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10x42 vs 15x50 - which view contains more stars?

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#1 Saturninus

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Posted 28 January 2015 - 04:00 PM

I remember once reading an article about a way to calculate which eyepiece in a telescope would show the most stars. The article was discussing the “majesty factor” that is used to characterize widefield eyepieces. It wasn’t the alder index, but something else that also accounted for differences in TFOV.

 

I’m thinking about getting a pair of Canon IS binoculars - either the 10x42 or the 15x50. So my question is, which one would produce a view with the highest total star count?

 

The 10x42 has a 6.5 degree field of view, and a 4.2 exit pupil, but only 10x mag and 42mm of aperture.

 

The 15x50 has a 30% less field of view at 4.5 degree and 20% smaller exit pupil at 3.3, but it has 50% more magnification at 15x and 20% more aperture at 50m (actually 40% if you measure by area)

 

Given those tradeoffs, which binocular would produce the view with the highest absolute star count? I know the 15x50 will have the higher alder index, but my impression is that at higher magnification, the distribution of stars would be less dense, so that even though I would be able to see fainter stars, there would be less of then in the narrower field of view.



#2 MartinPond

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Posted 28 January 2015 - 04:14 PM

You're sort of onto something there, but push it a bit more:

for shear quantity and width, a good old 7x35 @ 11-degrees fov trolling the Milky Way is more "majestic".

There are a lot out there used...if you don't want to do more than a simple outer cleaning, look for

ones that look like new, like they spent 40 yrs in the closet. They are all great, but some need an

overhaul inside.

 

There are fanatics here who have wider-feild binoculars than 11-degrees that, but those are extremely rare.

 

You can see majesty 'within' a nebula with those Canons, but total star-count is the domain of the extra or ultrawide.


Edited by MartinPond, 28 January 2015 - 04:16 PM.


#3 Mark9473

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Posted 28 January 2015 - 04:42 PM

Just a back-of-the envelope calculation.

Assume that the following simplistic formula is correct, and that both instruments have the same light throughput:

 

hypothesis: magnitude gain = 2.5x log M + 2.5x log D

where M is the magnification and D the aperture in cm.

 

10 x 42 goes 4.1 magnitudes deeper than the unaided eye;

15 x 50 goes 4.7 magnitudes deeper than the unaided eye.

 

Let's say the faintest magnitude reached is 9.5 in the 10x42 and 10.1 in the 15x50.

 

Now look at this site giving the number of stars at each magnitude step, and let's assume that this is accurate:

http://www.stargazin...wmanystars.html

 

There are 222054 stars up to mag 9.5 over the entire sky.

There are 418158 stars up to mag 10.1 over the entire sky.

(I used Excel for this: nr. of stars = 9.8663 x e^(1.0549 x mag.) calculated between mag 7.49 and 11.49).

 

The total sky is 41253 square degrees.

At mag 9.5 there are 5.38 stars per square degrees; in the 10x42 having 33.2 square degrees of FOV on average you'll see 178 stars.

At mag 10.1 there are 10.14 stars per square degree; in the 15x50 having 15.9 square degrees of FOV on average you'll see 161 stars.

 

In my opinion those numbers are close enough to say there is no difference.


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#4 Mark9473

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Posted 28 January 2015 - 05:00 PM

7x35

3.5 magnitudes gain over the unaided eye.

magnitude 8.9 reached in the above simulation.

117917 stars in the whole sky = 2.86 stars per square degree = 272 stars in an 11° FOV.

In my more mundane 7x35 having 9.3° TFOV that number becomes 194 stars.

 

You'll have figured out by now that AFOV is the primary driver within the range of "normal" binocular sizes and magnifications.



#5 Jon Isaacs

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Posted 28 January 2015 - 05:43 PM

7x35

3.5 magnitudes gain over the unaided eye.

magnitude 8.9 reached in the above simulation.

117917 stars in the whole sky = 2.86 stars per square degree = 272 stars in an 11° FOV.

In my more mundane 7x35 having 9.3° TFOV that number becomes 194 stars.

 

You'll have figured out by now that AFOV is the primary driver within the range of "normal" binocular sizes and magnifications.

