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Two related questions (visual magnitude + Uranus)



 
 
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  #1  
Old September 10th 11, 07:44 PM posted to sci.astro.amateur
[email protected]
external usenet poster
 
Posts: 17
Default Two related questions (visual magnitude + Uranus)

Hi group!

As you can imagine, a side-effect of the thread I initiated last week
on
"elusive Uranus", I got now extra-eager and impatient trying to see
Uranus.

So, I learned that under reasonably dark sky, it is a naked eye
object,
and very obvious with binoculars (should be the case with mine, a
10x50 pair).

Ok, but what about under full Moon? (well, or near-full, like in the
past couple of days?)

I normally would think not, but I ask because (and this leads me to
my other "related" question) on my front yard, in a heavily light-
polluted
area --- right on the border between white (big-city level light-
pollution)
and red, as per light-pollution maps http://www.jshine.net/astronomy/dark_sky/
I went out just to try and see the moon through the binoculars to see
if
it was worth the effort to take the telescope to shoot a couple of
pictures,
and *surprise* !!! I could see Andromeda (the galaxy!) *naked eye*
(I
had to see through the binoculars to double check that it wasn't my
imagination).

Even *yesterday*, I could see it (even though yesterday I would say
it was *truly barely* visible naked eye --- but through the
binoculars
it was like incredibly obvious. (I'm sure this is not at all news for
most
experienced astro-enthusiasts out there --- but it did take me by
surprise!!). I tried locating Uranus through binoculars, but without
success --- nothing seemed like a reasonable suspect.

So, I see that M31's visual magnitude is 3.4 (according to Wikipedia).
That's a lot, and actually I would expect a star of 3.4 to be visible
in
downtown NYC or some *heavily light-polluted* area (am I over-
estimating??). So, this suggests that the 3.4 is sort of the
integral
over the area, to obtain the "total amount of light" that the area
covered
by Andromeda emits (well, that we receive, I should say).

The funtion being integrated is somehow the "density" or the "true"
value of the luminous intensity, as in amount of light per unit of
surface;
with stars, the thing works more or less like the function being a
Dirac's delta, where its weight is precisely the visual magnitude of
the given star.

So, *finally* the question: is there a measurement of the brightness
of a celestial object that could be more directly related to the level
of
light pollution at which the object becomes visible? Intuitively, it
seems
like this would be closely related (if not being the exact same
measure)
to the "density" that I was describing above --- more specifically,
the
maximum value of this density of the given object (for objects that
occupy
an area, like M31)

IOW, I would like to be able to determine whether I *should* be able
to see some target (in this case Uranus, of course) simply by the
fact that some other target is visible --- I mean, it seems to me
(pure
intuition and rough estimate) that if Andromeda is visible under some
particular light-pollution conditions, then a 5.6 object *should* be
visible as well (possibly even obvious). But then, I don't think I
can
conclude that from the number 3.4 that I see for Andromeda; I do
not think the 3.4 reflects how faint Andromeda really is.

Hope I'm making some sense and that you will be able to understand
what I'm asking, and shed some light (I know, I know, this must be
the
second lamest pun, after the one we all know for Uranus :-\ :-) )

Thanks,

Carlos
--
  #2  
Old September 10th 11, 09:04 PM posted to sci.astro.amateur
Ken S. Tucker
external usenet poster
 
Posts: 740
Default Two related questions (visual magnitude + Uranus)

On Sep 10, 11:44 am, wrote:
Hi group!

As you can imagine, a side-effect of the thread I initiated last week
on
"elusive Uranus", I got now extra-eager and impatient trying to see
Uranus.

So, I learned that under reasonably dark sky, it is a naked eye
object,
and very obvious with binoculars (should be the case with mine, a
10x50 pair).

Ok, but what about under full Moon? (well, or near-full, like in the
past couple of days?)

I normally would think not, but I ask because (and this leads me to
my other "related" question) on my front yard, in a heavily light-
polluted
area --- right on the border between white (big-city level light-
pollution)
and red, as per light-pollution mapshttp://www.jshine.net/astronomy/dark_sky/
I went out just to try and see the moon through the binoculars to see
if
it was worth the effort to take the telescope to shoot a couple of
pictures,
and *surprise* !!! I could see Andromeda (the galaxy!) *naked eye*
(I
had to see through the binoculars to double check that it wasn't my
imagination).

