•

# Problems with Calculating True Field of View

29 replies to this topic

### #1 HotRod217

HotRod217

Explorer 1

• topic starter
• Posts: 59
• Joined: 04 Jan 2021
• Loc: Germany

Posted 14 May 2021 - 12:19 PM

Why is there such a huge difference in the calculated Real FOV when using Method 1 or 2?

My scope has a focal length of 1650mm. The eyepiece is a Orion LHD 80 degree, 9 mm, field stop diameter 33.4mm.

What is my real True Field of view?

### #2 Virtus

Virtus

Ranger 4

• Posts: 319
• Joined: 09 Mar 2021
• Loc: SE NC

Posted 14 May 2021 - 12:24 PM

Per Don Pensack's spreadsheet the 9mm Orion LHD has an effective field stop of 12.6mm, not 33.4 so method 1 is approximately correct.

• Ernesto.Nicola, areyoukiddingme and ECP M42 like this

Vostok 1

• Posts: 173
• Joined: 01 Mar 2019
• Loc: Northern California (Mt. Tam)

Posted 14 May 2021 - 12:24 PM

How can a 9mm EP have such a big field stop? Impossible, I think you are measuring the wrong thing -- the field stop is the little circle you see down inside when you look into the nosepiece end.

If you enter the right number, the two methods should be approximately equal.

-EDIT- and also, with a widefield you don't always see the field stop from the nosepiece side, there may be elements that come before the effective field stop.

Edited by rblackadar, 14 May 2021 - 12:26 PM.

• Voyager 3 and ECP M42 like this

### #4 skysurfer

skysurfer

Viking 1

• Posts: 932
• Joined: 05 Oct 2009
• Loc: EU, N 52 E 6

Posted 14 May 2021 - 12:25 PM

The first method is correct.

The second one is only when the eyepiece has maximum FOV which is not always the case.

### #5 AstroBrett

AstroBrett

Ranger 4

• Posts: 370
• Joined: 26 Jan 2009

Posted 14 May 2021 - 01:44 PM

HotRod is quoting the correct diameter for the field stop as given in the specs for the eyepiece.  But I agree with the others, that may be the actual physical dimension, but it is not the effective field stop diameter.

If you really want to know the actual FOV on your eyepieces, turn off the tracking on your scope and time the number of seconds it takes a start to transit the field of view. That is all that really matters, not what a particular formula gives. An don't forget that most formulas have assumptions built into them, and when those assumptions are not met, the results is usually incorrect.

Brett

• SteveG likes this

### #6 TOMDEY

TOMDEY

Cosmos

• Posts: 9,714
• Joined: 10 Feb 2014
• Loc: Springwater, NY

Posted 14 May 2021 - 01:49 PM

Both are generally correct but make different assumptions regarding linear and angular distortions. So you have to take all such with a grain of salt. They are only meant to be approximations; never intended as gospel. Dennis di Cicco recently wrote a great article on eyepieces, including a picture of his AFOV lens bench --- I'm building a bench to that same approach right now!  Or you can use what I show below there to get AFOV that takes into account angular distortion.  Tom

#### Attached Thumbnails

• vkhastro1 and george tatsis like this

### #7 Starman1

Starman1

Vendor (EyepiecesEtc.com)

• Posts: 50,657
• Joined: 23 Jun 2003
• Loc: Los Angeles

Posted 14 May 2021 - 02:58 PM

You can calculate an approximate field stop diameter when it is unknown:

Calculated field stop = (manufacturer's apparent field claim / 57.296) x eyepiece focal length.

So the 9mm Orion LHD 80° should have a field stop diameter of 80/57.296 x 9 = 12.57mm (12.6mm rounded off).

the manufacturer doesn't state the field diameter on their website, and Orion's data is accidentally copied from another focal length.

• Jon Isaacs, george tatsis, Ernesto.Nicola and 2 others like this

### #8 GUS.K

GUS.K

Viking 1

• Posts: 996
• Joined: 13 Aug 2014
• Loc: Australia.

