Easy Bellows Extension Factor

[@cweeks5211]

Literally, one of the top-five photo channels on KZbin!

[@jpcgordon]

I tried to get answers about bellows compensation on forums and got confused very quickly. This video is fantastic and clear, thanks so much!

[@rredd7777]

A nice, clear explanation. I was always confused by this concept before, but now I think I got it. Thanks!

[@mikenorton2727]

Like the channel. Also in these calculations is film repocracy and any additional filter to offset colour imbalances. Maths work to a point. Measurement on the ground glass and at subject helped for bellows extension.

[@AtlantaTerry]

Buy a cheap cloth tape measure or cut off a piece of gift wrap ribbon. Extend the tape/ribbon to the maximum length away from your ground glass. Cut the excess off. Calculate several bellows extensions then mark the tape/ribbon. Store one of these with each lens. To use: Compose and focus. Extend the tape/ribbon. Adjust your exposure. Expose the film. Terry Thomas... the photographer Atlanta, Georgia USA

[@Dadbros420]

Can you give us a pdf to that cheat sheet you made? Would be super helpful to have on the field :)

[@carlhale5405]

Thank you. I’m going to begin with a 240mm lens on a 4x5 Calumet camera bought on eBay. Will figure out the bellows extension cheat sheet option.

[@troystaten5633]

There is a fun cheat that I used to use back in the day. Using a tape measure and figuring the length of the lens in inches and the bellows draw in inches you put your thumbs on both the length of the bellows draw and the length of the lens then slide both thumbs the same amount so they line up with f stops and you have your exposure compensation. Example an 8" lens (210mm) 16 inches of draw is 2 stops of compensationn IE F8, F11 F16 the inches translate to F stops. . It works with any length of lens or bellows draw. It might not be accurate to a tenth of a stop but more than accurate enough for work.

[@jimpurcell]

Great explanation! Do you have kodak dark green safelight on the back of that shirt? Would be cool!

[@user-lu8mc3wo3i]

I worked as a professional photo finisher for about 15 years from 1980-95 and changed careers to Cisco networking in which I’ve been working the last 25. I’m getting near retirement and was thinking about opening a small photo studio/lab and wanted to get your advice as to the state of the industry and if you consider it a reviving market (kind of like vinyl in music). I’d love to get back into it. Thanks and great videos.

[@strega-nil]

I think there's an easier way than this that gives you a direct number of stops to change, which is: 2 * log2(bellows / focal) which ends up being: 2 * ln(bellows / focal) / ln(2) or approximately 2.9 * ln(bellows / focal) Thus, if you have an iphone and flip it on its side, you can get the stops to change by entering in: bellows (/) focal (=) (ln) (*) 2.9 with the measurements from the video, you type in 220 / 135 = ln * 2.9 and get an answer of 1.4 stops.

[@TheNakedPhotographer]

Another formula is (v * f)/F where v is the bellows extension, f is the aperture the lens is set to, and F is the focal length of the lens. The resulting number is the effective aperture. So if the lens is 150mm but extended to 300mm and set to f/11: (300 * 11)/150 = 22. The lens may be set at f/11, but acts like it’s set to f/22. The formula I discuss in the video is the widely used bellows extension formula, but all three work to arrive at the correct answer.

[@Larpy1933]

At 3:37, you measure “from the film to the front” as 22 cm. Good. That’s fine. My point here: it would be wise to focus that lens on infinity and measure “from the film to the front” to check if that corresponds to your lens’s focal length. (I think you know how to deal with any discrepancy.) Thanks. I found this very helpful. I have just begun using a Mamiya C220. It has a built-in “bellows extension corrrctin factor scale” that pulls out of the camera body with the “lens board”. I’ve concluded the OEM scale is incorrect. I’m using a 55/4.5 lens. At 4 or 5 metres subject distance, the correction factor shown is 1.5. This makes no sense. My Rolleicord w/ 75/3.5 lens focuses to 0.9 metres and needs no correction factor ever. I think I’ll make a new scale (self adhesive) and apply it over the OEM engraved metallic one.

