S ~ 8250/(F*E)

where S is the error ("slop") in "
F is the focal length in mm
E is the amount of enlargement of the print (3x is the standard for 35-mm film)

The slop is derived from the formula Theta = k*(h/F), with k = 206256 (the number of seconds in a radian) and h = 0.04 mm of image-drift tolerance (an empirical value from astrophotographs).

CONVERSION OF PLATE SCALE TO EFFECTIVE FOCAL LENGTH

EFL = mm per degree * 57.3 = 206265/" per mm

where EFL is the effective focal length in mm
57.3 is the number of degrees in a radian
206256 is the number of " in a radian

MAXIMUM RESOLUTION FOR A PERFECT LENS

Maximum Resolution = 1600/f

where Maximum Resolution is the maximum resolution for a perfect lens
f is the f-number (f/) of the lens

Most films, even fast ones, resolve only 60 lines/mm; the human eye resolves 6 lines/mm (less gives a "wooly" appearance). 80 lines/mm for a 50-mm lens is rated excellent (equal to 1 minute of arc); a 200-mm lens is rated excellent with 40 lines/mm. 2415 films yields 320 line pairs (160 lines)/mm (equal to 1 second of arc); Tri-X yields 80 lines/mm.

MINIMUM RESOLUTION NECESSARY FOR FILM

Minimum Resolution = Maximum Resolution * Print Enlargement

where Minimum Resolution is the minimum resolution necessary for film
> Maximum Resolution is the maximum resolution for a perfect lens
> Print Enlargement is the amount of enlargement of the print (3x is the standard for 35-mm film)

SIZE OF IMAGE (ANGULAR)

h = (Theta*F)/k

Theta = k*(h/F)

F = (k*h)/Theta

where h is the linear height in mm of the image at prime focus of an objective or a telephoto lens
Theta is the object's angular height (angle of view) in units corresponding to k

F is the effective focal length (focal length times Barlow magnification) in mm
k is a constant with a value of 57.3 for Theta in degrees, 3438 in minutes of arc, 206265 for seconds of arc (the number of the respective units in a radian)

The first formula yields image size of the sun and moon as approximately 1% of the effective focal length (Theta/k = 0.5/57.3 = 0.009).

The second formula can be used to find the angle of view (Theta) for a given film frame size (h) and lens focal length (F). Example: the 24 mm height, 36 mm width, and 43 mm diagonal of 35-mm film yields an angle of view of 27o, 41o, and 49o for a 50-mm lens.

The third formula can be used to find the effective focal length (F) required for a given film frame size (h) and angle of view (Theta).

LENGTH OF A STAR TRAIL ON FILM

Length = F*T*0.0044

where Length is the length in mm of the star trail on film
F is the focal length of the lens in mm
T is the exposure time in minutes
0.0044 derives from (2*Pi)/N for minutes (N = 1440 minutes per day)

EXPOSURE TIME FOR STAR TRAIL ON 35-MM FILM

T = 5455/F

where T is the exposure time in minutes for a length of 24 mm (the smallest dimension of 35-mm film)
F is the focal length of the lens in mm

MAXIMUM EXPOSURE TIME WITHOUT STAR TRAIL

T = (1397/F)

where T is the maximum exposure time in seconds without a star trail
1397 derives from 1' at reading distance (254 mm), the smallest angular quantity that can be perceived by the human eye without optical aid ("limiting resolution") and is equal to < 0.1 mm. This quantity also applies to the moon. 2-3x yields only a slight elongation. Use 20x for a clock drive.
F is the focal length of the lens in mm

The earth rotates 5' in 20 s, which yields a barely detectable star trail with an unguided 50-mm lens. 2-3' (8-12 s) is necessary for an undetectable trail, 1' (4 s) for an expert exposure. Divide these values by the proportional increase in focal length over a 50-mm lens. For example, for 3' (12 s), a 150-mm lens would be 1/3 (1' and 4 s) and a 1000-mm lens would be 1/20 (0.15' and 0.6 s). Note that to compensate for these values, the constant in the formula would be 1000 for a barely-detectable trail, 600 for an undetectable trail, and 200 for an expert exposure.

N.B. The above formulae assume a declination of 0o. For other declinations, multiply lengths and divide exposure times by the following cosines of the respective declination angles: 0.98 (10o), 0.93 (20o), 0.86 (30o), 0.75 (40o), 0.64 (50o), 0.50 (60o), 0.34 (70o), 0.18 (80o), 0.10 (85o).

SIZE OF IMAGE (LINEAR)

i = (h/D)*F
h = (D*i)/F
D = (h*F)/i
F = (D*i)/h

where i is the linear image size in mm of the image at prime focus of an objective or telephoto lens (for terrestrial objects, equal to 24 mm divided by the amount of enlargement of the print [3x is the standard for 35-mm film] for the smallest dimension of 35-mm film])
h is the linear height of the object in units corresponding to D
D is the distance of the object in units corresponding to h
F is the effective focal length (focal length times Barlow magnification) in mm

The last formula gives the focal length necessary to photograph a recognizable celestial (Linear Width in km) or terrestrial (Linear Width in m).

EXPOSURE DURATION FOR EXTENDED OBJECTS

E = f^2/(S*B)

where e is the exposure duration in seconds for an image size of >= 0.1 mm
f is the f-number (f/) of the lens
S is the film's ISO speed
B is the brightness factor of the object (Venus 1000, Moon 125, Mars 30, Jupiter 5.7)

Thus, a 2-minute exposure at f/1.4 is equivalent to a 32-minute exposure at f/5.6 (4 stops squared times 2 minutes), ignoring the effects of reciprocity failure in the film, which would mean that the 32-minute exposure would have to be even longer.

SURFACE BRIGHTNESS OF AN EXTENDED OBJECT ("B" VALUE)

B = 10^0.4(9.5-M)/D^2

where B is the surface brightness of the (round) extended object
M is the magnitude of the object (total brightness of the object), linearized in the formula
D is the angular diameter of the object in seconds of arc (D^2 is the surface area of the object)

EXPOSURE DURATION FOR POINT SOURCES

e = (10^0.4(M+13))/S*a^2

where e is the exposure duration in seconds for an image size of >= 0.1 mm
M is the magnitude of the object
S if the film's ISO speed
a is the aperture of the objective