Posted 14 May 2021 - 03:37 PM

Moving to Eyepieces forum for a better fit.

### #9 SeattleScott

SeattleScott

Voyager 1

• Posts: 11,610
• Joined: 14 Oct 2011

Posted 14 May 2021 - 04:58 PM

The 14mm has a calculated field stop of 19.6mm. I find it is closer to 19.3mm. So if the calculated field stop of the 9mm is 12.6, I suspect the actual effective field stop is around 12.4mm

Scott
• Starman1 and Jon Isaacs like this

### #10 RASutton

RASutton

Lift Off

• Posts: 24
• Joined: 23 Jan 2008

Posted 14 May 2021 - 08:50 PM

Here is another method that you can read at this link which takes you to the ISO current definitions of the terms involved in finding true field of view and a lot of other measures in telescopic systems. All of that are discussed in the Preview and in some other ISO standards that are mentioned in the annex. It is interesting reading.

https://www.iso.org/...dard/66244.html

### #11 Jon Isaacs

Jon Isaacs

ISS

• Posts: 92,505
• Joined: 16 Jun 2004
• Loc: San Diego and Boulevard, CA

Posted 15 May 2021 - 05:18 AM

Both are generally correct but make different assumptions regarding linear and angular distortions.

TFoV = AFoV / Mag  does make assumptions about the distortion of the eyepiece.  One can measure AFoV directly and TFoV directly and see an error of 10% or more.

TFoV = 180°/π  x  Field Stop (effective) / Focal length (telescope) makes no assumptions about the distortion of the eyepiece, it just  considers the angular size of focal plane that the eyepiece is magnifying which depends on the focal length of the scope and the diameter of the field stop. It's an isosceles triangle whose base is the field stop and whose height is the focal length of the telescope.

The one caveat is that some eyepieces have magnifying optics before the focal plane/field stop and this magnification affects the effective focal length of the telescope. Such eyepieces are call negative-positive because they have a Barlow-like tele-negative section as well as positive magnifying section. This why the Effective Field Stop must be used. The Orion 80 degree eyepiece is a negative-positive eyepiece so it's possible the actual field stop after the tele-negative section is 33.4 mm but I suspect it's just a webpage error, not uncommon out there on the web.

Jon

### #12 gnowellsct

gnowellsct

Hubble

• Posts: 18,686
• Joined: 24 Jun 2009

Posted 15 May 2021 - 06:05 AM

If you really want to know the actual FOV on your eyepieces, turn off the tracking on your scope and time the number of seconds it takes a start to transit the field of view. That is all that really matters, not what a particular formula gives. An don't forget that most formulas have assumptions built into them, and when those assumptions are not met, the results is usually incorrect.

Brett

At the celestial equator aka zero declination. Otherwise you'll need to correct the measured time with some trig. Timing a transit of Polaris is futile. Everything from Dec 90 to Dec 0 falls on a continuum from zero degrees per hour to 15 degrees per hour when you measure transit times.

Strictly speaking it all moves 15 degrees an hour. But the distance traversed changes, the same way the center of a car wheel barely moves compared to the outer circumference, but the rpm is constant for both.
• ECP M42 likes this

### #13 TOMDEY

TOMDEY

Cosmos

• Posts: 9,714
• Joined: 10 Feb 2014
• Loc: Springwater, NY

Posted 15 May 2021 - 06:45 AM

Yep, the factor is the cosine of the declination. The star's angular speed slows down by that factor.    Tom

• gnowellsct likes this

### #14 ECP M42

ECP M42

Ranger 4

• Posts: 318
• Joined: 28 Apr 2021
• Loc: central Europe 45°N

Posted 16 May 2021 - 08:14 AM

TFoV = AFoV / Mag  does make assumptions about the distortion of the eyepiece.

I am also of this idea (if I understand correctly). I am not sure, but I believe that eyepiece manufacturers assign an AFOV value adapted so that the calculation above takes into account the distortion and that the astronomer can simply calculate the field of view (FOV) without resorting to trigonometry every time.

