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 1a. Focal Ratio Focal Ratio = Focal Length / Aperture f = F / d Where: f = focal ratio F = focal length D = aperture F and D should be in the same units, such as millimeters or inches. Example: A Celestron C11 has an aperture of 11 inches and a focal length of 110 inches f = F / d f = 110 / 11 f = 10 A C11 has an aperture of 11 inches, a focal length of 110 inches, and a focal ratio of f/10.

 1b. Aperture from Focal Length and Focal Ratio Aperture = Focal Length / Focal Ratio D = F / f Where: f = focal ratio F = focal length D = aperture F and D should be in the same units, such as millimeters or inches. Example: A Celestron C11 has a focal length of 110 inches and a focal ratio of f/10 D = F / f D = 110 / 10 D = 11 A C11 has an aperture of 11 inches, a focal length of 110 inches, and a focal ratio of f/10.

 2. The Linear Diameter of the Airy Disk Airy Disk Diameter = 2.44 * Wavelength * Focal Ratio d = 2.44 * λ * f Where d is the linear diameter of the first interspace (the middle of the dark zone that separates the Airy disk from the first ring) in microns 2.44 is a constant λ is the wavelength of the light in millimeters f is the Focal Ratio of the telescope Example: What is the linear diameter of the Airy disk for a telescope with a focal ratio of f/10 at a wavelength of 550 nanometers (0.00055 mm) ? d = 2.44 * 0.00055 mm * f d = 0.001586 mm * f d = 0.01342 mm or about 13 microns A telescope with a focal ratio of f/10 can form an Airy disk with a diameter of about 13 microns at a wavelength of 550 nanometers.

 3. The Angular Diameter of the Airy Disk Airy Disk = (2.44 * Wavelength / Diameter of the Scope) * 206,265 θ = (2.44 * λ / D) * 206,265 Where θ is the angular separation in arcseconds 2.44 is a constant. (1.22 for the Bessell function that calculates the radius of the first interspace times 2 to get the diameter) λ is the wavelength of the light in millimeters. D is the diameter of the aperture of the telescope in millimeters. 206,265 is the conversion factor to convert radians to arcseconds. 206,265 is the number of arcseconds in a radian. A radian is a unit of angle, equal to an angle at the center of a circle whose arc is equal in length to the radius. One radian equals about 57.296 degrees, so 57.296 * 60 minutes in a degree * 60 seconds in a minute = 206, 265. Example: What is the angular diameter of the Airy disk for a telescope with an aperture of 279.4 mm (11 inches) working at a wavelength of 550 nanometers (0.00055 millimeters)? θ = (2.44 * λ / D) * 206,265 θ = (2.44 * 0.00055 mm / 279.4)* 206,265 θ = 0.99 arcseconds A telescope with an aperture of 279.4 mm (11 inches) can form an Airy disk that has an angular size of about 1 arcsecond.

 4. FWHM Angular Diameter of the Airy Disk Airy Disk at FWHM = (1.02 * Wavelength / Aperture) * 206,265 FWHM = (1.02 * λ / A) * 206,265 Where FWHM is the angular FWHM in arcseconds 1.02 is a constant λ is the wavelength of the light in millimeters A is the diameter of the aperture of the telescope in millimeters 206,265 is the conversion factor to convert radians to arcseconds. 206,265 is the number of arcseconds in a radian. Example: What is the angular diameter of the Airy disk at FWHM for a telescope with an aperture of 279.4 mm (11 inches) working at a wavelength of 550 nanometers (0.00055 millimeters)? FWHM = (1.02 * λ / A) * 206,265 FWHM = (1.02 * 0.00055 mm / 279.4)* 206,265 FWHM = 0.414 arcseconds A telescope with an aperture of 279.4 mm (11 inches) can form an Airy disk that has a FWHM angular size of about 0.4 arcseconds.

 5. FWHM Linear Diameter of the Airy Disk Airy Disk FWHM = 1.02 * Wavelength * Focal Ratio FWHM LD = 1.02 * λ * FR Where FWHM LD is the Full Width Half Maximum Linear Diameter of the Airy disk in microns 1.02 is a constant λ is Lambda, the wavelength of the light in millimeters FR is the Focal Ratio of the telescope Example: What is the linear diameter of the Airy Disk to its Full Width Half Maximum in a 5 inch f/8 telescope for light at a wavelength of 550nm. FWHM LD = 1.02 * Lambda * FR FWHM LD = 1.02 * 550nm * 8 Where 550nm = 0.00055 millimeters FWHM LD = 1.02 * 0.00055 * 8 where 1 micron = 0.001 mm FWHM LD = 4.5 microns The FWHM linear diameter of the Airy Disk in a 5 inch f/8 telescope for light at 550nm is 4.5 microns.

