Secondary Obstruction question

[Buqibu]

If a reflector has a primary mirror diameter (aperture) of lets say 20 inches, and a secondary mirror obstruction of 30%, does that mean you effectively only have a usable aperture of 14 inches? (Because 70% of 20 is 14). Sorry if this is a very basic question, just wondering.

[Robindonne]

No its the percentage of the scopes diameter. Not the overall percentage of the scopes aperture

in your example i think it must be around 10-15% obstructed of the initial aperture

Edited August 2, 2021 by Robindonne

[rl]

No...the obstruction percentage refers to the linear dimension. We need to be clear not to confuse linear and area numbers. 

Using your example, the area of the primary would be Pi* radius squared...= 314 square inches.

The obstruction is 30%, or a circle of 30% * 20 inches = 6 inches. 

The area of the obstruction is Pi * radius squared, = 28.3 square inches.

Thus the secondary obstructs 28.3/314 of the incoming light = 9%

This is a fairly small fraction which you will hardly notice. 

The effective area is 314 - 28.3 square inches = 285.7 square inches.

Working the circle area formula backwards, this equates to an unobstructed diameter of 19.1 inches. So the effective diameter changes very little for quite a big central obstruction. 

The theoretical resolution remains the same as a 20 inch unobstructed, but you would notice a loss of contrast since diffraction off the edges of the secondary causes a small amount of light spillage..a small amount goes where it should'nt so dark areas in the image aren't as dark as they should be, and light areas aren't quite as bright

[Buqibu]

Alright, thanks! I have been enlightened😀

[jetstream]

30% obstruction is getting up there IMHO.My truss dobs are 21% or less.

[Louis D]

Resolution is dependent on primary diameter regardless of central obstruction.  That's why very long baseline interferometry works.

You can read up on the effects of central obstructions here.