Mike,
For the 80/4 bellows lens, the equations also need similar modification,
but they are different:
(1) When using the telescoping auto-tube, add 51mm to the tube marking
Ae = A * [(51 + x + y) / 80]
Ae = effective lens aperture
A = lens aperture marking
x = telescoping auto-tube index mark
y = further extension from lens focusing helical
Note: this *only* works for the telescoping tube with the 80/4
bellows lens!
Thus, if the the tube is set to 69mm and the lens helical is fully
retracted (system of tube plus lens focused at infinity), the effective
aperture wide open is:
Ae = 4 * [(51 + 69 + 0) / 80]
Ae = 4 * (120 / 80)
Ae = 6
If the telescoping auto-tube is fully extended to 116mm and the lens
helical is also fully extended (approx. 9mm on the 80/4), this is maximum
magnification for the system of tube plus lens and the effective aperture
wide open becomes:
Ae = 4 * [(51 + 116 + 9) / 80]
Ae = 4 * (176 / 80)
Ae ~= 8.8
If using the bellows, the equation becomes:
Ae = A * [(18 + x + y) / 80]
x = bellows rail index mark
Note: this *only* works for the bellows with the 80/4 bellows lens!
Exposure corrections:
If:
Ae = A * [(51 + x + y) / 80], then
A = Ae / [(51 + x + y) / 80]
This finds the lens aperture marking to use to obtain a specific
effective aperture.
To adjust shutter speed (exposure time), multiply the non-TTL metered
shutter speed by
[(51 + x + y) / 80]^2
This is the square of the factor used to adjust aperture.
Magnification of subject at critical focus distance on film is:
M = [(51 + x + y) / 80] - 1
Thus, the maximum magnification you can achieve at minimum focus distance
with telescoping auto-tube fully extended to 116mm and the lens helical
completely extended (approx. 9mm) is:
M = [(51 + 116 + 9) / 80] - 1
M = (176 / 80) - 1
M = 1.2X
This is approx. 120 0fe-size.
-- John
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