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Binoculars and Dawes

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#1 azure1961p

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Posted 06 July 2025 - 03:02 PM

It's so much plainer with a telescope, dawes res, load up the power look through the ocular.  You either see that classic notch or you dont.  Binoculars though working from 7 to 10x and 20mm to 50mm aperture would require a different metric to discern between apertures and power.  You can't run 100x through your 7x50s to find out.

 

So is there then a real-world metric for angular resolution for comparative apertures and powers that can be worked out like a Rayleigh and Dawes?  There should be anyway.

 

Your thoughts?

 

Pete



#2 Astrojensen

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Posted 06 July 2025 - 03:28 PM

You can't run 100x through your 7x50s to find out.

You absolutely can. Just place a small (homemade) 20-30mm scope with interchangeable eyepieces behind one of the binocular's eyepieces. Or just a finderscope, perhaps with the crosshairs removed. A 7x50 with a 6x30 held behind it will work at 56x. 

 

It can be very revealing. 

 

Details can be found in Brandt's "Himmelswunder im Feldstecher". 

 

 

Clear skies!

Thomas, Denmark



#3 azure1961p

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Posted 06 July 2025 - 04:29 PM

Thanks Thomas.

 

From checking online 35 arc seconds seems to be a "limit" of sorts but then I recall 15" for some people.  I'm curious what some call a "split".  Is it a merged "split" etc.

 

Pete



#4 Bortle9

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Posted 06 July 2025 - 07:36 PM

A 7x50 with a 6x30 held behind it will work at 56x. 

 

Just curious, what formula is required to derive the 56x? I hobbled together the ratio of the exit pupils multiplied by the product of the magnification (7.14286/5)x(7x6) to get 59.99.

 

My kludge is surely useless, but my middle-school son just got a fancy new calculator so...



#5 Mark9473

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Posted 06 July 2025 - 08:28 PM

It's much easier than that. Just avoid writing a typo and then 7 x 6 = 42 or 7 x 8 = 56.
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#6 Bortle9

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Posted 07 July 2025 - 01:26 AM

It's much easier than that. Just avoid writing a typo and then 7 x 6 = 42 or 7 x 8 = 56.

But 50/7 is 7.14286. A new scientific calculator is making me pedantic.

 

Edit: Where does your "8" come from?


Edited by Bortle9, 07 July 2025 - 01:29 AM.


#7 quilty

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Posted 07 July 2025 - 03:32 AM

the only diff I find between small binos and true scopes is, you can easily split even single stars.
Just tilt one prism a bit and make the bino squint

Most normal 30-60 mm binos are not up to pows like 100x You can try but they won't reward.

Honestly, serious starsplitting I'd never try with a binocular

Mizar is a good one for small binos

Edited by quilty, 07 July 2025 - 03:36 AM.


#8 JoeFaz

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Posted 07 July 2025 - 07:35 AM

So is there then a real-world metric for angular resolution for comparative apertures and powers that can be worked out like a Rayleigh and Dawes?  There should be anyway.

 

Your thoughts?

The Rayleigh and Dawes limits carry the same meaning for binoculars as they do for telescopes, since they're very simply derived from the aperture alone. Those figures don't account for magnification in any way. You can't observe Rayleigh and Dawes doubles with fixed mag binos because of their relatively low magnification.

 

I'm not entirely sure what the question here is, but I assume it's something along the lines of how close of a double can you observe with a given binocular of a certain aperture and magnification?

 

A brief aside before trying to answer your question:

The Dawes limit for an 50mm binocular is about 2.32 arc seconds. Even if you were using a 20x50 binocular, the apparent separation (discussed in the next paragraph) of a 2.32" double is 46.4". The theoretical resolving ability of the human eye is about 60" (this number is too low for our case, and exceptionally few if any people can split a double star with a 60" separation to their naked-eye). The point is, fixed mag binoculars, in the formats actually available to consumers, won't make Dawes or Rayleigh splits even in theory.

