- Exposure time according the f/ratio

- Amplification factor and focuser in-travel of a Barlow

- Optimal focal ratio for a CCD camera

- Sky coverage by a CCD camera

- CCD Resolution

Theoretical performances of your scope

By a clear and dark night, the object being near overhead you can win over 1 magnitude on the values below. In a urban or suburban area these occasions are exceptional. On the contrary when the seeing is not perfect, you will reach with difficulty the values indicated. Check the virtual magnitude calculator for other data.

Scope Resolution (PR) in arc-seconds

PR : Scope resolution (arc sec)

D : Distance of the subject (mm)

d : Size of the subject (mm)

PR = inv sin (d / D) * 3600

From the top of a valley, 250m of altitude, at daytime a NexStar 5 with a 6 mm Radian eyepiece (208x) is able to see a 10 cm diameter symbol placed on a building located at ~20 km. The scope resolution is 1.03", near its theoretical resolution of 0.9".

The limit magnitude

A. The magnitude gain

Dm : Magnitude gain

D :  Objective diameter (mm)

d : Exit pupil diameter (mm)

Dm = 2.5 log₁₀ (D²/d²) = 5 log₁₀ (D) - 5 log₁₀ (d).

For a NexStar5 scope of 127mm using a 25mm eyepiece providing an exit pupil of 2.5mm, the magnitude gain is 8.5.

To check : Limiting Magnitude Calculations

B. The limit visual magnitude of your scope

lm_t : Limit magnitude of the scope

lm_s: Limit magnitude of the sky

lm_t = lm_s+5 log₁₀(D) - 5 log₁₀ (d)

or

lm_t = lm_s + Dm

For a NexStar5 scope of 125mm using a 25mm eyepiece providing a exit pupil of  2.5mm and observing under a sky offering a limit magnitude of 5, the limit visual magnitude of your optical system is 13.5.

Check my eyepieces worksheet EP.xls which computes this value in the last column according your scope parameters.

NB. To this value one have to substract psychological and physiological factors of everyone.

Amplification factor and focuser in-travel of a Barlow

A 150 mm scope opened at f/10 uses a 75 mm Barlow lens placed 50 mm before the old focal plane. What is the amplification factor A of this Barlow and the distance D between this lens and the new focal plane ?

F : Focal lenght of the objective , 150 mm * 10 = 1500 mm

f  : Barlow focal lenght , 75 mm

d : Distance between the Barlow and the old focal plane, 50 mm

D : Distance between the Barlow and the new focal plane

The resulting focal length R is,

R = (F * f) / (f - d)

R = (1500 * 75) / (75-50)

R = 4500 mm

If the amplification factor A = R/F,

A = 4500 / 1500

The amplification factor is 3

The focuser in-travel distance D (in mm) is,

D = f * (A -1)

D = 75 * (3 - 1)

The distance between the Barlow lens and the new focal plane is 150 mm.

Untracked exposure time

Using a SLR with a 35mm f/2 objective you want to know how long you can picture a conjunction between the Moon and Venus at 40° of declination before stars trails are visible on your film ?

T : Time (seconds)

F : Focal length of your optic (mm)

D : Declination of the subject (degrees)

T = 1000 / (F * cos D)

For this conjunction the longest exposure time is 37 sec.

NB. This is probably too long both for such a subject and because of the angular coverage of this wide-angle objective. An exposure time from 10 to 15 sec is preferable. For a 200mm used in the same conditions the exposure time is 6 times shorter (6 sec).

Exposure time according the f/ratio

You want to picture the Moon, no more at the resulting focal ratio f/30 but at f/10. What will be the new exposure time if it was of 1/10^th sec at f/30 ?

T_o : Old exposure time

T₁ : New exposure time

f_o : Old resulting f/ratio

f₁ : New resulting f/ratio

T₁ = T_o / ( f_o / f₁ )²

A subject pictured at f/30 an requesting 1/10^th of exposure, will only require 1/111^th sec at f/10; the scope is became 9 times faster !

Optimal focal ratio for a CCD camera (planetary imaging)

What focal ratio must I use to reach the resolution of my CCD camera which photodiods (pixels) are 10 microns wide ?

f   : focal ratio,

T  : pixel size in microns

l  : working wavelength in microns

f = 2 x T / l

The Nyquist's sampling theorem states that the pixel size must be equal to half the diameter of the Airy diffraction disk. Knowing this, for a 10 microns pixel and a maximum spectral sensitivity near l = 0.7 microns, we get a focal ratio of about  f/29, ideal for planetary imaging.

Exposure Calculator

A software from Michael A. Covington

Sky coverage by a CCD camera

For a focal length of 1250 mm, using a MX516c which chip size is 4.9x3.6mm, the sky coverage is 13.5x9.9', a good reason to use a focal reducer to open the scope aperture and fasten the exposition time.

NB. This formula is an approximation based on the equivalence between the tanget of an angle and its measurement in radians, that allows to write for a very small A angle : A(rad) = sin(A) = tg(A). But as soon as A > 5°, the approximation becomes rough and the resultat is no more correct. We can thus not use this formula to calculate the coverage of objectives of digital cameras. In this case we have to use the relation :

S = 2 * Arc Tg ( D / 2F )

For a focal length of 1250 mm, using a MX516c which pixel size is 9.8x12.6m, the resolution is ~1.6"/pixel.

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