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star magnitude and binoculars

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Ravensong

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Jul 10, 2003, 8:59:34 PM7/10/03
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i'm new to this group, and have very little experience with astronomy. was
hoping someone might be able to help with a recommendation...

if i want to view stars with a 12-12.5 magnitude, what is the size binoculars i
need (given that i'm not too greatly concerned about number of stars in my
field of vision)?

unfortunately, the only binoculars i ever use are for bird-watching, and we're
talking two different animals here. *ahem*

thanks,
Ravensong

Bernhard Rems

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Jul 11, 2003, 3:24:26 AM7/11/03
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I'd say you will have to be in the 4" range for that - and that makes pretty
heavy binos then.

Bernhard

"Ravensong" <ravens...@aol.com> schrieb im Newsbeitrag
news:20030710205934...@mb-m04.aol.com...

edz

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Jul 11, 2003, 10:13:31 AM7/11/03
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> > if i want to view stars with a 12-12.5 magnitude, what is the size
> binoculars i
> > need (given that i'm not too greatly concerned about number of stars in my
> > field of vision)?

Not only would you need large aperture, 100mm might be about right,
but also you would need pretty high magnification.

You could get a pretty good idea of what would be needed by checking
the limiting magnitude of telescopes for various size scopes and
modifying it according to the low magnification of binoculars. Best
bet would be to reduce what the lens is capable of by about 1
magnitude. That should give you a pretty good idea of what could be
seen with typical binocular magnification.

Limiting magnitude for a given lens size can only be reached at
optimum magnification. As magnification decreases, reachable
magnitude decreases. For a 6" scope, stars seen at 150x are about 1
magnitude fainter than what can be seen at 30x.

The faintest stars I've ever seen with my 5" scope at high power is
13.1mag. A 5" scope could not see 13th mag stars at 30x to 50x. I
cannot see 13th mag stars with my 90mm scope at any power.

If I had to guess, I's say 100mm binoculars at 60x to 80x or 125mm
binoculars at 40x to 60x.

bwhiting

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Jul 11, 2003, 12:42:29 PM7/11/03
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13.1 mag in a 5 inch glass is remarkable...you must have excellent
vision (and probably under age 35). I read where a 6 inch glass
is typically good down to about 12.9 mag (Olcott's Field Book of
the Skies, 1954)....so the guy with binoculars probably needs more
like the range of 120-150 mm of glass to see 12th magnitude stars.
Note...stars, not galaxies or other extended objects, where the
critical item is surface brightness, as the light is spread
out.
Tom W.
Message has been deleted

/\/\R /\/\

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Jul 12, 2003, 1:38:53 AM7/12/03
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PrisNo6 wrote:
> ravens...@aol.com (Ravensong) wrote in message
> news:<20030710205934...@mb-m04.aol.com>...

>> i'm new to this group, and have very little experience with
>> astronomy. was
>> hoping someone might be able to help with a recommendation... [snip]

>> unfortunately, the only binoculars i ever use are for bird-watching,
>> [snip]
>
> Ravensong -
>
> The following table is the short answer to your question, "What is the
> limiting magnitude of my binocular (or small telescope)." I have also
> added other advice for the beginner binocular backyard-amateur
> astronomers who start during the summer observing season:
>
> Enjoy - Kurt
>
> P.S. - Corrections from experienced newsgroup members are welcome.

Aren't Binos classed as Two Telescopes so that the light collected by the
two apertures has to be taken into consideration when working out the
limiting magnitude?

Just asking. Mr M.


Paul Schlyter

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Jul 12, 2003, 4:08:57 AM7/12/03
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In article <beo6ti$71oec$1...@ID-129199.news.uni-berlin.de>,

/\\/\\R /\\/\\ <mr...@ntlworld.com> wrote:

> Aren't Binos classed as Two Telescopes so that the light collected by the
> two apertures has to be taken into consideration when working out the
> limiting magnitude?

It's a bit more complex than that.

The limiting magnitude will be fainter that if you used just one of
those scopes on one eye. But it won't be quite as faint as if you
used a larger scope, with the same entrance pupil area as those
two scopes combined, on one eye. It will be somewhere in between.

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://www.stjarnhimlen.se/
http://home.tiscali.se/pausch/

/\/\R /\/\

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Jul 12, 2003, 4:46:09 AM7/12/03
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Paul Schlyter wrote:
> In article <beo6ti$71oec$1...@ID-129199.news.uni-berlin.de>,
> /\\/\\R /\\/\\ <mr...@ntlworld.com> wrote:
>
>> Aren't Binos classed as Two Telescopes so that the light collected
>> by the two apertures has to be taken into consideration when working
>> out the limiting magnitude?
>
> It's a bit more complex than that.
>
> The limiting magnitude will be fainter that if you used just one of
> those scopes on one eye. But it won't be quite as faint as if you
> used a larger scope, with the same entrance pupil area as those
> two scopes combined, on one eye. It will be somewhere in between.

Are you saying that my binos with two 60mm objectives will give me a
limiting magnitude similar to a single telescope with an objective of
approx. 85mm (given the same eyepieces)?
I'm sorry I'm being thick about this.
Mr M.


Roger Hamlett

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Jul 12, 2003, 5:07:26 AM7/12/03
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"/\/\R /\/\" <mr...@ntlworld.com> wrote in message
news:beohsj$7balf$1...@ID-129199.news.uni-berlin.de...
No, it'll be slightly less.
Take the limiting magnitude of a 60mm 'scope' = 'X'.
Work out the equivalent scope size for the two tubes = just under 85mm, and
this then has a limiting magnitude 'Y'. The limiting magnitude for the pair
of binoculars, will be _between_ these two figures. Not quite as good as the
85mm scope, but better than a single 60mm scope (so X < M < Y).
The exact 'figure', will depend on the observer, but if you say that the
60mm scope has a limiting magnitude of perhaps 11.9, and the 85mm scope a
limiting magnitude of perhaps 13.2, the binoculars will give a figure around
perhaps 12.5. However the observer, and the conditions, make so much
difference, that accurate comparison is very difficult...

