Midnight Culmination

[GrampianStars (Rob)]

G'day Y'all
I'm at a loss as to how to calculate the date of "Midnight Culmination"
of stars and DSO's.
i.e. the date of highest elevation above the local horizon
I know how to do the culmination;
i.e. "A simple example: I live at Pomonal, 37⁰ degrees south of the equator, at my location the celestial equator would be 90⁰ -37⁰ = 53⁰ degrees above the horizon. A star at +15⁰ degrees declination (north) would then culminate at 53⁰ -15⁰ = 38⁰ degrees altitude, (above horizon)."
What is bugging me is how to calculate the best date for crossing the meridian
i.e. for m31 Andromeda galaxy.

Any takers

[chrisp9au (Chris)]

G'day Robert,

I have a cheat for doing this!

Free software called HNSKY, a fast simple planetarium program for Windows.
Download it from http://www.hnsky.org/software.htm

Setup your location, set the time to midnight, get the display to show the meridian, and then use the F8 and F7 keys to advance/retreat by a day.
Shows me that M31 culminates at midnight on October 22nd here in Kyneton.
And in Pomonal, October 24th

Or, you could spend a bit of cash and get Sky Tools V3.0 which gives you optimum viewing dates and times for all objects.
Check it out at http://www.skyhound.com/ One of my better investments!

Cheers
Chris

[Gas Giant (Andrew)]

If you can get some way to read the Local Sidereal Time; many astro software packages give LST, or the Argo Navis computer has a built in sidereal clock, then:

Sidereal Time = Right Ascension on the Meridian:

that is, when an object's RA is the same as Sidereal Time, the object is on the Meridian.
Andrew

[Robh (Rob)]

I think you're restricting yourself too much. Any culmination after 10 pm would be fine (even earlier in Winter months). You might wait for a midnight culmination then find the weather is lousy on that night or the Moon is out in full. According to my software, the Andromeda Galaxy will culminate in Sydney at midnight around 17th Oct under a Gibbous Moon. Hardly ideal!
Here's a rough way to calculate the date of a culmination at some time, say 11 pm.
From your site, locate a star on the meridian at 11 pm. Get its RA.
e.g. Fomalhaut is roughly on the meridian for my location at 11 pm tonight. RA = 22.9 hours.
For the object you're waiting for e.g. Andromeda Galaxy, find its RA = 0.7 hours.
Find the number of hours difference to say one decimal place.
(24-22.9)+0.7 = 1.1+0.7 = 1.8 hours.
Multiply by 15 degrees (as 1 hour corresponds to 15 degrees) = 1.8x15 = 27 degrees.
The Earth moves 360 degrees in its orbit about the Sun every 365 days. This is pretty close to one degree per day. Therefore, the Andromeda Galaxy will culminate here at 11 pm in about 27 days i.e. about Nov 1st.

Regards, Rob

[glenc (Glen)]

The approx. RA at midnight on the 21st of the month is:

Month, RA
1, 8
2, 10
3, 12 [equinox]
4, 14
5, 16
6, 18
7, 20
8, 22
9, 0 [equinox]
10, 2
11, 4
12, 6

The RA on the meridian at 6 pm on the 21st of the month is equal to the month number times two.
At 8 pm RA = 2M + 2 e.g. For Jan, RA = 2x1+2 = 4 at 8 pm
At midnt RA = 2M + 6 e.g. For Jan, RA = 2x1+6 = 8 at midnt
At 4 am RA = 2M + 10 e.g. For Jan, RA = 2x1+10 = 12 at 4 am