Here's the one I made:Originally Posted by Doremus Scudder
Here's the one I made:
seems easier to just be familiar with the third stops (actual f numbers )
2
2.2
2.5
2.8
3.2
3.5
4
4.5
5.0
5.6
6.3
7.1
8
9
10
11
13
14
16
18
20
22
when i compensate ( which isn't often )
i convert the focal length to inches ( as the OP pretty much did )
8"
i measure the bellows extended ( as the OP pretty much did )
9"
think f8 >> f9 which is 1/3 stop
just open the lens up 1/3 a stop ( to the smaller f# )
if you aren't familiar with the inbetween fstops if you google F-number you will get the list of all of them between 0 and 22
in both full stops, half stops, as well as third stops and quarter stops ..
just keep this list in a camera bag with your tape measure and you won't have to do any fancy math
and it seems to work every time.
of course if you like doing fancy math, bringing magnification disks and calculators to use have a blast.
i'd rather just measure, convert to a closeby F# and compensate ... and i'm less apt to make an error
I just add a stop for the bellows and usually add a whole bunch more for the reciprocity failure. Works with most of my B&W film. All the more power to those who think in 1/3 stops!
But I have also cheated. During a studio session I have applied a bit of windage to the exposure and then developed the film (11x14) and then apply my guesstimation for the next exposure. I was using some Efke and the reciprocity failure was extreme! It was not really cheating, of course, but I do not do much studio work, and compared to the field, having the darkroom right there to develop sheets as I exposed them was sort of like having ULF Polaroids...a lot of fun.
A question -- would using tilt (front or rear) ever cause significant fall-off due to the there being greater distance from the lens to the film at the one end of the negative compared to the other?
Rhetorical question of the day: If there is fall-off at one end of the image, does that mean the other end has fall-on?
"Landscapes exist in the material world yet soar in the realms of the spirit..." Tsung Ping, 5th Century China
I think what the formula/calculator users are overlooking in the discussion is that the extension compensation is not difficult to determine if you know the f/stop sequence. You only have one measurement to make and it it linear distance from the lens to the film plane. You could then go nuts with multiplying this by that and using square roots and blah, blah, blah, if you wished. But, as you point out, knowing that 9 is 1/3 stop from 8 and that 10 is a third from 9, etc., gives you the correction with the least amount of effort. Really, the only ciphering needed is to be able to count arithmetically.
If you also memorize only three numbers in the sequence, (e.g., 4.0, 4.5, 5.0) you can easily generate the rest (e.g., 8.0, 9.0, 10.0 is twice the previous and 4.5 x 1.4 = 6.3, etc.).
No wonder Tink is always on Mark's ass...
I just use an iPhone app to do it.
Lachlan.
You miss 100% of the shots you never take. -- Wayne Gretzky
A question -- would using tilt (front or rear) ever cause significant fall-off due to the there being greater distance from the lens to the film at the one end of the negative compared to the other?
Difficult question, but certainly, yes.
If various parts of the image are located at various distances from the exit pupil of the lens, each square inch or square centimeter of the image receives different illuminations.
Moreover, in addition to the effect of the distance (inverse square law) there is an additional effect due to the fact that light can be spread on a slanted area. And this occurs in the edges of the image projected by a wide-angle lens without any tilt.
The basic explanation and photometric formula is simple but I do not see how to derive simple rules applicable in the field.
It is important to notice that the position of the exit pupil is crucial for the distribution of light in the image. The mean ray connecting the center of the exit pupil to a given image point gives the distance, and defines two angles between this ray and the plane of the exit pupil and the image plane.
In quasi-symmetrical lenses like wide-angle or standard view camera lenses, the exit pupil is located close to the iris, close to the lens board.
But with a telephoto, the exit pupil is not located close to the lens board, and the distances for the inverse square law cannot be measured from the lens board. For this reason the excellent Quick Disc fails to tell you what happens in your telephoto. You would have to re-compute a correction scale, taking into account the particular optical formula in use (actuall only the pupillar magnification is required to re-compute a dedicated telephoto Quick Disc).
Rhetorical question of the day: If there is fall-off at one end of the image, does that mean the other end has fall-on?
Difficult to say since there are two effects, the effect of the distance to the center of the exit pupil and the effect of the pupil plane and the image section being slanted with respect to the mean ray coming out of the exit.
In the case of extreme tilts combined with shifts, one can imagine situations where both effects add or subtract. Both effects can probably subtract, a longer distance at some place, but a piece of image perfectly perpendicular to the mean ray at the same place (needs to draw some sketches to see if this compensation effect is possible or not).
With an optical fiber connected to an exposure meter (such an accessory existed in for the Gossen Luna-pro series), it should be in principle easy to map the distribution of light; I would take the ground glass off to do this. And may be cover the entrance of the fiber with a piece of frosted plastic acting as a diffusor (like for a luxmeter).
low tech but seems to work fine for me:http://www.salzgeber.at/disc/index.html
Thanks,
Kirk
at age 73:
"The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep"
I use this also on 95 % of my closeup work. The only time I don't is when placing the disc in the scene might somehow disturb it. I carry about 5 or 6 and have laminated them so they last quite awhile.
Listen to everybody about the QuickDisc, it's great. Unless the math is why you're getting into this.
Wow, thank you everyone for the replies!!
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