Even *yesterday*, I could see it (even though yesterday I would say
it was *truly barely* visible naked eye --- but through the
binoculars
it was like incredibly obvious. (I'm sure this is not at all news for
most
experienced astro-enthusiasts out there --- but it did take me by
surprise!!). I tried locating Uranus through binoculars, but without
success --- nothing seemed like a reasonable suspect.

So, I see that M31's visual magnitude is 3.4 (according to Wikipedia).
That's a lot, and actually I would expect a star of 3.4 to be visible
in
downtown NYC or some *heavily light-polluted* area (am I over-
estimating??). So, this suggests that the 3.4 is sort of the
integral
over the area, to obtain the "total amount of light" that the area
covered
by Andromeda emits (well, that we receive, I should say).

The funtion being integrated is somehow the "density" or the "true"
value of the luminous intensity, as in amount of light per unit of
surface;
with stars, the thing works more or less like the function being a
Dirac's delta, where its weight is precisely the visual magnitude of
the given star.

So, *finally* the question: is there a measurement of the brightness
of a celestial object that could be more directly related to the level
of
light pollution at which the object becomes visible? Intuitively, it
seems
like this would be closely related (if not being the exact same
measure)
to the "density" that I was describing above --- more specifically,
the
maximum value of this density of the given object (for objects that
occupy
an area, like M31)

IOW, I would like to be able to determine whether I *should* be able
to see some target (in this case Uranus, of course) simply by the
fact that some other target is visible --- I mean, it seems to me
(pure
intuition and rough estimate) that if Andromeda is visible under some
particular light-pollution conditions, then a 5.6 object *should* be
visible as well (possibly even obvious). But then, I don't think I
can
conclude that from the number 3.4 that I see for Andromeda; I do
not think the 3.4 reflects how faint Andromeda really is.

Hope I'm making some sense and that you will be able to understand
what I'm asking, and shed some light (I know, I know, this must be
the
second lamest pun, after the one we all know for Uranus :-\ :-) )

Thanks,
Carlos


Your question(s) are good ones.
Have a read of this paragraph, note visual magnitudes and how
amateurs can use 'the little dipper' as a seeing reference,
http://en.wikipedia.org/wiki/Ursa_Minor#Stars

The four stars constituting the "bowl" of the little dipper are
unusual in that they are of second, third, fourth and, fifth
magnitudes. Hence, they provide an easy guide to determining what
magnitude stars are visible, useful for city dwellers or testing one's
eyesight.

It's very convenient as Polaris is easy to find.
If you keep notes, log the dimmest star you can see,
inverted vision is cheating ;-).
Cheers
Ken
  #3  
Old September 10th 11, 10:01 PM posted to sci.astro.amateur
Chris L Peterson
external usenet poster
 
Posts: 10,007
Default Two related questions (visual magnitude + Uranus)

On Sat, 10 Sep 2011 11:44:40 -0700 (PDT),
wrote:

Ok, but what about under full Moon? (well, or near-full, like in the
past couple of days?)


Even with moonlight, Uranus is no problem to see telescopically.
However, as a starhopper, you might have problems actually finding it
under bright skies.

So, I see that M31's visual magnitude is 3.4 (according to Wikipedia).
That's a lot, and actually I would expect a star of 3.4 to be visible
in
downtown NYC or some *heavily light-polluted* area (am I over-
estimating??). So, this suggests that the 3.4 is sort of the
integral
over the area, to obtain the "total amount of light" that the area
covered
by Andromeda emits (well, that we receive, I should say).


Correct. The magnitude of extended objects is determined by the total
light emitted, so large objects often have low magnitudes, when in
fact, they are very hard to see. For extended objects, you can pretty
much ignore published magnitude values. What you care about is surface
brightness, which is usually harder to find information about. Of
course, with some experience, you learn to look at the magnitude and
size, and mentally compute just how bright that object really is. But
the important point is that you really care about surface brightness,
not integrated intensity.

