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How to calc max field of view with maximum aperture.

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#1 Neptune

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Posted 18 January 2020 - 01:29 PM

Hi guys, I am new to the ATM area and have a question.

What is the formula for determining the maximum field given the below criteria.

 

Maximum aperture of a telescope = X  

 

F number of scope = Y

 

Exit pupil could go 5mm or maybe 6mm.

 

Eyepiece used 41mm Panoptic (68 afov)

 

Thanks,
David



#2 bridgman

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Posted 18 January 2020 - 01:39 PM

IIRC FOV is just AFOV divided by magnification, or 68 / ((X * Y) / 41), which simplifies to (68 x 41) / (X * Y).

 

Assumes aperture is in mm as well (to match 41mm) or you need to include conversion constants.

 

Exit pupil gets calculated separately - aperture divided by magnification, which IIRC ends up simplifying to eyepiece FL over F number or 41 / Y.

 

Let's check that... aperture / magnification -> X / ((X * Y) / 41) -> 1 / (Y / 41) -> 41 / Y. Yep.

 

For 5-6mm exit pupil with 41mm EP you want F number in the 7-8 range (41/5 = 8.2, 41/6 = 6.83).


Edited by bridgman, 18 January 2020 - 01:45 PM.

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#3 junomike

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Posted 18 January 2020 - 01:47 PM

Seeing as you already know the desired exit pupil and eyepiece you can calculate the F-Ratio as 41mm/5mm exit pupil  = F8.2  & 41mm/6mm exit pupil  = F6.83

Using F7 yields an exit pupil of 5.86mm.

Now your max FOV will be dependent on the Aperture as a 120mm F7 OTA will yield a larger FOV than a 200mm F7 OTA.

 

120 F7 =840mm FL

41 pan = 46mm Field Stop

46/840 X 57.3 = 3.14° FOV

 

200mm F7 = 1400mm FL

41 pan = 46mm Field Stop

46/1400 X 57.3 = 1.88° FOV


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#4 Bob4BVM

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Posted 18 January 2020 - 02:47 PM

Virtually ANY optics, telescope, AP calculation you could ever dream of :

 

 https://www.nexstars...RAC/form.html#1

 

Enjoy.

 

CS

Bob


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#5 jtsenghas

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Posted 18 January 2020 - 04:05 PM

Virtually ANY optics, telescope, AP calculation you could ever dream of :

https://www.nexstars...RAC/form.html#1

Enjoy.

CS
Bob


Thanks, Bob! I've bookmarked that one. There are several formulas there I hadn't had and could use.

#6 Jon Isaacs

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Posted 18 January 2020 - 04:16 PM

Thanks, Bob! I've bookmarked that one. There are several formulas there I hadn't had and could use.

 

There are a number of useful equations.   There are some mistakes. 

 

For example, the max field of view equations use 50.8 mm and 31.7mm rather than the actual maximim possible field stop diameters.

 

The handling of Afov ~ TFoV/field stop is not adequate explained. There is no direct relationship, only approximations.

 

Jon


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#7 mconnelley

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Posted 18 January 2020 - 04:22 PM

Hello:

  

   You need the eyepiece to demagnify the entrance pupil to fit within the ~6 mm pupil of your eye.  This demagnification factor is equal to how much the image is magnified.  So, if you have an 8" (200 mm) telescope, the minimum magnification where you can use the full aperture is 200/6 = 33x.  Note that this has nothing to do with the focal length of the telescope, f/ratio of the telescope, or focal length of the eyepiece.

 

 The true field of view is just apparent field divided by magnification.  If you have a Panoptic with 68 degrees apparent field, then the largest true field you can get with an 8" is 68/33 = 2.1 degrees.  

 

   Now, the telescope's mechanical design may not allow you to get this much true field.  For example, in an 8" SCT the primary mirror baffle tube limits the fully illuminated field to well under an inch.  However, with an 80" focal length, a 2 degree field would be nearly 3" across.  

Cheers

Mike



#8 CeleNoptic

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Posted 18 January 2020 - 04:45 PM

Hello:

  

   You need the eyepiece to demagnify the entrance pupil to fit within the ~6 mm pupil of your eye.  This demagnification factor is equal to how much the image is magnified.  So, if you have an 8" (200 mm) telescope, the minimum magnification where you can use the full aperture is 200/6 = 33x.  Note that this has nothing to do with the focal length of the telescope, f/ratio of the telescope, or focal length of the eyepiece.

