North from Polaris

[Brillig]

This will be a basic question, but I'm not used to using astronomical coordinates to find terrestrial coordinates.

Suppose I sight Polaris at local midnight. Using this sighting, I draw a line on the ground pointing North. Since Polaris isn't due north but is about 3/4 degree away from true north, the line I just drew will point to one side or the other unless Polaris happens to be just above or below north. The question is, how do I know which direction and by how much?

Wikipedia states that Polaris' coordinates are RA: 2h 31m 48.7s, Dec: 89 degrees, 15' 51". It seems like simple trigonometry is called for here, but I don't know where RA: 0 is from a terretrial perspective.

Or maybe I should ask the question a different way. Given the RA of Polaris, the current date, and my longitude, at what time should I observe Polaris for it to be due North? There should be two times of day that satisfy these criteria. There's a good chance one of them will be at night, but not necessarily.

Victor

[Brillig]

Never mind. This page:
http://www.ephemeris.com/space-time.html

with a rough way to calculate azimuth from RA and Dec gets me what I want.

Victor

On Thu, Oct 15, 2009 at 4:41 PM, Victor Engel <brillig@...> wrote:

This will be a basic question, but I'm not used to using astronomical coordinates to find terrestrial coordinates.

Suppose I sight Polaris at local midnight. Using this sighting, I draw a line on the ground pointing North. Since Polaris isn't due north but is about 3/4 degree away from true north, the line I just drew will point to one side or the other unless Polaris happens to be just above or below north. The question is, how do I know which direction and by how much?

Wikipedia states that Polaris' coordinates are RA: 2h 31m 48.7s, Dec: 89 degrees, 15' 51". It seems like simple trigonometry is called for here, but I don't know where RA: 0 is from a terretrial perspective.

Or maybe I should ask the question a different way. Given the RA of Polaris, the current date, and my longitude, at what time should I observe Polaris for it to be due North? There should be two times of day that satisfy these criteria. There's a good chance one of them will be at night, but not necessarily.

Victor

[Tom Peters-6]

Op 15-okt-2009, om 23:54 heeft Victor Engel het volgende geschreven:

> Never mind. This page:
> http://www.ephemeris.com/space-time.html
>
> with a rough way to calculate azimuth from RA and Dec gets me what  
> I want.
>
> Victor

   There is a simpler way.  If Polaris' R.A. is 2h31m48.7s (in 2000 I  
presume; its coordinates change quite fast so close to the pole), it  
means it follows the first point of Aries (aequinox) by that much  
time.  When the aequinox passes the meridian in the South, it is 0h  
sidereal time.  This is distinct from mean solar time, which runs a  
factor 366.25/365.25 slower: in one orbit around the Sun, there are  
365.25 solar days (revolutions of the Earth around its axis as seen  
from the Sun), but 366.25 sidereal days (actual revolutions of the  
Earth around its axis in a non-rotating reference frame).

   Now the (mean) Sun is in the aequinox around 21 March, so around  
noon local time on 21 March, when both the Sun and the aequinox are  
in the meridian, the sidereal time is 0; Polaris will then be on the  
local meridian about 2:32 PM; and of course 12h before and after  
that.  Around 23 September the (mean) Sun is in the other aequinox  
point, so when they cross the meridian in the North around local  
midnight, the spring aequinox is in the meridian in the South and it  
is 0h sidereal time again.  Polaris will then cross the meridian  
about 2:32 AM.  Interpolate for other days.

   You have to adjust for your actual geographic longitude to switch  
between clock time and local time.  The actual Sun may lead or lag by  
max. 16 min. because of its variable apparent speed over the year  
(the so-called equation of time, which shows itself in the so-called  
analemma sometimes drawn on sundials); however, (local) clock time  
follows a steady mean Sun, and that is what you use to compute the  
equally steady sidereal time.

   The American Ephemeris and Nautical Almanac used to have pole star  
tables to solve a similar problem: when shooting the altitude of  
Polaris to find geographical latitude, you need to correct for its  
daily motion of almost a degree around the pole.

> On Thu, Oct 15, 2009 at 4:41 PM, Victor Engel <brillig@...>  
> wrote:
> This will be a basic question, but I'm not used to using  
> astronomical coordinates to find terrestrial coordinates.
>
> Suppose I sight Polaris at local midnight. Using this sighting, I  
> draw a line on the ground pointing North. Since Polaris isn't due  
> north but is about 3/4 degree away from true north, the line I just  
> drew will point to one side or the other unless Polaris happens to  
> be just above or below north. The question is, how do I know which  
> direction and by how much?
>
> Wikipedia states that Polaris' coordinates are RA: 2h 31m 48.7s,  
> Dec: 89 degrees, 15' 51". It seems like simple trigonometry is  
> called for here, but I don't know where RA: 0 is from a terretrial  
> perspective.
>
> Or maybe I should ask the question a different way. Given the RA of  
> Polaris, the current date, and my longitude, at what time should I  
> observe Polaris for it to be due North? There should be two times  
> of day that satisfy these criteria. There's a good chance one of  
> them will be at night, but not necessarily.
>
> Victor

--
Tom Peters

[Brillig]

Thanks, Tom. That is helpful. As a practical matter, I've decided to use the R.A. indicated in Skyview Cafe. I figure it's more likely that will be a good figure than that I'll do everything correctly the first time. Or I can use it to check my work.

