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quick math question. (1 Viewer)

aha! second dimension as against first! with a touch of third thrown in. now that i understand intuitively, thank you david. i guess we dont need to add the perception of depth of field as third dimension but it is what i would be also calculating, if again intuitively, if comparing the 'real' subjective perception of view as against objective measurement. 'objective' being pov not fov. bloody autism, expands in some senses of awareness, detracts in others.
 
aha! second dimension as against first! with a touch of third thrown in. now that i understand intuitively, thank you david. i guess we dont need to add the perception of depth of field as third dimension but it is what i would be also calculating, if again intuitively, if comparing the 'real' subjective perception of view as against objective measurement. 'objective' being pov not fov. bloody autism, expands in some senses of awareness, detracts in others.

In simple optical terms depth of field is dictated by magnification. It would be the same for all 8x binoculars, so can be disregarded.

However, that ignores the user interaction and perceptions. There is a much more complicated answer which would include, variable effective aperture, visual acuity and focal accommodation, with consequential changes in circles of confusion, all modified by angular and magnification distortion and subject to psychological processing. Unfortunately no one here understands that lot so it's best to disregard it as well, and just go with what looks best ;)

David
 
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In simple optical terms depth of field is dictated by magnification. It would be the same for all 8x binoculars, so can be disregarded.

However, that ignores the user interaction and perceptions. There is a much more complicated answer which would include, variable effective aperture and visual acuity with consequential changes in circles of confusion, all modified by angular and magnification distortion and subject to psychological processing. Unfortunately no one here understands that lot so it's best to disregard it as well, and just go with what looks best ;)

David

im sorry but now you have stated it i am unable to disregard it. having expanded my consciousness to include a third dimension, the propinquity to, not the fourth but the infinite variables of possibility, i find a consensus of my possible selves wishes to request that you define further the phrase 'circles of confusion'. you may also if it interests you, include enlightenment on the relevant 'psychological (as opposed to 'physiological') processing'. you may include mathematics, autistic perception sometimes intuits from symbolic language and equation shapes beyond the linear logic.
 
"Any sufficiently advanced technology is indistinguishable from magic." - Arthur C. Clarke
circles of confusion, occultism
this is a dark science. i can see where wjc got led astray but i fell off my visual (and perceptual) cliff halfway through the explanation of shadow stereopsis and may yet be saved.
 
136m at 1000m

How does one convert that to ft. at 1000 yds?
It's a simple RATIO:

136m/1000m = 136yds/1000yds

136yds = 408'

136yds/1000yds = 408'/1000yds

Once you start MIXING or OMITTING units of measurement (feet/yds) you may get confused.
FOV=408 sounds better than FOV=136. Actually, it means nothing without units of measurement.
FOV = 408' implies 408'/1000yds or 136'/1000' which, as we know, is equal to 136m/1000m

Quiz: Which is the larger FOV 408' or 136m?
Hint: You must assume units of measurement and correct ratios.
Answer: Given the previous data these FOV's are equal. 408'/3000' = 408'/1000yds = 136yds/1000yds = 136m/1000m
 
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I think I've got it.:t:

So here's another one. Sometimes they only put the Fov in degrees. 7.1,8.1 etc. How do you get the Ft. Fov from that?
 
Pirsquared?

No, pies are round.

Seriously, it isn't difficult, just multiply by 3 for feet at 1000 yards, or if one wants yards at 1000 yards, it is the same as metres at 1000 metres.

Anyway, the stated figures are often way off, sometimes by 10% or more, so the only real way is to measure the binocular field or compare two binoculars side by side.

In addition, the magnification is often way off also.

If the fields are pretty similar then the area difference is just double the linear difference.

For instance, if one binocular has a 10% larger linear field then the area increase is about 20%.
Don't waste your time working out square degrees.

(Actually 1.1 squared is 1.21 so the area field is 21% bigger, but 20% is near enough).

(It is only with something like an EWA Minolta Standard MK 8x40 with about 9.4 degree field, compared with say a 8x42 with 7.5 degree field that the increased area field is more than just double the linear field difference).

With the 9.4 degree field compared to 7.5 degree field the linear field is 25% bigger.
The area field is 57% bigger, so considerably more than double the linear increase.
 
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For something like 7.8 or 8.1 degrees, one goes to radians.

A radian is nearly 57.3 degrees.
1/10th radian or 0.1 radian is 5.73 degrees.

So if the field is 5.73 degrees it is 100 metres at 1000 metres.

A 7.8 degree field is just 7.8 divided by 5.73 times 100 metres at 1000 metres.
That is 136 metres at 1000 metres.
 
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I think I've got it.:t:

So here's another one. Sometimes they only put the Fov in degrees. 7.1,8.1 etc. How do you get the Ft. Fov from that?

Use the link in #16. Click the boxes for Size and Degrees, then enter 3000 (ft) for distance and the FoV angle in degrees. Click calculate.

7.1° would be 372.2ft and 8.1° is 424.8ft.

David
 
i am going to petition the World Parliament for a ban on maths. we need simple statements like GOOD and GOODER. oh hold on, no World Parliament, wrong timeline again. Black crow, we are lost. i tried.
 
Use the link in #16. Click the boxes for Size and Degrees, then enter 3000 (ft) for distance and the FoV angle in degrees. Click calculate.

7.1° would be 372.2ft and 8.1° is 424.8ft.

David

Oh thanks friend. That's great. I'm very good at the kind of math where you put in some numbers and then click on calculate. :t:
 
FoV in
Yards at 1000 yards = Meters at 1000 meters
Feet at 1000 yards = 3 x the above

(To convert
Feet at 1000 yards TO Meters at 1000 meters
/ by 3.)
FoV in
Feet at 1000 yards = Figure in Degrees x 52.4
(To convert
Feet at 1000 yards TO Degrees
/ by 52.4)
That is all! No pies, but optional if you wish to celebrate this.
 
Yes, correct.
Is that a Nikon E 8x30? Then pie, and B :)
Nikon's spec page (link) for this model gives
Angular field of view (Real): 8.8 degrees, and
Field of view at 1000 meters: 154 meters.
154 meters at 1000 meters is
154 yards at 1000 yards, or
on x by 3
462 feet at 1000 yards.
 
At the risk of pouring some confusion over a moment of clarity, Nikon actually list the AFoV as 63.2° not, as most of us would, 8.8°x 8 =70.4°. They explain in Adhoc's link that the ISO standard committee decided the AFoV for an 8x should be calculated as if the viewer was 8x fold closer. The calculation looks quite daunting. Using the calculator from #16 click the angle and degree boxes, and enter the 462ft for size and 375ft for distance (3000ft÷8).and you will get the 63.3° ISO answer.

David
 
Hi adhoc,

That is all! No pies, but optional if you wish to celebrate this.

Great concise summary! :)

Let me try to cover the background:

The 360 degree horizon at 1000 m distance has a length of 2 * pi * 1000 m.

So if we have x degrees of field of vision, it's x / 360 deg * 2 * pi * 1000 m wide.

With that in mind, you don't even have to memorize any numbers. (Other than pi, which has an infinite number of digits ;-)

Regards,

Henning
 
Henning,
To further clarify, and for Black crow's enjoyment,
1 / 360 x 2 x Pi x 1000 x 3 = 52.39

David,
The real and apparent angles of the FoV are, of course, another subject,
which too I am sure has been covered well in the Forum before!
Real FoV x 8 is quite valid if the image is (or approximated to be)
curved as upon a perfect sphere.
Zeiss uses this simpler formula in their consumer lit.
as for example here (link) for the Victory FL 8x32.
 
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