IOW, I would like to be able to determine whether I *should* be able
to see some target (in this case Uranus, of course) simply by the
fact that some other target is visible...


For Uranus it's simple, since its brightness is effectively equivalent
to a stellar value, and there are lots of well described methods of
estimating how both the Moon and light pollution affect the stellar
limiting magnitude.
  #4  
Old September 11th 11, 04:31 AM posted to sci.astro.amateur
[email protected]
external usenet poster
 
Posts: 17
Default Two related questions (visual magnitude + Uranus)


The four stars constituting the "bowl" of the little dipper are
unusual in that they are of second, third, fourth and, fifth
magnitudes. Hence, they provide an easy guide to determining what
magnitude stars are visible


Huh ... Interesting. However, the caveat (working against
me, as per Murphy's Law :-) ) ... I can see only two of them
with zero effort, the third one *barely*, and not at all the fourth
one (the one with magnitude 5, that is). With the binoculars,
however, all of them are extremely obvious. But that's looking
to the north, which is opposite to where the Moon is, and also,
more or less opposite to where the light pollution is coming
(as I'm to the north-west of the city, north/north-west is the
darkest side of the sky from where I live --- the Moon is now
being added to the light pollution, and Pisces and Uranus
are precisely around there ...)

*sigh* ... anyways, next weekend!!! However, the Moon
shots, prime focus with a 1000mm focal distance f/5
Newtonian, are not half bad --- I'm pretty sure they're my
personal best; maybe some of the past eclipse pictures
might have come out a bit more impressive, not being so
excessively bright.

If you're curious, two of them are he

http://www.mochima.com/personal/DSC_5578.jpg
http://www.mochima.com/personal/DSC_5580.jpg

Thanks, and thanks Chris for your message!

Carlos
--
  #5  
Old September 11th 11, 06:14 AM posted to sci.astro.amateur
jwarner1
external usenet poster
 
Posts: 156
Default Two related questions (visual magnitude + Uranus)



wrote:

Hi group!

As you can imagine, a side-effect of the thread I initiated last week
on
"elusive Uranus", I got now extra-eager and impatient trying to see
Uranus.

So, I learned that under reasonably dark sky, it is a naked eye
object,
and very obvious with binoculars (should be the case with mine, a
10x50 pair).

Ok, but what about under full Moon? (well, or near-full, like in the
past couple of days?)

I normally would think not, but I ask because (and this leads me to
my other "related" question) on my front yard, in a heavily light-
polluted
area --- right on the border between white (big-city level light-
pollution)
and red, as per light-pollution maps
http://www.jshine.net/astronomy/dark_sky/
I went out just to try and see the moon through the binoculars to see
if
it was worth the effort to take the telescope to shoot a couple of
pictures,
and *surprise* !!! I could see Andromeda (the galaxy!) *naked eye*
(I
had to see through the binoculars to double check that it wasn't my
imagination).

Even *yesterday*, I could see it (even though yesterday I would say
it was *truly barely* visible naked eye --- but through the
binoculars
it was like incredibly obvious. (I'm sure this is not at all news for
most
experienced astro-enthusiasts out there --- but it did take me by
surprise!!). I tried locating Uranus through binoculars, but without
success --- nothing seemed like a reasonable suspect.

So, I see that M31's visual magnitude is 3.4 (according to Wikipedia).
That's a lot, and actually I would expect a star of 3.4 to be visible
in
downtown NYC or some *heavily light-polluted* area (am I over-
estimating??). So, this suggests that the 3.4 is sort of the
integral
over the area, to obtain the "total amount of light" that the area
covered
by Andromeda emits (well, that we receive, I should say).

The funtion being integrated is somehow the "density" or the "true"
value of the luminous intensity, as in amount of light per unit of
surface;
with stars, the thing works more or less like the function being a
Dirac's delta, where its weight is precisely the visual magnitude of
the given star.

