11 replies to this topic

[eblanken]

Hello All,

 

I am seeking a better understanding of the method to calculate the real image that is formed in my Refractor: I have a Skywatcher 100mm x 900mm, f/9 ED refractor and want to know what the actual image size is in real units. More specifically, I want to know what the image size is of the Full Moon (Luna) or the Sun (Sol) when the Refractor is aimed and focused on either of these two objects at the curved image surface (and please approximate that to a simple, flat image, if that is easier and assuming 0.5 Deg for the disk is fine).

 

Best,

 

Ed

 

P.S. See my signature and use Math as necessary, but a simple equation would be fine, if there is a handy approximation is available.

Edited by eblanken, 06 December 2024 - 11:04 PM.

[triplemon]

In short - the image size h with a focal length of fl is

 

h = 2 * fl * tan(angular size / 2)

 

 

If you don't have an scientific calculator, you can aproximate this for most cases involving small angles (less than a few degrees) by

 

   = fl * 2 * Pi  / 360 degrees * angular size in degrees

   = fl * 0.0175 * size in degrees

   = fl / 57.3     * size in degrees

 

Makes 7.8mm in your case.

 

Extra points if you know the trivia on 57.3 (mandatory for astro nerd status):

 

Televue's "Radian" eyepieces have a FOV of (approximately) 57.3 degress. Yep, the same 57.3 as in the above formula. Its all about circles, angles, degrees and Pi. This mathematical constant is called one radian ... and Al Nagler thought its cool to pick just that for his FOV and then name the whole series after it.

Edited by triplemon, 07 December 2024 - 08:01 PM.

[eblanken]

Thank you neighbor !!!

 

In short - the image size h with a focal length of fl is

 

h = fl * tan(angular size)

 

If you don't have an scientific calculator, you can aproximate the tan() for most cases involving small angles (less than a few degrees) like this by
 

   = fl * 2 * pi  / 360 degrees * angular size in degrees

   = fl * 0.0175 * size in degrees

   = fl / 57.3     * size in degrees

 

Makes 7.8mm in your case {THE MOON'S DIAMETER, RIGHT ???}.

 

Extra points if you recognize the 57.3 {Degrees per Radian}.

Televue's "Radian" eyepieces {I HAVE 6mm, 8mm, 10mm,12mm, 14mm, 16mm} have a FOV of 57.3 degress. Yep - the name of that eyepiece series and the FOV is a direct reference to the same 57.3 in the above formula. Its all about circles, angles, degress and PI. This mathmatical constant is called one radian ...

 

{I EDITED TO ADD}

 

Very Best Regards,

 

Ed

 

P.S. We should connect for coffee or a meal sometime !!! P.M. Me please !!!

 

P.P.S. TOMDEY's F * 9/1000 gets me 8.1 mm for Moon's diameter in my 900mm OTA, so good agreement with my neighbor (aka triplemon) . . .  

Edited by eblanken, 07 December 2024 - 02:15 AM.

[TOMDEY]

Hi Ed!

 

All three of these three approximations give the same answer to 4 significant figures, presuming that the sun or moon is 30 arc-min across. And that's really good enough... because field curvature and distortion of your particular lens design will also come into play. Seidel lateral field curvature goes linear with aperture and quadratic with field. Seidel lateral distortion goes cubic with field.

 

Image Diameter of a half-degree object at infinity approximations:

 

> radian subtent x F                 D = (π/360) x F          = 0.00872665 x F

> tangent of angle x F              D = tan(0.5^o) x F        = 0.00872686 x F

> 2 tangent of half-angle x F    D= 2 tan(0.25^o) x F    = 0.00872670 x F

 

PS: I came up with this show and tell on that subject quite some time ago >>> The moon is imaged on a piece of 3M Scotch magic mending Tape.    Tom

Attached Thumbnails

• 

Edited by TOMDEY, 07 December 2024 - 06:56 PM.

[eblanken]

Hi TOMDEY,

 

Thank you (once again) for your valuable inputs into my education !!!

 

All-the-Very-Best-Regards,

 

Ed

 

P.S. I can easily remember "Moon = Nine Thousandths of Focal Length (F.L.)", so thanks a bunch !!!

Edited by eblanken, 07 December 2024 - 01:34 AM.

[eblanken]

But wait !!!

 

TOMDEY: 0.009 * F {PHOTO IS CORRECT} NOT 0.0009 F ?!?!?!?

 

Ed

 

P.S. I did some editing, so re-reading carefully the above posts is advised . . . 

