#11
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46P, can't see
Mike Collins wrote:
StarDust wrote: On Sunday, December 9, 2018 at 6:36:56 AM UTC-8, Chris L Peterson wrote: On Sun, 9 Dec 2018 02:18:24 -0800 (PST), wrote: I live in the city, major light pollution, could not see 46P with my 80mm APO, mounted on NexStar GT goto. I know my GOTO points to the right location. I slew the scope in auto between Aldebaran and other 2 stars, like Mira back and forth, all came in to the center of my eyepiece. Alt/Az kind of suck, coordinates all ways changing. Day before tried it with my 50 mm bino, no luck to see it either. Is it still hard to see this comet? I saw it again last night, with a very thin haze in the sky. No problem with 8x50 binoculars, but the haze prevented me from seeing it with averted vision. Not obviously brighter than a few nights ago, but less contrast because of the poorer sky. But still a dark sky. It's very big, so I think it will be easily lost in any light pollution. I'd guess your best chance is with the lowest possible magnification, or with high enough magnification that you just get the core part of the coma (which requires accurate pointing, of course). How big is the comet in arc minutes? I used my Celestron Onyx ED f6.2 and Pentax XL40 mm eyepiece, I know this combination gives me a very wide field at low power. I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. It took me a while to find it. Eventually I used a combined high/low tech method and held my iPad mini running Luminos against the objectives of my 8x50 binoculars (on a tripod) and moving the pan/tilt until the comet was in the centre of the screen. At least in the back yard there are no Christmas lights. This is the British not US use of back yard, A small paved area not a big area of grass. It was not that dark so you should be able to find it from a city sky. But you have to stare at it for a long time to convince yourself that it’s there. I kept moving the binoculars away and then back again to convince myself that it really was the comet. But it definitely matched the position shown in Luminos in the straight line part of an asterism like an inverted question mark (or a tiny version of Leo’s head). Apple’s spell check decided I wanted USA instead of Ursa. I cant see any of the USA from longitude 1 degree East. |
#12
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46P, can't see
On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins
wrote: I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? |
#13
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46P, can't see
On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins
wrote: I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. Aren't you confusing seeing with transparency? Even with horrible seeing the Milky Way will be easily visible if only the transparency is good and the sky is dark. |
#14
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46P, can't see
On Sunday, December 9, 2018 at 2:26:42 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
StarDust wrote: ^^^^^^^^ Please post using your real name(, too). How big is the comet in arc minutes? Good (and free) astronomy software like Stellarium[1][2] can give you the numbers (be sure to use the latest version, currently 0.18.2). For now: Distance (D) (from Terra) ≈ 0.087 AU ≈ 1.3 × 10⁷ km (and closing; perigee on 2018-12-16) Radius of coma (estimated) ≈ 36'000 km (0°09'30") Tail length (estimated) ≈ 0.00812 × 10⁶ km (0°02'08") Core diameter (d) = 10.0 km If you know the diameter d and the distance D in SI units (or any other non-angular measure), you can calculate the diameter of a celestial object in degrees (of arc) using the equation φ = d/D/(2π) × 360° because sin(φ) ≈ φ for small angles φ (otherwise calculate φ = arcsin(r/D)/π × 360°): φ/2 _______ __ \ _.-': |PE _.-\ .: r -:'--------:--- d = φ `-._ : `-._:______ :-- D ---: This gives you the diameter in degrees (of arc), so for minutes of arc you must multiply by 60. For example, you can get Celestia’s 0°09'30" for the estimated radius of the coma with φ ≈ (36'000 km)/(1.3 × 10⁷ km)/(2π) × 360(°) × 60 ≈ 9.09' [arcsin(18'000 km/(1.3e7 km))/π × 360° × 60 ≈ 9.5' is the exact value as displayed by Stellarium]. So ignoring the uncertainty of the coma, the comet’s core diameter in minutes of arc is approximately only φ ≈ (10 km)/(1.3 × 10⁷ km)/(2π) × 360(°) × 60 ≈ 0.0026'. See also: https://www.wolframalpha.com/input/?i=(10+km%2F1.3e7+km)%2F(2pi)*360*60 Thus, in combination with a visual brightness of only 9.05 mag (9.25 mag with extinction; both according to Stellarium), and light pollution, it might be very hard to see with an amateur telescope. OTOH, as it should become brighter when it approaches Sol, observing it might be possible. However, the simulation with Stellarium indicates to me that even if it is observable, it might not be easily distinguishable from the fixed stars, because it is not moving sufficiently fast at this point. And ISTM that an equatorial mount is highly recommended, otherwise, once found, the comet is already out of the field of view after ca. 30 s (due to Terra’s rotation). I used my Celestron Onyx ED f6.2 and Pentax XL40 mm eyepiece, I know this combination gives me a very wide field at low power. Good luck. (It has been raining here all week, so no luck for me yet.) [1] https://stellarium.org/ [2] http://dslr-astrophotography.com/add-comets-stellarium/ -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail. I use C2A! http://www.astrosurf.com/c2a/english/ But to remind you, not every one is professor in math! |
