When do stars change direction?

[Galah (Patrick)]

This concerns circumpolar stars.

Is anyone aware of a calculation or software which can find the point (azimuth) where a circumpolar star "changes" direction from left-to-right (or right-to-left).
That is, the point where a tracking telescope changes from moving to the left, to moving to the right (or from moving to the right to moving to the left).

This can be found using (say) Starry Night and the step forward/backward in time function, but is a pain.

Regards,

Patrick.

[MrB (Simon)]

Isn't it just the stars declination transposed to azimuth? Just guessing here.
For example, if the stars declination is 3 degrees from the pole(-87degrees), then the point it changes direction ( which is due east or west of the celestial pole) would be at 3degrees east or west, so 177degrees and 183 degrees azimuth respectively? (all assuming SCP)

So, again assuming SCP..
Change from left to right(east): Azimuth = 90 - Declination
Change from right to left(west): Azimuth = Declination + 270
Declination as a negative value, eg 90 - -87 = 177 and -87+270 = 183)

Many self-corrections later, I think that is right
Happy to be corrected.

[mithrandir (Andrew)]

I'm not sure this answers your question.

The time an object crosses the meridian (and achieves maximum altitude) is when the local sidereal time is equal to its RA. Just about any planetarium program will tell you the UTC or civil time when this happens. At that time (assuming the southern hemisphere) its azimuth is 180 (ie due south). That is when it changes from moving up to down.

Around the SCP, objects move clockwise, so when crossing the meridian it will be moving right.

[pdalek (Patrick)]

Using the ancient notation
d : declination of star, -south +north
L : latitude of observer, -south +north
LHA : local hour angle of star, -east +west
Hc : altitude
Zn : azimuth

Hc = asin( sin(d) sin(L) + cos(d) cos(L) cos(LHA) )
Z = acos( (sin(d) - sin(L) sin(Hc)) / (cos(L) cos(Hc)) )
If sin(LHA) < 0 then Zn = Z else Zn = 360 - Z

Other ways to write formula but his gets signs right without having to think.

Images 1-3 show azimuth for three objects at differing declinations as observed from Melbourne

Max and min Zn (i.e. direction changes) when partial derivative of Zn wrt LHA is zero.
Messy to work out closed form solution. Easier numerically.

Image 4 shows maximum azimuth vs declination for Melbourne observer. Minimum = 360 - maximum.

Generating a plot or table of values takes four lines of code using Mathematica.

[OzEclipse (Joe Cali)]

I think some of the posters misread your question. The question is not very clear. It seems you want to know either when (time) or where(location) the azimuth reverses direction. Using the word "point" doesn't make your goal very clear. You did make it clear that you wanted the "point" when the azimuth reverses which is when the star is +/- 90 deg from transit. Most replies described the transit or culmination time when the altitude reverses not the azimuth.

The time of culmination or transit is the highest point is when the stars RA is equal to the local siderial time. Starry night shows this as transit time when you hover over the object with the mouse it's the 4th row down on my version - Starry night Pro.

If it is times you are looking for it is when the star is most east and most west. This is 5hrs 59m before and 5hrs 59m after the transit time.

The altitude and azimuth are
altitude = your latitude
azimuth (east most point) = 90+declination
azimuth (east most point) = 270-declination

This is the same formula as Simon just written a different way.

The way I've written these formulae, just use the declination as a positive number even though it is south ie negative. So if the dec is 65 south

azimuth (east most point) = 90+65 = 155
azimuth (east most point) = 270-65 = 205

cheers

Joe

[pdalek (Patrick)]

Turn on both alt-az and equatorial grids in your planetarium software and look south. Follow a point around dec circle and check the azimuth.
Max/min azimuth are not where you may intuitively guess.
Flat maps of spherical things are tricky.

[pdalek (Patrick)]

Maybe this is clearer:

Consider as a specific example a star 10 degrees from the south celestial pole, i.e. dec = -80.

Stand at the south pole, i.e. lat = -90. The star will rotate at constant altitude 80 through a full 360 deg of azimuth.

Stand at lat -80. At some point the star will be directly overhead so azimuth is not defined. At all other times the azimuth will be somewhere between 90 and 270.

Stand exactly at the equator. The azimuth will be as Joe or Simon described.

For latitude between the equator and -80, the max/min values must be calculated for that specific latitude.

[MrB (Simon)]

At all other times the azimuth will be somewhere between 90 and 270.

A -80 dec star is never further than 10 degrees from the celestial pole(nor any closer) therefore it's position when transposed to azimuth can never be further than 10 degrees from true south, or 180 degrees, so will only ever be between 170 and 190 degrees azimuth.

Stand exactly at the equator. The azimuth will be as Joe or Simon described.

And at any other latitude. Ones latitude has no effect on the stars azimuth, it only shifts the celestial poles altitude.

[pdalek (Patrick)]

Here is Rigil Kent, dec -61, as seen from McMurdo Station, at azimuth 0, i.e. North. Observer position matters.

[MrB (Simon)]

Ah sorry, yes I see what you are saying.

It's all too much thinking for my brain at 3:30am. I have to get up for work in 2.5 hours

[Galah (Patrick)]

Thank you for all your responses. However I am just about at wits end.

All the following were used / obtained using Starry Night Pro 6:
Home Location: Longitude: 145 degrees 58 minutes East
Latitude: 36 degrees 30 minutes South
Time Zone: AEST.
Star: Acrux. (Jnow) RA: 12H 27m 25.4s. Dec: -63 degrees 10 minutes 42 seconds.
Start Date: 17th March 2014. Time: 8:31:45 pm.
At azimuth approx. 145 degrees 51.211 minutes (altitude was 41 degrees 48 minutes 3 seconds) Acrux starts changing from moving to the east, to moving to the west (azimuth changes from decreasing to increasing).


Continuing on, at azimuth 214 degrees 8.789 minutes (also at altitude 41 degrees 48 minutes 3 seconds) Acrux starts changing from moving west, to moving east ( azimuth changes from increasing to decreasing. Date: 18th March 2014. Time: 5:34:29 am.


I have been unable to reach these same values using any of the provided methods including many hours with a spreadsheet using "the ancient notation" calculations provided (obviously doing something wrong there but cannot fault the calculations.).


Perhaps someone could please attempt to validate their suggestions using their methods with the above details and / or explain what might be happening.


Regards,



Patrick.

[Galah (Patrick)]

Sorry, forgot to include the LHA data used in the calculations suggested by Patrick (pdalek).


Used three:
SNP Information Pane - Hour Angle (Jnow) : 19h 27m 53.1s
Hour Angle Lines - display from the SNP pop-up menu for Acrux: 7h 27m 53
From C2A - Hour Angle (which Philippe Deverchere advised is the same as LHA)– 4h 31m 17s
Don’t know if any of these are the correct values to use for LHA.

Patrick.

[mithrandir (Andrew)]

It might be easier to determine the culmination time and subtract or add 1/6 sidereal day (5h59'1.023") to get the time for the max and min azimuth.