Low Power Wide Angle Inexpensive Eyepiece for my Mak? Ha! I should be so lucky!

[Mr Spock]

It may be you can fit a 2" visual back but that aperture is going to restrict the field of view. It's better to stick with 1.25" I think.

[Splodger]

55 minutes ago, Mr Spock said:

It may be you can fit a 2" visual back but that aperture is going to restrict the field of view. It's better to stick with 1.25" I think.

I'm with you. The Meade eyepiece I have is about as wide field as I can hope to get without spending a lot of money. At 40mm with 52° it's not at all bad.

[John]

52 minutes ago, Splodger said:

I'm with you. The Meade eyepiece I have is about as wide field as I can hope to get without spending a lot of money. At 40mm with 52° it's not at all bad.

In the 1.25 inch fitting, the maximum apparent field of view you can get with a 40mm focal length eyepiece is 43 degrees. That is limited by the internal diameter of the 1.25" barrel.

 

[Louis D]

2 hours ago, Splodger said:

I think I would be getting to the point of diminishing returns if I were to go down the route of converting it.

If I didn't already have a bunch of 2" eyepieces and 2" diagonals, I probably wouldn't have bothered with them on my 127mm Maks, but since I did, it was low risk to shell out $50 for a 2" SCT visual back and the Mak to SCT thread step ring.  Luckily, it worked out for me.

If I were in your situation, I'd get an 80mm to 100mm ED refractor to take in the wide true fields you're after.

[Splodger]

4 minutes ago, John said:

In the 1.25 inch fitting, the maximum apparent field of view you can get with a 40mm focal length eyepiece is 43 degrees. That is limited by the internal diameter of the 1.25" barrel.

 

Interesting. I had a hunt around and found that about half of the sites selling, or were selling as they are all sold out, the eyepiece have it at 52° and the other half, including meade.com, at 44°. Perhaps I bought it a telescope.com https://tinyurl.com/meade-40mm who are marketing it at 52°. Anyway Ive been bamboozled one way or another.

[John]

50 minutes ago, Splodger said:

Interesting. I had a hunt around and found that about half of the sites selling, or were selling as they are all sold out, the eyepiece have it at 52° and the other half, including meade.com, at 44°. Perhaps I bought it a telescope.com https://tinyurl.com/meade-40mm who are marketing it at 52°. Anyway Ive been bamboozled one way or another.

If you click on the "Specs" tab for that particular eyepiece it does say that the apparent field of view (AFoV) is 44 degrees. It is rather confusing that they mention 52 degrees in the "More Info" tab but I think that is a generic description of the Meade 4000 plossl range. It is only the 40mm focal length of the 1.25 inch 4000 series eyepieces that has that smaller AFoV.

You are not the first to be confused by this description:

https://www.cloudynights.com/topic/462043-meade-4000-plössl-40mm-does-it-really-have-52fov/

 

 

 

Edited January 3, 2022 by John

[Louis D]

I suppose it's entirely possible for a 40mm 1.25" eyepiece to have a 52° apparent field of view (AFOV) if it has enough edge distortion to stretch the outer regions far enough to fill the extra 8°.  After all, the 25mm ES-100 and 26mm (really 25mm) Meade MWA have roughly the same true field of view (TFOV), but the ES shows it in a ~100° AFOV while the Meade shows it in a ~82° AFOV.  That's a massive difference in edge distortion for the same TFOV.

However, I don't think the 40mm Meade Plossl in question has that amount of edge distortion.

[Zermelo]

Mr. Pensack's spreadsheet lists the Meade 40mm as having 40° AFOV.  

[John]

1 hour ago, Zermelo said:

Mr. Pensack's spreadsheet lists the Meade 40mm as having 40° AFOV.  

That may well be the true value. Manufacturer specs have been known to be a little on the optimistic side from time to time 

 

[Don Pensack]

22 hours ago, Splodger said:

Interesting. I had a hunt around and found that about half of the sites selling, or were selling as they are all sold out, the eyepiece have it at 52° and the other half, including meade.com, at 44°. Perhaps I bought it a telescope.com https://tinyurl.com/meade-40mm who are marketing it at 52°. Anyway Ive been bamboozled one way or another.

If they quote 52°, it is because they merely copied the information from a shorter focal length.  The maximums are 43° on a 40mm, 49.5° on a 32mm, so if you see figures quoted that are larger, they are advertising figures, not actual.

[Don Pensack]

On 04/01/2022 at 05:38, Louis D said:

I suppose it's entirely possible for a 40mm 1.25" eyepiece to have a 52° apparent field of view (AFOV) if it has enough edge distortion to stretch the outer regions far enough to fill the extra 8°.  After all, the 25mm ES-100 and 26mm (really 25mm) Meade MWA have roughly the same true field of view (TFOV), but the ES shows it in a ~100° AFOV while the Meade shows it in a ~82° AFOV.  That's a massive difference in edge distortion for the same TFOV.

However, I don't think the 40mm Meade Plossl in question has that amount of edge distortion.

The maximum without rectilinear distortion is 37.8° with a 27.4mm field stop.

