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How do you calculate apparent field of view for binoculars (and what is the largest?) (1 Viewer)

S

Saturninus

Active member
I’m most interested in binoculars that maximize “apparent field of view” – widening the subjective circle of vision even if the area of viewing is smaller due to magnification.

But I’m having trouble converting linear field of view to apparent field of view. I know that you can convert linear field of view to angular field of view by dividing the field of view in feet by 52. And I had thought that you can convert the angular field of view to apparent field of view by multiplying the degrees by the magnification.

Thus, the Swarovski EL 10x42 shows 336ft at 1,000 yds, or a 6.4 degree angular field of view. But the official spec shows only a 60 deg apparent field of view, so mag X angular field of view must not work the way I thought.

Then the Zeiss Victory SF 10x42 shows 360ft at 1,000 yds, but the spec sheet states only a 6.5 deg angular field of view??? I thought it should be 6.9 deg. And what would the apparent field of view be?

And how about the Kruger Caldera 10x42 – it claims 367ft at 1,000 yds, and an angular field of view a 7.0 deg!!!! Is that for real? And if so, how come I never hear more about them?

Anyway, of those three binoculars, which one has the widest circle of vision, and how would I calculate it accurately? What is the largest you have seen in a binocular?
 
Saturninus said:
I’m most interested in binoculars that maximize “apparent field of view” – widening the subjective circle of vision even if the area of viewing is smaller due to magnification.

But I’m having trouble converting linear field of view to apparent field of view. I know that you can convert linear field of view to angular field of view by dividing the field of view in feet by 52. And I had thought that you can convert the angular field of view to apparent field of view by multiplying the degrees by the magnification.

Thus, the Swarovski EL 10x42 shows 336ft at 1,000 yds, or a 6.4 degree angular field of view. But the official spec shows only a 60 deg apparent field of view, so mag X angular field of view must not work the way I thought.

Then the Zeiss Victory SF 10x42 shows 360ft at 1,000 yds, but the spec sheet states only a 6.5 deg angular field of view??? I thought it should be 6.9 deg. And what would the apparent field of view be?

And how about the Kruger Caldera 10x42 – it claims 367ft at 1,000 yds, and an angular field of view a 7.0 deg!!!! Is that for real? And if so, how come I never hear more about them?

Anyway, of those three binoculars, which one has the widest circle of vision, and how would I calculate it accurately? What is the largest you have seen in a binocular?
Click to expand...

If you type that exact question into Google, you will have many answers to choose from. Most will be the same. However, that will merely justify your answer.

Bill
 
The biggest apparent field of view I have seen is the new Zeiss SF 8x42.

I always multiply the angular FOV by the magnification.
 
Nikon binocular specs in their catalogues always always show a smaller value than the simple"multiply the angular FOV by the magnification". Nikon calculate apparent field of view using ISO standards, ref ISO 14132-1-2002. I think this is a laudable approach by Nikon but gives a smaller value than their rivals and it is harder to calculate requiring basic trigonometry. It is however the correct approach if you accept ISO Standards which many international organisations use.
 
Saturninus said:
I’m most interested in binoculars that maximize “apparent field of view” – widening the subjective circle of vision even if the area of viewing is smaller due to magnification.

But I’m having trouble converting linear field of view to apparent field of view. I know that you can convert linear field of view to angular field of view by dividing the field of view in feet by 52. And I had thought that you can convert the angular field of view to apparent field of view by multiplying the degrees by the magnification.

Thus, the Swarovski EL 10x42 shows 336ft at 1,000 yds, or a 6.4 degree angular field of view. But the official spec shows only a 60 deg apparent field of view, so mag X angular field of view must not work the way I thought.

Then the Zeiss Victory SF 10x42 shows 360ft at 1,000 yds, but the spec sheet states only a 6.5 deg angular field of view??? I thought it should be 6.9 deg. And what would the apparent field of view be?

And how about the Kruger Caldera 10x42 – it claims 367ft at 1,000 yds, and an angular field of view a 7.0 deg!!!! Is that for real? And if so, how come I never hear more about them?

Anyway, of those three binoculars, which one has the widest circle of vision, and how would I calculate it accurately? What is the largest you have seen in a binocular?
Click to expand...
I have read the Nikon 8x30 EII with a 70.2 Apparent FOV or 8.8 degrees at 462 feet at 1000 yards has the biggest FOV of any 8x binocular.
 
henry link said:
Here's another thread on measuring AFOV from Cloudy nights.

http://www.cloudynights.com/topic/493167-any-volunteers-to-measure-fujinon-fmt-sx-10x50-afov/

AFOV has to be measured. It could be accurately calculated, but only if a value for distortion is known and included in the formula. Neither the ISO method nor simple multiplication of the real field by the magnification takes distortion into account.
Click to expand...

Henry
Surely the important thing is that all binocular manufacturers follow the same approach? I am not qualified to say how good or bad the ISO standard is but a consistent or standard approach is helpful when comparing specifications.
Robert
 
The Kruger Caldera is for real regarding fov. My 8x42 Caldera checks in over 450 ft, with the listed fov being 438'. However this one illustrates Henry's post regarding distortion. I set up a tape measure at as precisely as as I can measure it 10 yards from the objective of a tripod mounted binocular. The conversion to 1,000 yards is then pretty simple math. As Henry points out, the killer in the deal is distortion. I could stretch the fov on mine to 460', but the edge distortion on any binocular makes it kind of difficult to measure precisely. However according to mfr specs the Caldera is a class leading wide fov.