Best Wishes


bwhiting

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Jul 12, 2003, 5:36:01 AM7/12/03
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I personally think you can add anywhere from 0.5 to 1 whole
magnitude to that table below....IF you are in very dark skies...
we have typically split the gravitationally lensed quasar up
in Ursa Major at CSSP with the 30 inch...mags 17.1 and 17.4,
having to go averted for the dimmer one. (and obviously those
are point sources).
IMHO,
Tom W.


>
> Telescope Limiting Magnitude
> Theoretical limits - extended down to 35mm
>
> Aperture Aperture Faintest point
> Inches mm source observable -
> theorectical
>
> 1.4 35mm 10.2
> 2 51 10.3
> 3 76 11.2
> 4 102 11.8
> 6 152 12.7
> 8 203 13.3
> 10 254 13.8
> 12.5 318 14.3
> 14 356 14.5
> 16 406 14.8
> 18 457 15.1
> 20 508 15.3
> 24 610 15.7
> 30 762 16.2
>

Roger Hamlett

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Jul 12, 2003, 7:02:15 AM7/12/03
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"bwhiting" <bwhi...@stargate.net> wrote in message
news:3F0FD681...@stargate.net...
Yes. Your comments agree, with a lot of other people's opinions on this.
There is a standard 'formula', which runs:
Mag=6.5 - (5 log d) + (5 log D)
where 'd' is the eyes exit pupil, and 'D' is the scopes apperture. However
it is allmost invariably accompanied by a 'coda', reading something like
'observed magnitude limits often exceed these figures'. The figures above,
are a little low, even from this formula (corresponding to an exit pupil of
about 8mm, rather than the normal 7mm max). Really dark/stable skies, and a
good observer, can often add at least half a magnitude to the theoretical
figures.
Like most such things, these figures are 'guides'. You know full well, that
there is no point trying to observe a mag 15 object, with a 150mm scope,
but, it may well be worth trying for a mag 13 object with the same scope.
All these formulae, start with assumptions (the primary one on the above, is
the visual limiting magnitude without any 'aid', at around mag 6.5). On a
lot of nights, seeing this 'deep', may well be impossible, but just
occasionally, when everything conspires to help, it can be possible to far
exceed this expectation, and this 'moves the goalposts'.
I remember on one amazing night, when the sky was both stable, and clear,
managing to count thirteen seperate stars in the Pleides, naked eye. The
very next night, I was straining to see six, which is much more 'normal'.

Best Wishes


Paul Schlyter

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Jul 12, 2003, 1:11:11 PM7/12/03
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In article <beohsj$7balf$1...@ID-129199.news.uni-berlin.de>,
No, I'm saying that the limiting magnitude of 60mm binoculars will be
somewhere between the limiting magnitudes for 60mm monoculars and 85mm
monoculars (where "monoculars" = normal single-tube scopes).

Brian Tung

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Jul 12, 2003, 7:09:40 PM7/12/03
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Kurt wrote:
> Telescope Limiting Magnitude
> Theoretical limits - extended down to 35mm

It should be emphasized that only the *differences* between the limits
are in any sense theoretical. There is also an empirical element to
the formula.

Specifically, these "theoretical" values are derived from some formula
that is similar to

M = 1.8 + 5 log A

where M is the limiting magnitude, A is the aperture in mm, and log is
to the base 10. The fact that this logarithm is multiplied by 5 does
have a basis in theory: To increase the log of the aperture is to
multiply it by 10, which multiplies the area by 100, and 100 times as
bright is 5 magnitudes.

However, the 1.8 is entirely empirical. It undoubtedlly varies quite
considerably from observer to observer, from site to site, and from
night to night. One should not refrain from attempting a target simply
because it falls outside these limits.

Personally, I think 1.8 is terribly conservative. It corresponds to a
limiting magnitude of about 6.0 for the unaided eye with a pupil of
7 mm, and only 5.3 with a pupil of 5 mm. This is routinely exceeded in
many sites by many observers.

> Practical _point source_ (individual stars) magnitude limits are
> perhaps 1-2.5 mags less than those listed above. You'll probably get
> many different opinions based for every amatuer astronomer you talk
> to.

I disagree with this also. Individual stars are as good as point
sources for this purpose, even taking into account diffraction effects.
Even with the same integrated magnitude, increasing magnification and
therefore increasing apparent size at the cost of decreasing apparent
surface brightness often *enhances* visibility rather than detracting
from it. See, for example, Mel Bartels's excellent site on this matter.

> Each binocular or small telescope, when paired with its owner's
> individual eyes, is unique. Practice with a star chart and your
> personal binoculars will quickly give you a feel at what magnitude of
> stars printed on star charts that you can ignore because they are too
> dim.

I agree with this advice wholeheartedly--the empirical approach is
best for any individual person. For a large group of people (say, SAA
readers), a general rule is often more useful, but it should, I think,
be taken with a grain of salt.

> [snipped lots of good observing advice and targets]

Brian Tung <br...@isi.edu>
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

PrisNo6

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Jul 14, 2003, 12:19:28 AM7/14/03
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br...@isi.edu (Brian Tung) wrote in message news:<beq4fk$ve2$1...@zot.isi.edu>...

> Kurt wrote:
> > Telescope Limiting Magnitude
> > Theoretical limits - extended down to 35mm
> > [snip table]

> > Practical _point source_ (individual stars) magnitude limits are
> > perhaps 1-2.5 mags less than those listed above. You'll probably get
> > many different opinions based for every amatuer astronomer you talk
> > to.

bwhiting noted that:

> I personally think you can add anywhere from 0.5 to 1 whole
> magnitude to that table below....

and Brian responded that:
> I disagree with this [Kurt's table] also. [big snip and rearrange.]

> Personally, I think 1.8 is terribly conservative. It corresponds to a
> limiting magnitude of about 6.0 for the unaided eye with a pupil

> of 7 mm, . . . This is routinely exceeded in many sites by many observers.

As always, Brian, thank you for catching any fuzzy-thinking in my
posts.

Harrington's table is based on a purely mathematical model (like M =
1.8 + 5 log A), and as J.B. Sidgwick noted in his _Amateur
Astronomer's Handbook_, that this curve "does not agree well with the
results of observation, yielding results which are consistently high
by about 1.5 mag throughout the aperature range of 2 to 20 ins."
citing Sidgwick's own experience and work done in the early 1900s by
the British astronomer W.H. Steavenson.

