equinox question

Today was the vernal equinox.
According to my understanding, this is a day on which the sun rises
and sets approximately due east and west and takes approximately 12
hours to make that trip. But Starry Night pro and a couple other
databases tell me that Sol rose at approx. 05:55 and set at approx.
18:04. The closest to a 12-hour day was on March 16.
My intuitive sense of this is that latitude should not be a factor. Am
I wrong in that assumption?
What am i missing?

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Heron stone posted:

Today was the vernal equinox.
According to my understanding, this is a day on which the sun rises
and sets approximately due east and west and takes approximately 12
hours to make that trip. But Starry Night pro and a couple other
databases tell me that Sol rose at approx. 05:55 and set at approx.
18:04. The closest to a 12-hour day was on March 16.
My intuitive sense of this is that latitude should not be a factor. Am
I wrong in that assumption?
What am i missing?

The vernal equinox occurs when the center of the sun's disk is directly over a
point on the sky where the ecliptic (the plane of the Earth's orbit) and the
plane of the Celestial equator cross (the point is called "the first point of
Aries (actually is located in Pisces). The sun, from our vantage point,
appears to be "moving" into the northern half of the celestial sphere, so the
vernal equinox is generally indicated as the time when spring begins in the
northern hemisphere. The date for equal times of sunrise and sunset differ
slightly from the date of the equinoxes, as the sun appears to be moving
constantly against the star background as the day goes on. This apparent
motion (along with the fact that the path the sun takes is tilted) is enough
that it will tend to cause the "equal day/night" date to be a bit different
from the exact date for the equinox. Clear skies to you.
--
David W. Knisely
Prairie Astronomy Club: http://www.prairieastronomyclub.org
Hyde Memorial Observatory: http://www.hydeobservatory.info/

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Barry Schwarz wrote in message ...
On Sat, 20 Mar 2004 19:36:28 -0800, heron stone
wrote:

Today was the vernal equinox.
According to my understanding, this is a day on which the sun rises
and sets approximately due east and west and takes approximately 12
hours to make that trip. But Starry Night pro and a couple other
databases tell me that Sol rose at approx. 05:55 and set at approx.
18:04. The closest to a 12-hour day was on March 16.
My intuitive sense of this is that latitude should not be a factor. Am
I wrong in that assumption?
What am i missing?

The question gets asked semi-annually. If you google the alt.astro*
and sci.astro* newgroups, you will probably find more comprehensive
explanations and links to sites with all the detail you could want.

I am sure there are other factors but consider just these two:

The motion of the sun on the ecliptic is continuous. On average,
the sum moves slightly more than .986 degrees per day. If the vernal
equinox occurred at exactly sunrise at your location, then 12 hours
later the sun would have moved .498 degrees and would be north of the
equator. The sun will therefore set a little later, increasing the
amount of daylight.

Atmospheric refraction causes the sun to be visible even when it
actually is slightly below the horizon at that location.
Consequently, "visible" sunrise occurs before "geometrical" sunrise
and "visible" sunset after "geometrical" sunset.




Remove the del for email

Two other possible issues: 1. I assume that outside of the refractive
effect of the atmosphere giving us an early peek at the sun (or a
slightly later sunset) another issue is at work. I believe that
sunrise and sunset are measured when the solar disk is first and last
visible (the edge). On March 20th daylight was 12:10. Perhaps it takes
about 5 minutes after the disk edge pierces the horizon for the center
of the disk to do so...and at sunset the center of the disk "sets" 5
minutes before actual sunset. This would shorten the "daylight" to
exactly 12 hours.

2. Using your example from above, with one slight modification..What
if the equinox occurs at your location when the sun is transiting your
meridian (local apparent noon) I would assume that here the daylight
(as measured by the center of the disk method) would be exactly 12
hours. Your thoughts?

Without consulting the equations, I would say the sun's refraction in the
atmosphere is about 4-5 minutes at each of sunrise and sunset accounting for
the extra daylight. In an atmophere-less planet the day/night would indeed
be equal.


"emarks" wrote in message
om...
Barry Schwarz wrote in message
...
On Sat, 20 Mar 2004 19:36:28 -0800, heron stone
wrote:

Today was the vernal equinox.
According to my understanding, this is a day on which the sun rises
and sets approximately due east and west and takes approximately 12
hours to make that trip. But Starry Night pro and a couple other
databases tell me that Sol rose at approx. 05:55 and set at approx.
18:04. The closest to a 12-hour day was on March 16.
My intuitive sense of this is that latitude should not be a factor.
Am
I wrong in that assumption?
What am i missing?