So, *finally* the question: is there a measurement of the brightness
of a celestial object that could be more directly related to the level
of
light pollution at which the object becomes visible? Intuitively, it
seems
like this would be closely related (if not being the exact same
measure)
to the "density" that I was describing above --- more specifically,
the
maximum value of this density of the given object (for objects that
occupy
an area, like M31)

IOW, I would like to be able to determine whether I *should* be able
to see some target (in this case Uranus, of course) simply by the
fact that some other target is visible --- I mean, it seems to me
(pure
intuition and rough estimate) that if Andromeda is visible under some
particular light-pollution conditions, then a 5.6 object *should* be
visible as well (possibly even obvious). But then, I don't think I
can
conclude that from the number 3.4 that I see for Andromeda; I do
not think the 3.4 reflects how faint Andromeda really is.

Hope I'm making some sense and that you will be able to understand
what I'm asking, and shed some light (I know, I know, this must be
the
second lamest pun, after the one we all know for Uranus :-\ :-) )

Thanks,

Carlos
--


Rule: Any object is in the *position* it occupies at a given
date & time - period. Whether you see it or not in its position
is another matter entirely. Many people find objects by star
hopping and then confirm the object.

You are makign this waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay
too complicated.



  #6  
Old September 11th 11, 07:21 AM posted to sci.astro.amateur
Paul Schlyter[_3_]
external usenet poster
 
Posts: 1,344
Default Two related questions (visual magnitude + Uranus)

On Sat, 10 Sep 2011 15:01:46 -0600, Chris L Peterson
wrote:
Correct. The magnitude of extended objects is determined by the

total
light emitted, so large objects often have low magnitudes, when in
fact, they are very hard to see.


Correct. Two examples: the integrated magnitude of the Andromeda
Galaxy (M31) is around +3 and the magnitude of the Milky Way is
around -7 which is brighter than Venus. A stellar object of mag -7
would be easily visible even under the most light polluted skies and
even in daytime, while the Milky Way, which extends across the sky,
can quickly be washed out even by moderate light pollution.
  #7  
Old September 15th 11, 12:47 AM posted to sci.astro.amateur
Brian Tung[_5_]
external usenet poster
 
Posts: 205
Default Two related questions (visual magnitude + Uranus)

Carlos wrote:
So, I learned that under reasonably dark sky, it is a naked eye
object, and very obvious with binoculars *(should be the case
with mine, a 10x50 pair).

Ok, but what about under full Moon? *(well, or near-full, like in the
past couple of days?)


It should be quite possible to *see* Uranus with a full Moon
(unless the Moon is right next to Uranus--dang, the 13-year-old
in me is giggling like a madman), but it's non-trivial to identify
which dot you're looking at is Uranus. Uranus's apparent size
is 3-4 arcseconds, and when magnified 10x is indistinguishable
from a star. (At least, I certainly can't tell the difference.)

One really wants a chart with Uranus's position plotted day by
day, or the computer equivalent thereof. Something that goes
down to seventh magnitude or so is nice, as those stars are
generally easily visible through binoculars. Tilt the map so that
it aligns with what you see in the sky; it may help to draw out
circles on the map that match the binoculars' field of view
(usually about 5 or 6 degrees at that magnification), to give you
an idea of what is likely to appear in the binoculars along with
Uranus.

So, *finally* the question: *is there a measurement of the brightness
of a celestial object that could be more directly related to the level
of light pollution at which the object becomes visible?


As others have indicated, surface brightness, measured in
magnitudes per solid angle (square arcsecond or square
arcminute), is the beast you want. It should be pointed out that
it's not exactly what you're describing, because the visibility of
an object depends not only on the light pollution, but also
(among other things) on the instrument you use to observe
it--both aperture and magnification. But for general purposes,
surface brightness is useful.

You may find the discussion at this page

http://mysite.verizon.net/vze55p46/id18.html

somewhat useful. It also discusses the notion of peak
brightness, which as its name implies tells you how bright the
brightest part of the object is.

--
Brian Tung (posting from Google Groups)
The Astronomy Corner at http://www.astronomycorner.net/
Unofficial C5+ Page at http://www.astronomycorner.net/c5plus/
My PleiadAtlas Page at http://www.astronomycorner.net/pleiadatlas/
My Own Personal FAQ at http://www.astronomycorner.net/reference/faq.html
 




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