[@billhackley3540]

very helpful, would this formula be applicable to my mamiya rb67 with such a reduction in overall bellows size, the chart/graph on the camera just confuses me

[@tedsmith6795]

I understand how to calculate the bellows extension factor until it comes to trying to understand telephoto lenses. Am I correct that the infinity distance on a 300mm telephoto could be say 200mm. Therefore the BEF is calculated by not using the focal length of the lens but the infinity measurement. I am a little confused and need some advice please.

[@natereed3409]

Aperture = Focal Length / Lens Diameter. A 135mm lens at F8 will have a aperture with an 16.875mm diameter opening (135/8=16.875). You can find the diameter of your aperture for all F stops and then refer to your aperture in millimeters rather than "stops" (F8=16.875mm, F22=6.136mm, etc.). This way if I'm focusing very close up with that 135mm lens, I set the desired aperture for depth of field but all I need to do is measure my focal length and then divide it by whatever the opening diameter is. Doing that you get your exact F Stop for your lens at that bellows extension. Easy peasy only one thing to measure in the field.

[@TheNakedPhotographer]

You’re using the method of effective aperture, but it seems you’re doing it the hardest way possible. I’ll post the formula here when I get home. The aperture is measured by the size of the entrance pupil when viewed from the front of the lens. I mention that because you don’t say what exactly you’re measuring and other people reading this may not know.

[@TheNakedPhotographer]

The effective aperture formula is (v x f)/F, v is the lens to film distance, f is the original f stop, and F is the focal length of the lens. So for a 150mm lens at f/8 extended to 300mm it would be (300x8)/150= (2400)/150= 16, so the effective f/stop changes from f/8 to f/16.

[@natereed3409]

@@TheNakedPhotographer I don't see how it's harder to get the math done at home and just look at a "cheat sheet" in the field. When I take an exposure all I have to do is measure my working focal length. I take that measurement and divide it by whatever my lens diameter is (that I have already written on my lens for each F Stop). That's it, I divide one number by one number. How is that harder?

[@TheNakedPhotographer]

You do you

[@natereed3409]

@@TheNakedPhotographer Just wanted to share what I think is a faster way to get to the same outcome. I apologize.

[@gerardodalchielelueiro6818]

I made my solution with a carsboard rule that have some distances and the ractor calculated for any one lens I use Best regards from Argentina

[@vijaykarve736]

what calculations would you use if you are using a modern crop sensor camera with bellows?

[@TheNakedPhotographer]

If you are using the internal meter then none. It will calculate it for you. If you are using a separate meter then you do the same formula I presented here.

[@TaylerAleks]

Thanks for your videos, dude

[@UlrichBrinkmann]

After 2 minutes into your video I had to stop. And now I'm writing this. First let me say I'm using exactly the formula you explain her for the occasional macro shots (which I got from Ansel Adams). So everything fine. Second, though, explaining the need for compensation by the increasing image circle -- which supposedly distributes the same amount of light over a bigger target area - irresistibly provoked my 'defeat confusion' reflex. I immediately thought of my 120mm Super Angulon which I use for 4x5 architecture. There we have a large image circle at infinity and no compensation necessary. So - as the formula suggests, I always found it helpful for understanding the compensation to look at the lens end rather than the film end and see that there the relevant change happens. The aperture opening gets *smaller* in relation to the lens-film plane distance. So what you are doing when extending the bellows is extending the focal length, whereas the aperture remains the same. Since the f-stop is the relative measure of lens opening to focal distance that's what you are decreasing. So when you're doubling the bellows length, you're doubling the effective f-stop number (e.g. 11→22). Which is four times less light. So I don't find the light distribution area explanation wrong, but rather inviting some confusion. (Took so many words to explain, omg. )

[@TheNakedPhotographer]

I understand how you are thinking, but the effect I described also plays into studio lighting where no aperture exists between light and subject. With bellows extension, the amount of light entering the camera remains unchanged, but less of it falls on the film since it is spread differently. Changing the aperture would change the amount of light entering the camera. In the studio, if the light is moved further back, it is also spread out and reduces exposure on the subject.

