How to calculate EP's apparent (angular) FOV and true FOV from EP's field stop

[Bigzmey]

Why use an inaccurate formula if you have an accurate one: TF = (field stop / telescope focal length) x 180 / pi or (FS/TFL) x 57.2958
That formula doesn't even require knowledge of apparent field or distortion, yet yields an accurate number for field stop and true field.

You agree that TFV = (FS/TFL) x 57.2958 is an accurate formula, right?

Now, angular FOV of EP = TFV x magnification. Magnification = TFL/EFL. We substitute TFV with (FS/TFL) x 57.2958.

Angular FOV = (FS/TFL) x 57.2958 x TFL/EFL = FS/EFL x 57.2958.

[notFritzArgelander]

Pardon the intrusion, but....

Why use an inaccurate formula if you have an accurate one: TF = (field stop / telescope focal length) x 180 / pi or (FS/TFL) x 57.2958
That formula doesn't even require knowledge of apparent field or distortion, yet yields an accurate number for field stop and true field.

You agree that TFV = (FS/TFL) x 57.2958 is an accurate formula, right?

Now, angular FOV of EP = TFV x magnification. Magnification = TFL/EFL. We substitute TFV with (FS/TFL) x 57.2958.

Angular FOV = (FS/TFL) x 57.2958 x TFL/EFL = FS/EFL x 57.2958.

The accuracy of the formula is limited to orthoscopic eyepieces with a little "o".

[Bigzmey]

Pardon the intrusion, but....

Why use an inaccurate formula if you have an accurate one: TF = (field stop / telescope focal length) x 180 / pi or (FS/TFL) x 57.2958
That formula doesn't even require knowledge of apparent field or distortion, yet yields an accurate number for field stop and true field.

You agree that TFV = (FS/TFL) x 57.2958 is an accurate formula, right?

Now, angular FOV of EP = TFV x magnification. Magnification = TFL/EFL. We substitute TFV with (FS/TFL) x 57.2958.

Angular FOV = (FS/TFL) x 57.2958 x TFL/EFL = FS/EFL x 57.2958.

The accuracy of the formula is limited to orthoscopic eyepieces with a little "o".

Which of the two: TFV = (FS/TFL) x 57.2958 or angular FOV of EP = TFV x magnification?

[notFritzArgelander]

Pardon the intrusion, but....

You agree that TFV = (FS/TFL) x 57.2958 is an accurate formula, right?

Now, angular FOV of EP = TFV x magnification. Magnification = TFL/EFL. We substitute TFV with (FS/TFL) x 57.2958.

Angular FOV = (FS/TFL) x 57.2958 x TFL/EFL = FS/EFL x 57.2958.

The accuracy of the formula is limited to orthoscopic eyepieces with a little "o".

Which of the two: TFV = (FS/TFL) x 57.2958 or angular FOV of EP = TFV x magnification?

They’re of equal validity. Small “o” orthoscopy is required.

[SkyHiker]

The flashlight test:
with illustrations:
https://www.cloudynights.com/topic/5744 ... ?p=7958975
https://www.cloudynights.com/topic/5744 ... ?p=7959408
Glasses/no glasses: doesn't matter as long as your tape measure is accurate and you know a tiny bit of trigonometry.

From that discussion I can't tell how the test works. The text in the post with the diagram says: "The distance from the eye point (the eye relief point, the narrowest point in the beam) to the surface and the diameter of the beam are measured and using trig, the AFoV can be calculated. ".

That suggests that it will give you the AFOV only for that particular "eye point". In other words, the value of y and the AFOV is a function of how far away the "eye point" is from the lens, isn't it? I could be wrong but without math that shows how a unique value can be obtained for arbitrary setups with different "eye point" locations, for all I know the outcome is arbitrary. The "eye point" is essentially an infinitesimally small light source. Put it in front of a lens, move it and you will see the size of the light cone change.

[Don Pensack]

The flashlight test:
with illustrations:
https://www.cloudynights.com/topic/5744 ... ?p=7958975
https://www.cloudynights.com/topic/5744 ... ?p=7959408
Glasses/no glasses: doesn't matter as long as your tape measure is accurate and you know a tiny bit of trigonometry.

From that discussion I can't tell how the test works. The text in the post with the diagram says: "The distance from the eye point (the eye relief point, the narrowest point in the beam) to the surface and the diameter of the beam are measured and using trig, the AFoV can be calculated. ".