Those results are resummarized here, with H_t as Harrington's purely
mathematical computation (whose math is almost identical to Sidgwick's
mathematical model computation), Si_p as Sidgwick's experience with
refractors, and St_p as Steavenson's experience with refractors, where
all sources have reported common data rows for the same aperature
size:

Telescope Limiting Magnitude by Aperature Size
(in inches and millimeters)
Theoretical (H_t) and
Practical (Si_p and St_p)

D D
" mm H_t Si_p St_p
2 51 10.3 12.1 12.2
3 76 11.2 13.0 13.1
4 102 11.8 13.6 13.7
6 152 12.7 14.5 14.6
8 203 13.3 15.1 15.2
10 254 13.8 15.6 15.7
20 508 15.3 17.1 17.2

Sources:

H_t column:
Phillips S. Harrington, Star Ware (2d ed. 1994) (John Wiley & Sons,
Inc.) republished at
http://www.stargazing.net/david/constel/howmanystars.html

Si_p and St_p columns:
J.B. Sidgwick. Amateur Astronomer's Handbook. (3rd ed. 1980 Dover) at
pp. 27-28. See graphic of above data at p. 28.

The above table is consistent with bwhiting's comment of "add anywhere
from 0.5 to 1 whole magnitude to that table . . ." and Roger's comment
of in "[r]eally dark/stable skies, and a good observer, can often add


at least half a magnitude to the theoretical figures."

I was aware of this difference when writing my prior post, but
Ravensong appeared to be an absolute beginner who typically will not
seek out dark skies. The observing experience of such beginners,
including myself, is that less stars are perceived than the
theoretical equation, either as result of light pollution or of
inexperience.

I should have made that distinction clear in my prior post.

(Although we do not know what size binoculars Ravensong was using,
assuming 35mm (1.4") field binoculars, the theorectical limiting
magnitude is 10.2 and practical adds Sidgwick's "practical" 1.5 mag,
for a total of 11.7 mag. That's near to, but less than her target of
12.0-12.5 mag.)

> For a large group of people (say, SAA readers), a general rule is often more useful, . . .

The imprecision of my writing drew responses from you, bwhiting and
Roger on the important distinctions, which all intermediate amateur
astronomers should be aware. Had I written more clearly, that could
have been avoided.

> Even with the same integrated magnitude, increasing magnification and
> therefore increasing apparent size at the cost of decreasing apparent
> surface brightness often *enhances* visibility rather than detracting
> from it.

Or as Sidgwick says the same thing (I think -:) at p. 32) with respect
to the apparent brightness of point source image: "Magnification may
neverthless make a stellar image appear to be brighter by darkening
the background, when the increased contrast between the star and the
sky is wrongly interpreted [by the human brain] as a brightening of
the stellar image."

Although I don't get how anything that enhances the contrast of a
faint star is "wrongly interpreted". -:)

Thanks again - Kurt

P.S. to Mr M. "/\\/\\R /\\/\\" -

Brian Tung and Roger Hamett provided two equations:

M = 1.8 + 5 log D_mm
Mag=6.5 - (5 log d_ep) + (5 log D_obj)

where mag 6.5 is the faintest star seen with the unaided human eye.

Take Roger's recital of the basic limiting magnitude equation,

Mag_limiting = 6.5 - 5*log(D_ep) + 5*log(D_obj)

Assuming, the exit-pupil size produced by a combination of the
telescope's objective and eyepiece is optimized, so that the
exit_pupil just fits the relaxed pupil of the human eye, let's say 7
mm or 0.3 inches, you get, evaluating 5*log(D_ep)(in inches):

Mag_limiting = 9.1 + 5*log(D_inches)

Brian's version is similar, but is expressed in millimeters and
assumes an observer with good eyesight who can see stars unaided down
to mag 6.0, instead of mag 6.5 (e.g. - "It corresponds to a limiting
magnitude of about 6.0 for the unaided eye with a pupil of 7 mm"):

Mag_limiting = 1.8 + 5*log(D_mm)

bwhiting

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Jul 14, 2003, 1:33:55 AM7/14/03
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I'd be very curious what your numbers for H_t, Si_p, and St_p
are for a 24 (610 mm)and a 30 inch (762 mm) glass, if you have them,
or can extrapolate them from the data.
Thanks ahead of time,
Tom W.

bwhiting

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Jul 14, 2003, 1:38:43 AM7/14/03
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Oh wait...that's easy...since a 30 is about
2.5 times a 20 (almost) and a 24 is about
midway, just add 0.5 mag to the 20 inch numbers for a 24",
and 1 full mag. (almost) for the 30 inch numbers,
or about 18.0 mag....but I think that would be for young
eyes only.
QED!
TW.

PrisNo6

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Jul 14, 2003, 2:03:56 AM7/14/03
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"/\\/\\R /\\/\\" <mr...@ntlworld.com> wrote in message news:<beohsj$7balf$1...@ID-129199.news.uni-berlin.de>...

> > /\\/\\R /\\/\\ <mr...@ntlworld.com> wrote:
> >> Aren't Binos classed as Two Telescopes so that the light collected
> >> by the two apertures has to be taken into consideration when working
> >> out the limiting magnitude?

I don't know the answer to your question. I have never seen a
separate set of limiting magnitude tables published or referenced by
published binocular amateur or professional astronomers, or a
discussion of how the limiting magnitude of binoculars are measured by
different equations than monoculars. All amateur binocular guides that
I have read refer the monocular-telescope limiting magnitude tables
and the limiting magnitude equation:

Mag_limiting = 6.5 - (5 log d_ep) + (5 log D_obj)

That's why I used that table as a rough guide. It seems to compare
with my own personal experience.

I would appreciate any references to books or articles discussing the
difference in binocular limiting magntidues that lurkers could
provide.

I suspect that the limiting magnitude for true point sources would not
be different for a two-eyeball vs. one-eyeball scope. True point
sources occur when the image that falls on the retina only causes one
or two rods to fire. See the diagram of the fovea at:

http://hyperphysics.phy-astr.gsu.edu/hbase/vision/rodcone.html#c2

With true point sources, having one or two extra rods discharging in
the other eyeball probably is not going to make much of a difference
to the brain's perception response and image formation.

Extended images are spread out over a larger portion of the fovea,
discharge many more rods outside the fovea centralis, and follow the
general rule for limiting magnitude stated by Roger:

Mag_limiting = 6.5 - (5 log d_ep) + (5 log D_obj)

But the faintest star a telescope can see (its limiting magnitude) is
really another way of expressing its light grasp:

Lightgrasp = ( ( D_obj / D_exitpupil) ^ 2 ) x transmission factor

Ignore the transmission factor (loss of light during passage through
the telescope) for a minute.

Roger's computations and your assumption concerning the doubling in
the size of the scope's aperature ignores that in a binocular there
also is a second exit pupil.

In otherwords, there is no difference between the light grasp results
of:

( D_obj / D_exitpupil ) ^ 2 one-eyeball scope

( ( 2*D_obj ) / (2*D_exitpupil) ) ^ 2 two-eyeball binos

The same reasoning would apply to the results of limiting magnitude
equation, m = 6.5 - 5*log(d_ep) + 5*log(D_obj), although the math is
different and more complicated.

Nonetheless, IMHE (in-my-humble-experience), extended images do appear
brighter in binoculars than in an equivalent-aperature-sized
telescope. (Equivalent in the sense that 2*Area_left_bino =
Area_telescope.) That's why one buys them.

But the brighter image implies an overall performance with a lower
limiting magnitude, that mathematically, doesn't appear to be there.

Kitchin's _Telescope and Techniques (Springer-Verlag 1995) gives some
hint at an answer to the contradiction. As Kitchin (and other basic
amateur astronomy books) note, above a telescope's minimum useful
magnification, a telescope presents a dimmer image of an extended
object than that seen by the naked eye. Kitchin continues:

"This result is, to most people, very surprising, and perhaps
contradicts their experience of using a telescope. . . . [F]aint
extended objects, like comets or the Andromeda galaxy, M31, do appear
brighter through a telescope, and sometimes binoculars are advertised
for night vision, whereas in fact their images cannot be brighter
than those seen with the unaided eye. The answer to this paradox lies
in the structure of the eye . . . . [O]n looking at the same image
through a telescope, it will be magnified and some portions of it will
have to fall on to parts of the retina away from the fovea centralis,
and thus be detected by the higher-sensitivity rods. Thus faint
extended sources do appear brighter through a telescope, not becuase
they are actually brighter, but because more sensitive parts of the
eye are being used to detect them. This property of the eye can also
be used deliberately . . . through the use of averted vision."
Kitchin at pp. 30-31.

Most binoculars are built with eyepieces above the minimum useful
magnification.

With the two-eyeball binocular, the image of an extended object is
spread out over two fovea (one in each eye), and the number of
discharging rods increases by 2 times, as compared to the one-fovea,
one eye-piece telescope. That would make a difference to the brain's
perception response and image formation and result in the perceived
brighter image. But perhaps as a matter of physics and mathematics,
the binocular does not escape the constraints of the limiting
magnitude equation.

But again, I really don't know the answer to your question. Maybe
there is different equation for estimating the limiting magnitude of a
binocular and I just don't know about it.

- Regards Kurt

Message has been deleted
Message has been deleted

Harald Lang

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Jul 16, 2003, 7:12:58 AM7/16/03
to

Hi Kurt,

<<
P.S. - if anyone has a copy of:

Schaefer, B. "Telescopic limiting magnitudes", Publ. Astron. Soc.,
102:212 (1990)

I'm still trying to get ahold of one.
>>

It can be generated and downloaded from the proper website -- I
have done so, but I have forgotten how. It seems to me it can't
be a terrible infringement of copyright law or whatver if you
download my copy:

<http://www.math.kth.se/~lang/schaefer.pdf>

As for the issue under discussion, my rule-of-thumb formula is

m = 3*Log(A) + 2*Log(X) + 0.6 + v

where
m = magnitude of faintest star (point source) visible in binocular
A = aperture in cm
X = magnification
v = magnitude of faintest star visible to the naked eye

This formula is somewhat more elaborated than the usual

m = constant + 5*Log(A)

since it takes into account that also magnification, not only
aperture, matters -- higher magnification darkens the sky
background. It is derived from Clark's "Visual Astronomy..." and
the data by Blackwell that he refers to there. I have seen that
Nils Olof Carlin has suggested basically the same formula for the
same reason. I wish to think that "great minds think alike" (Nils
is extremely smart and clever.)

Cheers -- Harald

edz

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Jul 16, 2003, 2:44:54 PM7/16/03
to
>
> The following table is the short answer to your question, "What is the
> limiting magnitude of my binocular (or small telescope)." I have also
> added other advice for the beginner binocular backyard-amateur
> astronomers who start during the summer observing season:
>
> Telescope Limiting Magnitude
> Theoretical limits - extended down to 35mm
>
> Aperture Aperture Faintest point
> Inches mm source observable -
> theorectical
>
> 1.4 35mm 10.2
> 2 51 10.3
> 3 76 11.2
> 4 102 11.8

Although this thread has resulted in excellent discussion and info
regarding limiting magnitudes, it has traveled a long way from
answering the original question. What is limiting magnitude in
binoculars. That answer cannot be found directly from the various
tables and formulae presented for limiting magnitude.

Limiting magnitude for a given lens size can only be reached at
optimum magnification. As magnification decreases, reachable

magnitude decreases. For a 6" scope, stars seen at a magnification of

150x are about 1 magnitude fainter than what can be seen at 30x.

All the information that has been posted for limiting magnitude is
valid only when optimum magnification is utilized in a diffraction
limited lens, neither of which exist in binoculars except maybe the
best which might have diffraction limited optics. Binoculars, which
operate at magnifications substantially below optimum for limiting
magnitude and maximum resolution, can never reach the limiting
magnitude or maximum resolution based on calculations dependant on
aperture. Binoculars are optimized for light gathering. For all
other limitations, aperture in binoculars is underutilized.

So, limiting magnitudes (and maximum resolution) for binoculars are
substantially less than what can be reached with a telescope lens of
the same diameter, primarily due to the lack of sufficient
magnification.

edz

edz

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Jul 16, 2003, 3:13:40 PM7/16/03
to
Harald Lang <la...@deleteme.math.kth.se> wrote in message news:<bf3bvq$5dq$1...@news.kth.se>...

>
> As for the issue under discussion, my rule-of-thumb formula is
>
> m = 3*Log(A) + 2*Log(X) + 0.6 + v
>
> where
> m = magnitude of faintest star (point source) visible in binocular
> A = aperture in cm
> X = magnification
> v = magnitude of faintest star visible to the naked eye
>
> This formula is somewhat more elaborated than the usual
>
> m = constant + 5*Log(A)
>
> since it takes into account that also magnification, not only
> aperture, matters -- higher magnification darkens the sky
> background.


This last post presents a formula that, although I have not used, at
least agrees that Lim Mag is not only dependant on aperture but also
dependant on magnification. This is very much true in binoculars.

Take the same aperture binocular and increase the magnification, you
will see more. I have done these types of tests. I have recorded a
great deal of testing data that shows the significance of
magnification to produce increases in limiting magnitude and
resolution in binoculars. I published the results of those tests in
an article "Binocular Performance" on CN. You might find it
interesting reading.

edz

PrisNo6

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Jul 17, 2003, 1:49:10 AM7/17/03
to
ezar...@gilbaneco.com (edz) wrote in message news:<90c65221.03071...@posting.google.com>...

> Harald Lang <la...@deleteme.math.kth.se> wrote in message news:<bf3bvq$5dq$1...@news.kth.se>...
> > [snip]

> I have recorded a great deal of testing data that shows the
> significance of magnification to produce increases in
> limiting magnitude and resolution in binoculars. I published
> the results of those tests in an article "Binocular Performance" on CN.
> You might find it interesting reading.

Ed - Your article, "How-to Understand Binocular Performance", at:

http://www.cloudynights.com/howtos2/binocular-performance.htm

was very helpful, including your reference to "Adler Index" for
ordering the performance of differing binoculars:

Adler Index = Sqrt(A) * magnification

PrisNo6

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Jul 17, 2003, 3:46:39 AM7/17/03
to
Harald Lang <la...@deleteme.math.kth.se> wrote in message news:<bf3bvq$5dq$1...@news.kth.se>...
[snip]

Prof. Lang, thank you for taking your time to post and add your and
Carlin's formulae. To summarize and to post a rule-of-thumb guideline
table for the limiting magnitude of bincoluars:

Carlin-Langs equation for binocular limiting magnitude:

m = 3*Log(A) + 2*Log(X) + 0.6 + v

where
m = magnitude of faintest star (point source) visible in binocular
A = aperture in cm
X = magnification
v = magnitude of faintest star visible to the naked eye

Adler binocular performance index:

Adler# = Sqrt(A) * A

Data to equation:
v_lim - Bortle
Bino Adler# A M 6 5 4 3
7x35 41 35 7 5.1 5.6 6.1 6.6
10x50 71 50 10 5.1 5.6 6.1 6.6
12x50 85 50 12 5.1 5.6 6.1 6.6
15x70 125 70 15 5.1 5.6 6.1 6.6
16x70 134 70 16 5.1 5.6 6.1 6.6
16x80 143 80 16 5.1 5.6 6.1 6.6
20x80 179 80 20 5.1 5.6 6.1 6.6

Results from Carlin-Lang equation:

Proposed "rule-of-thumb"
guideline table for
limiting magnitude for binoculars
by Bortle scale sky conditions
and binocular types

Limiting magnitude where faintest
naked-eye visible star is
5.1 5.6 6.1 6.6
Bino Adler#
7x35 41 12.0 12.5 13.0 13.5
10x50 71 12.8 13.3 13.8 14.3
12x50 85 13.0 13.5 14.0 14.5
16x70 134 13.6 14.1 14.6 15.1
16x80 143 13.8 14.3 14.8 15.3
20x80 179 14.0 14.5 15.0 15.5

Bortle class 6 5 4 3

A few quick wrap-up questions, if you have the time:

1) Has your and Carlin's equation ever been tested with a group of
binocular users to assess it's empirical accuracy?

2) Has it ever been published (for proper attribution)?

3) Since you are a professional scientist, what should be your
equation's attribution? I used Carlin-Lang based on your email.

4) By "v = magnitude of faintest star visible to the naked eye," did
you mean zenth limiting magnitude, or a more informal measurement of
the light pollution?

5) Does the equation makes predictions about limiting magnitude of
point sources (individual stars) and not extended objects (globular
clusters, open clusters, nebulae)?

Thanks again for contributing.

- Kurt

Harald Lang

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Jul 17, 2003, 6:22:21 AM7/17/03
to

Hi Kurt,

you wrote


<<
Proposed "rule-of-thumb"
guideline table for
limiting magnitude for binoculars
by Bortle scale sky conditions
and binocular types

Limiting magnitude where faintest
naked-eye visible star is
5.1 5.6 6.1 6.6
Bino Adler#
7x35 41 12.0 12.5 13.0 13.5
10x50 71 12.8 13.3 13.8 14.3
12x50 85 13.0 13.5 14.0 14.5
16x70 134 13.6 14.1 14.6 15.1
16x80 143 13.8 14.3 14.8 15.3
20x80 179 14.0 14.5 15.0 15.5
<<

These values are all 3 magnitudes too high. Note, A in the
formula (m = 3*Log(A) + 2*Log(X) + 0.6 + v) is aperture in *cm*
not mm.

If we write the Adler formula by the same token, it would be

m = 2.5*Log(A) + 2.5*Log(X) + 0.5 + v

so the difference is rather small -- Adler puts somewhat higher
weight on magnification relative to aperture compared to "my"
formula. (As for the number 0.6 and 0.5: these assume an eye
pupil of 6.3mm -- see below).

<<
A few quick wrap-up questions, if you have the time:

1) Has your and Carlin's equation ever been tested with a group of
binocular users to assess it's empirical accuracy?
>>

Not that I know of, but see comment later.

<<
2) Has it ever been published (for proper attribution)?

3) Since you are a professional scientist, what should be your
equation's attribution? I used Carlin-Lang based on your email.
>>

You take this "professional scientist" much too seriously. I'm a
scientist, and so are many people here, but in this context I'm a
moderately (less than most) experienced amateur astronomer. As
for published -- Nils Olof has it on his web page at:
<http://w1.411.telia.com/~u41105032/visual/limiting.htm>
I suggest you refer to that page.

