
Practical Formulae
You will find hereunder some formulae that can be useful to estimate various optical values in preparing your night session, like your scope or CCD camera resolution, the sky coverage by a CCD, etc.  Theoretical performances of your scope  Focusing tolerance and thermal expansion  Exposure time according the f/ratio  Amplification factor and focuser intravel of a Barlow  Optimal focal ratio for a CCD or CMOS camera  Sky coverage by a CCD or CMOS camera 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. The instrumental resolution is calculed from Rayleigh's law that is similar to Dawes' law but based on diffraction : with l, the wavelenght in millimeters D, the instrument diameter in millimeters 206265 is deduced from the parallaxe (1 pc/1 UA) In practice, in white light we can use the simplified formula : PS = 0.1384/D, where D is the instrument diameter expressed in meters. We get the below table :
Scope Resolution (PR) in arcseconds 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" (1.1" using Rayleigh's law). A. The magnitude gain Dm : Magnitude gain D : Objective diameter (mm) d : Exit pupil diameter (mm) Dm = 2.5 log_{10 }(D^{2}/d^{2}) = 5 log_{10 }(D)  5 log_{10 }(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 log_{10}(D)  5 log_{10 }(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. Focusing tolerance and thermal expansion Exposed in full Sun, an optical tube assembly sustains a noticeable thermal expansion. So, a Pyrex mirror known for its low thermal expansion will take more than two hours to reach the equilibrium (cf. this software to dowload from Cruxis). In the same time, the OTA will expand of a fraction of millimeter. So, from 10 to 25°C, an aluminium tube (coefficient of linear thermal expansion of 23x10^{6} K) 1000 mm long will extend of 0.345 mm or 345 microns. It is 100 times more than a fiber carbon tube (with a CLTE of 0.2x10^{6} K, a high reistant fibe rcarbon tube expands of 0.003 mm or 3 microns). Calculator : Calculation of the thermal expansion of solids This expansion has an impact on the focal length, and the focusing distance will be extended of a fraction of millimeter as well. It is thus necessary to check the tube distorsion and to compare it with the focusing tolerance (T_{foc}) using the next relation : T_{foc} = ± 8 * (F/D)^{2} * l_{550} * Dl with F/D, the optical system focal ratio, l_{550} the working wavelength and Dl the accuracy of the mirror polishing. For a telescope opened at F/D=6, l_{550} = 0.00055 mm and Dl = l/10, T_{foc} = ± 0.0158 mm or 16 microns. It means that in full Sun, the expansion coefficient of an OTA made of aluminium will be at least 20 time higher that the optical focusing tolerance ! Note that the tolerance increases with the focal ratio (for the same scope at F/D=20, T_{foc} = ± 0.176 mm) and pictures will be much less sensitive to a focusing flaw if you use a longer focal ratio, with of course a smaller field of view. For a DSLR equipped with a 16 mm f/2.8 wideangle lens, the focusing tolerance T_{foc} = ± 0.0034 mm against ± 0.028 mm at f/8. If you want to picture the total solar surface or the Moon in all its sharpnes, being a sphere, in some conditions it is impossible to get a perfect focusing in the optical axis, on the foreground, and in the same time on the limb. But according a small calculation, we can get it. The size of the sharpness field along the optical axis depends in the focal ratio F/D according to the next formula : Radius of sharpness field (°) = arctg (0.0109 * F^{2}/D^{3}) If parameters are expressed in millimeters, the radius of the sharpness field is expressed in degrees. Note that on hand calculators, arc tangent is the tan^{1} key. You can also use this online calculator. For example, for a 200 mm f/6 scope, the radius of the sharpness field is 0.112° or 6'44", or less than the half of the Sun or Moon radius (the Sun diameters is varying from 31'27" to 32'32" and the one of the Moon between 29'23" and 33'28"). To guarantee a sharpness across all the field, you need to increase the focal length of the same scope up to 2000 mm or F/D=10 (radius of sharpness field = 0.312° or 18'44") and even a but more if you wxant to picture a large prominence developping on the limb over a few arc minutes. Amplification factor and focuser intravel 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) / (7550) R = 4500 mm If the amplification factor A = R/F, A = 4500 / 1500 The amplification factor is 3
The focuser intravel 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. 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 wideangle 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 T_{1} : New exposure time f_{o} : Old resulting f/ratio f_{1} : New resulting f/ratio T_{1} = T_{o} / ( f_{o} / f_{1} )^{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 or CMOS 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. A
software from Michael A. Covington Sky coverage by a CCD or CMOS camera FOV : Field of View (arc minute) D : CCD size or diagonal (mm) f : Focal length of your scope (mm) FOV = 3438 * D / f or FOV = Arctan D * 10^{3} / f For a focal length of 1250 mm, using a MX516c which chip size is 4.9x3.6 mm, 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 FOV : FOV(rad) = sin(FOV) = tg(FOV). But as soon as FOV > 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 : FOV = 2 * Arc Tg ( D / 2f ) To download : CCD Calculator v1.4 de Ron Wodaski (et v1.5) FieldofView Simulator, Cloudmakers Field of view calculator, 12 Dimensional String R : CCD or CMOS resolution (arc sec/pixel) P : Pixel size (microns) f : Focal length of your scope (mm) R = 206 * P / f For a focal length of 1250 mm, using a MX516c which pixel size is 9.8x12.6m, the resolution is ~1.6"/pixel. 