Edited by eblanken, 07 December 2024 - 02:07 AM.

[TOMDEY]

But wait !!!

 

TOMDEY: 0.009 * F {PHOTO IS CORRECT} NOT 0.0009 F ?!?!?!?

 

Ed

 

P.S. I did some editing, so re-reading carefully the above posts is advised . . . 

Yep --- I just noticed and fixed that. Thanks!

 

90+% of the errata that I discover --- are indeed my own. And I'm sure there are hundreds or thousands buried in my old stuff here and there. Anytime a satellite that I worked on actually performs as required --- I breath a gargantuan (yet silent) sigh of relief... me and a thousand other guys and gals. When JWST got to L2, deployed, initialized --- and works --- we all wiped our sweaty brows, hugged, kissed... and passed out, with smiles on our faces.  Tom

[Kitfox]

^^^^ I love this post. Responsibility has a huge cost. But equally large benefits, if you are worthy. 

[eblanken]

Hi All,

 

Thank you each and every one of you for your VERY HELPFUL posts on my thread . . . 

 

Hello All,

 

I am seeking a better understanding of the method to calculate the real image that is formed in my Refractor: I have a Skywatcher 100mm x 900mm, f/9 ED refractor and want to know what the actual image size is in real units. More specifically, I want to know what the image size is of the Full Moon (Luna) or the Sun (Sol) when the Refractor is aimed and focused on either of these two objects at the curved image surface (and please approximate that to a simple, flat image, if that is easier and assuming 0.5 Deg for the disk is fine).

 

Best,

 

Ed

 

P.S. See my signature and use Math as necessary, but a simple equation would be fine, if there is a handy approximation is available.

 

 

In short - the image size h with a focal length of fl is

 

h = 2 * fl * tan(angular size / 2)

 

 

If you don't have an scientific calculator, you can aproximate this for most cases involving small angles (less than a few degrees) by

 

   = fl * 2 * Pi  / 360 degrees * angular size in degrees

   = fl * 0.0175 * size in degrees

   = fl / 57.3     * size in degrees

 

Makes 7.8mm in your case.

 

Extra points if you know the trivia on 57.3 (mandatory for astro nerd status):

 

Televue's "Radian" eyepieces have a FOV of (approximately) 57.3 degress. Yep, the same 57.3 as in the above formula. Its all about circles, angles, degrees and Pi. This mathematical constant is called one radian ... and Al Nagler thought its cool to pick just that for his FOV and then name the whole series after it.

 

 

Hi Ed!

 

All three of these three approximations give the same answer to 4 significant figures, presuming that the sun or moon is 30 arc-min across. And that's really good enough... because field curvature and distortion of your particular lens design will also come into play. Seidel lateral field curvature goes linear with aperture and quadratic with field. Seidel lateral distortion goes cubic with field.

 

Image Diameter of a half-degree object at infinity approximations:

 

> radian subtent x F                 D = (π/360) x F          = 0.00872665 x F

> tangent of angle x F              D = tan(0.5^o) x F        = 0.00872686 x F

> 2 tangent of half-angle x F    D= 2 tan(0.25^o) x F    = 0.00872670 x F

 

PS: I came up with this show and tell on that subject quite some time ago >>> The moon is imaged on a piece of 3M Scotch magic mending Tape.    Tom

 

 

^^^^ I love this post. Responsibility has a huge cost. But equally large benefits, if you are worthy. 

 

Now it is time for some deeper insight: I want to know more about the diameter as a function of length along the Axis of my Refractor of the light cone.

 

Say, I fully illuminate the Refractor with a lighted screen that fully illuminates the light cone and not just the Moon's Half Degree or Sun's Half degree.

 

If I call the "Thin Lens Approximation" for my 100mm x 900mm, F/9 lens Z = 0, what is the diameter of the f/9 light cone as it propagates down my Refractor from Z = 0mm to Z = 850 to 900mm to 950mm. I know that for much of the distance (Z < 850mm), the light cone can be modelled by a tri-angular cone such as (aka triplemon) shows in his drawing, but as the light cone gets near the focal plane, the diameter can no longer be modeled by the triangle that converges to a point at Z = 900mm (focal plane "image spreading"), but rather widens out as a de-focused "anti-cone" as the light cone approaches the focal plane spreading that has been discussed already or hinted at by the above. I am aware of vignetting issues . . . but would like a deeper understanding . . . 

 

Anybody willing to explore this topic deeper with me ?