Manufacturers know the value of eyepiece distortion, the astronomers who use them don't. You astronomers are lucky enough to be able to simplify things, so I don't understand why you have to rack your brains with complications. Also because the visual field will be seen precisely during the observations in the field (pardon the pun).

### #15 Starman1

Starman1

Vendor (EyepiecesEtc.com)

• Posts: 50,657
• Joined: 23 Jun 2003
• Loc: Los Angeles

Posted 16 May 2021 - 08:54 AM

I am also of this idea (if I understand correctly). I am not sure, but I believe that eyepiece manufacturers assign an AFOV value adapted so that the calculation above takes into account the distortion and that the astronomer can simply calculate the field of view (FOV) without resorting to trigonometry every time.

I wish that were true, but it is not.

The apparent field is a set thing determined by eye relief and lens diameters, and the advertised apparent field  is NOT chosen to fit into that formula.

First, in a line of eyepieces advertised to have the same apparent fields, the actual apparent fields will vary.

Second, apparent field cannot be used to derive the true field of the eyepiece unless the distortion figure is known, and that figure is never published.

Many manufacturers, though, do quote a field diameter for the eyepiece which CAN be used to calculate a true field in a particular scope if the telescope's focal length is known.

Another simple formula will result in a more accurate determination of true field.  The actual apparent field need not be known:

TF = EPFS/TFL x 57.296

where EPFS = eyepiece field stop, TFL = telescope focal length

Example of a difference in the formulae:

31mm Nagler in my scope:

The TF = AF/M formula gives a true field of 1.392°

The TF = EPFS/TFL x 57.296 formula gives a true field of 1.318°, 5.3% smaller.

Since a star timing yields the lower figure, you can see why a formula that bypasses apparent field and distortion is more useful for the amateur astronomer.

Manufacturers who quote field stop diameters are many.  Among them are:

TeleVue

Explore Scientific

Pentax

Nikon

William Optics (some)

Orion (some, and some are incorrect)

• Jon Isaacs likes this

### #16 Jon Isaacs

Jon Isaacs

ISS

• Posts: 92,505
• Joined: 16 Jun 2004
• Loc: San Diego and Boulevard, CA

Posted 16 May 2021 - 09:54 AM

I am also of this idea (if I understand correctly). I am not sure, but I believe that eyepiece manufacturers assign an AFOV value adapted so that the calculation above takes into account the distortion and that the astronomer can simply calculate the field of view (FOV) without resorting to trigonometry every time.

Manufacturers know the value of eyepiece distortion, the astronomers who use them don't. You astronomers are lucky enough to be able to simplify things, so I don't understand why you have to rack your brains with complications. Also because the visual field will be seen precisely during the observations in the field (pardon the pun).

To add to what Don wrote, the AFoV is an actual angle and one that can be directly measured.  David Knisely has documented a couple of methods, I use his Projection Method.

https://www.cloudyni...p/#entry6604421

By measuring the diameter of the projected beam, the distance from the eye relief point to the Projection surface the AFoV can be calculated.

AFoV = 2 x Arctangent ( D/(2 x L)) where D is the projected beam diameter and L the distance.

I've measured numerous eyepieces. The TeleVue Wide Fields are a good example that shows the importance of distortion. For example, I measured the 32 mm AFoV at 65 degrees, the actual specified AFoV.

I measured the field stop at 33.0 mm, the specified field stop.

But using the AFOV = MAG x TFoV, the AFoV would be 59.1 degrees rather than the actual 65 degrees. That's a 10% difference, the result of field distortion.

Some years ago I tried promoting the effective AFoV:

eAFoV  = Mag x TFoV = 57.3 deg/rad x Field stop/Focal length eyepiece. For the 32 mm WF, the eAFoV would be 59.1°

The idea was that if you knew the eAFoV, you could accurately calculate the TFoV using TFoV = eAFoV /Mag

It never caught on and with good reason.

