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brucepiano
super member
Reged: 12/02/07
Posts: 137
Loc: New England
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How do I calculate the lowest declination of a sky object I will be able to see at my latitude (42 degrees)?
Bruce
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AlanK
professor emeritus
Reged: 01/26/07
Posts: 510
Loc: Auckland, New Zealand
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Hi Bruce
Just subtract 90 from the latitude which gives a lowest declination of -48 degrees for sky objects in your case (that assumes your southern horizon is flat (no hills or other obstacles in the way).
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bbmagic
member
Reged: 09/18/08
Posts: 54
Loc: 40°30 111°53
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I'll second that. The zenith from your location is +42°, so 90° to the southern horizon is correct. Incidentally, Polaris is 42° from your northern horizon!
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MessierScott
sage
Reged: 06/18/07
Posts: 294
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Yes, if your southern horizon is flat, the lowest declination will be -48, but don't expect to see too much below about -45 declination except for the REALLY bright stuff.
Don't be looking for a 15th mag galaxy at declination -47!
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Scott Kranz
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Hrundi
Pooh-Bah
Reged: 02/06/08
Posts: 1159
Loc: Estonia
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Quote:
Yes, if your southern horizon is flat, the lowest declination will be -48, but don't expect to see too much below about -45 declination except for the REALLY bright stuff.
Don't be looking for a 15th mag galaxy at declination -47!
I think even these figures are generous. The lowest I've seen was M4 with my 12". It was at ~3 degrees, and quite faint. Not quite averted, but not a steady object.
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acey
member
Reged: 01/09/09
Posts: 39
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Hills, atmospheric extinction and light pollution all make it harder to get to the theoretical limit. From my 55 deg latitude I've had views of M4 and NGC253, both about 10 degrees above my horizon, but neither was particularly inspiring - and they're showpiece objects. I've seen them with smaller scopes from more southerly sites and there was no comparison.
Can anyone offer a rule of thumb for the magnitude loss close to the horizon? (i.e. something more rough-and-ready than the extinction formula, which isn't much fun to work out in your head on a dark night).
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BillFerris
Carpal Tunnel
Reged: 07/17/04
Posts: 2900
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Quote:
Can anyone offer a rule of thumb for the magnitude loss close to the horizon? (i.e. something more rough-and-ready than the extinction formula, which isn't much fun to work out in your head on a dark night).
It's heavily dependent on the climate where you observe. From the humid Midwest U.S., I generally stayed at least 15 degrees above the horizon. The exceptions were objects that never rose more than 10-15 degrees. Going lower just meant getting stuck in the muck. From the high desert sites I've used in Arizona, the sky near the horizon generally isn't an issue. Dry air and excellent transparency allow you to get very nice views of objects at just 5 degrees elevation.
I decide how low to go based on the appearance of the sky near the horizon. If the sky looks milky and hazy down low, I'll stay at least one fist's width (~10 degrees) above the muck. If the sky remains uniform in appearance as it drops low and there is a sharp cutoff where the sky meets a black horizon, then I don't worry about going too low.
Bill in Flag
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Tony Flanders
Post Laureate
Reged: 05/18/06
Posts: 3369
Loc: Cambridge, MA, USA
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Quote:
Can anyone offer a rule of thumb for the magnitude loss close to the horizon? (i.e. something more rough-and-ready than the extinction formula, which isn't much fun to work out in your head on a dark night).
There's an Excel spreadsheet for computing this in the online article Transparency and Atmospheric Extinction by me and Phil Creed.
According to the spreadsheet, at sea level in absolutely pristine air, extinction is 1.7 magnitudes 5 degrees above the horizon, 0.9 magnitudes at 10 degrees, and 0.6 magnitudes at 15 degrees -- all these at the peak wavelength for human night vision.
An aerosol optical depth of 0.10 -- average for the desert Southwest in the summer, and better than average in most parts of the U.S., raises the extinction to 2.9 at 5 degrees, 1.6 at 10 degrees, and 1.1 at 15 degrees.
Even at high altitudes in pristine air, there's an extremely heavy penalty for observing objects less than 15 degrees above the horizon.
