Jump to content

  •  

CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.

Photo

Crayford focuser for Cats

  • Please log in to reply
26 replies to this topic

#26 mvas

mvas

    Viking 1

  • -----
  • Posts: 795
  • Joined: 27 May 2015
  • Loc: Eastern Ohio

Posted 17 February 2018 - 10:58 AM

 

I think it is extremely important for us to discuss how we can get the best images possible from our SCT's.

We know that it is best to keep the image near the Back Focus as specified by Celestron.

The engineers at Celestron have performed the "hard" calculations for us.

 

The optimal Back Focus for the Best Image is simply:

  • 5.25" = EdgeHD 800
  • 5.75" = EdgeHD 925
  • 5.75" = EdgeHD 1100
  • 5.75" = EdgeHD 1400

 

Significant deviation from the above will degrade your image.

From: https://www.innovati...ocus-tolerance/

Conclusions
In a nutshell the tolerance (offset from the NBF distance) is tighter for good seeing and a large chip (diagonal), as we would expect.
The table below summarizes the maximum tolerances (rounded values) for a 10% spot diameter (encircled energy) increase v.s the diffraction and/or scope optical limits:

APS-C chip Full frame chip
1″ FWHM seeing +/-  20 mm +/- 20 mm*
2″ FWHM seeing +/- 50 mm +/- 20 mm

 

Those graphs clearly show there is one, and only one, Optimal Back Focus distance - at exactly "0.0 mm Offset from NBF" as shown in the top 2 graphs.

But let's be clear ... the graphs are the Worst Case location on each chip ( APS-C & FF) - in the extreme outer (4) corners only.

FF       =  21   mm off-axis from the optical center line ( upper curve )

APS-C = 13.5mm off-axis from the optical center line ( lower curve )

None of those graphs indicate how much better the spot size actually is near the center of the imaging chip.

Cropping can remove much of the outer edge error.

And if the "seeing" is random noise then stacking many images may reduce the bad effects due to the atmosphere.

 

It would be great to see a 3rd line in those top 2 graphs showing the the error at location 0.0 mm off-axis ( ie the center of the imaging chip ), too.

This 3rd line would certainly be below ( better than ) both blue curves.


Edited by mvas, 17 February 2018 - 11:12 AM.


#27 mvas

mvas

    Viking 1

  • -----
  • Posts: 795
  • Joined: 27 May 2015
  • Loc: Eastern Ohio

Posted 17 February 2018 - 11:01 AM

 

 

Here's some reading for you:

 

"Strictly talking, commercial SCT using primary mirror focusing are misaligned every time the two mirrors are not at a single optimum separation."

 

"The relation implies that, for nominal primary mirror shift ∆, error induced by extending back focus is independent of aperture. It only depends on secondary magnification and primary's focal ratio. For F1=2, m=5, and K2=0, every mm of reduction in mirror separation (∆=1), or nearly for every inch of focus extension, induces ~1/23 wave P-V wavefront error of over-correction, and as much of under-correction for widening the separation."

 

http://www.telescope...cs.net/SCT2.htm

 

if the focuser moves the focal plane 3 inches from the optimal location,  it would add about 1/8th wave of overcorrection. 

 

Jon

 

I tried plugging in their example numbers ( as given in your text above ) into their P-V Wave-Front Error formula ...

 

W = (1 - m^2 ) x ∆ / ( 512 x m^3 x FRpri^4 )

 

But I did not get the same answer of ~1/23 P-V Wave-Front Error, as stated in their text.

 

W = ( 1 - 5^2 ) x 1 / ( 512 x 5^3 x 2^4 ) 

W = 24 / 1,024,000

W = 1 / 42,667

 

I thought, "W" represented the P-V Wave-Front Error.

So, how does ( W = 1/42,667 ) convert to ~1/23 P-V Wave-Front Error?

 

I added the parentheses around the denominator to indicate order of operation.

 

In the above, I made the assumptions ( which I think was their intentions ) ...

a) Everything to the left of the divide symbol was the numerator

b) Everything to the right of the divide symbol was the denominator

 

Or computing from left to right  ...

W = (1 - m^2 ) x ∆ /  512 x m^3 x FRpri^4

W = 24 x 1 / 512 x 125 x 16

W = 48,000 / 512

W = 93 Huh?

 

 

 

That would be something to discuss with Vlad.  I think its well established that there is one optimal back focus for an SCT and deviating from it causes spherical aberration.

 

My goal here was just to point this out so those adding a Crawford would be aware of it.  

 

https://www.cloudyni...primary-on-sct/

 

jon

 

I agree with you goal.

I have sent an email to vlad regarding his formula, his example values and his result of "1/23 Wave Error", that does not compute.

No reply ...




CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.







Cloudy Nights LLC
Cloudy Nights Sponsor: Astronomics