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Mounting a Lunt B1800 Cak BF onto a 140mm refractor

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#1 SeymoreStars

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Posted 28 May 2024 - 06:52 PM

Are there ways to work around the 100mm limitation?

 

Can an ERF blocking all but CaK wavelength accomplish this?

 

I sent Lunt this question this morning and got no reply.

 

Thanks

Steve



#2 Jim Waters

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Posted 28 May 2024 - 07:22 PM

It may be best to call them and chat.



#3 SeymoreStars

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Posted 28 May 2024 - 07:24 PM

It may be best to call them and chat.

I did, someone besides Faye told me to send an email.



#4 MalVeauX

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Posted 28 May 2024 - 07:53 PM

Hi,

 

With a full aperture D-ERF that passes 393nm wavelength, yes, it will work fine.

 

This is precisely how I do it with my 150mm refractor with a tri-band D-ERF that passes 393nm, 430nm and 656nm.

 

150mm_cak_DERF.jpg

 

That said, I don't think you want 140mm aperture with calcium. That demands seeing around 0.6~0.7 arc-seconds peaks just to hope to get close to critical sampling of such angular resolution. The angular resolution is 59% higher than h-alpha and it is highly effected by seeing. 100mm in Calcium critically sampled is equivalent to more than what you see from critically sampled 150mm h-alpha systems. 140mm Calcium angular resolution is equivalent to around 222mm H-alpha angular resolution, ie, nearly a 9 inch aperture. You need incredible seeing for this.

 

Very best,


Edited by MalVeauX, 28 May 2024 - 08:00 PM.

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#5 SeymoreStars

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Posted 28 May 2024 - 08:58 PM

My other option is an 80mm refractor. Is it even worth the effort. The seeing range here is generally 1.2 - 2.5 arc seconds. Marty can you throw out some of the math that's leads you to your conclusion or point me to a thread that devles into it?

 

Thanks

Steve


Edited by SeymoreStars, 28 May 2024 - 09:54 PM.

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#6 hornjs

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Posted 28 May 2024 - 09:52 PM

As someone working on CaK imaging, it is extremely difficult at 1000 FL.  I am using an Omni 120xlt and rarely get seeing to support any barlow use at all.  But without the barlow I think it gives great views at 1000.  Granted this is stopped down to 57 aperture.   While the WL imaging for me has become very satisfying, I still put CaK in the "frustratingly mediocre" category.  


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#7 SeymoreStars

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Posted 28 May 2024 - 10:35 PM

Marty I wish I could understand what you wrote above.

 

Given the definition below - angular resolving power (or resolution) of a telescope

----------------------------------------------------------------------------------------------------------------------------------------------------

The angular resolving power (or resolution) of a telescope is the smallest angle between close objects that can be seen clearly to be separate. Resolution is limited by the wave nature of light. For a telescope having an objective lens or mirror with diameter D and operating at wavelength λ, the angular resolution (in radians) can be approximately described by the ratio λ/D.

----------------------------------------------------------------------------------------------------------------

Given that ...

nm = 1/1,000,000,000 of a meter

mm = 1/1,000 of a meter

 

I want to calculate the following...  393.(nm)/ 80(mm)

 

Convert 393nm to milimeters...and compute

 

.0000000393 / 80 = 0.00000000049125 (radians)

 

Convert radians to degrees...

0.00000000049125 (radians) = 0.0000000281465517 Degrees or in exponentiasl notation  (2.81465517e-8)

 

or

 

Convert radians to Arc Seconds...

0.00000000049125 (radians) = 0.000101327586 arc seconds

 

Now I don't know what to do with those result and am lost.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

 

To get back to some form of reality that I can comprehend...

 

Let say my seeing is a constant 2 arc seconds. How does that above relate to that?

 

Thanks!!

