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FrancisF
12-Aug-2017, 14:54
I am setting up a shot I have tried before…unsuccessfully. Here are the specs. I am foggy on the bellow extension factor and reciprocity failure. The set up is for a head shot in the outdoors. It is about a 2X enlargement.

20 x 24 Ilford FP4 film (125 ASA)

About 50 inch subject to lens distance

24 inch Goerz Artar Red Dot lens apochromat (American Optical) (no shutter) at f11

Gossen Luna Pro meter reads 17 which indicates a speed of 1/1000 at f11

What exposure should I use?

Thanks,

Francis

blue4130
12-Aug-2017, 15:14
Are you using flash or daylight?

Tim Meisburger
12-Aug-2017, 15:20
Measure the opening in the lens (the aperture) on its widest f stop. Divide 90 inches by that measurement. Assume for a moment that the opening is one inch, then your actual f stop is f90. Use that to calculate exposure and you should be right on. So in your example, if the meter wants 1/1000 at f11, and you actually have f90, then you need to reduce your shutter speed to 125th of a second. Reciprocity is not relevant.

FrancisF
12-Aug-2017, 15:38

Dan Fromm
12-Aug-2017, 15:51
Since you know the magnification, deciding what to do is easy. Effective f/# = (1 + magnification) * f/# set. At 1:1, effective aperture = 2 * aperture set. Set f/16, expose for f/32. At 2:1, set f/16, expose for f/45. And so on.

In your case, if the meter says 1/1000 @ f/11, you want a shutter speed three stops slower. 1/500, 1 stop slower; 1/250, 2 stops slower; 1/125, 3 stops slower. And there you are. Tim nailed it.

David Lobato
12-Aug-2017, 15:54
If the meter is reading an EV value of 17 at ISO 125, it's is too high for a shaded area. Shade would be close to 12 EV, 1/30th at f11 before bellows compensation. Adding 4 stops exposure for the 90 inch bellows is 1/2 second at f11. Someone correct me if I'm wrong on the 4 stops for the 90 inch bellows draw.

Dan Fromm
12-Aug-2017, 16:56
David, thanks for making me revisit the first post.

Francis, with a 24" lens 90" extension, 50" lens-to-subject distance (where did you measure from?) and 2x don't seem quite right. Would you please check your measurements?

2x wants 72" extension (rear node, which will be close to the diaphragm, to film distance) and 36" lens to subject distance (front node, which will be close to the diaphragm, to subject). This isn't your setup.

90" extension with a 24" lens means 2.75 x, lens to subject distance 33".

Jerry Bodine
12-Aug-2017, 18:30

The formula for correction due to bellows extension with ANY lens focal length is:
Exposure FACTOR (i.e., a multiplier) = [extension/focal length]^2. The ^2 means the number in brackets is squared.

Using your info, given Extension=90 and FL=24, the resulting EF=14 actually amounts to 3.8 stops and can be rounded to 4 stops. Since the EF is a FACTOR used to increase your exposure time, the new exposure time becomes 1/40 times 14, or about 1/3 second; so 1/3s at f11. This is approaching the reciprocity region for most films, and since you don't have a shutter to control short exposures anyway you may want to close the aperture down another two stops (giving 1-1/3s) to lengthen the exposure time enough to allow you to use a flat black card/darkslide/hat in front of the lens to time the exposure by hand. Of course, you’ll be limited by how long the subject of your “head shot” can hold still during the longer exposure. Reciprocity would not be severe.

Rich14
12-Aug-2017, 19:18

Rich

Johnny LaRue
12-Aug-2017, 19:20
Just a noob guess:
If your bellows is extended by almost 4x the lens' focal length, add 4 stops to the metered reading.
So f11 at 1/1000s becomes f11 at 1/60s.
Too easy? Too wrong? I'm missing something, right?

