A few questions occasioned by observing the Vernal Equinox

Factual Questions
1

  1. The Equinox was supposed to have happened at around 2:30 PM yesterday. What exactly does this mean? I was given to understand that the Equinox is the day on which both day and night are 12 hours long. What exactly was occurring at 2:30? If it occurred at noon, would day and night be exactly equal? Is it different from time zone to time zone? Can you compute how far off of equal day and night will be based on what time the Equinox occurs?

  2. How do you calculate the altitude and azimuth of the sun at a given time and date at a given latitude?

2

The equinox is a moment in time pertaining to the Earth’s rotation around the sun not it’s revolving on its axis. It’s the exact point (more or less) in the Earth’s orbit when its axis is perpendicular to its orbital plane.

3

The solstices are those moments in time when the Earth’s axis of rotation is pointed exactly to/from the Sun (well, exactly meaning as close as it ever gets). The equinoxes are those times when the Earth’s axis is exactly sideways to it, in other words when the Sun is directly over the equator. It’s a moment of time, not a day.

By the way, it’s not true that the solstices/equinoxes are not the “official” boundaries of the seasons. The Master Speaks.

4

I believe the equinox ocurred at approx UT Mar 20 19:15

At that time the declination of the Sun was zero as the sun crossed from the Southern hemisphere to the northern hemisphere. The moment when the sun reaches the maximum declination is called the solstice and then it starts moving in the opposite direction.

5

With a computer. You might think there is a simple formula for this. There isn’t. I made an Excel spreadsheet that does this, and it took a lot longer than I thought.

BTW Hail Ants is right, but confused the words “rotation” and “revolution”.

6

Er, no… The Earth’s axis is never perpendicular to the Earth’s orbital plane, is it? On the equinox, the sun lies in Earth’s equatorial plane, I think… is that what you were thinking of? That would then (to answer the OP’s question) be what occurred at 2:30 p.m. in Matt’s time zone.

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This question has probably been answered well enough already but let me throw in a couple of definitions from my “Practical Astronomy with Your Calculator” book:

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>> How do you calculate the altitude and azimuth of the sun at a given time and date at a given latitude?

>> There isn’t. I made an Excel spreadsheet that does this, and it took a lot longer than I thought

Yep, I have done this too. Just to give you an idea of what is involved… Assume you do not need terribly exact results (in which case you better be the US Naval Observatory and have a gazillion Gigabertz of computing power at your disposal plus some guys who know about astronomy)… just results which will give you a good approximation… let’s work backwards:

1- Assume you have the coordinates of the celestial body (extracted from the Nautical Almanac tables or other source) and your geographical position. Then calculating Height and azimuth is pretty straightforward. Let GHA (Greenwich Hour Angle) and Dec (Declination) be the coordinates of the celestial body and Lat (Latitude) and Lon (Longitude) the geographical coordinates of the observer. Then H (Height above the horizon) and Z (Azimuth) can be easily obtained using the following formulas:


Local Hour Angle (LHA)= GHA - Lon

Sin(Hc)= Sin(lat)*Sin(dec)+Cos(lat)*Cos(dec)*Cos(LHA)

                             Sin(LHA)                 
 Tan(Z) = ------------------------------------------- 
           Sin(lat) * cos(LHA) - Cos(lat) * Tan(dec)  

           Cos(dec) * Sin(LHA)  
 Sin(Z) = --------------------- 
                  Cos(Hc)       

Both formulas for (Z) are equivalent. 
Both return a value between -90° and +90° 
so Z needs to be adjusted to the correct 
quadrant in order to obtain Zn: 0° < Zn < 360°:
  If sign(sin(Z)) = sign(sin(LHA)) then Zn = Z + 180
      Else, If Z is negative then Zn = Z + 360
          Else Zn = Z

The first line is equivalent to: 
  If sin(Z)*sin(LHA) > 0 then Zn = Z + 180
or:   Z = Z + 90 * (1 + sign(sin(LHA) * sin(Z) ) )

You would then add a few corrections for parallax, atmospheric refraction, height of observer above the horizon, etc. and that would give you the Apparent H and Z. This is valid for any celestial body.

2- So how do you calculate the celestial coordinates of a given body? Some are easier than others. The Moon is very complex. With the Planets you would first calculate the heliocentric coordinates of the planet and of the Earth and, from these, calculate the geocentric coordinates of the planet. Pretty complex too. If you do not need great precission the stars are the easiest as they are pretty much fixed in the celestial sphere which in turn rotates at a pretty constant rate. The Sun’s geocentric coordinates are not difficult to calculate with a few formulas but not as straightforward as the stars.

Conclusion: If you want to know the times of the equinoxes and the solstices with 10 - 15 minutes precission, a few formulas in a spreadsheet or computer program will do. If you need to know it to the minute, then ask the USNO. Also, as they say: “A man with one watch always knows the time, a man with two watches is never sure”. Different observatories arrive at (slightly) different predictions.

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But that accuracy is nice to have, and the USNO is happy to share it with the general public.

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And just to confuse things a bit I will say that, while wind direction is given from where it blows from, current set (direction) is given in the direction where it goes to. A current which sets NW comes from the SE and goes to the NW. And current speed is called drift. I knew you’d want to know that.

11

Shoot! my apologies. that post was meant for this thread Can a moderator clean up the mess? Sorry about that.