Using Skyview Cafe, I determined that Polaris was on the Austin, Texas meridian at 2:39 AM two days ago. My daughter and I took a sighting at about midnight. I calculate that our sighting was off by 31' 16" due to the time difference between midnight and 2:39, plus any observational error. If it's clear enough, I'll try another sighting this weekend. For this excercise I'm using my own eye and a metal sphere (about the size of my pupil). The ball has a fixed location. I move my eye until Polaris is obscured by the ball. Then Helen marks the location of my pupil. I think that's about as pricise as I can manage observationally, using Polaris.

I could make two observations at midnight, 6 months apart (+/- adjustment for Earth's orbit not being a circle), and take their average. Then I wouldn't need to know where Polaris was.

Victor

On Fri, Oct 16, 2009 at 2:48 PM, Tom Peters <tpeters@...> wrote:

Op 15-okt-2009, om 23:54 heeft Victor Engel het volgende geschreven:

Never mind. This page:
http://www.ephemeris.com/space-time.html

with a rough way to calculate azimuth from RA and Dec gets me what I want.

Victor

 There is a simpler way.  If Polaris' R.A. is 2h31m48.7s (in 2000 I presume; its coordinates change quite fast so close to the pole), it means it follows the first point of Aries (aequinox) by that much time.  When the aequinox passes the meridian in the South, it is 0h sidereal time.  This is distinct from mean solar time, which runs a factor 366.25/365.25 slower: in one orbit around the Sun, there are 365.25 solar days (revolutions of the Earth around its axis as seen from the Sun), but 366.25 sidereal days (actual revolutions of the Earth around its axis in a non-rotating reference frame).

 Now the (mean) Sun is in the aequinox around 21 March, so around noon local time on 21 March, when both the Sun and the aequinox are in the meridian, the sidereal time is 0; Polaris will then be on the local meridian about 2:32 PM; and of course 12h before and after that.  Around 23 September the (mean) Sun is in the other aequinox point, so when they cross the meridian in the North around local midnight, the spring aequinox is in the meridian in the South and it is 0h sidereal time again.  Polaris will then cross the meridian about 2:32 AM.  Interpolate for other days.

 You have to adjust for your actual geographic longitude to switch between clock time and local time.  The actual Sun may lead or lag by max. 16 min. because of its variable apparent speed over the year (the so-called equation of time, which shows itself in the so-called analemma sometimes drawn on sundials); however, (local) clock time follows a steady mean Sun, and that is what you use to compute the equally steady sidereal time.

 The American Ephemeris and Nautical Almanac used to have pole star tables to solve a similar problem: when shooting the altitude of Polaris to find geographical latitude, you need to correct for its daily motion of almost a degree around the pole.

On Thu, Oct 15, 2009 at 4:41 PM, Victor Engel <brillig@...> wrote:
This will be a basic question, but I'm not used to using astronomical coordinates to find terrestrial coordinates.

Suppose I sight Polaris at local midnight. Using this sighting, I draw a line on the ground pointing North. Since Polaris isn't due north but is about 3/4 degree away from true north, the line I just drew will point to one side or the other unless Polaris happens to be just above or below north. The question is, how do I know which direction and by how much?

Wikipedia states that Polaris' coordinates are RA: 2h 31m 48.7s, Dec: 89 degrees, 15' 51". It seems like simple trigonometry is called for here, but I don't know where RA: 0 is from a terretrial perspective.

Or maybe I should ask the question a different way. Given the RA of Polaris, the current date, and my longitude, at what time should I observe Polaris for it to be due North? There should be two times of day that satisfy these criteria. There's a good chance one of them will be at night, but not necessarily.