Luis-F-S
12-Aug-2017, 19:50
Funny thing, in open shade, I'd use 1 sec at f/22, by the time you stop 4 stops for the bellows and 2 stops to get you to optimum aperture!

Drew Bedo
13-Aug-2017, 06:01
So, lets see . . . .

The focal length of the lens is 24 inches.
The bellows extension is 90 inches.
Each focal length equivalent of bellows extension will require an additional two stops of exposure, either by opening up the lens or lengthening the exposure time.

So: 90 inches divided by 24 inches gives 3.75 focal lengths of extension.

With two stops of additional exposure required for each additional focal length of extension that is 7.5 stops.

(90/24)2=7.5

Do I have that right?

I don't know much about reciprocity issues, but this will be a long exposure.

Dan Fromm
13-Aug-2017, 08:50
A few useful magic formulas for closeup/macro:

These apply to symmetrical lenses. The OP’s Apo Artar is symmetrical.

f = lens’ focal length
m = magnification
Practically speaking the OP’s lens’ nodes are at its diaphragm.

Extension (distance from the lens’ rear node to film) = f*(m + 1)

Whence m = (extension/f) - 1

Distance from the lens’ front node to the subject = f*(m + 1)/m

Effective f/ number = f/ number set * (m + 1)
Exposure time increase factor = (m + 1)^2

If the OP tells his meter he is shooting at his lens’ effective aperture the meter will suggest the right exposure time.

Jerry Bodine
13-Aug-2017, 10:36
All good info, Dan. If someone's kit contains a lens that is known to be un-symmetrical it would be useful to know how the exposure correction is determined. For that I refer to an explanation published in my Leica Manual, excerpted here:

EF = (1 + m/mp)^2 where mp is the pupil magnification; mp is the ratio of the diameter of the exit pupil to the diameter of the entrance pupil (seen from in front of the lens). It can be determined with sufficient precision by looking first into the rear of the lens and then into the front. Obviously, mp=1 for a symmetrical lens but will be less than 1 for an un-symmetrical lens, thus causing a higher EF due to reduced illumination at the film plane. Many lenses for MF or 35mm are likely un-symmetrical designs, and if the camera has through-the-lens metering EF would automatically be accounted for; a rangefinder user would not have that benefit. So the EF for an LF lens that is not known to be symmetrical or un-symmetrical can be determined after measuring its mp.

Rich14
13-Aug-2017, 12:13
So, lets see . . . .

The focal length of the lens is 24 inches.
The bellows extension is 90 inches.
Each focal length equivalent of bellows extension will require an additional two stops of exposure, either by opening up the lens or lengthening the exposure time.

So: 90 inches divided by 24 inches gives 3.75 focal lengths of extension.

With two stops of additional exposure required for each additional focal length of extension that is 7.5 stops.

(90/24)2=7.5

Do I have that right?

I don't know much about reciprocity issues, but this will be a long exposure.

No.

(90/24)^2 Means to square the quantity (90/24)

(90/24) = 3.75

3.75 squared = 14.06

That's the Bellows Factor. You have to increase the exposure 14 times. That's the equivalent of 3.8 stops exposure increase.

To get the number of stops from the bellows factor, either estimate in your head:
1 stop = 2x.
2 stops = 4x.
3 stops = 8x.
4 stops = 16x.
Since 14x is needed, it's a little less than 4 stops. You can safely round to 4 stops.

Or, the number of stops = Ln Bellows Factor/Ln 2.
You can also use Log base10. It doesn't matter.

Ln 14 = 2.6.
Ln 2 = 0.69
2.6/0.69 = 3.78 stops

Log base10 of 14 = 1.15
Log base10 of 2 = 0.3
1.15/0.3 = 3.8 stops

Rich

lungovw
13-Aug-2017, 14:28
My reasoning would be: f/11 with 24 inches focal length means an entrance pupil of 2.18 inches (24 / 11). Then I would measure the new lens/ground glass distance and calculate the actual diaphragm in this situation based on the entrance pupil that remained the same. If the 90 inches (bellows extension) is representative of that conjugated foci, then 90 / 2.18 = f/41.28, that I would round to 45 and get the same fours stops as Jerry and Johnny calculated above. I guess it is all the same. When distance doubles, area quadruples in the base of a cone light.