 6. Image Scale Per Pixel Image Scale per Pixel = (Pixel Size / Focal Length) * 206.265 IS = (PS / FL) * 206.265 Where IS is the image scale per pixel in arcseconds PS is the size of the pixel in microns FL is the focal length in millimeters 206.265 converts to arcseconds Example: Calculate the image scale per pixel for a Canon T2i (550D) camera with 4.3 micron pixels when used on an 11 inch f/10 telescope with 2,794 millimeters of focal length. IS = (PS / FL) * 206.265 IS = (4.3 / 2,794) * 206.265 IS = 0.00154 * 206.265 IS = 0.317 arcseconds per pixel Each pixel in a Canon T2i camera when used on a 11 inch f/10 telescope will cover about 0.3 arcseconds of the sky.

 7. Critical Sampling Focal Length = ((Aperture * Pixel Size) / (wavelength * 1.02)) * 1,000 * Sampling Factor FL = ((D * PS) / (λ * 1.22)) * 1,000 * SF This is based on the angular radius of the Airy disk. Where FL is the effective focal length in mm D is the diameter of the aperture of the telescope in mm PS is the pixel size in microns λ is Lambda, the wavelength of light in millimeters 1.22 is a constant for the angular radius of the Airy disk 1,000 converts millimeters to microns SF is the sampling factor (usually from 3.5 to 7) Discussion: The Airy disk can be smaller than the theoretical formula indicates. The FWHM angular size is more appropriate for high-resolution planetary photography. Nyquist requires a sampling rate of an absolute minimum of 2x, but a minimum of 3.5x is more appropriate for astrophotography. Example: Calculate the effective focal length needed to critically sample an image with a sample factor of 3.5x with a scope of 279.4 mm diameter aperture and a camera with 4.3 micron pixels for a wavelength of 550 nanometers. For high-resolution planetary work, the critical sampling factor should be a minimum of 3.5x for mediocre seeing up to 7x for excellent seeing. FL = ((D * PS) / (λ * 1.22)) * 1,000 * SF FL = ((279.4 * 4.3) / (550 * 1.22)) * 1,000 * 7 FL = (1201 / 671) * 1,000 * 7 FL = ((5164) / (671)) * 1,000 * 7 FL = 1.790 * 1,000 * 7 FL = 1790 * 7 FL = 12,530 mm To critically sample at 7x on a night of great seeing, for a telescope with an aperture of 279.4 mm, and a camera with 4.3 micron pixels, at a wavelength of 550 nanometers, we would need 12,530 mm of focal length. The native focal length of the scope is 2,794 mm. We would need a Barlow to reach 12,530 mm. 12,530 / 2,794 = 4.48 So we would need about a 4.5x Barlow to reach 12,530 mm of focal length to critically sample at 7x.

 8. Focal Length Needed for a Desired Image Sampling Focal Length = Pixel Size * 206 / Image Sampling FL = PS * 206 / IS Where FL is the focal length needed in mm PS is the pixel size in microns 206 converts to arcseconds IS is the Image Sampling Example: Lets say the seeing is 1 arcsecond on a given night. We need to sample this at a minimum of 3.5x. So we divide 1 / 3.5 = 0.29. 0.29 arcseconds per pixel is the image sampling we want to use. Calculate the focal length needed to sample at 0.29 arcseconds per pixel with 4.3 micron pixels. FL = PS * 206 / IS FL = 4.3 * 206 / 0.29 FL = 4.3 * 206 / 0.29 FL = 886 / 0.2 FL = 3,055 mm 3,055 mm is the focal length needed to sample 0.29 arcseconds with 4.3 micron pixels.

 9. Angular Size Based on Distance Angular Size = (Linear Size * 206,265) / Distance θ = (LS * 206,265) / D Where: θ is the angular size in arcseconds LS is the linear size in kilometers 206,265 is the conversion factor from radians to arcseconds D is the distance to the object in kilometers Size and distance can be in any units as long as they are the same. Example: What is the angular size of a Cassini's Division, which is 4,800 km wide, when Saturn is at a distance of 1,320,000,000 km? θ = (LS * 206,265) / D θ = (4,800 km * 206265) / 1,320,000,000 θ = 990072000 / 1,320,000,000 θ = 0.75 At opposition, Cassini's division subtends about 0.75 arcseconds as seen from the Earth. Encke's division is 325 km. That means Encke's division subtends about 0.05 arcseconds as seen from Earth.

