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Apparent FOV? A quick way to measure!

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#1 GlennLeDrew

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Posted 15 October 2010 - 01:56 AM

In the spirit of the objective effective aperture test using a flashlight...

I've been inspired to provide this after all the discussion here regarding AFoV, and the great amount of guesswork going on. It's not quite as simple as the objective aperture test, but not arduous at all.

This method also involves shining a flashlight into the bino, but in reverse. First, test things out to get a feel for how big an illuminated circle you can get before getting too faint. The bigger this circle, the more accurate the measurement.

If you can't see the full circle at a time, simply wiggle the flashlight a bit so that opposite sides of the circle can be seen alternately. In such case, you can make small pencil marks and measure their separation afterward.

Set the focus at least close to infinity focus. Some binos have field stops which are fixed with respect to the body; as the eyepiece is focused in/out AFoV actually changes somewhat! (If the field stop is fixed within the eyepiece itself, AFoV doesn't ever vary. If you know this to be the case, eyepiece position will not matter.)

Some binos suffer significant edge-of-field darkening due to a too-small rear prism aperture, which can result in a non-sharp field edge. Simply measure to the farthest visible 'edge' of the illuminated circle.

1) Set the bino up on a tripod, near a white wall and aimed so that the eyepices are facing the wall as prependicularly as you can obtain. (The wider the AFoV, the more critical is perpendicularity. Any tilt will result in an elliptical circle of light due to the geometry of projection.)

2) Shine a flashlight into an objective, from up close so that you get no spillage of light onto the wall.

3) Measure the diameter of the illuminated circle of light on the wall.

4) Measure the distance between the wall and the location of the eyepiece's eyepoint (exit pupil). If the eye relief is stated as, e.g., 18mm, measure right to the eye lens itself and then subtract the 18mm.

5) AFoV = 2 * ARCTAN((circle diam. / 2) / distance to eye point)


An example:

- The illuminated circle of light measures 300mm in diameter.
- The distance from wall to eye lens = 280mm.
- The eye relief is 15mm; the wall-to-eye point distance is therefore 280 - 15 = 265mm.

AFoV = 2 * ARCTAN((circle diam. / 2) / distance to eye point)
AFoV = 2 * ARCTAN((300 / 2) / 265)
AFoV = 2 * ARCTAN(150 / 265)
AFOV = 2 * ARCTAN(0.566)
AFoV = 2 * 29.5
AFoV = 59.0 degrees

With care you can surely derive the AFoV to within 1/2 degree via this method. And as you can see, a larger projection distance lessens individual measurement error.

Lastly, this value will conform to the real AFoV, which is the apparent angle subtended on the retina by the illuminated field. For larger AFoVs particularly, there will usually be some discrepancy when compared to the 'fictitious AFoV' simply calculated from TFoV and magnification. This is normal.

Let's see some results!

#2 Jawaid I. Abbasi

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Posted 15 October 2010 - 03:04 AM

Forgive me, if I understood this way.

Attached Thumbnails

  • 4116950-ArcTan.jpg


#3 Jawaid I. Abbasi

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Posted 15 October 2010 - 03:07 AM

The formula you used is:
a=c/b
a=arctan(c/b)

#4 Rich V.

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Posted 15 October 2010 - 10:46 AM

No, Jawaid, by the triangle you labeled above:

Side "c" should be 1/2 of the projected circle diameter where AFOV = 2 x angle "C".

AFOV = 2 x arcTan [(circle/2)/ b]

Some test results using this method:

Fuji 16x70s (64° simple AFOV) have a measured AFOV of 60°
Nikon 10x35E2 (70° simple) -- 66.25°
Nikon 7x35E (52.5° simple) -- 49.68°
Pentax 10x50 PCF-V (50° simple) -- 46.82°
Pentax 10x43 DCF-SP (60° simple) -- 57.1°

Thanks once again for a simple explanation of true AFOV measurement, Glenn!

Rich V

#5 GlenM

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Posted 15 October 2010 - 11:13 AM

What would we do without you Glenn?

Always excellent threads and posts.

#6 KennyJ

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Posted 15 October 2010 - 11:22 AM

Congratulations to both Glenn and Glen .

The former for the well written article , the latter in celebration of another birthday !

Kenny

#7 GlenM

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Posted 15 October 2010 - 11:24 AM

Thank you Kenny.

You could never accuse me of a well written article :grin:

#8 Obx

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Posted 15 October 2010 - 11:57 AM

This is all Greek to me. But I appreciate that some people understand it.

#9 captain11

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Posted 15 October 2010 - 12:26 PM

Glen,
Your magical flashlight is once again so very illuminating. Thank you for your insight and bountiful contribution.

#10 GlennLeDrew

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Posted 15 October 2010 - 12:44 PM

Jawaid,
From your diagram, the correct formula is:

C = ARCTAN(c/b)

This gives the semi-angle (half-angle), which then is multiplied twice to get the full angle. The semi-angle must be determined first because of the requirement for a right triangle with trig functions.