The simple way is to multiply angular fov by magnification, as you have done. The angular field in degrees is figured by dividing the feet of field by 52.5, also as you have done. In my experience, the ISO standard reduces this by about 5*. Someone posted here once, but a search can't find the thread, a handy dandy ISO afov calculator. I have that on my computer, but the file type will not allow itself to be attached.

If all you want is a reference for the widest fov, not a mathematically correct one, the old fashioned way works just fine. It is just not "correct". Both the ISO method and the SLOW (Simple, Logical, Old fashioned Way) still rank the field of view the same way, one just reduces all the numbers without changing the rank order of narrow to wide, it just gives a smaller number.
 
Last edited:
Petrus82 said:
The biggest apparent field of view I have seen is the new Zeiss SF 8x42.

I always multiply the angular FOV by the magnification.
Click to expand...
If you use your calculation method. The Zeiss 8x42 SF has 8.5 degree angular FOV, 68 degree Apparent FOV and a 444 foot FOV@1000 yds. The Nikon 8x30 EII has an 8.8 degree angular FOV, 70.4 degree Apparent FOV and a 462 foot FOV@1000 yds. The EII is still king of the wide fields at 8x. Long live the king.
 
Last edited:
You fellows are talking SLOW (Dennis) vs ISO (Robert & Nikon) afov computations. Both are correct for the method quoted.
 
By the plain definition, apparent field of view is the actual field of view times the power.

In the past, many designs had 11-degrees actual, 77 degrees apparent,
and some had up to 13 degrees actual fov at 7x, for 91 degrees apparent.


Her is an extra-wide at 10x:
http://www.ebay.com/itm/Bushnell-Xt...751?pt=LH_DefaultDomain_0&hash=item27fcf0a87f
70 degrees apparent..

Here is a super-wide at 5x made today:
http://www.ebay.com/itm/Visionking-...399?pt=LH_DefaultDomain_0&hash=item3aa685d347
....that's 77 degrees apparent...



Saards are famous for actual fov, but have lower powers...

The cost is outer field fuzzing.
Some are better compensated than others.
The most well-known extra-wide with better outer field traits
is the Bushnell 7x35 Rangemaster.

Telescope buffs regularly get up to and over 100 degrees apparent,
and sometimes in custom super-binoculars. It's a matter of the
eyepiece and an extra-low-dispersion objective. Not cheap.

Because extremely advanced eyepieces are rather massive
in size and weight, you don't see them in normal binocular
product lines. Portability is a key thing in handheld binoculars.
You can easily find superwide EPs in the telescope world.
They may weigh 1-2 lbs each and cost a lot.

Most lens were hand-final-finished long ago, and could be moved
towards the more ideal parabolic shape. They were more expensive
in adjusted dollars, too. There are some aspherical elements added now.
Amateur telescope makers still hand-work parabolic refractor lenses.
Nice if you have a lot of time on your hands.
There are also some with mid-barrel "field flatteners", but that's usually
used now for perfection of the field rather than more field.
The flattener reduces the max. field you can attain.
It's most useful with smaller objectives for binoculars.
 
Last edited:
Zeiss's literature declares a 450' FOV @ 1000 yards or 150 Meters @ 1000 Meters for their injudiciously discontinued 7x42 Victory FL.

Bob
 
ceasar said:
Zeiss's literature declares a 450' FOV @ 1000 yards or 150 Meters @ 1000 Meters for their injudiciously discontinued 7x42 Victory FL.

Bob
Click to expand...

They might actually be judicious, given their
"princess and the pea" clients.
If you're an Alpha maker, you will take more hits to your reputation
these days for any defects in the field than for missing field width.
Any binoculars
over 500ft have a decent amount of blur and coma near the edges,
so 450' seems like a very difficult and tortured decisions for Zeiss...

Given the boast-fests Alpha owners get into, your general reputation
could get a black eye if serioous aberration showed up anywhere.
And it would at 500ft//7x, unless you made a beast of the eyepieces.
Then....suddenly you're too heavy.

Extra-wides aren't big at any price-point now, though all makers are
doing a decent job creeping up to 420-430 ft. I enjoy that width when it's
clean almost all the way.

In the end, your eyes don't even have much acuity that far out..
...it's just people's perceptions. I see trouble at the edge of Aculons,
but only if I look for that, not the birds and the fishers.
 
I've been using the simple formula outlined above by Optic Nut for two decades, and I'm so used to it that downsizing to ISO standards, even if they are more accurate, is difficult to get used to at this point (old snoopy dawg, new snoopy dawg tricks).

As to how the level of distortion enters the picture, unless the distortion is extreme, I don't think that's going to have a significant impact on how much AFOV most people actually see. Plus what I see doesn't always match the distortion level. For example, looking through a 10x42 SE and 10x42 HG side by side, the HG appears to have a larger apparent FOV to me than the SE even though in theory, the SE should have the larger AFOV because it has more distortion (although both are relatively low in pincushion, the HG is very low, enough to start the "ball" rolling).

So as with other metrics, I think much of what we see through binoculars lies in perception rather than mathematical formulas, light transmission numbers, or ISO standards. The proof is that Bob sees almost everything I see through binoculars differently, in many cases, the opposite way I see it. If math ruled, we'd see the same thing through the same binoculars, but we don't.

Brock
 
Hi,
I do not use to calculate AFOV, but have one rule for binocular selection:

If 8x needs to be good enough then multiply it with FOV in meters and result must be at least 1000 /125m is minimum for 8x.../
If 10x then at least 1100
If 12x then at least 1150
If 7x then at least 950

And the sweetspot at least 75% of course...
 
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