4% distortion yields 39.3°

6% distortion yields 40.0° (the average 40mm Plössl)

10% distortion (very large for this type of eyepiece) yields 41.6°.

You can see how unlikely even 43° is.

Edited January 5, 2022 by Don Pensack

[vlaiv]

9 minutes ago, Don Pensack said:

The maximum without distortion is 37.8° with a 27.4mm field stop.

What sort of distortion?

They are both distortions of some sort. You can't map sphere onto flat plane without some type of distortion.

AMD vs Rectilinear distortion.

Two edge cases are:

y = f * tan(angle)

and

y = f * angle

where y is distance from center of the field to field stop and f is focal length of EP. If we take 27.4mm, half of that is 13.7 so we have:

13.7/40 = tan(angle) => angle = ~18.9 or AFOV = 37.8° for zero rectilinear distortion

13.7/40 = angle = ~19.6238 or AFOV of ~39.25° for zero AMD

 

 

[Don Pensack]

18 hours ago, vlaiv said:

What sort of distortion?

They are both distortions of some sort. You can't map sphere onto flat plane without some type of distortion.

AMD vs Rectilinear distortion.

Two edge cases are:

y = f * tan(angle)

and

y = f * angle

where y is distance from center of the field to field stop and f is focal length of EP. If we take 27.4mm, half of that is 13.7 so we have:

13.7/40 = tan(angle) => angle = ~18.9 or AFOV = 37.8° for zero rectilinear distortion

13.7/40 = angle = ~19.6238 or AFOV of ~39.25° for zero AMD

 

 

You're right.  I was referring to rectilinear distortion.  I should have specified.  I edited.

Edited January 5, 2022 by Don Pensack

[vlaiv]

8 minutes ago, Don Pensack said:

You're right.  I was referring to rectilinear distortion.  I should have specified.  I edited.

You mention percent of distortion. Can you elaborate a bit on that? I'm not quite familiar of what that could mean.

I understand two edge cases - AMD and rectilinear. Those two can easily be explained with a bit of geometry.

Zero rectilinear is first image - each angle is mapped to its "projection" on field stop. Same angles near the center are smaller on focal plane than angles near the edge (see segments on straight line in first image).

Second image is zero AMD. Each angle is mapped onto equal part on focal plane and focal plane is actually made up of joined segments along the arc (drawn in the image). This keeps all angles the same (zero angular magnification) - but bends straight lines.

When you say 10% distortion that gives 41.6° while max that I can imagine is case of zero AMD - and it is ~39.25° - what the mapping looks like in that case?

[Splodger]

On 04/01/2022 at 13:08, Don Pensack said:

6% distortion yields 40.0° (the average 40mm Plössl)

A little disappointing I must say. But at least 40° on my scope encompasses my major point of focus

[Don Pensack]

On 05/01/2022 at 08:28, vlaiv said:

You mention percent of distortion. Can you elaborate a bit on that? I'm not quite familiar of what that could mean.

I understand two edge cases - AMD and rectilinear. Those two can easily be explained with a bit of geometry.

Zero rectilinear is first image - each angle is mapped to its "projection" on field stop. Same angles near the center are smaller on focal plane than angles near the edge (see segments on straight line in first image).

Second image is zero AMD. Each angle is mapped onto equal part on focal plane and focal plane is actually made up of joined segments along the arc (drawn in the image). This keeps all angles the same (zero angular magnification) - but bends straight lines.

When you say 10% distortion that gives 41.6° while max that I can imagine is case of zero AMD - and it is ~39.25° - what the mapping looks like in that case?

Positive rectilinear distortion stretched the field radially.  It allows a certain field stop to hold a larger apparent field than its size indicates.

As an example, the TeleVue 24mm Panoptic has a 27mm field stop and a 68° apparent field.

With zero RD, the apparent field would be 58.7°  With a 15.8% distortion, the field becomes 68°.  Observers have noted for years that the design has a lot of RD and many observers remark that it is quite noticeable.

In contrast, take the 24mm APM Ultra Flat field, which has a 27.3mm field stop (from timing) and a 63° apparent field (from the flashlight test).

The zero RD figure for that focal length and field stop would be 59.3°.  A 6.3% radial distortion gets to the 63° measured.  

Both eyepieces have pincushion distortion, but at very different levels.

Tests of visibility for distortion in a moving field shows that roughly a 7% RD is seen by the eye as distortionless in a moving field, which is why many reviewers of the 24mm APM point to seeing almost no distortion.

Alas, the eyepiece is not completely without lateral astigmatism.

 

With zero RD, any widefield eyepiece will have significant AMD, and this leads to "globe" or "rolling ball" distortion when panning a field.

In practice, in the hundreds of pairs of binoculars I've looked through, no designer opts for zero RD (straight lines always seem to curve very slightly near the edge), but RD is minimized while AMD is usually the predominant issue.

 

[vlaiv]

9 hours ago, Don Pensack said:

With zero RD, the apparent field would be 58.7°  With a 15.8% distortion, the field becomes 68°.  Observers have noted for years that the design has a lot of RD and many observers remark that it is quite noticeable.

How can we be sure that all additional AFOV is due to distortion and not perhaps change in focal length?

