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General Astronomy >> General Observing and Astronomy

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Monte Porche
member


Reged: 01/23/08

Loc: Haughton, LA
Calculating the position of Polaris
      #5813923 - 04/21/13 08:00 PM

I have an interesting puzzler....

I am trying to figure out a way that, given a known latitude, longitude, and time....you can mathematically calculate the position of Polaris.

I know that the elevation will be equal to the Latitude.

I'm trying to figure out how to write a formula to calculate the number of degrees from magnetic north to rotate.

Any suggestions?

Thanks.


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Pharquart
sage
*****

Reged: 11/11/09

Loc: Southwest Minneapolis Metro
Re: Calculating the position of Polaris new [Re: Monte Porche]
      #5814002 - 04/21/13 08:38 PM

The formula would involve 2 components. One varies over many years and the other varies over the course of a day. The first is the location of the celestial pole. The north celestial pole is above a flat horizon by an amount equal to the location's latitude, and it's a fixed distance east or west of magnetic north, depending on location. The amount is known as the location's "magnetic declination." I'm sure there's a mathematical formula that can calculate it, otherwise they'd have a hard time producing the charts that show it. Google "magnetic declination" and you'll get the map.

The second component would predict Polaris's position as it rotates around the celestial pole. Polaris is less than 1 degree away from the celestial pole; over the course of 24 hours it draws a small circle around the pole. There used to be paper wheel calculators (like a circular slide rule) showing the location of the pole relative to Polaris given a date and local time. People used them (and may still) to determine where to put Polaris on their polar alignment scopes so their GEM was pointing right at the celestial pole. This formula to create this is probably easier than calculating magnetic declination. You'd have to find a specific date/time when Polaris was directly "above" (relative to the horizon) the pole, then figure one full counter-clockwise circle plus about 1 degree (1/365.25 of a circle) around the pole each day.

So your formula would be something like this:

Polaris's location = location of the celestial pole (elevation = latitude, east/west location = magnetic north offset by magnetic declination based on exact location) + deviation of Polaris from celestial pole (about 0.7 degrees in a direction determined by date and local standard time)

Brian


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wirenut
Pooh-Bah


Reged: 09/21/06

Loc: m'dale Pa
Re: Calculating the position of Polaris new [Re: Pharquart]
      #5814190 - 04/21/13 09:52 PM

I Know NOAA has a magnetic deviation chart and magnetic deviation calculator program, if that of any help. I believe there is some random movement in our magnetic poles as well as the ability to flip polarity. this may prove a problem in making future predictions based on magnetic N location.

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GlennLeDrew
Postmaster
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Reged: 06/18/08

Loc: Ottawa, Ontario, Canada
Re: Calculating the position of Polaris new [Re: wirenut]
      #5814493 - 04/22/13 12:48 AM

Polaris is currently within about 3/4 of a degree of the celestial pole. Can one use a magnetic compass to reliably obtain better pointing certainty than this?

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buddyjesus
Carpal Tunnel
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Reged: 07/07/10

Loc: Davison, Michigan
Re: Calculating the position of Polaris new [Re: GlennLeDrew]
      #5814651 - 04/22/13 04:53 AM

a planetarium program will tell you the altitude and azimuth for any target you get details on.

I use this jig for polar setup, though I don't do imaging. http://www.iceinspace.com.au/63-499-0-0-1-0.html


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Kraus
Pooh-Bah


Reged: 03/10/12

Loc: Georgia.
Re: Calculating the position of Polaris new [Re: buddyjesus]
      #5814969 - 04/22/13 10:33 AM

When your local sidereal time equals Polaris' right ascension, Polaris is exactly above the celestial pole-your meridian.
When the two are different by twelve hours, Polaris is exactly beneath the celestial pole-your meridian.

At six hours either side, Polaris is to the right or left of the celestial pole-your latitude.