<<
4) By "v = magnitude of faintest star visible to the naked eye," did
you mean zenth limiting magnitude, or a more informal measurement of
the light pollution?
>>

Ideally it should be the faintest star visible to the naked eye
in the same area as for the binocular observation.

<<
5) Does the equation makes predictions about limiting magnitude of
point sources (individual stars) and not extended objects (globular
clusters, open clusters, nebulae)?
>>

Point sources!! Roger N Clark considers in his book "Visual
Astronomy..." the visibility of extended objects (however, there
are a few mistakes there, which have been discussed extensively
between him, me, Mel Bartels and Nils Olof Carlin a few years ago.
I had hoped a new edition would come out -- the old one is out
print since a long time -- it's a wonderful book, but it hasn't
happened yet AFAIK.)

There is a big problem in practise in trying to use any formula
predicting the visibility of extended objects. Clark uses angular
size and magnitude as input relevant for the object. But objects
differ in other respects: some (galaxies) have a very bright
core, which is relatively small, so the brightness falls off
gradually. It's not even clear where boundary is (at what
isophote?) hence the "size" is a fuzzy number. Other objects,
have a more even brightness.

Clark's work relies on a very thorough empirical study made by
H.R Blackwell during WWII (1946 "Contrast Thresholds of the
Human Eye") which had nothing to do with amateur astronomy; it
had some military use IIRC. From Blackwell's study one can see
that for small objects (close to point sources) one can see 0.4
magnitudes dimmer objects if the sky brightness gets 1 magnitude
dimmer. Nils Olof's and my formula is a simple consequence of
that relation -- that's all. In this sense, it has been
empirically tested, but in a different context -- not by applying
it directly to observation of faint stars in binoculars (AFAIK).

Indeed, we can argue as follows: Assume first that we have a
binocular whose aperture is A and magnification is Xo=A/y, where y
is the (current) diameter if our eye pupil. Then the exit pupil
of the binocular exactly matches our eye pupil, so the sky
background (neglecting internal light losses in the binocular --
it should be small in modern binoculars, I assume) in the
binocular is the same as naked eye. The star, however, appears
5*Log(A/y) magnitudes brighter. That means that we can see
5*Log(A/y) magnitudes dimmer stars than naked eye.

However, assume now that we crank up magnification to X=Xo*k
for some k greater than 1. This will have no impact on the
brightness of the star in the binocular; it will however darken
the sky background by 5*Log(k) magnitudes (per unit of angular
area), since the diameter of the exit pupil will be reduced by a
factor 1/k. According to Blackwell-Clark this means that we can
see yet 0.4*5*Log(k) magnitudes fainter, i.e., in total

5*Log(A/y) + 0.4*5*Log(k) magnitudes.

Now we just clean up that expression:

5*Log(A/y) + 0.4*5*Log(k)
= 5*Log(A/y) + 2*Log(X/Xo)
= 5*Log(A/y) + 2*Log(X) - 2*Log(Xo)
= 5*Log(A/y) + 2*Log(X) - 2*Log(A/y)
= 3*Log(A/y) + 2*Log(X)
= 3*Log(A) + 2*Log(X) - 3*Log y

So the faintest star we can see is of magnitude

3*Log(A) + 2*Log(X) - 3*Log y + v

where v is that faintest star we can see naked eye. Now I have
put y=0.63 cm for some reason I have forgotten; this makes
-3*Log(y) = 0.6, and we must be consistent and measure A also in
cm. That's Nils-Olof's and my formula.

If the exit pupil is larger than the eye pupil, the formula
should be

m = 5*Log(X) + v

As you can see, it wouldn't be right to attribute it to any of us
-- the major work comes from Blackwell's article and Clark's
discovery of its relevance for amateur astronomy. For reference,
give Nils-Olof's web page, he attributes both Clark and
Blackwell.

BTW, did you download Schaefer's article? I put it temporarily on
my web site, and I'll take it away if you are done with it.

Cheers -- Harald

Paul Schlyter

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Jul 17, 2003, 6:24:03 AM7/17/03
to
In article <9865fa0b.0307...@posting.google.com>,

PrisNo6 <fish...@csolutions.net> wrote:

> Harald Lang <la...@deleteme.math.kth.se> wrote in message news:<bf3bvq$5dq$1...@news.kth.se>...
> [snip]
>
> Prof. Lang, thank you for taking your time to post and add your and
> Carlin's formulae. To summarize and to post a rule-of-thumb guideline
> table for the limiting magnitude of bincoluars:
>
> Carlin-Langs equation for binocular limiting magnitude:
>
> m = 3*Log(A) + 2*Log(X) + 0.6 + v
>
> where
> m = magnitude of faintest star (point source) visible in binocular
> A = aperture in cm
> X = magnification
> v = magnitude of faintest star visible to the naked eye

Why not replace this formula with:

m = 3*Log(A) + 2*Log(X) - 3*Log(a) + v


where
m = magnitude of faintest star (point source) visible in binocular
A = aperture in cm
X = magnification
v = magnitude of faintest star visible to the naked eye
a = size of your eye pupil in cm

Note that -3*Log(0.63) = 0.6 thus the Carlin-Lang formula assumes an eye
pupil size of 6.3 mm

Harald Lang

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Jul 17, 2003, 11:30:36 AM7/17/03
to

Paul Schlyter wrote:

<<
Why not replace this formula with:

m = 3*Log(A) + 2*Log(X) - 3*Log(a) + v

where
m = magnitude of faintest star (point source) visible in binocular
A = aperture in cm
X = magnification
v = magnitude of faintest star visible to the naked eye
a = size of your eye pupil in cm

Note that -3*Log(0.63) = 0.6 thus the Carlin-Lang formula assumes an eye
pupil size of 6.3 mm
>>

You are absulutely right. But if we want to compare different
binoculars, it is a good idea to plug in some number for a. I
don't remember exactly why I choose 6.3mm, but I think it was
because I had an old estimate of my own pupils that were close to
that, and Log 0.63 = -0.20 is a nice round number.