[Louis D]

4 hours ago, vlaiv said:

How can we be sure that all additional AFOV is due to distortion and not perhaps change in focal length?

As far as I'm concerned, they're the same thing.  If you increase image scale to stretch the image over a larger area, then you must have locally decreased the eyepiece focal length.  Vice versa, if you decrease image scale to squeeze more image into the same or a smaller area, then you must have locally increased the eyepiece focal length.  Is there another explanation as to how image scale can be changed in the apparent field of view without changing the local magnification power of the eyepiece across the field of view?

[Don Pensack]

6 hours ago, vlaiv said:

How can we be sure that all additional AFOV is due to distortion and not perhaps change in focal length?

Normally, I would ask the same question.

But TeleVue's stated design philosophy is to reduce angular magnification distortion to as close to zero as possible, leaving RD wherever it falls.

So in this particular case, the extra field is due to pincushion distortion, not a change in focal length/magnification.

[Don Pensack]

1 hour ago, Louis D said:

As far as I'm concerned, they're the same thing.  If you increase image scale to stretch the image over a larger area, then you must have locally decreased the eyepiece focal length.  Vice versa, if you decrease image scale to squeeze more image into the same or a smaller area, then you must have locally increased the eyepiece focal length.  Is there another explanation as to how image scale can be changed in the apparent field of view without changing the local magnification power of the eyepiece across the field of view?

Yes.  Magnification can stay the same and the object be distorted in shape.  

Take the example of a double star, which, with AMD, changes its apparent separation as the star nears the edge of the field.

With RD, it does not, though the position angle may change.

[Mr Spock]

I made a post about these issues earlier.

https://stargazerslounge.com/topic/378099-a-few-eyepiece-tests/

You can see from the 42mm LVW it's field of view is 72°, but due to distortion it only covers 65° actual. Overall it's not actually 42mm, it's only that in the centre.

[John]

2 minutes ago, Mr Spock said:

I made a post about these issues earlier.

https://stargazerslounge.com/topic/378099-a-few-eyepiece-tests/

You can see from the 42mm LVW it's field of view is 72°, but due to distortion it only covers 65° actual. Overall it's not actually 42mm, it's only that in the centre.

Doesn't the 42mm LVW have 65 degrees marked on the barrel ?

 

[vlaiv]

1 hour ago, Louis D said:

As far as I'm concerned, they're the same thing.

I don't think they quite are.

Imagine following scenario, you have two eyepieces.

You look thru one eyepiece with your left eye and thru other eyepiece with your right eye and let your brain try to merge image.

Image consists out of full moon in the center of the field and surrounding sky up to field stop (sky is bright enough against field stop so that you can easily distinguish it in each eye).

If focal lengths are the same - then moon will overlap perfectly from both eyes - there will be no double image of it.

If AFOV is the same - then field stops will align perfectly - there will be no double image.

You can have the same moon image and different surrounding sky images and you can have different moon sizes for same sky/field stop image. There are third and fourth cases - when both are the same and both are different.

This shows that AFOV and magnification are not effectively the same thing.

(you don't need the moon to compare AFOVs of two different eyepieces - you just need blank well lit wall - hold two eyepieces - each against one eye and let brain try to merge the image - if you have trouble and can't align field stops - that means AFOVs are different - and in fact, you can judge which one is larger by favoring each eye in turn).

30 minutes ago, Don Pensack said:

Normally, I would ask the same question.

But TeleVue's stated design philosophy is to reduce angular magnification distortion to as close to zero as possible, leaving RD wherever it falls.

So in this particular case, the extra field is due to pincushion distortion, not a change in focal length/magnification.

This is the reason I asked in the first place about percent of distortion. Televue has same formulae about relationship of AFOV, field stop and focal length on one of their pages:

https://www.televue.com/engine/TV3b_page.asp?id=113

Now, as you mentioned:

16 hours ago, Don Pensack said:

TeleVue 24mm Panoptic has a 27mm field stop and a 68° apparent field.

If that eyepiece has zero AMD, then we need to use

and if we use that formula on 24mm FL and 27mm field stop we get:

beta = 27 / 24 = 1.125 radians = 64.46°

But if we do something else - and assume that AFOV is indeed ~68° then we can do following:

68° = 27 / actual_fl => actual_fl = 27mm / 68° = ~22.75

So maybe Panoptic 24mm is actually 22.75mm FL eyepiece.

Or maybe there is some middle ground and FL is something like 23.4mm and AFOV is from there 66° and for marketing purposes it is declared as 24mm 68° to be in line with rest of Panoptic EPs.

In the end - maybe field stop is not precisely 27mm but a bit more?

 

[Mr Spock]

16 minutes ago, John said:

Doesn't the 42mm LVW have 65 degrees marked on the barrel ?

No, 72° 

It is also physically measured at 72°, it's just not 42mm... See the images for comparison - all taken with the same iPhone. You see how much larger the fov is over the 22mm 65° LVW.

[vlaiv]

2 minutes ago, Mr Spock said:

No, 72° 

At least they no longer sell it as 72° EP - it's listed as 65°: