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1a. Focal Ratio
Focal Ratio = Focal Length / Aperture
f = F / d
Where:
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.
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1b. Aperture from Focal Length and Focal Ratio
Aperture = Focal Length / Focal Ratio
D = F / f
Where:
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.
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2. The Linear Diameter of the Airy Disk
Airy Disk Diameter = 2.44 * Wavelength * Focal Ratio
d = 2.44 * λ * f
Where
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.
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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
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.
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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
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.
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5. FWHM Linear Diameter of the Airy Disk
Airy Disk FWHM = 1.02 * Wavelength * Focal Ratio
FWHM LD = 1.02 * λ * FR
Where
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.
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6. Image Scale Per Pixel
Image Scale per Pixel = (Pixel Size / Focal Length) * 206.265
IS = (PS / FL) * 206.265
Where
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.
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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
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.
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8. Focal Length Needed for a Desired Image Sampling
Focal Length = Pixel Size * 206 / Image Sampling
FL = PS * 206 / IS
Where
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.
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9. Angular Size Based on Distance
Angular Size = (Linear Size * 206,265) / Distance
θ = (LS * 206,265) / D
Where:
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.
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10. Linear Size Based on Distance
Linear Size = (Angular Size * Distance) / 206,265
LS = (θ * D) / 206,265
Where
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.
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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
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.
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12a. Effective Focal Length from Eyepiece Projection
Effective Focal Length = Scope Focal Length * Magnification
EFL = FL * M
Where
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.
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12b. Effective Focal Ratio from Eyepiece Projection
Focal Ratio = Effective Focal Length / Aperture
F/# = EFL / A
Where
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.
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12c. Magnification from Eyepiece Projection
Magnification = (Eyepiece Distance - Eyepiece Focal Length) / Eyepiece Focal Length
M = (ED - EF) / EF
Where
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.
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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
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.
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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
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
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13c. Magnification from Afocal Projection
Magnification = Camera Lens Focal Length / Eyepiece Focal Length
M = CF / EF
Where
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.
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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
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.
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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
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.
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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
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.
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16. Field of View
Field of View = ((57.3 / scope focal length) * frame size)
FOV = ((57.3 / FL) * FS)
Where
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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