Victor

--
Tom Peters

[Tom Peters-6]

Op 16-okt-2009, om 22:13 heeft Victor Engel het volgende geschreven:

> Thanks, Tom. That is helpful. As a practical matter, I've decided  
> to use the R.A. indicated in Skyview Cafe. I figure it's more  
> likely that will be a good figure than that I'll do everything  
> correctly the first time. Or I can use it to check my work.
>
> Using Skyview Cafe, I determined that Polaris was on the Austin,  
> Texas meridian at 2:39 AM two days ago. My daughter and I took a  
> sighting at about midnight. I calculate that our sighting was off  
> by 31' 16" due to the time difference between midnight and 2:39,  
> plus any observational error. If it's clear enough, I'll try  
> another sighting this weekend. For this excercise I'm using my own  
> eye and a metal sphere (about the size of my pupil). The ball has a  
> fixed location. I move my eye until Polaris is obscured by the  
> ball. Then Helen marks the location of my pupil. I think that's  
> about as pricise as I can manage observationally, using Polaris.
>
> I could make two observations at midnight, 6 months apart (+/-  
> adjustment for Earth's orbit not being a circle), and take their  
> average. Then I wouldn't need to know where Polaris was.

You could use the Egyptian method.  Find a circumpolar star that has  
the same R.A. as Polaris, or 12h more (the Egyptians possibly used  
alfa Draconis as pole star, and delta Ursa Maioris - the point where  
the tail attaches to the body).  Use a plumbline to find the moment  
when the stars are aligned vertically - then they are due North and  
the plumbline gives the direction at the horizon.

But you do not need to use Polaris for this method.  Best pair  
probably still is alfa and beta UMa, at 11:03:44 and 11:01:50  
(2000).  When they align in the vertical (meridian), they indicate  
North.  The difference in right ascension (1:54) amounts to 28.5' ,  
about a solar or lunar disk, if you can tolerate this error.

These alignments conveniently happen before midnight around this time  
of year.
--
Tom Peters

[Brillig]

Dear Tom and Calendar People,

On Fri, Oct 16, 2009 at 6:23 PM, Tom Peters <tpeters@...> wrote:

You could use the Egyptian method.  Find a circumpolar star that has the same R.A. as Polaris, or 12h more (the Egyptians possibly used alfa Draconis as pole star, and delta Ursa Maioris - the point where the tail attaches to the body).  Use a plumbline to find the moment when the stars are aligned vertically - then they are due North and the plumbline gives the direction at the horizon.

I may try that technique to check my work.
 

But you do not need to use Polaris for this method.  Best pair probably still is alfa and beta UMa, at 11:03:44 and 11:01:50 (2000).  When they align in the vertical (meridian), they indicate North.  The difference in right ascension (1:54) amounts to 28.5' , about a solar or lunar disk, if you can tolerate this error.

I was hoping to get a little closer than that.
 

These alignments conveniently happen before midnight around this time of year.
--
Tom Peters

One problem I have in the city is light pollution. Polaris is visible in the middle of the night, but only just. I've been waiting for a night with low humidity, and I finally got one.

I ended up taking to sightings of Polaris. The first I took at midnight, and with my daughter's assistance, I got a mark on the driveway just below the pupil of my eye when the ball at the end of my radio antenna obscured Polaris.

The next sighting was done more carefully, using a lamp shade from a clip-on shop lamp. It just so happens that the angle of a line passing through one of the vent holes through the hole into which the light bulb normally is inserted matches my latitude. This allowed me to place the lamp shade flat on the ground and peer through the vent hole up to Polaris. I simply moved the lamp shade around until Polaris was obscured by the ball on the antenna. The ball is about 4mm across. The hole in the lamp shade is about 10mm across. The two were about 2.8 meters apart. So I think my precision using this technique is about 8-9 minutes of arc.

I used a plumb line from the string connecting the ball at top of the antenna through the lampshade vent hole down to the grown in order to get a second lay point for a horizontal north line. There was some uncertainty with this position due to the wind blowing the plumb line. I protected from the wind by passing the plumb line through a cardboard shipping tube used for wind protection.

I'm pretty confident that the error in end points (ball and vent holes) plus the error in plumb is less than a cm.

I was curious how much my approximately midnight sighting differed from my other sighting (taken at 2:28 AM local time). I haven't measured the difference yet, but it seemed to be about an inch. Running the numbers, I calculated Polaris moved 31 minutes of arc from midnight the day I sighted the first time to 2:28, the day I sighted the second time, which it should have been due north. So sine of 31 minutes times my distance of 2826 millimeters should be about an inch. My calculator says 25 mm. That's pretty good confirmation.

Victor

[Brillig]

Just a side note:

I knew before that alpha and beta UMa aligned with north. I didn't realize until this project that alpha and beta UMi also aligned with north.

I was thinking that to improve on the slight error in alignment of alpha and beta UMi, taking advantage of the fact that Polaris is actually very close to due north, one could simply make sure alpha and beta UMi are nearly vertical, and then simply use Polaris as the reference point. Beta UMi is used only to establish the time of observation. I'd rather just use my watch, though, since Beta UMi is so hard to see from this latitude in the city.

Victor