Tim Meisburger
13-Aug-2017, 19:15
There is a lot of unnecessary calculation going on in this thread. Just remember this formula: f number = focal length divided by internal lens diameter (more properly called entrance pupil). Lets call that F = fl/p. For most LF lenses you can simply measure the entrance pupil with a ruler and you will be pretty close, but it is also possible to calculate it if you have two of the three numbers, as lungovw has above. So

fl = 24, F=11, so p = 24/11 or 2.18.

Now the trick to all this is to remember that the fl stamped on the lens is its focal length at infinity. But in your case, you are not using the lens at infinity, you are using it at 90 inches, Consequently, your focal length for this image is 90 inches!

So, pop that in our equation (F = fl/p) and it becomes F = 90/2.18, or an actual F number of 41 (or 45, as suggested above). Just determine the correct shutter speed for the subject a F45, and the film will be correctly exposed. Then, if its a long exposure, you can deal with reciprocity normally.

13-Aug-2017, 19:57
Goodness. Forget the frigging math. Have the subject hold a Calumet bellows correction wafer on her cheek and measure it with the corresponding ruler on the GG. DONE. It does not matter what the lens is or the format or anything. The proof is in the pudding.

For the life of me the willingness and ability of people here to make this excessively move complicated that needs to be is mind boggling. The objective here is to make photographs. This is not math class. Onward!

Rich14
13-Aug-2017, 20:21
There is a lot of unnecessary calculation going on in this thread. Just remember this formula: f number = focal length divided by internal lens diameter (more properly called entrance pupil). Lets call that F = fl/p. For most LF lenses you can simply measure the entrance pupil with a ruler and you will be pretty close, but it is also possible to calculate it if you have two of the three numbers, as lungovw has above. So

fl = 24, F=11, so p = 24/11 or 2.18.

Now the trick to all this is to remember that the fl stamped on the lens is its focal length at infinity. But in your case, you are not using the lens at infinity, you are using it at 90 inches, Consequently, your focal length for this image is 90 inches!

So, pop that in our equation (F = fl/p) and it becomes F = 90/2.18, or an actual F number of 41 (or 45, as suggested above). Just determine the correct shutter speed for the subject a F45, and the film will be correctly exposed. Then, if its a long exposure, you can deal with reciprocity normally.

Jim,

Respectfully, I disagree. The gyrations that have been suggested over the years for understanding a very, very simple relationship are just amazing.

All to avoid dreaded MATH. (Shudder!).

What is so hard to comprehend? What is so hard to do?

The length of the bellows when extended.
Divided by the lens focal length.
Squared.
That's all there is to it. One silly measurement.
That's the Bellows Factor. Thats the amount the exposure needs to be increased.

1. Two.
2. Four
3. Eight
4. Sixteen
5. Thirty Two
Those are f/stop relationships. Pick the value closest to the Bellows Factor. Guess an in-between value when necessary. How many steps were necessary to get to that value? That's the number of f/stops to open up.

This is not higher math. It's not hard.

If the Bellows Factor is eight, we need 3 stops more exposure.
If the Bellows factor is 20, we need more than 4 stops, but less than 5 stops. Call it 4-1/3 stops. Take the shot.

It's a direct, basic embracing of the actual situation.

Just measure the darn bellows and everything else falls into place.

Anyone who does it two or three times will never forget it. The only special piece of equipment needed is a little tape measure. I carry a spring-loaded cloth one made for sewing. It doesn't weigh a thing. Takes up no room in the camera case. Very low tech. Works great.