And to all; thanks for the kind words! I'm happy this will be useful for some.

#11 Jawaid I. Abbasi

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Posted 15 October 2010 - 01:08 PM

Glenn,
My second question which I was about to ask is explained which is "2".
Since, you determined Base(b) and Perpendicular© then if you add both minus 180. Will not it be the full angle of "Hypotenuse" that is a=(b+c)-180

So, according to the numbers you have provided that is 60.5 degree (AFOV) I know you know all the numbers then I but it is what I calculated finally

Oh, Glen M; Congratulations your 90th birthday :)

#12 Jawaid I. Abbasi

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Posted 15 October 2010 - 01:21 PM

Forget my last post.

Yes the formula you derived is correct because using the 180 degree; of course will be half so need "2" to get full angle.

#13 Ad Astra

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Posted 15 October 2010 - 02:17 PM

Glenn,
This looks fascinating - I think I could turn this into an optics lab for astronomy students. If I get it worked up, I'll post it for everyone to see.

Would the same technique work for an isolated telescope eyepiece?

Dan

#14 Gordon Rayner

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Posted 15 October 2010 - 03:45 PM

A minor point:

The angular size of the flashlight , as seen from the objective lens , should be larger than the true field of the binocular.

Ideally, the flashlight is a source of only parallel rays. It could be considered the equivalent of a circular group of very bright stars at infinity.

If the flashlight were to recede from the binocular past the distance at which its angular size is equal to the true field of the binocular, the circle of light projected onto the wall would be smaller in diameter than the apparent field of the binocular or telescope.

#15 Gordon Rayner

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Posted 15 October 2010 - 04:35 PM

Another way seems to be:

Tape a laser pointer to a rotatable mount,such as a tripod head, or onto a jar lid nested in a horizontal wood, plastic, or metal vee supported on a table.

Aim the laser at the objective of the binocular or telescope. Rotate the binocular or telescope until the laserbeam exits from the eyepiece approximately parallel to the (cylindrical) eyepiece barrel, and perpendicular to the measuring wall. Reposition the laser as required.

Now,rotate the laser until the beam projected through the binocular onto the wall disappears. Mark that spot on tape attached to the wall.
Next, rotate the laser in the opposite angular direction, until the beam disappears. Mark that spot on the taped wall.

Now, use the trigonometric expression in Glenn's post above to calculate the apparent field.

The center of rotation of the laser must be close enough to the objective lens such that the laser can be rotated through an angle at least as large as the true field of the binocular, before the beam on the wall disappears.

It may be desirable that the axis of rotation of the laser mount be near the point at which the beam leaves the laser. But I have not yet tried this measurement method. That rotation axis should be 90 degrees to the laser beam.

#16 GlennLeDrew

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Posted 15 October 2010 - 05:37 PM

Dan,
Yes, this works for any positive eyepice, whether installed in a telescope (or bino, or microscope) or not.

As Gordon points out, the field stop needs to be fully illuminated if the entire circle is to be seen. This is why I mentioned 'wiggling' the light source about if this condition is not fulfilled. If the wiggling were do be done at high enough frequency (say, 10/second), the effect would be to pretty much see the full circle at any one time. But in practice, most 'normal' flashlights will at least come close to fully illuminating the field stop, and so minimal 'wiggling' will be required to sample opposite sides of the illumination pattern.

If testing an eyepiece in isolation, and if the light source is of larger diameter than the eyepice, I recommend some form of light blocker to keep extraneous light from illuminating the surface upon which the circle of light is projected. (We want good contrast so as to obtain a large circle for more accurate measurement.) This blocker can be a piece of cardboard with a 1.25" or 2" hole, depending on eyepiece barrel diameter.

#17 Man in a Tub

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Posted 15 October 2010 - 10:02 PM

Let's see some results!


OK, I just did two quickies (probably too darn fast!) with my Brunton Eterna 15x51 roof prism already mounted on my Parks Video tripod.

The Brunton Eterna blurb states: 3.8° TFOV (200 ft.@ 1000 yds.) with eye relief of 15mm.

Without all my figures, here's what I got:

First time: 58.47°
Second time: 58.44°

Per the simple formula: 3.8° x 15 = 57°

Close enough for folk? ;)

#18 Jawaid I. Abbasi

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Posted 16 October 2010 - 12:14 AM

Todd,
So its really worked. I will do 15x70 perhaps next week or two when I visit my sister house.

#19 Man in a Tub

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Posted 16 October 2010 - 12:15 AM

Here's two more.

I. My lovable Pentax 20x60 PCF WP II:

Stated TFOV 2.2° (114 ft. @ 1000 yds.) with eye relief of 21mm

I measured 44.45°.

Per the simple formula: 2.2° x 20 = 44°


II. Nikon 12x50 Action Extreme:

Stated TFOV 5.5° (288 ft. @ 1000 yds.) with eye relief of 16.1mm

I measured 65.98°.