That suggests that it will give you the AFOV only for that particular "eye point". In other words, the value of y and the AFOV is a function of how far away the "eye point" is from the lens, isn't it? I could be wrong but without math that shows how a unique value can be obtained for arbitrary setups with different "eye point" locations, for all I know the outcome is arbitrary. The "eye point" is essentially an infinitesimally small light source. Put it in front of a lens, move it and you will see the size of the light cone change.

OK, a bright light enters the telescope, passes through the eyepiece and exits, narrowing to the exit pupil, then expands past the exit pupil.
The cones have equal angles--in other words, the apparent field angle from exit pupil to eyepiece is exactly the same angle on the outside of the exit pupil.
The light cone continues to expand until it hits a wall.
The distance from the exit pupil to the wall is measured, the width of the light circle on the wall is measured, and simple trigonometry calculates the angle at the exit pupil, which equals the apparent field.
It's just that simple.
I even have made it simpler by stopping my refractor down so the exit pupil is smaller and the point where the exit pupil is is more easily identified because in the unstopped-down telescope, the exit pupil can be large and hard to identify as an exact point.

You have to remember, the exit pupil is that point behind the eyepiece where the field edge first becomes visible as you near the eyepiece. It is the narrowest point of light behind the eyepiece, and it is from this point the apparent field is determined in an eyepiece.
That would be difficult to measure without a special instrument to do so, so the flashlight test provides a way to measure the apparent field at home with no special tools needed.

[SkyHiker]

The distance from the exit pupil to the wall is measured, the width of the light circle on the wall is measured, and simple trigonometry calculates the angle at the exit pupil, which equals the apparent field.
It's just that simple.

That would be the least of my problems, lol.

I even have made it simpler by stopping my refractor down so the exit pupil is smaller and the point where the exit pupil is is more easily identified because in the unstopped-down telescope, the exit pupil can be large and hard to identify as an exact point.

You have to remember, the exit pupil is that point behind the eyepiece where the field edge first becomes visible as you near the eyepiece. It is the narrowest point of light behind the eyepiece, and it is from this point the apparent field is determined in an eyepiece.
That would be difficult to measure without a special instrument to do so, so the flashlight test provides a way to measure the apparent field at home with no special tools needed.

I did the test myself with an ES 80 Apo. What I did not get from the diagram was if you had to create a light source at a focal point before the eyepiece. That was my mistaken impression. After setting up my Howie Glatter with Tublug (it has a lens) as a remote light source it became pretty clear that that was the wrong idea. Then I put my hiking headlamp right in front of the objective, and sure enough that worked like a charm. It's like a flood light that makes the field stop disc function as its own light source. That way the projection on the ceiling does not depend on anything else, just the eyepiece itself.

To check if the FOV changes in function of wearing eyeglasses or not I went outside with my ED80 on the Nano EQ3. No such effect at all. Basically, for a sharp image the focuser puts the focal plane at the field stop and that's what you are observing. Without extreme eyeball gymnastics that focal plane cannot move much so the FOV is constant, eyeglasses or not AFAICT. So yea the flashlight test should be pretty accurate.

[Don Pensack]

A little point about the exit pupil and the image in the telescope:
Eyepieces are afocal. The image is in focus at all distances from the eyepiece, but as you move away from the eyepiece, though the image stays in focus, the apparent field shrinks because the aluminum edge of the eye lens progressively cuts off more and more of the view.
The pupil of the eye sees a smaller and smaller portion of the expanding cone of light behind the eyepiece.
The eye takes those rays and focuses them on the retina, so the image is in focus at all distances from the eyepiece as long as the image was in focus to begin with.
Think of a large flat mirror reflecting the stars. It doesn't focus the stars at all, yet we see them in focus because our eyes focus the image.
So if eyepieces are afocal, what is the exit pupil?
It is an image of the primary mirror, magnified. It is the point where the image from the primary focuses down to its smallest point. It just so happens that the primary mirror is reflecting an image of the sky, like the flat mirror, except it is reflecting a large image
and bringing it down to a smaller size.
This eyepiece duality reminds me of the photon as particle and wave duality. Eyepieces are afocal, but we need to be at a set distance from the eyepiece to see the whole field of view in the eyepiece.