2022-05-11: APP 2.0.0-beta 1 has been released !!!
Download links per platform:
How to remove satellites?
I gathered height hours on the Witch Head nebula, in 60 seconds exposures. There is satellites on nearly all of my subs, following nearly all the same path. I used automatic integration and on my final stack, I can clearly see their trails. I was wondering if there is something I can do to get rid of them. Normaly I'd just try different integration method but with this 500 frames, it took me about 24 hours.
Here is the starless version, we see it better.
Oeh that is annoying, especially when they're on the same track. If you have enough subs, outlier rejection should work. I'd suggest to go for a more agressive rejection in tab 6 (integrate), like kappa high of 2.0, just to see if that has enough effect. You can also start integrating, say, 100 frames and then combine the resulting 5 stacks later as well. This should also give a nice result and you can test the rejection a bit faster.
Thanks for your help!
I'll work from there and probably report back in a few days 🙂
Thanks for your help Vincent!
Even after using a kappa high of 2.0, the satellites was still showing very faintly in the starless, bright background image. They do not appear in the processed image however.
Beautiful image! Ok, so yeah I think that will be about the best you can do then. Dithering with big steps will also help, but too aggressive rejection will start to hurt the signal I think.
Even if dethering, wouldn't the satellites still be at the same apparent location against the nebula? I guess I should look at a batterie to power my laptop and guide cam all night. Guiding would have help me with this one 😉
Ah yes, I think you're right about that. Well, at least dithering will help no matter what for the rest. 🙂 Regarding the rejection, you can go even lower with the kappa or also play around with the kappa low, and setting that lower. But I'm afraid then it will start to hurt the signal and as you see in the final integration it looks good.