In an image, after subtracted by dark, divided by flat, sky still left.
I think up a good way to eliminate the sky.
Assume the distribution of gas in sky is equivalent everywhere.
So there is a simple relation between sky intensity and elevation.
I = I{0}sec\theta + A,
where I is the intensity of the sky you observe,
I{0} is the intensity of the sky in \theta = 0 degree,
A is real const
How to calibrate the sky intensity?
In an image, after subtracted by dark, divided by flat, sky still left. I think up a good way to eliminate the sky.
Assume the distribution of gas in sky is equivalent everywhere. So there is a simple relation between sky intensity and elevation. I = I{0}sec\theta + A, where I is the intensity of the sky you observe, I{0} is the intensity of the sky in \theta = 0 degree, A is real const