Steve



#8 MalVeauX

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Posted 29 May 2024 - 04:50 AM

Hi Steve,

 

Keep it simple, it's not the telescope that has the angular resolution, its just the wavelength of light. Angular resolution increases as frequency decreases. 393/656 = 59% approximately. This is just a quick and dirty way to do it. If we just calculate critical sampling for the wavelength on a system, you'll notice the focal-ratio increases significantly as the frequency decreases, and that's to account for angular resolution increase of the wavelength.

 

θ = 1.22 * λ / d

Focal length ≥ ((aperture x pixel size) / (wavelength x 1.22)) x 3

 

You can see the relationship here where wavelength is in the operation and how it will influence things.

 

If your seeing is 1.2 arc-second peaks, that's good enough for about 120mm aperture in 656nm, so about 70~80mm in 393nm Calcium but knowing the seeing will effect it even more, so error on the poorer side (ie use smaller aperture or just be patient with lucky imaging). If your last seeing was 2 arc-seconds, that's 80mm in 656nm, so 50mm in 393nm.

 

Very best,


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#9 ch-viladrich

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Posted 29 May 2024 - 05:03 AM

Hi Steve,

 

Years ago, a friend tested on Lunt Ca K on a 200 mm f/8 refractor. The filter was "burned".

 

I tested one on a 150 mm f/7 for a couple of sessions. There was no problem with the filter. But, I have no idea how the filter would have aged after days of observation.

Accordinly, to be on the safe side, and as said by Marty, you should use an ERF, either external (= full-aperture) or internal (= sub aperture).

 

On my side (using other Ca K filters), I have a 50 mm sub-aperture ERF filter on my 150 f/7 refractor :

 

http://astrosurf.com...-Astronomik.jpg

 

On top of that, there are two diffculties with Ca K imaging :

- turbulence increases when wavelength decreases (= going to the blue side of the spectrum),

- optical quality of most refractors sharply decreases when going to the blue side of the spectrum (this is why results are better when using an aperture stop).

 

Clear skies !

 

Christian


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#10 SeymoreStars

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Posted 29 May 2024 - 07:51 AM

Thank you Marty and Christian. I am composing a response and really appreciate your help.


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#11 SeymoreStars

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Posted 29 May 2024 - 07:57 AM

I understand longer FL = light cone has steeper angle
I will use an 80mm telescope
The mixing of nano meters and milimeters in an equation has me scratching my head, LOL

 

Givens ...

telescope      = 80mm
ASI533 pixels  = 3.76nm
CaK wavelength = 393nm

 

 

Focal length = ((aperture x pixel size) / (wavelength x 1.22)) x 3

 

((80*3.76) / (393*1.22)) * 3 = 1.8821173820548116631210111375297 (should the decimal point be moved LOL)

 

Someday I want to understand the equation, but not today ;-)

 

Thank you very much!!

Steve

 

Edit: forgive me nighttime, more aperture is better, orientation.


Edited by SeymoreStars, 29 May 2024 - 08:02 AM.

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#12 MalVeauX

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Posted 29 May 2024 - 09:49 AM

Hi Steve,

 

You have to account for units. Thankfully its all metric (except for degrees/radians). Once converted to all like-unit, the calculation is simple.

 

Here's a pre-made calculator for this, you can just change variables as needed to get a quick calculation. The formulas are shown and the work can be seen so logic can be followed:

 

Attached File  Angular Resolution, Seeing & Critical Sampling Calculator (Martin Wise, Wider Columns, 02072024).xls   12KB   10 downloads

 

Very best,


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#13 SeymoreStars

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Posted 29 May 2024 - 09:54 AM

Thanks Marty!!


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#14 SeymoreStars

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Posted 29 May 2024 - 11:42 AM

Ah the "Fried" parameter (page 64) of "Solar Astronomy", thank you Christian



#15 SeymoreStars

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Posted 20 June 2024 - 05:35 PM

I pulled the trigger on a Lunt B1800 Cak blocking filter.


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