Jon

• gnowellsct likes this

### #17 ECP M42

ECP M42

Ranger 4

• Posts: 318
• Joined: 28 Apr 2021
• Loc: central Europe 45°N

Posted 16 May 2021 - 10:29 AM

Hi Starman, you have listed many thorny points, to which I would like to ask 1000 questions, some of which are uncomfortable. I can only accept your experience on this, but I will try to support my hypothesis anyway, to see if it can hold up.

1 - The apparent field is a set thing determined by eye relief and lens diameters...

2 - In a line of eyepieces advertised to have the same apparent fields, the actual apparent fields will vary.

3 - Apparent field cannot be used to derive the true field of the eyepiece unless the distortion figure is known, and that figure is never published.

4 - Many manufacturers, though, do quote a field diameter for the eyepiece which CAN be used to calculate a true field in a particular scope if the telescope's focal length is known.

5 - Another simple formula will result in a more accurate determination of true field.  The actual apparent field need not be known: TF = EPFS/TFL x 57.296 ...

1 - are you 200% sure of this and for all producers?

I find the method you listed to be absurd, as manufacturers know quite precisely all the distortion, focal, ER, field diaphragm values, and other data that might play out here (I'm talking about manufacturers, not sellers).

Why use a formula that is not precise at all, based on ER and ocular lens diameter, when you have all the data you need to find the exact value adapted to the thousandth and the distortion?

2 - here it is possible that it is the focal length, the data that brings the error (?)

I have no experience with telescope eyepieces, but I know that it is certainly more difficult to assess the precise focal length of an eyepiece, than that of an achromatic doublet, due to the question of the optical thicknesses involved (I have experience).

3 - it is precisely for this reason that the manufacturer is the only one who can calculate AFOV adapted to the distortion.

4 - however we are at the same starting point: if the data of the field diaphragm is wrong or its value is nominal or the astronomer cannot measure it precisely (together with the exact focal length), even this result cannot be realistic.

Which is precisely why the OP opened the topic.

5 - a 5% error, not bad at all, for the amateur astronomer ... or am I wrong?

It is much more complicated to do complex calculations using data, that must be found on a case-by-case basis and many times impossible for most users to find.

FOV = AFOV / Mag   is more easier for everyone, accepting the 5% error ... or not?

Edited by ECP M42, 16 May 2021 - 10:34 AM.

### #18 Starman1

Starman1

Vendor (EyepiecesEtc.com)

• Posts: 50,657
• Joined: 23 Jun 2003
• Loc: Los Angeles

Posted 16 May 2021 - 12:25 PM

Hi Starman, you have listed many thorny points, to which I would like to ask 1000 questions, some of which are uncomfortable. I can only accept your experience on this, but I will try to support my hypothesis anyway, to see if it can hold up.

1 - are you 200% sure of this and for all producers?

I find the method you listed to be absurd, as manufacturers know quite precisely all the distortion, focal, ER, field diaphragm values, and other data that might play out here (I'm talking about manufacturers, not sellers).

Why use a formula that is not precise at all, based on ER and ocular lens diameter, when you have all the data you need to find the exact value adapted to the thousandth and the distortion?

Yes, it's physics.  The apparent field cannot be wider than a limit set by eye lens diameter and eye relief. though it can be narrower.  The apparent field beyond that is set by the refraction inside the eyepiece itself.

Apparent field is designed in but it is usually not exact.  A given range of eyepieces could easily be +/- 2° from the quoted figure.

2 - here it is possible that it is the focal length, the data that brings the error (?)

I have no experience with telescope eyepieces, but I know that it is certainly more difficult to assess the precise focal length of an eyepiece, than that of an achromatic doublet, due to the question of the optical thicknesses involved (I have experience).

Focal lengths rarely differ by more than +/- 0.1mm from the stated focal length.  Some are rounded off (a 4.9mm focal length might be sold as a "5mm", for instance), but this will make only a very minor difference in apparent field.

3 - it is precisely for this reason that the manufacturer is the only one who can calculate AFOV adapted to the distortion.

Calculate, perhaps.  But we can measure it fairly precisely.  Jon showed one method.  Other CNers have shown photographic methods of determining it, and S&T showed another method in a recent article on eyepieces.