 

To the question as I understand it: because binoculars lack the magnification to be diffraction limited, you can essentially toss aperture and the associated Rayleigh and Dawes limits out the window (sort of, obviously aperture has some relevance to observing double stars with binos in general). The remaining factors (ignoring magnitudes and delta magnitude of the double star in question, which are also obviously relevant, but I'll ignore for this discussion) are magnification and visual acuity of the observer. So, the real determining factor is the apparent separation of the double. Take a 70" double, and use a 10x bino. The apparent separation is 700" (70" x 10). This would be a very easy split for most people. How low of an apparent separation you can get to will depend on the specific double (we're ignoring that for now), potentially the quality of the bino (also going to ignore that for now), and the observers visual acuity. Making an assessment of how tight a double you can split with a given binocular should be done with them mounted. Assuming the double contains two stars of the same or similar magnitude (and they aren't excessively bright or faint), and that the binoculars used are of good quality, I would suggest most people are going to be capable of splitting a double with an apparent separation of around 180-200 arc seconds. Some people will be able to go closer, some won't be able to go that tight. What you can do is something you'll only discover with experience.

 

*To determine apparent separation, multiply the actual separation by the magnification of the bino.

*To determine the actual separation that yields a desired apparent separation with a given bino - divide the desired apparent separation by the magnification of the bino.


Edited by JoeFaz, 07 July 2025 - 07:37 AM.


#9 azure1961p

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Posted 07 July 2025 - 08:08 AM

Thanks guys!

 

Joe, my gratis here, you're spot on.  I've dabbled about online and the 10x50 claims I've read of were at the most extreme 6 miles for a lunar crater (which I highly doubt and it's probably a bright ejects blanket making it appear easier than it truly is.  Some folks have claimed down to 15 arc seconds on the closest doubles.   Which then leads me to believe it HAD to be an elongation not a clean split.

 

It's a curious thing  about binos that they don't have this established. There must be some kind of Dawes-like correlate that can be arrived at that atleast gives a fair idea of these limits.  It'd not be so carved in as a textbook Dawes or Rayleigh but a fair estimation instead.  

 

Pete


Edited by azure1961p, 07 July 2025 - 08:13 AM.


#10 selenograph

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Posted 07 July 2025 - 08:12 AM

Just curious, what formula is required to derive the 56x? I hobbled together the ratio of the exit pupils multiplied by the product of the magnification (7.14286/5)x(7x6) to get 59.99.

 

My kludge is surely useless, but my middle-school son just got a fancy new calculator so...

 

The magnification of the binoculars (7x50) with a finderscope (6x30) is 42. The calculation is simple: Magnification of the binoculars multiplied by the magnification of the finderscope: 7 x 6 = 42.
I built such a combination many years ago for fun. I measured the exit pupil at the time, and it was 1.2 mm, so the calculated magnification is correct.

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Edited by selenograph, 07 July 2025 - 08:15 AM.

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#11 Astrojensen

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Posted 07 July 2025 - 08:13 AM

But 50/7 is 7.14286. A new scientific calculator is making me pedantic.

 

Edit: Where does your "8" come from?

I simply made a typo. 7x6 is obviously not 56, but 42... I had previously been experimenting with an 7x binocular and a 8x monocular, so the "56x" had become stuck in my mind, for some reason, and Mark caught it right away. That's what you get for posting late in the evening.

 

The exit pupils are completely irrelevant here, it's the magnification of the respective instruments that count. 

 

 

Clear skies!

Thomas, Denmark 


Edited by Astrojensen, 07 July 2025 - 08:26 AM.

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#12 azure1961p

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Posted 07 July 2025 - 08:17 AM

The magnification of the binoculars (7x50) with a finderscope (6x30) is 42. The calculation is simple: Magnification of the binoculars multiplied by the magnification of the finderscope: 7 x 6 = 42.
I built such a combination many years ago for fun. I measured the exit pupil at the time, and it was 1.2 mm—so the calculated magnification is correct.

 

This is great and it'll prove the glass correction for better or worse, BUT, way back down at 7x I'm inclined to believed it'll be something more like a *fair estimate*.  I can say I've seen the limits of my 8x42s when the ISS flew overhead one dusky evening.  That'd be a threshold in itself.  It was clearly not stellar but rather an extended though ambiguous shape.   It had angles but they defined only an amalgam of something undefined.

 

Pete



#13 azure1961p

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Posted 07 July 2025 - 09:03 AM

OK, here's something to consider...

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#14 Bortle9

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Posted 07 July 2025 - 09:22 AM

I simply made a typo. 7x6 is obviously not 56, but 42...

Cheers Thomas! I was wondering if there was an optics formula that I was unfamiliar with, but this explains it; kludge not required.




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