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# Calculating the position of Polaris

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### #1 Monte Porche

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Posted 21 April 2013 - 07:00 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.

### #2 Pharquart

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Posted 21 April 2013 - 07:38 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

### #3 wirenut

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Posted 21 April 2013 - 08:52 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.

### #4 GlennLeDrew

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Posted 21 April 2013 - 11:48 PM

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?

### #5 buddyjesus

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Posted 22 April 2013 - 03:53 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.iceinspac...99-0-0-1-0.html

### #6 Kraus

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Posted 22 April 2013 - 09:33 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.

### #7 edosaurusrex

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Posted 22 April 2013 - 12:18 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

### #8 Monte Porche

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Posted 22 April 2013 - 12:54 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.

### #9 Kraus

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Posted 22 April 2013 - 01:55 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.

### #10 Monte Porche

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Posted 02 May 2013 - 08:23 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.

### #11 Pharquart

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Posted 02 May 2013 - 09:37 PM

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

### #12 Monte Porche

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Posted 02 May 2013 - 10:03 PM

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

### #13 Matt2893

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Posted 03 May 2013 - 09:08 AM

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....

### #14 Kraus

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Posted 05 May 2013 - 06:09 AM

### #15 salientbunny

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Posted 05 May 2013 - 12:02 PM

1 degree past Polaris in the direction of Kochab?

### #16 Monte Porche

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Posted 08 May 2013 - 08:50 PM

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.

### #17 rjhinton

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Posted 20 August 2013 - 02:11 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

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.

Robert

### #18 Thomas Karpf

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Posted 20 August 2013 - 03:00 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.

### #19 rjhinton

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Posted 21 August 2013 - 07:06 AM

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

### #20 brianb11213

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Posted 21 August 2013 - 07:20 AM

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.

### #21 blb

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Posted 21 August 2013 - 10:50 AM

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

Any suggestions?

Don't! For what reason would you try to do this? Even most old surveying compasses only measure to the nearest degree and modern hand held compasses measure to the nearest five degrees. Polaris only deviates by about one and a half degrees from one side of the north pole to the other side, so you are trying to determin something that is beyond the ability of normal measurement. Not to mention that magnetic declination varies depending on where you are on the surface of the earth (so you need to know your lat. long. for each observing location) and the magnatic pole moves over time and there are small daily wobbles in the magnatic pole to account for too. Then you have to take into account local anomalies like magnetic rocks, fences, ferrous metals in your pockets, proximity to cars, power lines, etc., etc., etc. Not trying to be ugly but unless you are interested in just doing the exercise, this is fools play.

P.S. even in colonial times when magnetic compasses were used, magnetic bearings were measured at each end of a line being measured due to magnetic anomolies in the ground. If you are only taking one measurement and expecting accuracy, the measurement can't be made with any certainty it is correct.

### #22 Tony Flanders

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Posted 21 August 2013 - 02:40 PM

Even most old surveying compasses only measure to the nearest degree and modern hand held compasses measure to the nearest five degrees.

I think that's pretty pessimistic; I can definitely get +-3 degrees with a regular hand-held compass, and probably quite a bit better. Five degrees is a long, long way!

Having said that, I agree with the rest of your post; I don't see any point in this particular exercise.

### #23 Tony Flanders

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Posted 21 August 2013 - 02:45 PM

[quote name="Tony Flanders"][quote]Even most old surveying compasses only measure to the nearest degree and modern hand held compasses measure to the nearest five degrees.[/quote]

I think that's pretty pessimistic; I can definitely get +-2 degrees with a regular hand-held compass, and probably better. Five degrees is a long, long way!

Now that the requirement to find Polaris has been removed, this becomes simply what people do most with compasses: attempt to determine true north.

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Posted 21 August 2013 - 05:56 PM

The altitude of the pole is EXACTLY the same as your latitude. If you live near the coast, the best source of current magnetic variation is a marine chart. (Variation is sometimes called magnetic declination. Magnetic deviation is another issue - caused by local magnetic effects from metal)
There is a handy calculator here

TVMDC is the rule sailors use. True corrected for Variation gives Magnetic and Magnetic corrected for Deviation gives Compass!

### #25 blb

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Posted 21 August 2013 - 09:14 PM

Even most old surveying compasses only measure to the nearest degree and modern hand held compasses measure to the nearest five degrees.

I think that's pretty pessimistic; I can definitely get +-3 degrees with a regular hand-held compass, and probably quite a bit better. Five degrees is a long, long way!

Having said that, I agree with the rest of your post; I don't see any point in this particular exercise.

I do have a hand held compass that measures to one degree, that I used when surveying but most compasses are graduated to five degrees or more, I have several of these in my collection. That being said, there is a big difference between precision and accuracy. You can make measurements very precisely and even do it repeatedly and yet not be accurate. This is true when making measurements that are smaller than what you are using to measure with is able to measure.

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