 

 The true field of view is just apparent field divided by magnification.  If you have a Panoptic with 68 degrees apparent field, then the largest true field you can get with an 8" is 68/33 = 2.1 degrees.  

 

 

That's approximation, but the precise formula  is based on Eyepiece Field Stop

 

TFOV=(EPFS/TFL) x 57.296

 

where
TFOV = True Field of View
EPFS = Eyepiece Field Stop
TFL = Telescope focal length

 

So, for 8" f/6     the 41mm Panoptic will provide the max TFOV of 2.19* at 29.3x.   (46/1200) x 57.296 = 2.19


Edited by CeleNoptic, 18 January 2020 - 04:53 PM.


#9 Jon Isaacs

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Posted 18 January 2020 - 04:58 PM

Hello:

  

   You need the eyepiece to demagnify the entrance pupil to fit within the ~6 mm pupil of your eye.  This demagnification factor is equal to how much the image is magnified.  So, if you have an 8" (200 mm) telescope, the minimum magnification where you can use the full aperture is 200/6 = 33x.  Note that this has nothing to do with the focal length of the telescope, f/ratio of the telescope, or focal length of the eyepiece.

 

 The true field of view is just apparent field divided by magnification.  If you have a Panoptic with 68 degrees apparent field, then the largest true field you can get with an 8" is 68/33 = 2.1 degrees.  

 

   Now, the telescope's mechanical design may not allow you to get this much true field.  For example, in an 8" SCT the primary mirror baffle tube limits the fully illuminated field to well under an inch.  However, with an 80" focal length, a 2 degree field would be nearly 3" across.  

Cheers

Mike

Mike et.al:

 

In this case, the eyepiece is specified so this means the focal ratio is specified if the exit pupil is specified and thus for a given aperture the focal length is known. The 41mm Panoptic has a 46 mm field stop.

 

The focal ratio:

 

Fr = FLeye / Exit Pupil  For 6 mm:  41mm /6 mm = F/6.83

 

The TFoV: 

 

TFoV = 180°/Pi x field stop / Fr x Aperture = (57.3 deg/radian x 46 mm)/(6.83 x Aperture) 

 

This is a more exact equation than TFoV = Afov/ Mag because of field distortion.  This equation is independent of magnification and eyepiece focal length. It just looks at the diameter of the image at the focal plane. 

 

For an 200mm  scope the focal length is 1366 mm. (6.83 x 200mm)

 

The TFoV = 57.3 x 46mm/1366mm = 1.93°

 

The 41 mm Panoptic has a 68° AfoV but there's distortion.

 

2.1° (actually 2.04°) versus 1.93° is a small but significant error of almost 6%.

 

One can measure AFoV directly as well as the TFoV or field stop but the relationship between the two is approximate.

 

AFOV Measuring .jpg

Jon



#10 BGRE

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Posted 18 January 2020 - 05:09 PM

That formula ( TFOV=(EPFS/TFL) x 57.296) is itself an approximation which is only exact if the distortion produced by the objective is such that there is a linear relationship between lateral position of the image and field angle. For most objectives there is a linear relationship between the lateral position of the image and the tangent of the field angle.

 

Also when using the approximation 1"=25mm one shouldn't quote the result to more than 2 significant figures.

 

Using 1"=25.4mm and the correct formula

 

TFOV = 2*ARCTAN(EPPFS/(2*TFL))

 

TFOV = 2.16 degrees. 

 

Eyepiece distortion is irrelevant if the entrance pupil diameter is determined by a physical stop that has no glass between it and the telesccope objective (e.g, Panoptic eyepiece) 


Edited by BGRE, 18 January 2020 - 05:17 PM.

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#11 Vic Menard

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Posted 18 January 2020 - 07:01 PM

Hi guys, I am new to the ATM area and have a question.

What is the formula for determining the maximum field given the below criteria.

I'm not sure if this is where you're heading--still, fascinating out-of-the box reading!