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edosaurusrex
sage
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Reged: 01/30/07

Loc: Ft Worth, TX
Re: Calculating the position of Polaris new [Re: Kraus]
      #5815265 - 04/22/13 01:18 PM

A quick "sloppy" formula for the local sidereal time at Greenwich for 2013 is
99.8+360.9856*DoY where Doy is the Day of the Year for 2013. So April 21, 2013 at 2100 (9pm CDT) is 111+21/24+5/24 = 112.0833, then 99.8+360.9856*112.0833 = 40560.27deg = 240.27deg
The local sidereal time is 240.27 + L where L is your longitude (+ = East, - = West). For me at W97 the answer is 240.27+(-97) = 143.27 = 9.55h RA
The Hour Angle = H = local sidereal - RA Polaris[expressed in degrees or 15*(h+m/60+s/3600)].

Once you have the Hour Angle, the latitude, and declination the Azimuth and Altitude can be found from standard formulas.

RA Polaris is 41.98deg, Dec Polaris is 89.32deg, and my Latitude is 33N

tan Az=sinH/(cosH*sin(Lat)-cos(Lat)*tan(Dec)) [add 180 deg to AZ to be reckoned from North]
sin Alt=sin(Lat)*sin(Dec)+cos(Lat)*cos(Dec)*cosH

H=143.27-41.98 = 101.32deg

tan Az=sin(101.32)/(cos(101.32)*sin(33)-cos(33)*tan(41.98))

Az=179.2+180=359.2deg

sin Alt=sin(33)*sin(41.98)+cos(33)*cos(41.98)*cos(101.32)

Alt=32.9deg

But I kinda agree with Glenn on this one.

Ed

Edited by edosaurusrex (04/22/13 08:37 PM)


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Monte Porche
member


Reged: 01/23/08

Loc: Haughton, LA
Re: Calculating the position of Polaris new [Re: edosaurusrex]
      #5815322 - 04/22/13 01:54 PM

Thanks for all of the information....It's plenty to process, but I think it's going to point me in the right direction.

What I am trying to do is program a microcontroller to help do a polar alignment when there is no visible sight line to Polaris due to obstruction.

It's partly for my own practical use, and mainly as a project for an electronics class.

So, it's not critical to be hyper accurate. If I can get in the ballpark, that will be good enough for the project...*and* I can always tweak it once I get it operational.


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Kraus
Pooh-Bah


Reged: 03/10/12

Loc: Georgia.
Re: Calculating the position of Polaris new [Re: Monte Porche]
      #5815488 - 04/22/13 02:55 PM


...no visible sight line to Polaris due to obstruction...

I take it you'll be mobile so star drift is out. It takes too much time from an observing session unless you get lucky to use a star bright enough to be seen just after sunset and before twilight ends.

Polaris is close enough for casual viewing-you will find objects.


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Monte Porche
member


Reged: 01/23/08

Loc: Haughton, LA
Re: Calculating the position of Polaris new [Re: Kraus]
      #5837282 - 05/02/13 09:23 PM

I just realized that I may have been making this more difficult than I needed to make it.

Since I'm trying to calculate the coordinates to do a polar alignment based on a known longitude, latitude, and time....

I don't need to calculate the location of Polaris. I need to calcualte the location of the celestial pole.

And, unless I'm mistaken, that would simply require an elevation roughly equivalent to the longitude, and then calculating the magnetic deviation of that location so that I can point the object at the north pole.


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Pharquart
sage
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Reged: 11/11/09

Loc: Southwest Minneapolis Metro
Re: Calculating the position of Polaris new [Re: Monte Porche]
      #5837389 - 05/02/13 10:37 PM

Quote:

And, unless I'm mistaken, that would simply require an elevation roughly equivalent to the longitude, and then calculating the magnetic deviation of that location so that I can point the object at the north pole.




One minor correction: you set the elevation (angle) roughly equivalent to the latitude of the location.

Brian


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Monte Porche
member


Reged: 01/23/08

Loc: Haughton, LA
Re: Calculating the position of Polaris new [Re: Pharquart]
      #5837429 - 05/02/13 11:03 PM

drat....I always get those two mixed up..

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Matt2893
super member


Reged: 09/18/12

Loc: ChicagoLand
Re: Calculating the position of Polaris new [Re: Monte Porche]
      #5837968 - 05/03/13 10:08 AM

Quote:

drat....I always get those two mixed up..




I was taught to remember LATitude as a Fat guy, and LONGitude as a Long tall skinny guy. LAT rhymes with FAT. Not PC, but works....