4 - however we are at the same starting point: if the data of the field diaphragm is wrong or its value is nominal or the astronomer cannot measure it precisely (together with the exact focal length), even this result cannot be realistic.

Which is precisely why the OP opened the topic.

If no field diameter is known, it can only be approximately calculated.  This may not be any more accurate than the TF = AF/M, though.

One additional problem with TF = AF/M is not only that AF may not be accurate, but that M is dependent on knowing the exact focal length of the scope, which is also in error on most scopes due to the variability there.

5 - a 5% error, not bad at all, for the amateur astronomer ... or am I wrong?

It is much more complicated to do complex calculations using data, that must be found on a case-by-case basis and many times impossible for most users to find.

I have found errors up to 15% when using the TF=AF/M formula.  I used the TeleVue eyepiece to illustrate the large error in a highly-accurate eyepiece of known apparent field and field diameter in a telescope where the focal length is known to a mm or two.

Since AF can vary in a line of eyepieces (one line of Plössls tested many years ago rand from 47-52°), and the focal length of the telescope is rarely known exactly (especially in SCTs and MCTs), but eyepiece field diameter is often published,

and since the field diameter formula bypasses distortion and apparent field figures entirely, it is the more accurate way to determine true field.  And it is not a more complex formula at all.

I also have experience at comparing figures derived from star timing versus the use of both formulas and the TF=EPFS/TFL x 57.296 comes the closest to the exact true field seen.

And imagers can substitute a 1 for EPFS in that formula and get the image scale on the focal plane of any scope, useful for figuring chip size for an intended image.

FOV = AFOV / Mag   is more easier for everyone, accepting the 5% error ... or not?

Perhaps if the error was always 5% or smaller, but that isn't the case, alas.

Edited by Starman1, 16 May 2021 - 12:26 PM.

• Jon Isaacs likes this

### #19 ECP M42

ECP M42

Ranger 4

• Posts: 318
• Joined: 28 Apr 2021
• Loc: central Europe 45°N

Posted 16 May 2021 - 01:47 PM

Thanks for the talk, Starman.

I have found errors up to 15% when using the TF=AF/M formula.

This seems to nullify my assumptions. But you who have direct experience with some producers, would it be possible to ask them the question to hear the answer?

### #20 Jon Isaacs

Jon Isaacs

ISS

• Posts: 92,505
• Joined: 16 Jun 2004
• Loc: San Diego and Boulevard, CA

Posted 16 May 2021 - 01:57 PM

4 - however we are at the same starting point: if the data of the field diaphragm is wrong or its value is nominal or the astronomer cannot measure it precisely (together with the exact focal length), even this result cannot be realistic.

Which is precisely why the OP opened the topic.

The TFoV of a telescope + eyepiece can be measured quite accurately by "drift timing". This involves measuring the time it takes for a star to cross from one edge of the field to the other. Then if the focal length of the telescope is known with reasonable accuracy, the "effective" field stop diameter can be determined.

Recently, there was a question of the effective field stop of the 30mm 70 degree APM UFF. I believe least 4 of us measured it, Don and I by drift timing, David Russel by a FoV measurement.

It took me a while to refine my technique and i eventually used a scope with a precisely known focal length.  My measurements were within less than 1 second for a drift time of about 4 minutes.

If I remember correctly, the range measured by the 4 of us was from 36.2mm to 36.7 mm, about 1% .

Jon

• ECP M42 likes this

### #21 ECP M42

ECP M42

Ranger 4

• Posts: 318
• Joined: 28 Apr 2021
• Loc: central Europe 45°N

Posted 16 May 2021 - 02:04 PM

Hi Jon, thank you for the link.

But here we are talking about finding the observed FOV, not the actual AFOV.

David Knisely has documented a couple of methods, I use his Projection Method.

I finally tried this "lamp method", with some binoculars that I am more than sure of the real AFOV I see.
But the result with the lamp is always greater than 3% of what I see with the eyes inside the binoculars.

So I prefer to keep using the eyes instead of the lamp. But I've tried and if any mistakes are accepted, that might be fine too.

