Yes it is a complicated topic.
For light gathering, obstructed area can be taken into consideration.e.g. A 20 cm lens/mirror has an area of 314 cm^2. If this was a Newt with a 5 cm secondary mirror (19.6 cm^2) then the resulting area for gathering light would be 294.4 cm^2. This works out to a 6.25% loss of light gathering power - a completely negligible amount in practical terms.
Light gathering comparisons of refractors and Newtonians must also take into consideration the % of light lost by passing through a refractor lens or light lost by inefficient reflectivity of the mirrors in a Newt...... Working out all these percentages is a very complex problem.
Resolution is quite a bit easier. An 8" refractor and an 8" Newt with 25% (by diameter) have the same resolving power for point sources like stars. Actually the Newt will do a few percent better than the refractor because the central obstruction makes the central spot of the diffraction pattern a wee bit smaller - the light is slightly diverted to the first diffraction ring. But in practical terms this difference can be ignored.
For contrast transfer the central obstruction reduces the ability of the telescope to discern fine, low contrast detail. Tiny central obstructions of, say 10% or 15% (by diameter), make no noticeable effect on contrast transfer. Even a 20% obstruction make so little a difference that it cannot be noticed, even by expert observers. Once we start to approach 25% and 30% or even 35% like many Schmidt-Cassegrains the unobstructed scope begins to noticeably pull ahead of the Newt.
The difficulty is trying to pin down a 'rule of thumb' which declares an 'equivalence' between a refractor of aperture x cm and a Newt of aperture y cm with a central obstruction of k cm. There exists a rule of thumb that states that one should simply subtract the central obstruction from the Newt's aperture to find the 'equivalent' refractor. I feel this 'rule' is far too harsh.
The MTF graphs below compare the contrast transfer of a 12 inch, 25% obstructed scope( 3inches) with a 10 inch unobstructed scope. At every spatial frequency the larger scope transfers more contrast to the eye and it also exceeds the resolving power of the smaller one. But the 'rule of thumb' which simply subtracts the central obstruction says the larger scope's 'equivalent' refractor is 12" - 3" =9". This is demonstrably false. A 12-nch Newt with a 3-inch secondary mirror exceeds the capabilities of a 10-inch refractor in every possible scenario - it's not even close. This also applies to any two scopes whose apertures are in the ratio of 6 to 5....
There are so many factors other than the above - air currents in the tube, the 'seeing' near your observing site, the seeing further up in the sky, the quality of the optics you are using....user experience, too.
Dave
Edited by Cotts, 19 January 2025 - 01:14 PM.