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mediopollito
13-Mar-2015, 17:59
Hello,

I did a quick calculation of the bellows extension factor for my 210mm lens in a specific situation. I got:

210mm = approx. 8"
Length of bellows when focused on subject = 9"

(9/8)2 = 1.265

Log2 of 1.265 is about 0.37, which would be the difference in stops.

I know that's not a big difference and probably wouldn't matter in this case, but my question is, once the extension factor is calculated and the difference in stops is determined, do I increase or decrease the stops?

An easier example: if I'm currently at f/8 and I calculated the extension factor to be 2, which would mean a 1 stop difference, do I correct it to an f/11 or an f/5.6?

Thank you!

Jim Becia
13-Mar-2015, 18:24
Hello,

I did a quick calculation of the bellows extension factor for my 210mm lens in a specific situation. I got:

210mm = approx. 8"
Length of bellows when focused on subject = 9"

(9/8)2 = 1.265

Log2 of 1.265 is about 0.37, which would be the difference in stops.

I know that's not a big difference and probably wouldn't matter in this case, but my question is, once the extension factor is calculated and the difference in stops is determined, do I increase or decrease the stops?

An easier example: if I'm currently at f/8 and I calculated the extension factor to be 2, which would mean a 1 stop difference, do I correct it to an f/11 or an f/5.6?

Thank you!

You would increase your exposure. In your example you would have to add approximately 1/3 stop, either by increasing the time, or opening the f stop up by a third (for example) from f11 to about f8.7

diversey
13-Mar-2015, 18:26
What do you think you should do in this case?:)

mediopollito
13-Mar-2015, 18:31
You would increase your exposure. In your example you would have to add approximately 1/3 stop, either by increasing the time, or opening the f stop up by a third (for example) from f11 to about f8.7

Thank you, Jim!!

mediopollito
13-Mar-2015, 18:32
What do you think you should do in this case?:)

I know I'm supposed to "increase" the stop, but that is confusing because I don't know if that means to increase the number (which would make the f-stop smaller, say from f/8 to f/11) or open the f-stop more, which would decrease the number (for example, from f/8 to f/5.6)

William Whitaker
13-Mar-2015, 19:28
Ya don't get nuthin' fer free. Gotta buy another stop's worth of light...

Joe Smigiel
13-Mar-2015, 21:36
I know I'm supposed to "increase" the stop, but that is confusing because I don't know if that means to increase the number (which would make the f-stop smaller, say from f/8 to f/11) or open the f-stop more, which would decrease the number (for example, from f/8 to f/5.6)

You need to increase the exposure as bellows extension increases. That means more light is needed so the increase would be towards the larger opening in the f/stop sequence, not the larger number. In your comment above, f/5.6 would represent the proper choice.

mediopollito
13-Mar-2015, 21:43
You need to increase the exposure as bellows extension increases. That means more light is needed so the increase would be towards the larger opening in the f/stop sequence, not the larger number. In your comment above, f/5.6 would represent the proper choice.

Got it. Thank you, Joe!

Doremus Scudder
14-Mar-2015, 04:13
Just to maybe simplify things for later:

F-stops are fractions. f/22 (notice the slash...) is an opening 1/22nd of the focal length of the lens; f/16 = 1/16th of the focal length, etc. 1/16th is obviously larger than 1/22nd... Remember this, and you'll never have to ask which way to go to add more exposure.

Best,

Doremus

Tracy Storer
14-Mar-2015, 09:16
Just to maybe simplify things for later:

F-stops are fractions. f/22 (notice the slash...) is an opening 1/22nd of the focal length of the lens; f/16 = 1/16th of the focal length, etc. 1/16th is obviously larger than 1/22nd... Remember this, and you'll never have to ask which way to go to add more exposure.

Best,

Doremus
To expand on Doremus excellent esplanation.....a 50mm f/1.0 lens(superfast) has infinity focus of 50mm and an entrance pupil of 50 mm, so 50/50 = 1. And THAT'S what f-stops are all about Charlie Brown.

