How to calculate minimum resolvable separation given aperture and magnification?

[The60mmKid]

There's a reason I became a social scientist 😞 My mathematical abilities are hopeless to the extent that I often forget how old I am. Please help me. I want to know the minimum separation between equal magnitude doubles that is resolvable in 18x70 binoculars. I understand how to figure this out using aperture alone, but when magnification becomes a variable, I feel like I'm 15 years old again, sitting in physics class praying to various premodern deities for answers.

[mikeDnight]

If you divide 116 by the aperture you'll find the theoretical resolution of two equally bright stars. 

In your case 116/70 =1.7 arc seconds, approximately.

Seeing conditions may prevent this on poor nights.

Edited February 8 by mikeDnight

[Mr Spock]

There's a useful calculator here: https://astronomy.tools/calculators/telescope_capabilities

 

[The60mmKid]

I appreciate the replies, but I don't think they address the question unless I misunderstand. The formula and tool calculate for a 70mm instrument, but my question also involves the fixed magnification of 18x. There's no way my 18x70 binoculars are going to split 1.7" doubles... as cool as that would be 😉

[vlaiv]

I think that you won't come close to theoretical resolution of 70mm aperture for several reasons.

First is quality of optics, but more important is magnification - that is too low.

If we assume perfect optics, then it's down to visual acuity of observer.

https://en.wikipedia.org/wiki/Visual_acuity

There is table on above page that lists MAR for different grades of visual acuity that is important factor - it is minimum angle of resolution measure and is expressed in arc minutes in said table.

20/20 vision equates to 1 MAR of resolution - which means that 20/20 person needs to see two equal doubles at one minute of arc separation to be able to just resolve them (see the gap).

Since you have binoculars that provide x18 magnification - that angle will actually be 1 minute of arc / 18 = 60 arc seconds / 18 = 3.33' separation.

This is for person having 20/20 vision and perfect optics.

Binoculars are often fast achromats that suffer from spherical aberration which will somewhat soften the view so the actual figure will be larger, and if you have less than 20/20 - this will add to separation needed.

For example 20/30 vision adds 50% to separation so you'll be able to resolve around 5'.

 

[Mr Spock]

Under perfect conditions, the human eye has a resolving power of 60 arc seconds (60"). So at a magnification of x18 you would split 3.3" given good seeing conditions and a steady base.

[The60mmKid]

Perfect! Not to brag, but I have excellent vision and a magnificent pair of binoculars 😏 3.33" doubles: Here I come! 

[The60mmKid]

1 minute ago, The60mmKid said:

Perfect! Not to brag, but I have excellent vision and a magnificent pair of binoculars 😏 3.33" doubles: Here I come! 

But maybe I'll round up to 3.34" to play it safe.

[The60mmKid]

17 minutes ago, vlaiv said:

I think that you won't come close to theoretical resolution of 70mm aperture for several reasons.

First is quality of optics, but more important is magnification - that is too low.

If we assume perfect optics, then it's down to visual acuity of observer.

https://en.wikipedia.org/wiki/Visual_acuity

There is table on above page that lists MAR for different grades of visual acuity that is important factor - it is minimum angle of resolution measure and is expressed in arc minutes in said table.

20/20 vision equates to 1 MAR of resolution - which means that 20/20 person needs to see two equal doubles at one minute of arc separation to be able to just resolve them (see the gap).

Since you have binoculars that provide x18 magnification - that angle will actually be 1 minute of arc / 18 = 60 arc seconds / 18 = 3.33' separation.

This is for person having 20/20 vision and perfect optics.

Binoculars are often fast achromats that suffer from spherical aberration which will somewhat soften the view so the actual figure will be larger, and if you have less than 20/20 - this will add to separation needed.

For example 20/30 vision adds 50% to separation so you'll be able to resolve around 5'.

 

After further reflection, my confusion has returned. You mention 3.33' (arc minutes) here, whereas @Mr Spock mentioned 3.3" (arc seconds). Based on my observing experience, the former strikes me as quite wide, and the latter strikes me as quite close.

Also, is aperture not a variable in this calculation? Would there not be a difference between the resolving ability of a 18x70 binocular and an 18x35 (hypothetically) binocular? That's surprising to me since aperture plays a clear role in the resolving ability of telescopes.

Edited February 8 by The60mmKid

[Zermelo]

The aperture is relevant to the initial answers you received. It sets a lower limit on the tightness of doubles that can be split, irrespective of eyesight. Then you need to take account of the limits of your eyes, and if the instrument is operating at a low magnification, the retinal cells may not be close enough to take advantage of the resolving power of the instrument. An eagle might be able to see finer splits with your binoculars than you can.