 10. Linear Size Based on Distance Linear Size = (Angular Size * Distance) / 206,265 LS = (θ * D) / 206,265 Where LS is the linear size in kilometers θ is the angular size in arcseconds D is the distance to the object in kilometers 206,265 is the conversion factor from radians to arcseconds Size and distance can be in any units as long as they are the same. Example: What is the actual size of a crater on the moon that has an angular size of 0.9381 arcseconds when the moon is at a distance of 403,635 km? LS = (θ * D) / 206,265 LS = (0.9381 * 403,635) / 206,265 LS = 378649 / 206,265 LS = 1.84 km The linear size of a crater that has an angular size of 0.9381 arcseconds when the moon is at a distance of 403,635 km is 1.84 kilometers.

 11. Effective Focal Length from an Object's Size in an Image Focal Length = (Object Size in pixels * Pixel Size * 206.265) / Object Size in arc sec FL = (OSP * PS * 206.265) / OSAS Where FL is the effective focal length in mm OSP is the object size in pixels PS is the pixel size in microns 206.265 converts to arcseconds OSAS is the object size in arcseconds in the sky Example: Calculate the effective focal length you are working at if you take an image of Jupiter that is 448 pixels wide in the image taken with a Canon T2i with 4.3 micron pixels when Jupiter subtends an angle of 49 arcseconds in the sky. FL = (OSP * PS * 206.265) / OSAS FL = (448 * 4.3 * 206.265) / 49 FL = (397348.896) / 49 FL = 8,109 mm The effective focal length is 8,416 mm if Jupiter is 49 arcseconds in the sky and 465 pixels wide in the image taken with a camera with 4.3 micron pixels.

 12a. Effective Focal Length from Eyepiece Projection Effective Focal Length = Scope Focal Length * Magnification EFL = FL * M Where EFL = Effective Focal Length in mm FL = Scope Focal Length in mm M = Magnification Example: what is the effective focal length of a telescope with a 125 mm aperture at f/8 with 1,000 mm of focal length with the 18mm eyepiece that is 90 mm from the camera sensor. The magnification is 4x, so: EFL = FL x M EFL = 1,000 x 4 EFL = 4,000 mm The effective focal length is 4,000 mm.

 12b. Effective Focal Ratio from Eyepiece Projection Focal Ratio = Effective Focal Length / Aperture F/# = EFL / A Where F/# = F/number or Focal Ratio EFL = Effective Focal Length in mm A = Aperture in mm Example: With a 125 mm aperture f/8 telescope with 1,000 mm of focal length and an 18mm eyepiece that is 90 mm from the camera sensor, the magnification is 4x and the effective focal length is 4,000 millimeters. What is the effective focal ratio? F/# = EFL / A F/# = 4,000 / 125 F/# = 32 The effective focal ratio is f/32.

 12c. Magnification from Eyepiece Projection Magnification = (Eyepiece Distance - Eyepiece Focal Length) / Eyepiece Focal Length M = (ED - EF) / EF Where M = Magnification ED = Eyepiece distance to camera sensor in mm EF = Eyepiece focal length in mm Example: What is the magnification with an 18mm eyepiece that is 90 mm from the camera sensor? M = (ED - EF) / EF M = (90 - 18) / 18 M = (72) / 18 M = 4 The magnification is 4x.

 13a. Effective Focal Length from Afocal Projection Effective Focal Length = (Camera Lens Focal Length / Eyepiece Focal Length) * Scope Focal Length EFL = (CF / EF) * FL Where EFL = Effective Focal Length in mm CF = Camera Lens Focal Length in mm EF = Eyepiece Focal Length in mm FL = Scope Focal Length in mm Example: What is the Effective Focal Length when a camera with a lens with 50mm of focal length is used afocally with an 18mm eyepiece in a scope with 2,794 mm of focal length? EFL = (CF / EF) * FL EFL = (50 / 18) * 2,794 EFL = (2.78) * 2,794 EFL = 7,767 mm The Effective Focal Length is 7,767 millimeters.