Rich

Tim Meisburger
14-Aug-2017, 03:25
Interesting. So by direct calculation I end up needing two stops, and you three. I hope the OP reports the final result! (its such a large sheet of film, I think I would mount a 4x5 reduction back and then test. But I'm cheap!).

Jerry Bodine
14-Aug-2017, 14:14
...I hope the OP reports the final result!...

I hope so too, Tim. We've not heard from him since he first posted his questions. Perhaps he's out actually shooting.

FWIW, years ago, I set up my Sinar Norma at home to check out where on the lens to measure for the bellows extension; I did this for each of my lenses (3 Symmars and 3 Super Angulons) focused on infinity. Found that all the Symmars could be measured to the aperture control ring and all the SA's could be measured to the front of the lensboard. Then I made up a small chart of exposure factors calculated at incremental bellows extensions out to the bellows limit. It includes all six lenses. I keep that little chart in my field notebook (that is always with me); it saves having to do the calcs in the field, avoiding possible errors. I, too, use a very small metal re-coiling tape measure. So I only need to measure the extension, look up the factor, then shoot. It's a very fast system. We all make choices for our workflow, that's mine.

BTW, I'm curious whether I'm the only one who's noticed that OP stated his lens has no shutter. I tried to help him a bit in dealing with that.

Jim Michael
14-Aug-2017, 20:08
From about 22 to 90 in f-stop inches is 4 stops.

Tim Meisburger
15-Aug-2017, 07:08
This tool (Quick Disk) also works well. I printed one and then laminated it with clear packing tape, and it lives in my camera bag. http://www.salzgeber.at/disc/index.html

docw
15-Aug-2017, 08:46
This is a helpful thread, but I would love to see a photo of a 90 inch bellows extension. It must be quite breath-taking.

Jerry Bodine
15-Aug-2017, 09:11
This tool (Quick Disk) also works well. I printed one and then laminated it with clear packing tape, and it lives in my camera bag. http://www.salzgeber.at/disc/index.html

Tim, that certainly is a useful convenient tool. I carry it as well, and it handles the bulk of any close-ups I do. But it is limited to only a two-stop correction.

Luis-F-S
15-Aug-2017, 15:38
From about 22 to 90 in f-stop inches is 4 stops.

FrancisF
15-Aug-2017, 19:43
Thanks to all for the insights and suggestions. I shall try Rich's formula first.

I will be happy to share the results here.....but that is a quandry. It will be a tiny image that can only really be appreciated in person.

Francis Fullam
Chicago, IL

Jim Galli
15-Aug-2017, 20:22
What Tim said in the second post. Measure the hole. Measure the length. Divide. Easy-peasy. 2" hole, 90" extension f45. 1 inch hole 90 inch extension f90. 1 1/2" hole, class? right, f64 No need to make it any harder.

Jim Becia
17-Aug-2017, 10:21

What Jim and Luis state. This is without a doubt, the easiest method. It's simply 4 stops going 22(24) inches to 32, 32 to 45, 45 to 64, 64 to 90.

CHELLM
19-Mar-2018, 16:05
Hullo FrancisF..
I strongly doubt that in a "shaded area" you have an answer from lightmeter of 1/1000 s. at f. 11. Just remember the f. 16 rule in full sun with shutter speed reciprocal of film ASA.. Therefore in full sun you should have this reading: f. 16 @ 1/125 s. Given you taking the picture during the midday hours..
I would check first how correct the reading of lightmeter is (battery ?)

If the subject is in an "open shadow" (i.e shaded by tall building w/out strong reflection from nearby water or glass building) I would open iris up to f. 4 or use 1/8 s. shutter speed if appropriate.. F. 4 @ 125 s. or f. 16 @ 1/8 s. are the starting points to THEN make all calculations advised above ..

Best

Mauro

jwallace68
21-Mar-2018, 19:11
Um.. what about the "no shutter" part of the description? A fractional shutter speed is going to be a challenge using the lens cap..

- JasOn