### #22 Jon Isaacs

Jon Isaacs

ISS

• Posts: 92,505
• Joined: 16 Jun 2004
• Loc: San Diego and Boulevard, CA

Posted 16 May 2021 - 02:20 PM

Hi Jon, thank you for the link.

But here we are talking about finding the observed FOV, not the actual AFOV.

I finally tried this "lamp method", with some binoculars that I am more than sure of the real AFOV I see.
But the result with the lamp is always greater than 3% of what I see with the eyes inside the binoculars.

So I prefer to keep using the eyes instead of the lamp. But I've tried and if any mistakes are accepted, that might be fine too.

This thread is about the relationship between the AFoV and the TFoV.

I'm not sure how you can accurately know the real AFoV you see. Binocular manufacturers provide TFoV and magnification and ignore field distortion of the eyepiece.

Jon

### #23 ECP M42

ECP M42

Ranger 4

• Posts: 318
• Joined: 28 Apr 2021
• Loc: central Europe 45°N

Posted 16 May 2021 - 02:46 PM

I'm not sure how you can accurately know the real AFoV you see.

With my eyes, Jon.

A meter and a pencil and eyes (and a scientific calculator), as I explained in Mark's thread.

I also measured some binoculars by removing the eyepieces from the body, but the result is identical when measuring the whole binocular. What you see in the eyepiece is the eyepiece's field diaphragm.

The method is precisely the reverse of that with the lamp. That is, it is the method to precisely measure what you see with your eyes. Is the "eye method"

### #24 Starman1

Starman1

Vendor (EyepiecesEtc.com)

• Posts: 50,657
• Joined: 23 Jun 2003
• Loc: Los Angeles

Posted 16 May 2021 - 03:18 PM

If you use any form of target to measure, you are not measuring the apparent field, you are measuring the true field.

And you cannot derive apparent field from the true field with any accuracy.  You can only derive the field diameter seen.

Binocular specs usually include the width of the field at a set distance.

You can calculate the field width in mm from this.

You cannot accurately calculate the apparent field of view with more than +/- 10% accuracy.

Now, that measurement has a lot of validity since you can derive a field diameter for the eyepiece that would be constant in all telescopes.

You just cannot back into the apparent field that way.

Except for one astronomy hack: at 57.296° of apparent field, the focal length of the eyepiece equals the field diameter.

This is not exact, and there are some exceptions.  But, it leads to another easy formula:

Field diameter = AF/57.296 x FL

The farther the apparent field is from 57.296°, the greater the error in this assumption because field distortion increases rapidly.

But, when using this formula on TeleVue eyepieces, whose field diameters are specified by TeleVue, the differences are in the neighborhood of 0.0 to 0.3mm except for a few whose apparent fields might be off.

Ernest Maratovich's measurements of some Tele Vue eyepieces do show some discrepancies (e.g. 11mm Type 6 Nagler, claimed 82°, actual 78°).

When the measured apparent fields are input, the differences between calculated and actual field diameters slim down.

• Jon Isaacs and rocketsteve like this

Messenger

• Posts: 431
• Joined: 17 Jan 2021
• Loc: Milano (Italy)

Posted 16 May 2021 - 03:39 PM

Why is there such a huge difference in the calculated Real FOV when using Method 1 or 2?

My scope has a focal length of 1650mm. The eyepiece is a Orion LHD 80 degree, 9 mm, field stop diameter 33.4mm.

What is my real True Field of view?

This has developed into a fascinating discussion but following skysurfer's  observation I'd like to tell the OP: if you start entering all the technical rabbit holes as a beginner you might get totally lost

Your first method is true and tested and for practical purposes it's enough for 99.9% of amateur astronomers. I venture to think that Don, Jon and the others who engaged in the discussion would not disagree. Stick to it. You don't need another – not yet

Edited by radiofm74, 16 May 2021 - 03:41 PM.

• BFaucett and ECP M42 like this

## Recent Topics

 Cloudy Nights LLC Cloudy Nights Sponsor: Astronomics