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# Can somebody please explain "Clear Aperture"?

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### #1 Astroforecast

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Posted 13 December 2009 - 11:36 PM

I just got my first non-refractor telescope which is a Celestron C6. I always thought that "clear aperture" was the objective diameter minus the diameter of any obstruction such as the secondary mirror. However when I look at specifications of the C6 they say clear aperture of 150mm. Shouldn't the clear aperture be 150mm minus the diamter of the secondary?

The reason I'm asking is because I'm trying to compare the light grasp of a C6 to something I am familiar with such as my 120mm refractor.

Regardless of the exact definition of "clear aperture", I calculate that the C6 will have a light grasp about 1.4 times greater than the 120mm refractor. Correct? I used a 50mm secondary diameter for that calculation because I don't know the true size in a C6.

Thanks

### #2 Karl_Bonner_1982

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Posted 14 December 2009 - 12:39 AM

Here's how I calculate light grasp. Rather than mess with the area of a circle, I simply calculate the area of the square that circumscribes the aperture; for a 100mm refractor it would be 10,000 sq mm. You then have to subtract the square of the central obstruction - if you had a 100mm Mak-Cass with a 30mm central mirror then you'd subtract 30^2, or 900, to get an effective square aperture of 9100 mm^2. Then you take the square root of 9100 to find the refractor aperture that gives the same effective square aperture as a 100mm Mak with 30mm C.O.

This method works because the ratio of a circle's area to the area of its circumscribing square is a constant pi/4. So to calculate true apertural area you just multiply the effective square aperture by pi/4.

### #3 David Knisely

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Posted 14 December 2009 - 01:03 AM

I just got my first non-refractor telescope which is a Celestron C6. I always thought that "clear aperture" was the objective diameter minus the diameter of any obstruction such as the secondary mirror. However when I look at specifications of the C6 they say clear aperture of 150mm. Shouldn't the clear aperture be 150mm minus the diamter of the secondary?

The reason I'm asking is because I'm trying to compare the light grasp of a C6 to something I am familiar with such as my 120mm refractor.

Regardless of the exact definition of "clear aperture", I calculate that the C6 will have a light grasp about 1.4 times greater than the 120mm refractor. Correct? I used a 50mm secondary diameter for that calculation because I don't know the true size in a C6.

Thanks

Clear aperture in a catadioptric telescope (SCTs, Maksutovs) is often considered to be the width of the corrector plate from one side to the other. In SCTs, the primary mirror tends to be just a bit larger than the corrector plate's diameter, as the corrector plate deviates the light a bit requiring a bigger primary to catch it all and make it converge towards the secondary mirror. The C6 has a clear aperture of 5.91 inches. It has a secondary obstruction diameter of 2.20 inches, so the total surface area that collects light is 26.6 square inches. This would be equivalent to the light gathering area of a 5.39 inch (137mm) aperture unobstructed telescope like a refractor. However, the resolution of the instrument would still be about that of a 5.91 inch aperture telescope. Clear skies to you.

### #4 MitchAlsup

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Posted 14 December 2009 - 01:48 PM

Clear Aperture is the "effective" aperture after taking all of the stops and obstructions into account.

So, Schmidt cameras have a aperture of (say) 48" and a main mirror size of 72", but a clear aperture of only 45.28" due to the holder for the 14" plates.

If the aperture is partially opaqued by a apodizing mask, then you end up having to compute how much light gets transmitted (T-stop), and then refer back to what unobstructied aperture would allow this same amount of light through.

### #5 Astroforecast

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Posted 14 December 2009 - 03:13 PM

Hmmm... the last two posts seem like two different definitions of clear aperture.

The description by Mitch sounds technically accurate to me. For an SCT with a 150mm corrector plate and a 56 mm central obstruction it seems like the clear aperture should be listed as 139mm (using the method described in the last line of Mitch's post) instead of 150mm as Celestron does.

### #6 David Knisely

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Posted 14 December 2009 - 05:49 PM

Hmmm... the last two posts seem like two different definitions of clear aperture.

The description by Mitch sounds technically accurate to me. For an SCT with a 150mm corrector plate and a 56 mm central obstruction it seems like the clear aperture should be listed as 139mm (using the method described in the last line of Mitch's post) instead of 150mm as Celestron does.

From an optical physics standpoint, the definition of "clear aperture" depends strongly on who you talk to (like various telescope manufacturers, including Celestron). The term "clear aperture" isn't even mentioned in many optics texts (it isn't in my old edition of OPTICS by Hecht & Zajac). Originally, it referred to the diameter of the part of a lens which was exposed to light after any aperture stops were applied. Some refer it to an effective light gathering aperture based on exposed area after the obstruction is taken into account, but this definition ignores the fact that the resolution of the instrument only depends on the diameter of that opening and not on any "effective" aperture. In other cases, the term does not refer to some numerical aperture diameter, but is used to describe an optical system that does not have any obstructions (i.e., a refractor or an off-axis or tilted-component reflector). To be fully correct, one should just use the word "aperture" or "entrance pupil" rather than mixing in some other word that may be confusing. The aperture of the C6 is 5.91 inches. The Rayleigh resolution for that aperture is about 0.92 arc seconds which is better than the resolution of an "effective aperture" of 5.39 inches that the C6 has for its light gathering ability (and finer than a 120mm aperture refractor's figure of 1.15 arc seconds). Clear skies to you.

### #7 turtle86

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Posted 14 December 2009 - 07:17 PM

Excellent explanation.

Clear aperture in a catadioptric telescope (SCTs, Maksutovs) is often considered to be the width of the corrector plate from one side to the other. In SCTs, the primary mirror tends to be just a bit larger than the corrector plate's diameter, as the corrector plate deviates the light a bit requiring a bigger primary to catch it all and make it converge towards the secondary mirror. The C6 has a clear aperture of 5.91 inches. It has a secondary obstruction diameter of 2.20 inches, so the total surface area that collects light is 26.6 square inches. This would be equivalent to the light gathering area of a 5.39 inch (137mm) aperture unobstructed telescope like a refractor. However, the resolution of the instrument would still be about that of a 5.91 inch aperture telescope. Clear skies to you.

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