[The60mmKid]

1 minute ago, Zermelo said:

The aperture is relevant to the initial answers you received. It sets a lower limit on the tightness of doubles that can be split, irrespective of eyesight. Then you need to take account of the limits of your eyes, and if the instrument is operating at a low magnification, the retinal cells may not be close enough to take advantage of the resolving power of the instrument. An eagle might be able to see finer splits with your binoculars than you can.

Ok, but here magnification and aperture (i assume, perhaps incorrectly) are both variables, in addition to the others mentioned (i.e., eyesight, optical quality). So where does that leave us?

[Zermelo]

Perhaps you are looking for a single formula that takes all the parameters into account? I've not seen one. The capabilities of the instrument and of your eyesight both impose limits on the resolving power, and either may be the limitation in practice. Usually, it's the instrument (aperture) that is relevant, and the usual formulae assume perfect instrument optics, normal visual acuity, and the absence of other limiting factors. If you're splitting doubles, then it would be more usual to be employing higher magnifications.

[lunator]

There is a simple relationship in my experience..

The Dawes limit means you will not cleanly split the pair. If you want to cleanly split the pair the Raleigh criteria is more relevant. For white stars e.g 5500nm light it is 138/D.

From experience if you substitute the magnification M for D you will get a pretty close result.

So it becomes 138/18 ~ 7.5" 

Also from experience you can split pairs with a delta M of upto 2 magnitudes. 

I hope this helps.

Cheers

Ian

[The60mmKid]

Just now, Zermelo said:

Perhaps you are looking for a single formula that takes all the parameters into account? I've not seen one. The capabilities of the instrument and of your eyesight both impose limits on the resolving power, and either may be the limitation in practice. Usually, it's the instrument (aperture) that is relevant, and the usual formulae assume perfect instrument optics, normal visual acuity, and the absence of other limiting factors. If you're splitting doubles, then it would be more usual to be employing higher magnifications.

I'm looking for a way to account for aperture (70mm) and magnification (18x) and that assumes perfect eyesight and optics, which is usually what we do when we talk about telescopes' theoretical ability to split doubles.

I'm aware that we usually use higher magnifications for splitting doubles 😉 I have experience with that. But sometimes one wishes to use a certain instrument!

[The60mmKid]

2 minutes ago, lunator said:

There is a simple relationship in my experience..

The Dawes limit means you will not cleanly split the pair. If you want to cleanly split the pair the Raleigh criteria is more relevant. For white stars e.g 5500nm light it is 138/D.

From experience if you substitute the magnification M for D you will get a pretty close result.

So it becomes 138/18 ~ 7.5" 

Also from experience you can split pairs with a delta M of upto 2 magnitudes. 

I hope this helps.

Cheers

Ian

This does help! Based on my observing experience, that seems like a realistic split with the given aperture and magnification. I will observe a relatively equal ~8" pair with my 18x70s and report back.

[vlaiv]

44 minutes ago, The60mmKid said:

After further reflection, my confusion has returned. You mention 3.33' (arc minutes) here, whereas @Mr Spock mentioned 3.3" (arc seconds). Based on my observing experience, the former strikes me as quite wide, and the latter strikes me as quite close.

Also, is aperture not a variable in this calculation? Would there not be a difference between the resolving ability of a 18x70 binocular and an 18x35 (hypothetically) binocular? That's surprising to me since aperture plays a clear role in the resolving ability of telescopes.

Ah, sorry - my bad, I pressed ' instead of " (which is just shift away).

Aperture is variable - but binoculars resolve independent of the eyes (they produce the image regardless if someone is actually looking thru them) so things don't really compound.

If binoculars do resolve and human eyes are able to resolve that resolved image - we have a separation, in other cases - we don't (if either binoculars or eyes can't do their part - or both).

[Richard N]

There are probably too many uncontrolled variables to give a precise answer to a question like this. More fun might be to compile a list of test doubles using Stelle Doppie. The advanced search will allow you to search by separation and magnitude difference and compile a list of test doubles that you could amuse yourself with. https://www.stelledoppie.it/index2.php?section=2&azione=ricerca_avanzata

I'm guessing that uncontrolled variables will include your state of mind on the day, the seeing, the direction, the airmass etc. Then there will be some unknowns related to the optics themselves.

[The60mmKid]

Why is that we talk about theoretical limits of telescope resolution all the time but uncontrolled variables suddenly plague us when it's binoculars 🤔

Am I missing something?

Edited February 20 by The60mmKid

[vlaiv]

2 minutes ago, The60mmKid said:

Why is that we talk about theoretical limits of telescope resolution all the time but uncontrolled variables suddenly plague us when it's binoculars 🤔

Am I missing something?

