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Is this not immediately intuitive and simple?

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No, not really. It's constructed to be simple only when the focal lengths match. How does the TFoV in a 12mm eyepiece with an RFOV of 80 degrees compare to a 15mm eyepiece with an RFOV of 65? But what if I had told you instead that the 12mm has an EFSD of 16.8mm and the 15 has an EFSD of 17.0mm? You know immediately which will have the larger true field, and that required no arithmetic at all.

Now suppose I want to calculate the TFoV in my spreadsheet and I know my scope has a focal length of 1200mm. All the field stop formula says is that true field is proportional to EFSD. That proportionality constant is (180/pi)/TFL, usually approximated by 57.3/TFL. So I can calculate that once and for all, in my example it's 57.3/1200 = 0.04775 . Any time I want to convert an EFSD to a TFOV for that scope, I just multiply by 0.04775.

Now contrast that with the way you use the RFOV. With the RFOV you have to divide by magnification, which means that I first have to divide the telescope focal length by the eyepiece focal length, and then I have to divide the RFOV by that. That's not less complicated.

Remember the whole point here was calculating TFOV. Whereas EFSDs are superior for that, the analogous RFOV advantage would be in comparing angular magnification distortion. However, for that purpose it would be even easier to supply a quantity to represent that directly, e.g. 1 - RFOV/AFOV.

A long time ago, Jon Isaacs and I were pushing this idea of RFOV, only we called it something different, but the more I think about it, the more I realize that Al Nagler was right all along--just report EFSDs. That's what every manufacturer should do, and when they don't, we should use our drift-test data to do it for them.

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Mike

Stuff that I use:

• 7" f/6.7 home-built Newt, eq platform, Pentax 40XW and 5XO, Tele Vue 13E and 2x Barlow, ZAO-II 6mm
• 120mm f/8.3 home-built grab-n-go Newt with 7-21mm Nikon Zoom
• Canon 15x50 IS and Eagle Optics 12x50 Ranger binoculars
  

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