 

 

To play the devil's advocate:

 

AFoV in the calculations is a major driver.  But the assumption in these calculations is that the fields are sharp enough that one can actually see those stars.  I think that's a shaky assumption..  7X binoculars with an 11 degree TFoV, that's a 77degree AFoV.. If the eyepieces were as sharp as the 31mm Nagler and the fields as flat as the 106FSQ or NP-101, that assumption might be reasonable.   But my experience with widefield binoculars, the outer portions of the fields are far from flat and the eyepieces show major amounts of astigmatism.  Add in the fact that the fields not fully illuminated and the gain in the number of stars one sees with the added AFoV would see to be much less than a calculation would suggest.  

 

Just my two cents.

 

Jon


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#6 Mark9473

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Posted 28 January 2015 - 05:55 PM

I couldn't agree more, Jon. I never pretended it was an exact calculation; we all know optical quality plays a big role. You're right in pointing that out more explicitly.

Do not forget about binocular summation though, which your precious Nagler 31 / NP-101 combination does not give you.



#7 Erik Bakker

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Posted 28 January 2015 - 06:02 PM

Just some food for thought: I have a 7x42 and 10x56 of the same design and make. In number of stars visible but especially on impression and brightness the stars leave on me, the 10x56 is miles ahead of the 7x42.



#8 Saturninus

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Posted 28 January 2015 - 06:21 PM

Wow that's a cool "back of the envelope" calculation. I'm not surprised that there isn't that much of difference, since the AFOV of those two specifications are essentially the same even though the AFOVs are different. I would have thought that a slight edge would go to the 15x50 since the same AFOV is supported by more aperture, but this suggests that the "rate of return" on magnification is lower than the "rate of return" on TFOV, if that makes sense.

 

In practical terms, I can see how that might be the case. A star that is suddenly included in the view because the 10x42 TFOV is larger would likely be more noticeable than a star that was previously just beyond visible detection in the 10x42 binocular but just crosses the threshold of visible detection in the 15x50 binocular. So then perhaps the 10x42 would feel like it has more stars that are brighter?



#9 Saturninus

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Posted 28 January 2015 - 06:28 PM

Just some food for thought: I have a 7x42 and 10x56 of the same design and make. In number of stars visible but especially on impression and brightness the stars leave on me, the 10x56 is miles ahead of the 7x42.

 

I would expect that to be the case since the gap in aperture is much wider. The gain from aperture probably overwhelms any efficiency that might have been gained or loss in the tradeoff between TFOV + exit pupil vs magnification.

 

If however it were a 8x42 vs 10x50, would it be much closer? I expect it would be indistinguishable, but from an academic point of view, which would yield more? 



#10 plyscope

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Posted 28 January 2015 - 06:31 PM

I remember once reading an article about a way to calculate which eyepiece in a telescope would show the most stars. The article was discussing the “majesty factor” that is used to characterize widefield eyepieces. It wasn’t the alder index, but something else that also accounted for differences in TFOV.

 

 

Maybe the article was this one by Mel Bartels;

 

http://www.bbastrodesigns.com/rft.html



#11 Mark9473

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Posted 28 January 2015 - 06:32 PM

The 10x42 would show more bright stars, but in the 15x50 the bright stars that are seen, are brighter than in the 10x42.

 

Do not forget the all-important factor of resolution at these relatively low magnifications.

The site I linked to giving the number of stars, does not take into account whether or not you can resolve them.

Quite a few of the stars up to mag 9 or 10 or whatever limit you're looking at, will be in star clusters and the difference in stars seen resolved at 15x is dramatically higher than at 10x.

 

Both instruments may theoretically show more or less the same number of stars, but the 15x50 will show more stars that matter.


Edited by Mark9473, 28 January 2015 - 06:32 PM.


#12 GlennLeDrew

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Posted 28 January 2015 - 08:48 PM

Mark has derived the fundamental driver here; AFoV.

And Jon points out the importance of image fidelity in actually seeing those stars, particularly the dimmer ones.

If we lived in a Universe which had the same space density of stars everywhere (as though tgere was an 'infinitely' large galaxy), *all* apertures at given AFoV would reveal similar star numbers within the field--if, moreover, space were also quite clear of dust.

But we live inside a very finite--and polluted--galaxy. For lines of sight well removed from the disk mid-plane, the space density of stars drops off pretty rapidly. And for views along or near the disk mid-plane, interstellar extinction generally induces an early fall-off in star counts. These two phenomena result in the situation whereby the star counts per unit solid angle begin to fall off at about 11m or so. This makes a smaller instrument a more impressive star number presenter than a large one. As aperture goes up, star fields tend to become ever sparser--except, of course, for regions of the Milky Way which present anomalously low extinction over very long lines if sight, such as the M24 star cloud.
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#13 Jon Isaacs

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Posted 29 January 2015 - 03:13 PM

I couldn't agree more, Jon. I never pretended it was an exact calculation; we all know optical quality plays a big role. You're right in pointing that out more explicitly.

Do not forget about binocular summation though, which your precious Nagler 31 / NP-101 combination does not give you.

 

Mark has derived the fundamental driver here; AFoV.

And Jon points out the importance of image fidelity in actually seeing those stars, particularly the dimmer ones.

If we lived in a Universe which had the same space density of stars everywhere (as though tgere was an 'infinitely' large galaxy), *all* apertures at given AFoV would reveal similar star numbers within the field--if, moreover, space were also quite clear of dust.

But we live inside a very finite--and polluted--galaxy. For lines of sight well removed from the disk mid-plane, the space density of stars drops off pretty rapidly. And for views along or near the disk mid-plane, interstellar extinction generally induces an early fall-off in star counts. These two phenomena result in the situation whereby the star counts per unit solid angle begin to fall off at about 11m or so. This makes a smaller instrument a more impressive star number presenter than a large one. As aperture goes up, star fields tend to become ever sparser--except, of course, for regions of the Milky Way which present anomalously low extinction over very long lines if sight, such as the M24 star cloud.

 

 

I tend to think of it as more variability with a larger scope.  The rich regions, clusters and dense star fields, are almost overwhelming in a larger scope.. And part of the equation is the brightness of the stars.. I think that's what's most different when viewing with a larger aperture.  

 

Jon



#14 GlennLeDrew

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Posted 30 January 2015 - 07:25 AM

The apparent star brightness function with aperture is an interesting discussion by itself. The question being: What is the fractional star count by apparent magnitude interval as aperture varies? Let's call this the apparent magnitude fraction, or AMF.

To start, we picture a random naked-eye view through a tube which restricts to some angle shared by the typical eyepiece AFoV. There will be so many stars seen in each magnitude interval, which we may bin as 0-0.99, 1-1.99, 2-2.99, etc. the resulting counts will set the naked-eye AMF.

Now we choose an aperture which collects 100 times more light, thus boosting star brightness by 5 magnitudes (at that exit pupil equaling our iris diameter.) A zero magnitude star is made to appear as -5m, a 4th magnitude star appears as -1m, a 9th magnitude star appears as 4m, and so on.

But of course the rather smaller FOV samples fewer of the brighter stars. Statistically we would hardly ever catch the zero and first magnitude stars, and not that often those of second or third magnitude. To first order, the telescopic AMF will shift downward by 5 magnitudes in terms if the actual apparent magnitude. But the 5-magnitude brightness gain compensates so as to yield a view much like the naked-eye view in terms of numbers by 'telescopic magnitude.' Indeed, if the space density of stars was everywhere the same to arbitrarily large distance, and interstellar extinction were not present, any random telescopic view at an iris-equaling exit pupil would reveal an AMF like that seen by the unaided eye.

But the real space distribution of stars and the presence of interstellar extinction truncate the stars counts as we look to ever fainter limits. The result is a decrease in numbers of the fainter stars seen as aperture increases; the AMF is trimmed at the faint end. And so the profusion of dim specks seen with smaller apertures is lost to larger apertures. The AMF becomes biased toward apparently brighter stars with increasing aperture.

Note, however, that this does not imply a compensating increase in *numbers* of the apparently brighter stars. The overall star count goes down at all apparent magnitude intervals beyond 11m or so, the slope of this decrease increasingly steepening with fainter limits.

So, does a *relatively* higher number of brighter stars (in spite of the smaller total star count) result in the impression of a brighter star field in the eyepiece on a larger scope? I find the comparative paucity to be the key factor, which results in a reduced impact; I prefer a *rich* star field, even if most are on the dim side.

Again, selected milky way fields and star clouds which provide little extinction to great distance are the anomaly here, these 'windows' exhibiting star counts which maintain a flatter AMF to fainter limits (and hence for larger apertures) than generally encountered. We should be wary of allowing these exceptions to color our impression of the more typical telescopic field.

To appreciate the truncation of star numbers at fainter limits, we need only look to W. Herschel's 'star gauges' he undertook in an attempt to determine our location within the sidereal system. His result placed us not far from the center, in large part because of the phenomenon of widespread interstellar extinction, which imposes an artificial horizon of sorts when we consider the dimmer stars in and near the band if the milky way. And on sight lines removed from the milky way band we quickly leave the more densely populated thin disk and sample the much sparser halo.

#15 Saturninus

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Posted 30 January 2015 - 02:57 PM

Wow great reading here. That article by Mel Bartel's that Plyscope provided the link for above was the one I remember, but I never really understood it until reading the explanations above. One thing I don't understand though - why would a small scope show more stars? Is that a function of its lower magnification? Wouldn't a larger scope with the same magnification still show more stars? Or are they assuming a fixed exit pupil, so that a smaller scope would have lower magnification / wider TFOV and thus show more stars?

#16 MartinPond

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Posted 30 January 2015 - 03:41 PM

Glenn:    Still not sure where you come down on the OP, but...heh, yeah, great reading!

 

Jon:  Point well taken on the edges of 11-degree binoculars, though your eyes fizzle outside of 20-30D afov anyway

         (so 'majesty' may not be so much about detail anyhow).

          So many 11-Deg binoculars are just a big, thick Plossl at the eye...nice but fading.

          When I think 'majesty', only the Empire Super Sport comes close to delivering.

 

 

         Here's a thought:   If there is so much varibility in the density, wouldn't choosing one power be a mistake?



#17 Mark9473

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Posted 30 January 2015 - 05:16 PM

One thing I don't understand though - why would a small scope show more stars? Is that a function of its lower magnification? Wouldn't a larger scope with the same magnification still show more stars?

 

Like Glenn explained, for the large part of the sky which is away from the galactic plane, you run out of stars. We're looking at a cloud of stars that are part of our home galaxy, but that cloud is not infinitely large. As you go to ever smaller FOV in ever larger scopes, you're looking deeper into the universe. Once you start looking beyond the "edge" of our galaxy, there's nothing much there until you get to the next galaxy.


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#18 GlennLeDrew

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Posted 30 January 2015 - 05:26 PM

To understand how star counts decrease with increasing aperture, first imagine the 'ideal' case where star counts *would* remain constant at all apertures. And that is if the Universe were uniformly populated by stars everywhere, with no interstellar extinction in effect.

Let's get this out of the way. Even though stars differ hugely in intrinsic luminosity, any random view would statistically contain a reasonably similar sampling of the various types.

In our hypothetical Universe, suppose we double our aperture. by retaining the same exit pupil and AFoV, the magnification is doubled and the TFoV is halved. The doubling of aperture collects 4X the light, which is a gain of 1.5 magnitudes, and permits to see twice as far for given star luminosity.

The doubling of distance probed is countered by the halving of the TFoV, the result being the same volume of space sampled. Because of the uniformity in space density of stars, an equal volume of space viewed will contain the same number of stars.

Now suppose we introduce a smooth distribution of obscuring dust throughout our Universe. The more distant stars sampled by a larger aperture will be more strongly attenuated than the nearer stars. And so in general the fainter stars are made to fade out more strongly than they would due to distance alone. A larger cone of space sampled by a larger aperture will appear to contain fewer stars than expected from a naive calculation based on just the volume.

As pointed out earlier, our real galaxy imposes a fall-off in star counts with distance due to both extinction (profound in and near the disk mid-plane) and decreasing star density not far removed from the disk system (profound for lines of sight greater than about 10 degrees in galactic latitude.)

Edited by GlennLeDrew, 30 January 2015 - 05:30 PM.


#19 MartinPond

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Posted 31 January 2015 - 12:29 AM

So, big eyes (and powers) done in by the dust, count-wise.

Small (or low power) eyes limited by the ability to even detect certain objects without the contrast/prying of the power.

 

Seems like there should be an optimum between those two. 



#20 GlennLeDrew

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Posted 31 January 2015 - 03:20 AM

The 'optimum' configuration for obtaining THE richest star field is that which reveals stars to 11-12m. An 80-100mm bino operating at a ~5mm exit pupil and having a large AFoV is hard to beat. Nirvana in this realm could be attained with a binoscope of such aperture that accepts 2" eyepieces, such as 22mm Naglers if the objective f/ratio is short, or 30mm 80 degree oculars of suitably slim barrel diameter if the f/ratio is long-ish.

Edited by GlennLeDrew, 31 January 2015 - 03:20 AM.


#21 Saturninus

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Posted 31 January 2015 - 08:11 PM

ok now I think I understand in my simplified (dumbed down) way. Thinking of the viewing area as a "volume of space" surveyed by the binocular helps a lot.

 

So I understand that if you had a 10x50 binocular (with an 65 AFOV) field of view, you have an area of observation than can be described as 6.5 TFOV wide and 10x deep, illuminated by a 5.0 exit pupil. If you were to stay at that aperture and directly trade TFOV for magnification, and use a binocular of 12x50, you area of observation described by 4.3 TFOV and 12x would show less stars, because the lower exit pupil is insufficient to reveal the stars in the distant end of the 12x dimension of the volume of space observed.

 

What was surprising to me is the notion that when you use a greater aperture to keep the same 5.0 exit pupil, say a 12x60 binocular, such that volume of space is described as 5.4 TFOV wide and 12x deep, there are sill less stars. But I understand the explanation above that this is because dust clouds and the distribution of stars are such that the stars in distant end of the deeper volume of space are more poorly illuminated and spread out relative to the stars at the near end of a wider TFOV view.  Am I understanding it right so far?

 

The question I am thinking about now is, how much more aperture would you have needed to reveal an equal or greater amount of stars in a volume of space described by 5.4 TFOV and 12x magnification deep vs the volume of space described by 6.5 TFOV and 10x magnifcation supported by 50mm aperture or 5.0 exit pupil. The discussion above suggest you need more than the 60mm required to keep an equal exit pupil between the two views. How much bigger of an exit pupil would you need? If you went to a 12x70 binocular such that the 5.4 TFOV by 12x mag view is supported by a 7.0 exit pupil, would that now reveal more stars than the 6.5 TFOV by 10x mag view that is supported by a 5.0 exit pupil? Is there a function that describes that equivalence? 



#22 Saturninus

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Posted 31 January 2015 - 08:12 PM

The 'optimum' configuration for obtaining THE richest star field is that which reveals stars to 11-12m. An 80-100mm bino operating at a ~5mm exit pupil and having a large AFoV is hard to beat. Nirvana in this realm could be attained with a binoscope of such aperture that accepts 2" eyepieces, such as 22mm Naglers if the objective f/ratio is short, or 30mm 80 degree oculars of suitably slim barrel diameter if the f/ratio is long-ish.

 By the way, what do you mean be "11-12m?"



#23 Mark9473

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Posted 01 February 2015 - 04:18 AM

11-12 magnitude



#24 pez_espada

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Posted 13 January 2017 - 01:55 PM

I would like to resurrect this topic as I find it extremely interesting and it seems that no definite conclusion has been reached in the thread. So there is a definite theoretical combination of the mentioned factors, namely aperture, TFoV, magnification,and exit pupil, that optimizes the star count for binoculars (and telescopes for that matter) under ideal dark skies?


Edited by pez_espada, 14 January 2017 - 09:07 AM.


#25 oldmanrick

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Posted 13 January 2017 - 03:12 PM

Fascinating topic!

 

I think it simply boils down to what part of the sky you are looking at.  The 10X42 will definitely show more stars in a field of uniform density containing mostly stars of a magnitude visible in these binoc's, but few that are slightly too small to see in them.  The 15X50 will show more stars in areas of the sky having many stars of a magnitude not visible in the 10X42, but visible in the 15X50. 

 

The heavens are awesomely variable.  That's one reason viewing the sky is so exciting and interesting.  Also the reason one needs different instruments for different views!


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