 

http://www.bbastrodesigns.com/rft.html

 

and maybe this one too:

 

http://www.bbastrode...ertureCalc.html


Edited by Vic Menard, 18 January 2020 - 07:14 PM.


#12 Neptune

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Posted 18 January 2020 - 07:58 PM

Thanks guys for the incredible info you all have provided.

 

I might have added to my initial criteria a field of view at the eyepiece of 1.2 to 1.3 degrees would be sufficient and

 

I would like a 16" dia. scope that provides an exit pupil of about 6mm using a 41mm Panoptic.  I don't think it is possible to accomplish this given any f ratio mirror.


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#13 Jon Isaacs

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Posted 18 January 2020 - 08:04 PM

That formula ( TFOV=(EPFS/TFL) x 57.296) is itself an approximation which is only exact if the distortion produced by the objective is such that there is a linear relationship between lateral position of the image and field angle. For most objectives there is a linear relationship between the lateral position of the image and the tangent of the field angle.

 

Also when using the approximation 1"=25mm one shouldn't quote the result to more than 2 significant figures.

 

Using 1"=25.4mm and the correct formula

 

TFOV = 2*ARCTAN(EPPFS/(2*TFL))

 

TFOV = 2.16 degrees. 

 

Eyepiece distortion is irrelevant if the entrance pupil diameter is determined by a physical stop that has no glass between it and the telesccope objective (e.g, Panoptic eyepiece) 

 

Whose post are you replying To?

 

What is the focal length of the scope you are using?  It looks to me like you're computing the field for the 8 inch F/6. This provides a 6.83 mm exit pupil, the OP was asking about 5mm and 6 mm exit pupils.

.

These are small angles, the small angle approximation is plenty accurate.

 

The half tangent provides 2.1615°  

 

The small angle approximation provides 2.1619°, that's an error of 0.015% 

 

For that to be significant, the field stop would need to be accurate to 8 microns on the field stop.. 

 

Jon



#14 Neptune

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Posted 18 January 2020 - 08:16 PM

Playing with the numbers, I see that my 16" scope cannot provide the criteria I am looking for.  If I reduce the scope to a 13" f6.75 it is very close to  do what I was thinking.  Hate to reduce the aperture though.

I am beginning to see the relationships between the diameter of the objective and the f number of said objective.



#15 Jon Isaacs

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Posted 18 January 2020 - 08:19 PM

Thanks guys for the incredible info you all have provided.

 

I might have added to my initial criteria a field of view at the eyepiece of 1.2 to 1.3 degrees would be sufficient and

 

I would like a 16" dia. scope that provides an exit pupil of about 6mm using a 41mm Panoptic.  I don't think it is possible to accomplish this given any f ratio mirror.

 

You could manage that with the 21 mm Ethos but you'd be at about F3.0, F/3.5 with a Paracorr. That would provide a 1.46° field.

 

The 31 mm Nagler in a 16 inch F/4.4 with a Paracorr provides 1.17° at about 66x with a 6.1 mm exit pupil. With the 21 Ethos, you're at 98x with a 1.00° field, I use that a lot.

 

That's where I'm at. I'm 6 ft tall, no ladder or stool needed.

 

Jon



#16 Neptune

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Posted 18 January 2020 - 08:38 PM

You could manage that with the 21 mm Ethos but you'd be at about F3.0, F/3.5 with a Paracorr. That would provide a 1.46° field.

 

The 31 mm Nagler in a 16 inch F/4.4 with a Paracorr provides 1.17° at about 66x with a 6.1 mm exit pupil. With the 21 Ethos, you're at 98x with a 1.00° field, I use that a lot.

 

That's where I'm at. I'm 6 ft tall, no ladder or stool needed.

 

Jon

 

Jon,  all very good points. As luck would have it I currently have on hand a 41mm Pan, 31 Nagler and 21 Ethos.  It really didn't occur to me that I could use an eyepiece other than the max field (in a 2" eyepiece) 41mm Panoptic.



#17 BGRE

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Posted 18 January 2020 - 08:51 PM

Whose post are you replying To?

 

What is the focal length of the scope you are using?  It looks to me like you're computing the field for the 8 inch F/6. This provides a 6.83 mm exit pupil, the OP was asking about 5mm and 6 mm exit pupils.

.

These are small angles, the small angle approximation is plenty accurate.