#15
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46P, can't see
Chris L Peterson wrote:
On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins wrote: I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? Yes I find it hard to format my brain in minutes and seconds of arc. I think of degrees and decimal degrees. When I wrote planetarium software in the 80s I used decimals and only converted to minutes and seconds for the final display. The stars, moon and planetswere plotted manually from Mercator projections of the zodiac. |
#16
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46P, can't see
On Sunday, December 9, 2018 at 9:13:17 PM UTC-8, Chris L Peterson wrote:
On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? That's very small? Few people here saying the comet is very large. Maybe the brightest part, the nucleus of the comet is 8 arc minutes? |
#17
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46P, can't see
On Monday, December 10, 2018 at 5:19:55 AM UTC-8, Mike Collins wrote:
I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? Yes I find it hard to format my brain in minutes and seconds of arc. I think of degrees and decimal degrees. When I wrote planetarium software in the 80s I used decimals and only converted to minutes and seconds for the final display. The stars, moon and planetswere plotted manually from Mercator projections of the zodiac. Think of the size of the Moon is 1/2 degree, or 30 arc minute, Saturn is about 45 arc seconds and compare? |
#18
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46P, can't see
On Mon, 10 Dec 2018 13:19:53 -0000 (UTC), Mike Collins
wrote: I find it hard to format my brain in minutes and seconds of arc. I think of degrees and decimal degrees. Do you feel the same about time? So you use hours and decimals of hours instead of hours, minutes and seconds? "I'll see you at 9.835" - such a statement would be wilder most people... When I wrote planetarium software in the 80s I used decimals and only converted to minutes and seconds for the final display. That's natural. You want to use one unit instead of mixing different units internally in the software. For angles that unit could be degrees. Or radians, so the built-in trig functions work without any need for unit conversion. For time, hours could be that unit. Or, perhaps even better, days counted from some reference date. All with fractions to full machine precision of course. For display purposes you convert angles to whatever you want: degrees with decimals, or degrees and minutes with decimals, or degrees, minutes and seconds perhaps with decimals. The day count is converted to the calendar date followed by hours, minutes and seconds. If there's any input, the opposite conversion needs to be done. |
#19
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46P, can't see
On Mon, 10 Dec 2018 05:23:56 -0800 (PST), StarDust
wrote: On Sunday, December 9, 2018 at 9:13:17 PM UTC-8, Chris L Peterson wrote: On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? That's very small? Few people here saying the comet is very large. Maybe the brightest part, the nucleus of the comet is 8 arc minutes? What you can see with the naked eye is 5-10 arcminutes. The actual coma is over a half a degree, but the outer part only shows up in images. It's too dim for the eye. I don't know that 8 arcmin is all that small. It's the size of large maria on the Moon. |
#20
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46P, can't see
On Mon, 10 Dec 2018 13:19:53 -0000 (UTC), Mike Collins
wrote: Chris L Peterson wrote: On Sun, 9 Dec 2018 23:15:38 -0000 (UTC), Mike Collins wrote: I just found it again from my back yard. Not perfect seeing. I can just make out the Milky Way but only one star in the bowl of USA minor is visible. I make it out to about 8 seconds of arc. 8 _minutes_ of arc? Yes I find it hard to format my brain in minutes and seconds of arc. I think of degrees and decimal degrees. When I wrote planetarium software in the 80s I used decimals and only converted to minutes and seconds for the final display. Likewise. I always use degrees and decimals for my work (and that includes right ascension). And I don't normally bother with the conversion to sexagesimal at all, even in the final output. There's seldom a need. Happily, I see professional astronomy abandoning that system, as well. It is common now to see coordinates given in decimal degrees on both axes of the sky. |
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