Thursday, August 27, 2015

Using the Celestron Field Flattener with an SBIG ST-8300M and FW8-8300

I purchased a used Celestron 9.25" XLT SCT intending to use it primarily for planetary imaging at f/10 or greater. There was no need for a focal reducer/flattener. I did purchase a used Celestron 94175 f/6.3 flattener/reducer so I could, if I wanted to, image deep sky objects. As it turns out, deep sky objects have been the C925's main use.

Reducer/Flatteners like the Celestron 94175 work best when at a specific distance from the sensor. Can this distance be known? Let's use the calculator at  Wilmslow Astro and find out.

For a standard f/10 SCT, the separation to give f/6.3 (where we presume the field will be flattest) should be 105mm.

Two nights ago I imaged NGC 7008, the Fetus Nebula, using my CCD and the focal reducer. The stars were nice and round all the way to the edge:

NGC 7008 (click to enlarge) reported that the image scale was 1.58" per pixel, which translates to f/6.0. According to the above calculator this should happen at a reducer/CCD separation of 113mm**. 

If the reducer is flattest at f/6.3 I need to decrease the separation from 113mm to 105mm, or by about 8mm.

Happily, I can do that. The nosepiece I was using was a 2" focuser-to-TeleVue IS adapter + IS to T-ring adapter. I also have a nosepiece that's just 2" focuser to T-ring, and amazingly it will let the CCD get about 8mm closer. Just about perfect!***

The next night I'm out I'll try this configuration and see if it gets the focal ratio correct and also produces round stars.

* This suggests that the performance of the reducer is rather insensitive to the separation. Here I'm 8mm off and the stars look good. This has limits, though. In a previous image a 12.7mm spacer was included and the stars at the corners were plainly distorted. That image had an image scale of 1.65" per pixel, which corresponds to a focal ratio of  f/5.75. The calculator says this happens at a separation of about 123mm. I measured the separation as about 125mm. That image suggests the separation reduction should be 20mm, which is roughly the same as 12.7 + 7.3 (the spacer plus the ~8 mm suggested by the other image).

** One thing that's usually left ambiguous in reducer explanations is the point on the reducer from which the CCD distance is measured. In the case of the system operating at f/6 I measured the CCD to be 125mm from the front of the reducer. which is the same thing as 114mm from the center and about 100mm from the rear thread base. Because the center position in only 1mm different from the formula's value of 113mm, and given the inexactness of my measurement, It's reasonable to conclude separation should be measured from the center of the reducer.

*** Is this an accident? The Antares 50mm long 2" focus tube on the back of the SCT makes this possible, but it predates the ST-8300 + filter wheel and couldn't be designed to work with their backfocus requirement. But it works out nicely, right?

What does this say about using a 0.5 focal reducer? The model I have is a SmartAstronomy 2", which is probably identical to the GSO 2". GSO says that the focal length is 106mm and the optimal separation is 53mm. The calculator puts it a bit larger at 56mm. Because this reducer screws into 2" nosepieces, the closest I can get it to my CCD is about 69mm (measuring from the center of the reducer). That puts it at about f/4. I'd guess that's too far from ideal to produce pleasing images, but it's worth a try. A Lumicon low profile nosepiece could trim 9mm off that, getting it close: f/4.7. Trying that will cost about $35.

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