrote:
> In article <44CE775A.7090003@jpl.nasa.gov>,
> Bill Owen <wmo@jpl.nasa.gov> wrote:
>
> > Paul Schlyter wrote:
> >> In article <44CE2FBB.50809@jpl.nasa.gov>, Bill Owen <wmo@jpl.nasa.gov> wrote:
> >>>The ecliptic is defined as the *mean* orbit of the Earth-Moon
> >>>barycenter. Short- and long-period perturbations from the other
> >>>planets move the EMB off the ecliptic. To put it another way, the
> >>>osculating orbit of the EMB will have a small but nonzero inclination,
> >>>and the EMB itself will never (well, hardly ever) be at the node.
> >>>
> >>>-- Bill
> >>
> >>
> >> The most important perturbations on the EMB comes from Venus and Jupiter.
> >> This will cause the EMB to cross the plane of the ecliptic a few times
> >> each Earth year. I don't think one can say that something "never" or
> >> "hardly ever" happens when it happens THAT often....
> >>
> >> The inclination of the EMB's osculating orbit is always very small of
> >> course, but the longitude of the node of the EMB's osculating orbit
> >> will vary a lot, quickly, and irregularly.
> >
> > You're right ... but I didn't write that the EMB will never (well,
> > hardly ever) *cross* the node, but that it will never (well, hardly
> > ever) *be at* the node. Yes, crossings happen quite often, but the
> > set of times when the EMB is *exactly* on the ecliptic is still a set
> > of measure zero.
> >
> > Sorry if I confused you.
> >
> > -- Bill
>
> From a strictly mathematical point you're of course right. Otoh this
> does not apply just to zero latitude, it applies to *any* given
> ecliptic latitude: the EMB is never (well hardly ever) there....
>
> So one might then wonder: if the EMB never (well hardly ever) is
> anywhere, where *is* it ???? And if the EMB is on one side of the
> ecliptic, and later on the other side of the ecliptic, without ever
> having been on the ecliptic in between -- how did it get through the
> ecliptic? Some tunnel effect? :-)
>
>
> Since we're dealing with measureable quantities which all have some
> measurement error, the purely mathematical point of view isn't
> particularly relevant. A more relevant point of view would be to e.g.
> consider when the EMB is close enough to the plane of the ecliptic to
> be, with our current measurement capabilities (or with some tolerance
> level), indistinguishable from being exactly on the ecliptic. And
> that will be a time interval longer than zero, and it will happen a
> few times each year.
>
> --
> ----------------------------------------------------------------
> Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
> e-mail: pausch at stockholm dot bostream dot se
> WWW: http://stjarnhimlen.se/
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Date: 05 Aug 2006 14:02:27
From: oriel36
Subject: Re: Summer Solstice Calculation Questions
|
Odysseus wrote:
> In article <44D36810.3050503@jpl.nasa.gov>,
> Bill Owen <wmo@jpl.nasa.gov> wrote:
>
> <snip>
>
> > So the two planets which perturb the earth's orbit the most were both
> > well south of the ecliptic when they were closest to us, moving the
> > earth-moon barycenter southward as well. Then in early November, the
> > moon was at its *northernmost* ecliptic latitude, so the earth (being
> > on the other side of the barycenter) was at the southern point of its
> > monthly motion. All three -- Venus, Jupiter, Moon -- moved the earth
> > southward. The result was that the sun appeared at a latitude of
> > +1.06 arcsec.
>
> Thanks for the detailed analysis! It certainly appears that there must
> be few occasions on which those factors would combine to greater effect
> than they did in 1999.
>
> --
> Odysseus
Do you clearly understand that he is using axial orientation to
dictate orbital conditions.
The one thing about 'perturbations',they wreck the ability to
appreciate the Roemerian insight which emerges as observed anomalous
motion in Io.Newton was of course exploiting this intricate Copernican
refinement and of course when he does comment on Io he does'nt even
recognise it as the innermost satellite of Jupiter .
" Some inequalities of time may arise from the Excentricities of the
Orbs of the Satellites; [etc.]... But this inequality has no respect to
the position of the Earth, and in the three interior Satellites is
insensible, as I find by computation from the Theory of their
Gravity." Optics 1704
This of of course the usual Newtonian rubbish but as nobody else
appears to have the neccessary intuitive faculties to light up the
original insight of Roemer and demolish these stupid Newtonian
conceptions,the great and exquisite astronomical insight withers and
is replaced by 'perturbations'.
Climatology desperately needs the correct local relationship between
axial and orbital motions and orientations.Carry on believing what you
need to ,even these 'detailed analysis' which go nowhere and do
nothing.Humanity cannot afford your kind.
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