The question gets asked semi-annually. If you google the alt.astro*
and sci.astro* newgroups, you will probably find more comprehensive
explanations and links to sites with all the detail you could want.

I am sure there are other factors but consider just these two:

The motion of the sun on the ecliptic is continuous. On average,
the sum moves slightly more than .986 degrees per day. If the vernal
equinox occurred at exactly sunrise at your location, then 12 hours
later the sun would have moved .498 degrees and would be north of the
equator. The sun will therefore set a little later, increasing the
amount of daylight.

Atmospheric refraction causes the sun to be visible even when it
actually is slightly below the horizon at that location.
Consequently, "visible" sunrise occurs before "geometrical" sunrise
and "visible" sunset after "geometrical" sunset.




Remove the del for email

Two other possible issues: 1. I assume that outside of the refractive
effect of the atmosphere giving us an early peek at the sun (or a
slightly later sunset) another issue is at work. I believe that
sunrise and sunset are measured when the solar disk is first and last
visible (the edge). On March 20th daylight was 12:10. Perhaps it takes
about 5 minutes after the disk edge pierces the horizon for the center
of the disk to do so...and at sunset the center of the disk "sets" 5
minutes before actual sunset. This would shorten the "daylight" to
exactly 12 hours.

2. Using your example from above, with one slight modification..What
if the equinox occurs at your location when the sun is transiting your
meridian (local apparent noon) I would assume that here the daylight
(as measured by the center of the disk method) would be exactly 12
hours. Your thoughts?

"John Galt" wrote in
:

Without consulting the equations, I would say the sun's refraction in
the atmosphere is about 4-5 minutes at each of sunrise and sunset
accounting for the extra daylight. In an atmophere-less planet the
day/night would indeed be equal.


"emarks" wrote in message
om...
Barry Schwarz wrote in message
...
On Sat, 20 Mar 2004 19:36:28 -0800, heron stone
wrote:

Today was the vernal equinox.
According to my understanding, this is a day on which the sun
rises
and sets approximately due east and west and takes approximately
12 hours to make that trip. But Starry Night pro and a couple
other databases tell me that Sol rose at approx. 05:55 and set at
approx. 18:04. The closest to a 12-hour day was on March 16.
My intuitive sense of this is that latitude should not be a
factor. Am
I wrong in that assumption?
What am i missing?

The question gets asked semi-annually. If you google the
alt.astro* and sci.astro* newgroups, you will probably find more
comprehensive explanations and links to sites with all the detail
you could want.

I am sure there are other factors but consider just these two:

The motion of the sun on the ecliptic is continuous. On average,
the sum moves slightly more than .986 degrees per day. If the
vernal equinox occurred at exactly sunrise at your location, then
12 hours later the sun would have moved .498 degrees and would be
north of the equator. The sun will therefore set a little later,
increasing the amount of daylight.

Atmospheric refraction causes the sun to be visible even when it
actually is slightly below the horizon at that location.
Consequently, "visible" sunrise occurs before "geometrical" sunrise
and "visible" sunset after "geometrical" sunset.




Remove the del for email

Two other possible issues: 1. I assume that outside of the refractive
effect of the atmosphere giving us an early peek at the sun (or a
slightly later sunset) another issue is at work. I believe that
sunrise and sunset are measured when the solar disk is first and last
visible (the edge). On March 20th daylight was 12:10. Perhaps it
takes about 5 minutes after the disk edge pierces the horizon for the
center of the disk to do so...and at sunset the center of the disk
"sets" 5 minutes before actual sunset. This would shorten the
"daylight" to exactly 12 hours.

2. Using your example from above, with one slight modification..What
if the equinox occurs at your location when the sun is transiting
your meridian (local apparent noon) I would assume that here the
daylight (as measured by the center of the disk method) would be
exactly 12 hours. Your thoughts?



Skymap Pro 7 shows lengthening daylight hours as we go further north. This
no doubt due to refraction effects. The sun rises and sets at a more acute
angle to the horizon as we go further north. That is, it takes longer for
the entire disk of the sun to appear and disappear at sunrise and sunset
respectivley thereby lenghtening the refraction effect and length of
daylight.

Here are the results for the equinotical sun rise/set for 20 March 2004

Equator
rise 6:04am set 6:11pm daylight = 12:07

45 deg north
rise 6:03am set 6:13pm daylight = 12:10

60 deg north
rise 6:01am set 6:15 pm daylight = 12:14

Martin


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