Per the simple formula: 5.5° x 12 = 66°


#20 Rich V.

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Posted 16 October 2010 - 02:07 AM

Todd, I'm a bit surprised your measurements/calculations work out to numbers so close to the simple AFOV calculation. In the few binoculars I've measured Glenn's way I get numbers smaller than the simple AFOV; numbers more in line with the ISO standard method of AFOV calculation that Nikon now uses.

Nikon's AFOV calculation

For instance, ISO standard calculation: 8x binocular w/ 7° TFOV

AFOV = 2 x tan¯¹ (mag x tan TFOV/2)
____ = 2 x tan¯¹ (8 x tan 3.5)
____ = 2 x tan¯¹ (8 x .061)
____ = 2 x tan¯¹ (.488)
____ = 2 x 26.01°
____ = 52.02°

Simple calculation:

AFOV = 8 x 7° = 56°

Is this an instance where any pincushion distortion added by the eyepiece would make the measured AFOV larger than the undistorted ISO AFOV calculation would indicate? :confused: Comments welcome.

Rich V

#21 Man in a Tub

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Posted 16 October 2010 - 03:21 AM

Todd, I'm a bit surprised your measurements/calculations work out to numbers so close to the simple AFOV calculation.

Rich V


I computed what I measured very quickly.

I'll redo the Nikon 12x50 Action Extreme measurements more meticulously. It's the easiest binocular to work with.

#22 Man in a Tub

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Posted 16 October 2010 - 04:30 AM

Redoing Nikon 12x50 Action Extreme with 5.5° TFOV.

ISO standard method of AFOV:

(I just plugged in the values for the Nikon.)

AFOV = 2 x tan¯¹ (mag x tan TFOV/2)
____ = 2 x tan¯¹ (12 x tan 2.75)
____ = 2 x tan¯¹ (12 x .0480)
____ = 2 x tan¯¹ (.5764)
____ = 2 x 29.96°
____ = 59.92°

Simple calculation:

AFOV = 12 x 5.5° = 66°

My second result with slower measurements for the quick way to measure: 65.05°

AFOV = 2 x ARCTAN ( (circle diam./2) / distance to eye point)*

AFOV = 2 x ARCTAN ( (269/2) / 210.9)

AFOV = 2 x ARCTAN (134.5 / 210.9)

AFOV = 2 x ARCTAN (.6377)

AFOV = 2 x 32.5274°

AFOV = 65.0548°

*In my calculation, the distance to the eye point is the distance from the wall to the eye lens less Nikon's stated eye relief of 16.1mm. See Glenn's opening post. I got that right, didn't I?

That's the best I can do, I think.

:shrug:

#23 Rich V.

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Posted 16 October 2010 - 12:28 PM

Looks good to me, Todd.

Like I mentioned above, the introduction of pincushion distortion to provide an "undistorted" view while panning may likely give the Nikon the wider AFOV you're seeing as measured with Glenn's method.

My experience with Pentax binoculars is that they don't add much pincushion to their design as I see the "globe effect" more readily in my Pentax binoculars than in others. There I would have expected the AFOV number to be smaller than the simple calculation. Obviously, there is something else going on here I don't understand! I'll measure my Pentax bins later today and add them to my list above.

Also, using the ISO formula usually relies on specs provided by the mfgr. which may or may not be accurate, frequently erring to a larger number based on what I've seen.

Rich V

#24 Rich V.

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Posted 26 October 2010 - 11:55 AM

Glenn, at least a couple of us are still thinking about this discussion. What is your take on measured AFOV using your method vs. the ISO calculation that takes angular distortion into account? The AFOV measurements I've taken using your method seem to be closer to the results using the ISO method demonstrated in posts above rather than the simple calculation.

It seems clear to me that as one views a target through a wide field instrument (whether it's the naked eye or a WF binocular) that a wider AFOV design will show more apparent angular distortion towards the edge of the field of view. The ISO standard that Nikon now uses takes this angular distortion into account giving a lower AFOV number than using the simple mag. x TFOV method.

I see it as being like looking at a picket fence from a close vantage point; the pickets appear to be x units apart right in front of you but the further down the fence you look, the closer the pickets appear. The wider the angle you take in, the more compressed the pickets appear to be; the image scale does not appear uniform across the field.

Based on Holger Merlitz's article on the "globe effect" it appears to me that adding pincushion distortion to an eyepiece design will counter this effect in a binocular but will also skew the AFOV to a greater angle (but not change the TFOV).

Since your measurement method is taking the eyepiece (and its distortion) into account wouldn't eyepieces with more pincushion distortion always give a wider AFOV measurement than an eyepiece with an undistorted design?

I'd appreciate any insight you can give as I'm really trying to understand the principle!

Rich V

#25 EdZ

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Posted 26 October 2010 - 12:35 PM

Redoing Nikon 12x50 Action Extreme with 5.5° TFOV.


the Nikon AE 12x50 actually has only a 5.1° TFOV, not the advertised 5.5°

edz


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