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Tony Flanders
First and foremost observing love: naked eye.
Second, binoculars.
Last but not least, telescopes.
And I sometimes dabble with cameras.
Edited by Tony Flanders (07/07/09 02:00 PM)
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JakeSaloranta
sage
Reged: 09/18/08
Posts: 229
Loc: Sisu, Sauna, Sibelius...
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Quote:
I think even these figures are generous. The lowest I've seen was M4 with my 12". It was at ~3 degrees, and quite faint. Not quite averted, but not a steady object.
One of the funny ones I have is Messier 8 at an altitude of 3° 49' with the sun only -11° below horizon the horizon. And it wasn't all that bad! 
/Jake
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Hrundi
Pooh-Bah
Reged: 02/06/08
Posts: 1159
Loc: Estonia
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Quote:
Quote:
I think even these figures are generous. The lowest I've seen was M4 with my 12". It was at ~3 degrees, and quite faint. Not quite averted, but not a steady object.
One of the funny ones I have is Messier 8 at an altitude of 3° 49' with the sun only -11° below horizon the horizon. And it wasn't all that bad!
/Jake
Wow  I was doubting if M8 was within reach at all. Must have been exciting to horizon hug. The viewing position like that tends to be quite comfortable.
This april, the lowest I got from that region was M17 at about 10 degrees, sun -11 as you had. I was pretty stunned, it was gorgeous.
You wouldn't happen to have a sketch of M8? 
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JakeSaloranta
sage
Reged: 09/18/08
Posts: 229
Loc: Sisu, Sauna, Sibelius...
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Quote:
Wow I was doubting if M8 was within reach at all. Must have been exciting to horizon hug.
All it needs is a clear sea horizon looking down south to Estonia
Quote:
You wouldn't happen to have a sketch of M8?
http://www.kolumbus.fi/jaakko.saloranta/Deepsky/Messier/M8.html
M22, M28 and most of the brighter deep sky objects in the Sagittarius-Scorpius are should be visible from Estonia as well. Just make sure you have a really good southern horizon!
The only ones I haven't been able to spot from here are M6, M7, M54, M55, M62, M69, M70 and M83.
My personal challenge has always been M83. It rises only about 30' above the horizon and I've been trying to hunt it down for years in vain 
/Jake
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Hrundi
Pooh-Bah
Reged: 02/06/08
Posts: 1159
Loc: Estonia
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Wow, that's a lot of detail!
M6, M7, M70, M69, M70 are completely impossible here.
M83 might rise 1 degree around here. If I go to the southern parts, I can gain another degree. It should be possible, but it'd probably be invisible :/
M55 however is even more fascinting. From where I am, it rises 5', and from my usual observing place, 9'.
Does that make it completely impossible?
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JakeSaloranta
sage
Reged: 09/18/08
Posts: 229
Loc: Sisu, Sauna, Sibelius...
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Quote:
M55 however is even more fascinting. From where I am, it rises 5', and from my usual observing place, 9'.
Does that make it completely impossible?
I'd say it does but never say never. Give it a go in a month!
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timokarhula
member
Reged: 01/30/06
Posts: 87
Loc: Sweden
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Last April, on an exceptionally transparent night, I got M4 for the first time in the telescopic field of view (I live at latitude +59°52'). The globular cluster's outskirts was easily resolved into 11 magnitude stars (and fainter?) with my 10-inch at 48x. Its famous bar of stars at the core was obvious at 125x. M4 was at the altitude of 3°.2 (including standard refraction). That night, Antares (altitude 3°) shone brightly naked eye which was an indication of the night's quality.
Last August, I got to see the nebulosity in M8 (not only the star cluster NGC6530) with 18x50 IS binoculars. The altitude of M8 was 3°.6 while the sun was 12°.9 down. That night, I observed all the southern summer Messier's that are theoretically visible from here. Last year, I saw all the (theoretically visible) 102 Messier's with binoculars in two nights, well in "1 1/2 nights", because all the summer Messier objects were seen before the end of astronomical twilight. 
/Timo Karhula
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