 

The half tangent provides 2.1615°  

 

The small angle approximation provides 2.1619°, that's an error of 0.015% 

 

For that to be significant, the field stop would need to be accurate to 8 microns on the field stop.. 

 

Jon

 

The false claim that the formula is exact is problematic.

Its bad practice to claim a formula to be exact when it isn't .

This only leads the inexperienced to apply it to situations where it is wildly inaccurate. (there are many examples on various websites where the author has assumed that a formula is exact when it isn't and even done a series expansion which is of little or no value because the formula from which the series was derived is inaccurate). 

The assumptions made should always be clearly stated so that an approximate formula can be used appropriately.

e.g. for half field angles of less than 0.1 radians (5.7 degrees) the error is less than 0.33% and increases approximately in proportion to the square of the semi field angle.

 

The actual difference in the example is approximately 0.012% not 0.015%.

 

The error that results from assuming 1"= 25mm is more significant.

 

Care should be taken when the field stop is internal such as for eyepieces employing a Smyth lens in front of the field stop.

Similarly where the exit pupil has significant SA, formulae using the exit pupil diameter can be wildly inaccurate.



#18 Neptune

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Posted 18 January 2020 - 09:24 PM

Thank you Jon for steering me in the right direction.  I like the idea now of a 16" scope with a paracorr making the final f ratio about 3.7 to 3.8

 

What f ratio mirror would be needed before the paracorr?



#19 Jon Isaacs

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Posted 19 January 2020 - 05:39 AM

The false claim that the formula is exact is problematic.

Its bad practice to claim a formula to be exact when it isn't .

This only leads the inexperienced to apply it to situations where it is wildly inaccurate. (there are many examples on various websites where the author has assumed that a formula is exact when it isn't and even done a series expansion which is of little or no value because the formula from which the series was derived is inaccurate). 

The assumptions made should always be clearly stated so that an approximate formula can be used appropriately.

e.g. for half field angles of less than 0.1 radians (5.7 degrees) the error is less than 0.33% and increases approximately in proportion to the square of the semi field angle.

 

The actual difference in the example is approximately 0.012% not 0.015%.

 

The error that results from assuming 1"= 25mm is more significant.

 

Care should be taken when the field stop is internal such as for eyepieces employing a Smyth lens in front of the field stop.

Similarly where the exit pupil has significant SA, formulae using the exit pupil diameter can be wildly inaccurate.

 

Show me a telescope with a field of view of 11.4°.

 

The best equation is the simplest that is not too simple.  This thread it turns out we're looking at a 16 inch with a 6 mm exit pupil. We're solving that problem. 

 

I solved it and added some practical experience.. The 31 mm Nagler is the eyepiece to be thinking about. The effective field stop is 42 mm.

 

Jon



#20 BGRE

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Posted 19 January 2020 - 05:52 AM

A  25mm aperture 5x finder with wide angle objective and 82 degree AFOV eyepiece with wide angle objective comes to mind.

 

Formulae are little use without an indication of their limitations and/or any assumptions made.

 

Practical experience alone is not always sufficient.

 

A false claim was made about an equation being exact.


Edited by BGRE, 19 January 2020 - 06:00 AM.


#21 Jon Isaacs

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Posted 19 January 2020 - 06:14 AM

A  25mm aperture 5x finder with wide angle objective and 82 degree AFOV eyepiece with wide angle objective comes to mind.

 

Formulae are little use without an indication of their limitations and/or any assumptions made.

 

Practical experience alone is not always sufficient.

 

A false claim was made about an equation being exact.

 

Did some one say there was an exact solution?  I certainly didn't.  This is the closest thing I can find:

 

"TFoV = 180°/Pi x field stop / Fr x Aperture = (57.3 deg/radian x 46 mm)/(6.83 x Aperture)

 

This is a more exact equation than TFoV = Afov/ Mag because of field distortion.  This equation is independent of magnification and eyepiece focal length. It just looks at the diameter of the image at the focal plane."

 

I said it was a more exact equation which it certainly is.  

 

So ask yourself, is this helping David understand how to resolve his problem.  I am not writing a text book, I am trying to help David understand how to build a 16 inch Dob with a 1.2 degree TFoV with a 6mm exit pupil.  

 

Jon


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