What does this formula mean?

[hi8ane]

As an enthusiatic beginner to imaging I'm soaking up all I can read but was puzzled by this formula which appeared in the March issue of S@N magazine on p64 relating to lunar imaging.

"... decide on your focal length. This can be calculated by the simple formula fl=3.6xDxP where D your scope's aperture in mm and P the size of the camera sensor's pixels in microns."

I'd welcome clarification as, for example, my own scope is 3.6x72x3.75 which implies a focal length of 972mm but what does his mean - max practical? I've used a x3 Barlow on my 432 fl scope with good results. 

Thanks in anticipation

[Owmuchonomy]

It is suggesting the optimal focal length to aim for based upon your set up.  So your 3x barlow is not far off to be honest.  The result is also substantially affected by the seeing so its only a guide.  I use a different approach and that is to aim for a f/ of around 5x my pixel size.  So for my ASI174mm that equates to 5 x 5.8 or somewhere between f/25 and f/30.  Hence I use a 2.5 x powermate in my C9.25 which is an f/10 scope.  That said, i have tried a 5x powermate too on the Moon so f/50 which wasn't great but not bad using an IR pass filter.

[hi8ane]

Thanks very much - appreciate your insight.  Do you know if there's any science to the formula? Seeems to be based on resolution which implies a degree of subjectivity. I know I'm pushing things a lot with my starter setup as I can can only get 40 pixels to cover the diameter of Venus and 70px for the others. So I need at least a fundamental 1200mm focal length really. Is that right?

[Owmuchonomy]

I'm sure there is some science to it but more importantly it is a tried and tested method. However, the S@N article is talking about the Moon, an object half a degree across. If you wish to image Venus then yes you are going to need a lot more FL. For Venus, I am using a FL of at least 5 metres.

[Owmuchonomy]

I like this presentation: 

 

[sharkmelley]

Another way of writing that formula is  fl/D = 3.6xP

Now fl/D is the focal ratio so the formula becomes: F_ratio = 3.6 x P

So as an example, for a typical pixel size of 3.75microns (ASI224 camera) it suggests using a focal ratio of F14 (3.6x3.75) - so you would add sufficient Barlows to your scope to extend the F-ratio to that level.

Technically, if you adhere to this formula then the Airy disc diameter is sampled by 5 pixels at the green wavelength.  The Airy disc is an indicator of the maximum resolution of the scope and if you sample it at 5 pixels you are pretty close to the theoretical physical limits of the aperture i.e. you won't improve resolution much further by increasing F-ratio.

You can push the F-ratio slightly further if you really want to:  at 5 x P you will be sampling one cycle of the MTF (modulation transfer function) extinction frequency at 3 pixels.  There's no subjectivity involved - these are hard limits imposed by physical optics and Shannon sampling.

Mark

[hi8ane]

Thank you Mark and Chris, very helpful and clear. When I was reviewing my Jupiter and Venus images I did wonder how to work out the various limits; none of the astrophotography guides I have go deep enough. Looks like the limits on my equipment are around 1000 mm focal length so the setup I have is the furthest I can go - REALLY good to know.