At 14:20 9/28/02, mike wrote:
John,
I had some time to go over your equations this weekend. Specifically
the corrected equation for effective aperture using the markings on the
lens. This is: (Effective A) = A*[(f+x+y)/f].
f = lens focal length
x = lens extension from tubes or bellows
y = lens extension from focusing helical (if other than infinity focus)
I am trying to compute the effective aperture of the 135/4.5 on the
telescoping auto tube when focused at infinity (65mm ext). I'm not coming
up with anything meaningful probably because I'm not sure I understand the
definition of the variable "y". You state, "y=lens extension from focusing
helical _if focused closer than infinity_" (emphasis is mine) Plugging in
"0" obviously is meaningless.
No . . . if the lens focusing ring is left at "infinity" then there's zero
additional extension from the focusing helical. It was intended for use
with the non-bellows type lenses when used with extension tubes or the
bellows. As an example using this equation with an 85/2 with the 25mm,
14mm and 7mm auto tubes stacked (total additional extension is 46mm), lens
aperture set to f/8, and lens focus set to the infinity mark:
Ae = 8 * (85 + 46 + 0) / 85 = 12.3
The effective aperture is f/12.3 if lens helical is at its infinity
focus position.
I must assume that y must be >0 or if I leave the focusing ring alone and
just use the telescoping tube i can eliminate the "y" variable altogether
(same as choosing a value approaching 0). This gives an effective aperture
of f6.7 at the very best for the 135/4.5. The 135/2.8 yields f4.1 albeit
in a bulky package.
You are using the 135/4.5 "bellows" lens specifically for use on the
bellows or telescoping auto-tube. These lenses cannot be used directly
mounted on the camera body's lens flange, but *must* be used with bellows
or telescoping auto-tube. I will make an educated guess the 135/4.5 was
designed for use with the telescoping tube, although it *can* be used with
the bellows. Even if it could be mounted directly to the camera body, it
would be too close to the film plane to even focus at infinity. Don't try
it; I suspect the lens extends rearward of the flange and would hit the
mirror, possibly damaging the camera body!
The equation I gave needs to be modified when using this lens (or its
sister bellows lenses) with the telescoping tube or bellows. The table for
this lens shows that it is focused at infinity when the lens helical is
fully retracted and the telescoping auto-tube is set to 65mm, or the
bellows is set so its index mark is a 98mm. The telescoping auto-tube
marks give true tube length (extension). The bellows index markings are
33mm longer than its true extension (65mm + 33mm = 98mm). Don't ask me why
this is, it just is (perhaps someone else knows why). To be focused at
infinity, the rear lens node must be extended 135mm from the film
plane. If a true extension of 65mm is required for this lens to be focused
at infinity, then another 70mm is required. Where is it? It's in the
lens, either by physical length, or effective length.
Therefore, when using the telescoping auto-tube, add 70mm to the tube
marking, and use that for "f + x" in the equation. Modifying it for the
telescoping auto-tube, the equation now becomes:
Ae = A * [(70 + x + y) / 135]
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 135/4.5
bellows lens!
Thus, if the the tube is set to 65mm and the lens helical is fully
retracted (system of tube plus lens focused at infinity), the effective
aperture wide open is:
Ae = 4.5 * [(70 + 65 + 0) / 135]
Ae = 4.5 * (135 / 135)
Ae = 4.5
If the telescoping auto-tube is fully extended to 116mm and the lens
helical is also fully extended (approx. 7mm on the 135/4.5), this is
maximum magnification for the system of tube plus lens and the effective
aperture wide open becomes:
Ae = 4.5 * [(70 + 116 + 7) / 135]
Ae = 4.5 * (193 / 135)
Ae ~= 6.4
If using the bellows, the equation becomes:
Ae = A * [(37 + x + y) / 135]
x = bellows rail index mark
Note: this *only* works for the bellows with the 135/4.5 bellows lens!
If you're using the 90/2, 50/2 or 50/3.5 Macro lenses, use the corrected
equation I originally gave. They are focused at inifinity if mounted
directly to the camera body with the lens helical at the infinity mark.
Exposure corrections:
If:
Ae = A * [(70 + x + y) / 135], then
A = Ae / [(70 + x + y) / 135]
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
[(70 + x + y) / 135]^2
This is the square of the factor used to adjust aperture.
Magnification of subject at critical focus distance on film is:
M = [(70 + x + y) / 135] - 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. 7mm) is:
M = [(70 + 116 + 7) / 135] - 1
M = (193 / 135) - 1
M ~= 0.43X
This is slightly over 40 0fe-size.
Hope this helps you out a little . . .
-- John
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