In that case, you really should calculate what aperture is required to achieve an unobstructed equivalent to 6.1, maybe 6.2"
Huh? Please forgive my lack of understanding. This old guy can be a bit slow sometimes, but you have lost me So... How would I calculate a 6.2" unobstructed aperture for a C14 with a 4.5" secondary obstruction?
Original poster: Apologies I'm not trying to hijack this thread.