 

Best,

 

Ed

Edited by eblanken, 08 December 2024 - 09:04 PM.

[dan_h]

Hi All,

 

Thank you each and every one of you for your VERY HELPFUL posts on my thread . . . 

 

Now it is time for some deeper insight: I want to know more about the diameter as a function of length along the Axis of my Refractor of the light cone.

 

Say, I fully illuminate the Refractor with a lighted screen that fully illuminates the light cone and not just the Moon's Half Degree or Sun's Half degree.

 

If I call the "Thin Lens Approximation" for my 100mm x 900mm, F/9 lens Z = 0, what is the diameter of the f/9 light cone as it propagates down my Refractor from Z = 0mm to Z = 850 to 900mm to 950mm. I know that for much of the distance (Z < 850mm), the light cone can be modelled by a tri-angular cone such as (aka triplemon) shows in his drawing, but as the light cone gets near the focal plane, the diameter can no longer be modeled by the triangle that converges to a point at Z = 900mm (focal plane "image spreading"), but rather widens out as a de-focused "anti-cone" as the light cone approaches the focal plane spreading that has been discussed already or hinted at by the above. I am aware of vignetting issues . . . but would like a deeper understanding . . . 

 

Anybody willing to explore this topic deeper with me ?

 

Best,

 

Ed

This is a good question. Important stuff if you want to calculate baffles. 

 

The taper of the light cone is different for different sizes of field at the image plane, and different again for different focal lengths.   You simply need to determine how much the cone tapers for a given length of the optical path. 

 

Consider the following example:

Clear aperture  =  A =  100mm

Image size = S = 20mm  (you get to choose this based of eyepiece field stop, camera size, personal preferences, etc)

Focal length of scope = Fl

 

In all cases the light cone tapers a total over the length of the cone, aperture minus the image size.  (A-S). 

 

For the scope in the example of 1000mm focal length, the light cone tapers 80mm over 1000mm or 80/1000= .08mm/1mm. Rate of taper = (A-S)/FL

 

To determine the size of the cone at any position along the length of the scope,  simply multiple the distance from the focal plane by the rate of change and add that to the image size. 

 

Size of the light cone at a distance d from the focal plane  is S + d(rate of taper) =  20mm + d(.08mm).  At 100mm distance this would be 20+8=28mm.   Size of the light cone 200mm from the image plane = 20mm + (200 x .08) = 20 + 16 =36mm

 

Working from the other direction, the size of the light cone from the clear aperture would be  A - d(rate of taper).

So in the example, the light come at 300mm would be  100mm - 300(.08mm)  =  100-24 = 76mm 

 

To summarize,  Size of light cone is determined as Aperture minus taper or image size plus the taper.  

 

Unfortunately, I can easily do the arithmetic but I don't know the industry standards for the symbols. 

 

hope this makes sense,

 

dan

Edited by dan_h, 09 December 2024 - 02:05 PM.

[eblanken]

Hi dan (aka dan_h),

 

Yes, this makes perfect sense to me conceptually. And yes, I was thinking about baffles and the adjustment necessary to account for the image size being a "non-zero" point, but the image is formed by what are "until the lens" parallel wavefronts that are "bent" into a light cone (in my example of the 100mm x 900mm, f/9). The image, as we have already seen earlier in the thread, of a solar or lunar disk of 0.5 deg is 0.009 *F. The image of the whole larger-angle screen that I am transitioning the thread toward, is slightly less clear to me. Obviously, the f/9 OTA can't see 180 Deg of field, but some portion of that gets to the image plane, so the image plane is larger than 0.009 *F. Any baffles or field stop will restrict what size is at the Z = 900mm in my example.

 

One thing to consider is the end of the focuser tube or a diagonal's free path diameter. In the case of MCT or SCT, there are baffles to consider as well.

 

Thanks for your contribution, dan !!!

 

Best,

 

Ed

Edited by eblanken, 09 December 2024 - 07:52 PM.

[dan_h]

 

 

 Any baffles or field stop will restrict what size is at the Z = 900mm in my example.

 

 

Best,

 

Ed

Precisely!  The total image size would be  much larger than we typically see if it wasn't for the restrictions in the light path be it a baffle, focuser tube, diagonal or the field stop in an eyepiece.   We use the restrictions to limit the image to the size we can work with in a camera or eyepiece.   This is not a bad thing because once you get further off axis the image quality deteriorates due to field curvature, astigmatism and coma  depending on the optical design. 

 

I have always wanted to build a camera obscura.  Truly wide field viewing.

 

dan