Not sure if that is true.

I'm sure that sky conditions play major part when talking about visual separation of doubles as well.

Sometimes talk about theoretical resolution of the telescope is had in context of planetary imaging for example. There we don't really entertain these variables as they are effectively excluded by the process of planetary imaging (lucky imaging where we discard subs that are too distorted by atmosphere).

[The60mmKid]

32 minutes ago, vlaiv said:

Not sure if that is true.

I'm sure that sky conditions play major part when talking about visual separation of doubles as well.

Sometimes talk about theoretical resolution of the telescope is had in context of planetary imaging for example. There we don't really entertain these variables as they are effectively excluded by the process of planetary imaging (lucky imaging where we discard subs that are too distorted by atmosphere).

Many of the conversations on splitting doubles that I've seen online make claims like, "___ telescope can theoretically split ___ doubles," while taking for granted that atmospheric conditions and eyesight are uncontrollable and important variables. (For example, I don't mind claiming that my FOA-60Q can split 2" equal doubles, and I don't feel the need to offer  caveats about seeing and eyesight since I assume most of us know about that already.) So, I anticipated a similar reply when asking about binoculars. I figure most people asking such questions on such a forum know about the impacts of seeing, etc., so that's partially why I'm surprised that we seem to be considering binoculars in a different way than we do telescopes in this instance.

Edited February 8 by The60mmKid

[vlaiv]

2 minutes ago, The60mmKid said:

Many of the conversations on splitting doubles that I've seen online make claims like, "___ telescope can theoretically split ___ doubles," while taking for granted that atmospheric conditions and eyesight are uncontrollable and important variables. So, I anticipated a similar reply when asking about binoculars. I figure most people asking such questions on such a forum know about the impacts of seeing, etc., so that's partially why I'm surprised that we seem to be considering binoculars in a different way than we do telescopes in this instance.

I think that it's down to two things:

- most people that use telescopes to split doubles are familiar with influence of seeing, and it's often omitted for that reason, but when wanting to be fully accurate in description - it is included

- eyesight is not that important variable if one can change magnification. You select magnification that allows you to easily see what the telescope is capable of. There is seldom discussion (but it does happen) - what can you split with say x40 power or similar. Most of the time, recommendation is to go with very high powers, even higher than one would use for planetary for example. That removes eyesight from the equation as at those magnifications - eye has no issues resolving things. With binoculars - it is a thing since one is tied to certain magnification - and that magnification tends to be on very low side of things - which is not suitable for splitting doubles because of eyesight issues.

[The60mmKid]

25 minutes ago, vlaiv said:

 eyesight is not that important variable if one can change magnification. You select magnification that allows you to easily see what the telescope is capable of. There is seldom discussion (but it does happen) - what can you split with say x40 power or similar. Most of the time, recommendation is to go with very high powers, even higher than one would use for planetary for example. That removes eyesight from the equation as at those magnifications - eye has no issues resolving things. With binoculars - it is a thing since one is tied to certain magnification - and that magnification tends to be on very low side of things - which is not suitable for splitting doubles because of eyesight issues.

Helpful. Thanks 👍

[Nik271]

It's also worth pointing out that angular  separation you can resolve depends on the magnitudes of the stars. Bright stars are the best, for example Castor AB  at 5'' will look 5x18=90'' or 1.5 arcminutes apart in your binoculars, so you should be able to split it with excellent vision.  On the other hand a similar double star pair  of say 9-th magnitude and separation of 5'' may be not resolvable. This is because human acuity depends on the brightness of the objects since the retina uses different cells for low levels of light.

[John]

I tend to use the terms illustrated here by David Knisely:

@Nik271 makes a very good point about the impact that uneven brightness of component stars has as well 👍

I'm sure I read somewhere about minimum magnifications needed to enable the human eye to see the resolution that the optics are capable of showing but I can't recall what it is 🤔

This CN thread covers this but it's quite detailed:

On Magnification for Resolving Double Stars - Double Star Observing - Cloudy Nights

Edited February 9 by John

[The60mmKid]

There have definitely been some good points made about observing doubles in general... But again, the specific question was about the theoretical limitations of a 70mm instrument with 18x magnification to resolve equal magnitude doubles, only considering the variables of aperture and magnification (and, thus, assuming perfect conditions and eyesight for the sake of the theoretical exercise). Some of the responses related to aperture, but not magnification. Others related to magnification, but not aperture. And others introduced other variables. If we don't know how to calculate using these two variables, that's totally fine. But I think it's helpful to keep the actual question in mind.

I found @lunator's reply most relevant and am eager to observe an 8" equal pair and report back.