Cheers -- Harald

Paul Schlyter

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Jul 17, 2003, 1:49:35 PM7/17/03
to
In article <bf6fes$mk8$1...@news.kth.se>,
OTOH we have to plug in a number for v (the naked-eye magnitude limit)
too....

So is -3*Log(a) + v pretty much a constant? To answer that question
properly would of course require asking a lot of people to measure
both their dark-adapted pupil size and their visual magnitude limit
under dark skies.

edz

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Jul 17, 2003, 3:01:12 PM7/17/03
to
Proposed "rule-of-thumb"
guideline table for
limiting magnitude for binoculars
by Bortle scale sky conditions
and binocular types

Limiting magnitude where faintest
naked-eye visible star is
5.1 5.6 6.1 6.6
Bino Adler#
7x35 41 12.0 12.5 13.0 13.5
10x50 71 12.8 13.3 13.8 14.3
12x50 85 13.0 13.5 14.0 14.5
16x70 134 13.6 14.1 14.6 15.1
16x80 143 13.8 14.3 14.8 15.3
20x80 179 14.0 14.5 15.0 15.5

I would have a real hard time believing that anyone, even the best
observor with the best eyes with the very best binoculars, could ever
even hope to reach results indicated in this chart.

I have mag 5.1 and sometimes mag 5.4 to mag 5.6 skies. Even under the
best conditions in mag 5.4 skies I could not even hope to see 13th
magnitude stars with any binocular, let alone a 50mm binocular. In
practice under mag5.4 skies, I would say 35mm binoculars could reach
maybe to mag 9.5 to mag 10. 70mm and 80mm binoculars might reach mag
12, but that would be a stretch. With mag 5.4 skies, I cannot see mag
13 stars with 16x70 Fujinons of 20x80 Deluxe binoculars.

Since I have't yet followed thru the formula to try and determine for
myself which factors are giving the most weight to the result, I am
not sure why the predicted results are so high. I can only say, from
all the test results I have recorded, the predicted results are
significantly too high.

edz

edz

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Jul 17, 2003, 3:39:24 PM7/17/03
to
Harald Lang <la...@deleteme.math.kth.se> wrote in message news:<bf5tct$jfs$1...@news.kth.se>...

> Bino Adler#
> 7x35 41 12.0 12.5 13.0 13.5
> 10x50 71 12.8 13.3 13.8 14.3
> 12x50 85 13.0 13.5 14.0 14.5
> 16x70 134 13.6 14.1 14.6 15.1
> 16x80 143 13.8 14.3 14.8 15.3
> 20x80 179 14.0 14.5 15.0 15.5
> <<
>
> These values are all 3 magnitudes too high. Note, A in the
> formula (m = 3*Log(A) + 2*Log(X) + 0.6 + v) is aperture in *cm*
> not mm.
>

This agrees with what I'm stating about the results my my testing.
The Binocular Performance article highlights the results of a year of
testing with nearly a dozen pair of binoculars. Although I have not
identified the exact lim mag of every binocular, I have spent enough
time in the field with all the binocs and charts and have taken enough
notes to be able to say these values are too high.


> so the difference is rather small -- Adler puts somewhat higher
> weight on magnification relative to aperture compared to "my"
> formula.

Adler does put more weight on magnification, which my testing has led
me to believe is a step in the right direction. I am inclined to put
even greater weight on magnification. The conclusions I reach in
Binocular Performance testing are aperture may be responsible for a
20% increase in seeing, while increasing magnification at a given
aperture may be responsible for an 80% increase in seeing. These
increases would need to be interpreted in the context of the equipment
I used in my tests and the magnification ranges that equipment
provided.

The most important conclusion I reached is that magnification has a
far greater effect on the ability to see thru a given size binocular
than does the aperture itself, because the higher the magnification in
use, the greater the potential of the aperture is utilized.

> There is a big problem in practise in trying to use any formula
> predicting the visibility of extended objects.

Contrast is what helps the most in seeing extended objects. The next
step beyond the Adler index, which already accounts for the weight
that should be given to magnification, is accounting for and applying
some factor for the quality of the coatings and baffels. Contrast is
the quality that improves as these components improve. That is why a
16x70 Fujinon binocular will outperform a 20x80 Oberwerk binocular.
Not only does this quality provide the needed performance for seeing
exteded objects but also it improves the performance for viewing point
source limiting magnitude.

Although these formulae do show a relative index of performance, just
the same as the Adler Index, they do not account for that quality that
determines the ultimate performance and will always fall short in that
judgement.

How to incorporate such a quality into a formula that will predict
limiting magnitude I am not sure. I have made my attempt at a much
simpler ranking in the hopes of providing a means to judge and rank
relative performance.

edz

Message has been deleted
Message has been deleted
Message has been deleted

Harald Lang

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Jul 17, 2003, 5:05:02 PM7/17/03
to

Paul Schlyter wrote:

>OTOH we have to plug in a number for v (the naked-eye magnitude limit)
>too....
>
>So is -3*Log(a) + v pretty much a constant? To answer that question
>properly would of course require asking a lot of people to measure
>both their dark-adapted pupil size and their visual magnitude limit
>under dark skies.

I was unclear. When I compare binoculars I compare the numbers


3*Log(A) + 2*Log(X) + 0.6

This is about as many magnitudes *deeper* I can see compared to
naked eye. Of course, if we just want to rank binoculars, we
could leave out the 0.6. I prefer to add the 0.6 though, for then
I can interpret the result as above. (Leaving it out would imply
an eye pupil of 10mm, if we want to stick to the same
interpretation, which is unrealistic, of course.)

BTW, Nils Olof writes the formula as you did, with eye pupil as a
parameter. It all depends on what you are using the forula for.
Anyone can adjust it for his own eyes and purposes.

Cheers -- Harald

Harald Lang

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Jul 17, 2003, 5:08:46 PM7/17/03
to

edz wrote:
<<
Limiting magnitude where faintest
naked-eye visible star is
5.1 5.6 6.1 6.6
Bino Adler#
7x35 41 12.0 12.5 13.0 13.5
10x50 71 12.8 13.3 13.8 14.3
12x50 85 13.0 13.5 14.0 14.5
16x70 134 13.6 14.1 14.6 15.1
16x80 143 13.8 14.3 14.8 15.3
20x80 179 14.0 14.5 15.0 15.5
[snip]

Since I have't yet followed thru the formula to try and determine for
myself which factors are giving the most weight to the result, I am
not sure why the predicted results are so high. I can only say, from
all the test results I have recorded, the predicted results are
significantly too high.
>>

Yes, as I pointed out in another post, they are three magnitudes
too high given the formula I gave.

