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46P, can't see



 
 
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  #11  
Old December 10th 18, 12:18 AM posted to sci.astro.amateur
Mike Collins[_4_]
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Default 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  
Old December 10th 18, 06:13 AM posted to sci.astro.amateur
Chris L Peterson
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Default 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  
Old December 10th 18, 07:20 AM posted to sci.astro.amateur
Paul Schlyter[_3_]
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Default 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  
Old December 10th 18, 07:53 AM posted to sci.astro.amateur
StarDust
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Default 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  
Old December 10th 18, 02:19 PM posted to sci.astro.amateur
Mike Collins[_4_]
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Default 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  
Old December 10th 18, 02:23 PM posted to sci.astro.amateur
StarDust
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Default 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  
Old December 10th 18, 02:32 PM posted to sci.astro.amateur
StarDust
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Default 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  
Old December 10th 18, 04:03 PM posted to sci.astro.amateur
Paul Schlyter[_3_]
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Default 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  
Old December 10th 18, 04:06 PM posted to sci.astro.amateur
Chris L Peterson
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Posts: 10,007
Default 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  
Old December 10th 18, 04:09 PM posted to sci.astro.amateur
Chris L Peterson
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Posts: 10,007
Default 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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