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Kraus
Pooh-Bah


Reged: 03/10/12

Loc: Georgia.
Re: Calculating the position of Polaris new [Re: Matt2893]
      #5841205 - 05/05/13 07:09 AM


Ahhh. Reverse navigation eh?


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salientbunny
member


Reged: 06/13/12

Loc: Southeast Georgia, US
Re: Calculating the position of Polaris new [Re: Kraus]
      #5841638 - 05/05/13 01:02 PM

1 degree past Polaris in the direction of Kochab?

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Monte Porche
member


Reged: 01/23/08

Loc: Haughton, LA
Re: Calculating the position of Polaris new [Re: Kraus]
      #5848836 - 05/08/13 09:50 PM

Quote:


Ahhh. Reverse navigation eh?




heh...not quite.

My end game is a 3 part process.

part 1: Prove that it's possible to calculate the proper elevation and direction to point a mount mathematically.

part 2: design and construct a circuit that will guide you in setting up a polar alignment (using a barn door tracker). Basically, on an LCD screen, it will tell you to rotate the tracker clockwise (or counterclockwise) until it is properly positioned. Then, it will tell you to elevate (or lower) the platform until the proper angle is reached.

Part 3: Modify the circuit using motors so that it will automatically position the barn door tracker properly, and will properly track the Earth's rotation.


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rjhinton
newbie


Reged: 05/03/13

Re: Calculating the position of Polaris new [Re: edosaurusrex]
      #6036726 - 08/20/13 03:11 PM

Quote:

A quick "sloppy" formula for the local sidereal time at Greenwich for 2013 is
99.8+360.9856*DoY where Doy is the Day of the Year for 2013. So April 21, 2013 at 2100 (9pm CDT) is 111+21/24+5/24 = 112.0833, then 99.8+360.9856*112.0833 = 40560.27deg = 240.27deg
The local sidereal time is 240.27 + L where L is your longitude (+ = East, - = West). For me at W97 the answer is 240.27+(-97) = 143.27 = 9.55h RA
The Hour Angle = H = local sidereal - RA Polaris[expressed in degrees or 15*(h+m/60+s/3600)].

Once you have the Hour Angle, the latitude, and declination the Azimuth and Altitude can be found from standard formulas.

RA Polaris is 41.98deg, Dec Polaris is 89.32deg, and my Latitude is 33N

tan Az=sinH/(cosH*sin(Lat)-cos(Lat)*tan(Dec)) [add 180 deg to AZ to be reckoned from North]
sin Alt=sin(Lat)*sin(Dec)+cos(Lat)*cos(Dec)*cosH

H=143.27-41.98 = 101.32deg

tan Az=sin(101.32)/(cos(101.32)*sin(33)-cos(33)*tan(41.98))

Az=179.2+180=359.2deg

sin Alt=sin(33)*sin(41.98)+cos(33)*cos(41.98)*cos(101.32)

Alt=32.9deg

But I kinda agree with Glenn on this one.

Ed




This is great information, but I need some additional info to understand what you're doing. Could you please identify what the number 99.8 represents and how you get 40560.27deg = 240.27deg. I have assumed that 5/24 is the timezone offset for CDT.

Thanks in Advance,
Robert


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Thomas Karpf
Carpal Tunnel


Reged: 02/09/09

Loc: Newington, CT
Re: Calculating the position of Polaris new [Re: rjhinton]
      #6036820 - 08/20/13 04:00 PM

I would expect that 99.8 and 240.27 fall under the category of 'starting point'.

The numbers get messy because a year is approximately 365.242159 synodic days long, so the beginning of any particular day in any particular location is unlikely to start with 'neat' numbers.


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rjhinton
newbie


Reged: 05/03/13

Re: Calculating the position of Polaris new [Re: Thomas Karpf]
      #6037852 - 08/21/13 08:06 AM

Ok. But how does this formula get from 40560.27deg to 240.27deg?

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brianb11213
Postmaster
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Reged: 02/25/09

Loc: 55.215N 6.554W
Re: Calculating the position of Polaris [Re: rjhinton]
      #6037864 - 08/21/13 08:20 AM

Quote:

Ok. But how does this formula get from 40560.27deg to 240.27deg?



You can throw away whole circles (360 degrees) & the angle is still the same. 40560.27 - 240.27 = 112 x 360.


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