Cheers -- Harald

Harald Lang

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Jul 17, 2003, 7:24:36 PM7/17/03
to

Kurt wrote:
<<
Clark's output from the input of "angular size and magnitude" is his
Object Detection Magnitude ("ODM")?
>>

I don't have his book here (I'm currently in my country house),
but as I recall, he is trying to determine two things:

- Is a given object (defined by size and brightness) at all
detectable in a particular telescope (i.e., aperture)
- What magnification "should" be used so as to see the object
"best"?

This latter question isn't as clear as it might seem at first.
Furthermore, Clark slips on the math here, so many of his tables
are out of whack. (For instance, he erroneously draws the
conclusion that a *smaller* aperture in some cases should use a
*higher* magnification.) However, his approach is innovative, and
his book was an eye opener for me. Too bad he is not about to give
out corrected edition, as it seems.

As an aside, Bill Ferris' article in the current S&T is based on
Mel Bartel's programme. He (Mel) is one of the persons who
spotted Clark's error, and his programme was written to correct
for that (his correction is not the same as the one Nils Olof
uses [in another context than this binocular formula], though. We
found that Mel and I had one idea of what was the best procedure,
Nils Olof and Roger -- when Roger eventually recognised his error --
had another.)

<<
In Clark-Blackwell-based equation and Carlin's discussion at:

http://w1.411.telia.com/~u41105032/visual/limiting.htm

I'm not tracking if there is any adjustment for the fact of binocular
vision through two light gathering devices. For any binocular size of
D_mm, the combined light area of the two tubes is sqrt(2)*D_mm or
1.414*D_mm.
>>

First, the data that is used, i.e., Blackwell's data, are all
based on binocular vision (i.e., with two eyes.) Second, for the
binocular formula, I don't think it matters, as long as we use
equally many eyes when we determine the naked eye limiting
magnitude as when looking through the binocular/monocular.

I wouldn't use the binocular formula for telescopes at high
magnification, though, for at even moderately high magnification
(small exit pupil) we come far outside the range of Blackwell's
data (too dark sky background), so we don't know if the relation
holds to a reasonable approximation any longer that it can be
used even as a rule of thumb.

<<
In the Clark-Blackwell-based equation,

1) is an adjustment for binocular seeing implicit in the model because
Blackwell's detection data was based on the ability of naked-eye
observers* (using their natural binoculars) to see faint objects?

2) is there a collecting area adjustment in the equation for the use
of the two fovea and two objectives?

3) is the fact that the collecting area of a binocular is 1.414*D_mm
the area of one monocular side not relevant to determining limiting
magnitude in the Clark-Blackwell model?
>>

I think the answer I gave above applies to all these questions.

IIRC Clark never adresses the question about monocular vs.
binocular vision. Schaefer (in the paper you downloaded) does,
though, and he claims that looking with two eyes increases the
"sensitivity" by 0.38 magnitudes, which corresponds to a factor
of square root of two [he refers to a study by Pirenne, 1943, in
Nature (152).] But as I said, if you use monocular vision when
you determine "v" in the equation

m = 3*Log(A) + 2*Log(X) + 0.6 + v

then m will obviously decrease by the same amount as v, so the
correction will be made automatically, whatever the magnitude of
the correction might be.

Cheers -- Harald
P.S. The discussion between me, Mel, Nils Olof and Roger Clark
took place spring 1999. Nils wrote his web page 1997, so our
discussion had no impact on that. I shouldn't have brought up
that discussion at all -- it had nothing to do with the issue at
hand.

Message has been deleted

edz

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Jul 17, 2003, 10:48:50 PM7/17/03
to
> Harald, thank you for help in nailing this down.
>
> With Harrington's _Touring the Universe through Binoculars_, Crossen's
> & Tirion's _Binocular Astronomy_, and Moore's _Exploring the Night Sky
> with Binoculars_, the popularity of binocular amateur astronomy has
> experienced some growth, at least as a pleasurable supplement to the
> telescope observing.
>
> The question "what is the difference between the limiting magnitude of
> binoculars and telescopes" seems to come up more often.
>
> - Kurt

I must admit there is an impressive array of figures here and fomula
to support those figures. However, I respectfully disagree that you
have nailed this down.

Unless you were to gather a collection of binoculars of varying
magnifications and aperture, all by the same manufacturer, then maybe
under those circumstances your tables might be accurate. However,
this is not reality.

Binoculars are of a tremendously wide variety of components and
quality of manufacture. These formula go so far as to take into
consideration the affects of magnification, aperture, sky background
and even eye pupil. And yet they fail to take into account one of the
single most inportant qualities of the binocular that helps provide
the ability to see and that is contrast.

Contrast provided by the optical system of binoculars may be equally
as important as aperture. I have already shown by repeated testing
and have publised results showing that a smaller aperture with better
contrast is capable of seeing more than a larger aperture with lesser
contrast. Your tables would never show this. I don't fully
understand why that is ignored, but it is my opinion that until it is
incorporated into the calculation, you will not have a representative
indication of the predicted outcome.

Of course, you could say that the tabulated data would give a good
indication of the performance of a given set of binoculars of
equivalent quality. However, that limits it's usefullness. It is
more likely to be useful data if the end user were able to apply a
factor allow the placement of their own binoculars into the field of
the data by some means that accounts for the endlessly varying levels
of quality and premium (or lack of) features incorporated into the
manufacture.

edz

Message has been deleted

PrisNo6

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Jul 18, 2003, 8:31:55 PM7/18/03
to

Thanks, this is a great plain english summary of governing principles.

Regards - Kurt

Message has been deleted

Anton Jopko

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Aug 4, 2003, 7:25:06 AM8/4/03
to
Hi Kurt,

one can also lose a third of a magnitude between the center of the field of
view and the edge. most people would focus on the centre of field though.
averted vision is not taken into account either.
Cheers

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