Joe Smigiel
14-Mar-2015, 10:35
Just to maybe simplify things for later:

F-stops are fractions. f/22 (notice the slash...) is an opening 1/22nd of the focal length of the lens; f/16 = 1/16th of the focal length, etc. 1/16th is obviously larger than 1/22nd... Remember this, and you'll never have to ask which way to go to add more exposure.

Best,

Doremus

I always remember the relationship by thinking about pizza. 1/8th slice of the pie is more than 1/11th.

Those darn shutter speeds are fractions too. 1/8th is more than 1/15th.

lecarp
14-Mar-2015, 10:45
I always remember the relationship by thinking about pizza. 1/8th slice of the pie is more than 1/11th.

Those darn shutter speeds are fractions too. 1/8th is more than 1/15th.

What an excellent excuse to always have pizza with me while working!

Mark Sawyer
14-Mar-2015, 10:50
Hello,

I did a quick calculation of the bellows extension factor for my 210mm lens in a specific situation. I got:

210mm = approx. 8"
Length of bellows when focused on subject = 9"

(9/8)2 = 1.265

Log2 of 1.265 is about 0.37, which would be the difference in stops...

I don't know why people use this ridiculous "bellows extension factor" formula. It's one of the Rube Goldberg methods of large format photography that gets repeated often enough that everyone figures "it must be right."

Here's what you do:

1.) Measure your bellows extension, aperture to film plane.
2.) Measure the diameter of your aperture though the front of the lens.
3.) Divide the bellows extension by the aperture diameter.

That gives you your actual f/stop. None of this "divide the square of this by the square of that and get a factor for changing something else" nonsense...

Jim Becia
14-Mar-2015, 11:16
Hello,

I did a quick calculation of the bellows extension factor for my 210mm lens in a specific situation. I got:

210mm = approx. 8"
Length of bellows when focused on subject = 9"

(9/8)2 = 1.265

Log2 of 1.265 is about 0.37, which would be the difference in stops.

I know that's not a big difference and probably wouldn't matter in this case, but my question is, once the extension factor is calculated and the difference in stops is determined, do I increase or decrease the stops?

An easier example: if I'm currently at f/8 and I calculated the extension factor to be 2, which would mean a 1 stop difference, do I correct it to an f/11 or an f/5.6?

Thank you!

Another simple method is this. If you are using a 210, that is approximately an 8 inch lens. If you extend your bellows out to 11 inches, that adds one stop. If you extend out to 16 inches, that adds another stop. By simply assigning the f stop numbers to your bellows extension, you can easily figure your bellows extension factor. The only difference is that you have to add to the exposure when your actual f stop numbers go up - 8, 11, 16, etc. Hope that helps.

Heroique
14-Mar-2015, 11:45
I always remember the relationship by thinking about pizza...

I'm a pizza guy too (in the field, not at my desk), so I thought about sharing this again from a recent K.I.S.S. thread:

I don’t worry about bellows correction unless the distance to my subject is 10x my focal length, or closer.

When this happens, I add 1/2 stop for every 25% increase in bellows extension (beyond infinity).

-----
Example: The OP is using a 210mm lens. 10 x 210mm = 2100mm (or 2.1 meters). So when my subject is 2.1 meters from the camera, or closer, I'll add a suitable amount of light in 1/2-stop increments.

DG 3313
14-Mar-2015, 12:27
I was told (and have done this for some time) to focus the lens at infinity. Measure from the film plane to a point at or near the shutter that is equal to the focal length of the lens...in this case 8 1/4" (210mm). Remember this point.

When shooting closer than infinity (after I focus the camera) I measure the distance between the same points again. Normally this distance is 11-16" for me.