 13b. Effective Focal Ratio from Afocal Projection Effective Focal Ratio = (Camera Lens Focal Length / Eyepiece Focal Length) * Scope Focal Ratio EFR = (CF / EF) * f Where EFR = Effective Focal Length in mm CF = Camera Lens Focal Length in mm EF = Eyepiece Focal Length in mm f = Scope Focal Ratio Example: What is the Effective Focal Ratio when a camera with a lens with 50mm of focal length is used afocally with an 18mm eyepiece in a scope with a focal ratio of f/10? EFR = (CF / EF) * f EFR = (50 / 18) * 10 EFR = (2.78) * 10 EFR = 27.8 The Effective Focal Ratio is f/27.8

 13c. Magnification from Afocal Projection Magnification = Camera Lens Focal Length / Eyepiece Focal Length M = CF / EF Where M = Magnification CF = Camera Lens Focal Length in mm EF = Eyepiece Focal Length in mm Example: What is the magnification when a camera with a lens with 50mm of focal length is used afocally with an 18mm eyepiece in a telescope? M = CF / EF M = 50 / 18 M = 2.78 The Magnification of the combined optical system of a 50mm camera lens used afocally with an 18mm eyepiece in a telescope is 2.78x.

 14a. Effective Focal Length Barlow Projection Barlow focal lengths are negative but use the absolute value of the focal length and input a positive number. EFL = scope focal length * ((Barlow distance - Barlow focal length) / Barlow focal length) EFL = FL * ((BD - BFL) / BFL) Where EFL = Effective Focal Length BD = Barlow Distance to sensor in mm BFL = Barlow Focal Length in mm FL = Scope Focal Length in mm Example: What is the effective focal length when a Televue 3x Barlow with a focal length of -52.3 mm when it is used at a distance of 150 mm from the sensor with a scope of 2,794 mm of focal length? EFL = FL * ((BD - BFL) / BFL) EFL = 2,794 * (157 / 52.3) EFL = 2,794 * 3 EFL = 8,382 The Effective focal length is 8,382 millimeters.

 14b. Effective Focal Ratio with Barlow Projection Barlow focal lengths are negative but use the absolute value of the focal length and input a positive number. EFR = scope focal ratio * ((Barlow distance - Barlow focal length) / Barlow focal length) EFR = FR * ((BD - BFL) / BFL) Where EFR = Effective Focal Ratio BD = Barlow Distance to sensor in mm BFL = Barlow Focal Length in mm f = Scope Focal Ratio Example: What is the effective focal Ratio when a Barlow with 18mm of focal length is used at a distance of 50mm from the sensor with an f/10 telescope? EFR = FR * ((BD - BFL) / BFL) EFR = 10 * ((50 - 18) / 18) EFR = 10 * (32 / 18) EFR = 10 * 1.78 EFR = 17.8 The Effective focal ratio is f/17.8.

 15. Maximum Exposure Length for Planetary Rotation Maximum Time = Smear Allowed / ((Pi * Planetary Diameter) / Rotation Period of the Planet) MT = SA / ((Pi * PD) / RP) Where MT = the maximum time of the video in minutes SA = Smear Allowed is the maximum amount of smear in arcseconds allowed Pi = the mathematical constant of the ratio between a circle's circumference to its diameter PD = the diameter of the planet in arcseconds RP = planet's rotational period in minutes Example: Jupiter's rotation period is 9 hours, 55 minutes and 30 seconds, or 595.5 minutes. At opposition, Jupiter's size is about roughly 44 to 50 arcseconds, say 47 for this example. Let's use a value of 0.5 arcseconds for the maximum smear allowed. Pi is, of course, approximately 3.14. MT = SA / ((Pi * PD) / RP) MT = 0.5 / ((147.58) / 595.5) MT = 0.5 / 0.25 MT = 2 min It is important to consider the quality of the seeing and the size of the detail that you are attempting to record on a given night. On nights of very good seeing around 1 arcsecond, using a value of 0.5 arcseconds for the maximum smear allowed is probably ok. On nights of excellent seeing, where you are trying to capture detail down around 0.5 arcseconds, you will need to lower the amount of maximum smear allowed and decrease the maximum time of the video. Likewise, if the seeing is only average at about 3 arcseconds, you can relax the value for maximum smear allowed.

 16. Field of View Field of View = ((57.3 / scope focal length) * frame size) FOV = ((57.3 / FL) * FS) Where FOV = Field of View in degrees FL = Scope Focal Length in mm FS = Frame Size in mm f = Scope Focal Ratio Example: A Canon T2i (550D) has a sensor that is 22.3 x 14.9 mm. What is the size of the field of view on the long side when used with a 1,040mm focal length telescope? FOV = ((57.3 / 1040) * 22.3) FOV = (0.055) * 22.3) FOV = 1.23 degrees The long side of the frame has a field of view of 1.23 degrees. Example: A Canon T2i (550D) has a sensor that is 22.3 x 14.9 mm. What is the size of the field of view on the short side when used with a 1,040mm focal length telescope? FOV = ((57.3 / 1,040) * 14.9) FOV = (0.055) * 14.9) FOV = 0.82 degrees The short side of the frame has a field of view of 0.82 degrees.

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