Since I've extended the bellows enough to have measurable light fall off. I look at the difference in distance as F-stops. 8" +/- lens with 11" of bellows draw is the same exposure as the 8" lens focused at infinity set at F-11. That is a one stop loss of light and must be compensated for some how. I just add one stop to what ever my base exposure would gave been.

I don't know how this works with a lens that focuses at infinity closer than the actual focal length....but it works great on my lenses.

photonsoup
14-Mar-2015, 13:02
Wellllll heres how Ive always done it.
Choose your lens, in this case 210mm. choose a f stop in this case f22.
divide 210 by 22 = lens opening of 9.5454mm.
now take your actual bellow extension lets say 485mm. divide that by lens opening of 9.5454. 485/9.5454=50.8
at that extension the marking of f22 is actually f51.

Ken Lee
14-Mar-2015, 14:16
You might find this brief article helpful: Formulas for Bellows Extension and Compensation (http://www.kenleegallery.com/html/tech/bellows.php)

Doremus Scudder
15-Mar-2015, 02:04
I hate doing calculations in the field; even easy ones. And, measuring the diameter of my aperture from the front of the lens would often endanger either me or my subject; unmounting it to measure is cumbersome. I can't do square roots in my head...

I have a number of excuses why I long ago made an easy-to-reference table of bellows extension factors for the various lenses I own out to the maximum bellows extension of the longest camera I own. I do carry a cloth tape measure in my kit, but I can usually estimate extension well enough for black and white film using my hands and eye. I just find (one way or the other) the total extension, consult the table and apply the factor. For obvious extensions I don't even have to consult the table.

Do the calculations once at your leisure, write them down and never bother with them again is my approach. I can then concentrate on important things when photographing, like not double-exposing...

Best,

Doremus

Emmanuel BIGLER
15-Mar-2015, 04:54
From Doremus :
I hate doing calculations in the field

Same for me: of course I love to discuss about calculations here on the forum, and preferably far beyond what is actually useful ;) (for example : what happens if my lens is a telephoto, and if I insist on using it for close-up shots?)
but in the field, of course no computer nor any mobile phone nor any kind of digital device are allowed for me. Even no quartz watch is allowed, still too much silicon inside!

Hence for calculations of exposure compensation, nothing beats the good old and 100% analogue methods like Herr Salzgeber's "Quick Disc"!!
http://www.salzgeber.at/disc/

And as Herr Salzgeber says: 3 • Don´t forget to remove the QuickDisc before taking the picture! (http://www.salzgeber.at/disc/manual.html)

This reminds me the legal disclaimer printed on a cardboard, accordion-shaped sun-shade bought once in California to put under you car's windscreen : Do NOT drive with the sun-shade in place under your windscreen!

Joe Smigiel
15-Mar-2015, 07:38
I long ago made an easy-to-reference table of bellows extension factors for the various lenses I own out to the maximum bellows extension of the longest camera I own. I do carry a cloth tape measure in my kit, but I can usually estimate extension well enough for black and white film using my hands and eye. I just find (one way or the other) the total extension, consult the table and apply the factor. For obvious extensions I don't even have to consult the table.

jnantz
15-Mar-2015, 09:29
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

Vaughn
15-Mar-2015, 09:56
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?

Joe Smigiel
15-Mar-2015, 10:51
seems easier to just be familiar with the third stops (actual f numbers )

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...

Lachlan 717
15-Mar-2015, 11:50
I just use an iPhone app to do it.

Emmanuel BIGLER
15-Mar-2015, 12:21
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).

Kirk Gittings
15-Mar-2015, 14:44
low tech but seems to work fine for me:http://www.salzgeber.at/disc/index.html

Jim Becia
15-Mar-2015, 16:53
low tech but seems to work fine for me:http://www.salzgeber.at/disc/index.html

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.

Kevin Crisp
15-Mar-2015, 17:05
Listen to everybody about the QuickDisc, it's great. Unless the math is why you're getting into this.

mediopollito
15-Mar-2015, 21:01
Wow, thank you everyone for the replies!!

appletree
17-Mar-2015, 09:52
Very helpful information. Love the table posted as well. The math doesn't bother me, but easier to have a quick reference.

So, obvious question...in the OP's example, I knew you needed to increase exposure (since I had been studying up on LF), but not sure why? I assume just due to the bellows being extended and the lens being further away from the film plane you literally need to let more light in to equate to the same exposure value?

Heroique
17-Mar-2015, 10:38
You're basically right, moving the lens away from the film is like moving a lamp away from a wall – the wall gets dimmer...

Through-the-lens metering would normally compensate for this automatically.

But people like us calculate a suitable amount of light to add – and provide it the fun way. ;^)

appletree
17-Mar-2015, 10:48
Great, thank you.

I printed that table posted as well. I guess it matters more on focal lengths and specifics of bellows, etc rather than brand/model of the lens, so it should work to keep handy. I might cut the ones I need and laminate it.

I also have an app for this, but hardcopy is always nice.

Emmanuel BIGLER
17-Mar-2015, 11:23
From appletree
but not sure why? I assume just due to the bellows being extended and the lens being further away from the film plane you literally need to let more light in to equate to the same exposure value?

Yes, exactly.
In fact you only consider what is going on at zero tilt and zero shift, the reference position being on-axis at the center of the image.
From this reference position, the required illumination decreases when you stretch the bellows, following and inverse square law.
When your bellows draw increases, to be very precise, the distance for the inverse square law has to be measured from the position of the exit pupil of the lens.
The situation in terms of illumination is exactly the same if you had a small, flat and diffuse light source located where the exit pupil is located and illuminating the film. In this situation, if you are not too close to the small light source, the amount of light detected per square area of film (at zero tilt), at the center of the field, decreases following an inverse square law versus the distance.

Most lenses we use in LF, with the exception of telephotos, have their exit pupil located very close to the lens board, so in practical use with quasi-symmetrical lenses simply consider the distance from the lens board to the image plane i.e. the total bellows draw. Divide this distance by the focal length and square the number, and you get your exposure factor in close-up photography
The method is even valid when the image is not sharp and grossly defocused!

The other approach implemented in Herr Salzgeber's Quick Disc is to measure the magnification ratio with a test target.
In fact the magnification ratio (image size ) / (object size), for a sharp image (and not for a grossly defocused image!), is directly related to the ratio computed previously
magnification = (image size) / (object size) = (total bellows draw - focal length) / (focal length) = ((total bellows draw) / (focal length)) -1

hence the exposure factor = ((total bellows draw) / (focal length))2 = (1 + magnification)2.

The nice feature in the QuickDisc is that you do not measure the image size in terms of inches or centimeters, but you directly read values in terms of increased exposure f-stops, on a non-linear scale, in the style of slide rules & other analogue calculators of good old days!

In other words, measuring the size of the image of a test target is an indirect way to measure the bellows draw, but with the Quick Disc system, you directly read your correction values without doing any maths!

And now for those who are interested in maths and useless rhetoric ;)

When the lens is a telephoto, the previous formula is only slightly changed and becomes

exposure factor = (1 + (magnification / PM) )2

where PM is the pupillar magnification factor of the lens. This is a fixed value characteristic of the lens formula, it is a fixed value like the focal length for a fixed focal length formula, but can change a lot in zoom lenses when operating the zooming ring (hopefully, we do not have zoom lenses in LF, a real blessing for this discussion ;) ).

For a quasi-symmetrical lens, this PM factor is close to 1 and both formulae are the same.
For a telephoto, the PM factor is smaller than unity, and can be as small as 0.5, for example in old Voigtländer Telomar lenses; the 360 Schneider tele-Arton has a PM factor equal to approx. 0.57.
But nobody uses large format telephotos for close-up, so the general formula has probably never been used by any LF photographer since one and a half century.

Halide
31-Dec-2015, 21:27