[@UlrichBrinkmann]

@@TheNakedPhotographer Yes but. The spreading out is more 'incidental' than the moving back. You can put blinders on your light source and so cut off the spreading out, but that doesn't mean there is more light on the subject (although the circle of coverage is greatly reduced).

[@rubenisedgars]

Another great video!! One question - shouldn’t this formula work also for extension tubes on a 35mm camera? The principle seems to be the same. However, the results are not. Using the formula for a 26mm tube and a 50mm lens give a factor of 0.2704 which takes the exposure into heavy underexposure.

[@TheNakedPhotographer]

Most 35mm cameras have built in meters that automatically compensate. But, the math would be 50+26=76mm extension so (76x76)/(50x50)= 5776/2500, which comes out to a factor of 2.3. That is just over one stop more exposure.

[@rubenisedgars]

Thanks for the response! I'm using a hand-held meter, that's why the question. Just to understand - in case of my 35mm example - why are you squaring the summed length of extension tube + lens (26+50) but in case of your bellows example just the length of the bellows was squared (the 135mm of the lens were not counted in for extension)?

[@TheNakedPhotographer]

You square the extension, in your case the extension is the lens plus the tube. In my large camera the measurement includes the same thing, it just looks different.

[@rubenisedgars]

Ok, I see. Thanks for this!

[@davyboyo]

Interesting. I understand the inverse square relationship with respect to the intensity of light, but your explanation got me wondering. Does the image circle of of a given lens affect the light intensity the film sees? For example I use a fujinon 180mm for 4x5 and the circle is much larger than some other lenses with the same focal length and max aperture ( I think it's close to 8x10 coverage from memory ). I can therefore only assume there is less total gathered light for a given area of film compared to the same spec lens with a smaller image circle? In which case using t stops would become somewhat of a necessity.. Or is there some inherent property of lenses that I'm not accounting for?

[@TheNakedPhotographer]

If think about it, the portion of the image circle that covers the 4x5 negative is giving the same amount of light when it is focused at infinity regardless of whether the edges extend further out to cover 8x10. Its when we extend the lens further from the film that we have a problem.

[@davyboyo]

@@TheNakedPhotographer that part makes sense. Now that I think about it more, I assume that a lens with a larger image circle has a larger image gathering circle, so two lenses of the same focal length and max aperture with different circles actually do transmit different quantities of light, but the same quantity for a given area of the image circle. In other words my question was pointless. While I'm here I just want to say your channel is among the best and most informative I have come across. Thanks so much, I've learned heaps from you.

[@manuelalvarez6566]

The light bulb came on. Thank you

[@jaywesselink8709]

About 30 years ago assisting multiple commercial table top photographers I was introduced to the 2 inch card with a scale marked index card to measure bellows factor. It works so easily with large format or even medium format cameras and lenses any focal length lens. You simply put the 2 inch card into the scene for a moment to then use the scale card to measure the 2 inch card projected on the ground glass. The scale is marked in thirds of a stop from 1/3 f-stop to three stops.

[@TheNakedPhotographer]

I have one of those made by Calumet. I have never used it.

[@jaywesselink8709]

The Naked Photographer I haven’t been doing any close-up work lately but it is so simple to just set the card in there and measured on the glass ground glass you don’t have to get your calculator out for it. I understand all the math but this is so quick.

[@jaywesselink8709]

The Naked Photographer Wayback in the day I simply copied the scale and have kept it on file and in my camera case. And for a 2 inch card you can just use a 35 mm slide mount.

[@fotolookconde]

Can we not use an App instead of measuring and working out maths?

[@TheNakedPhotographer]

How can an app know the extension of you don’t measure it?

[@nickfanzo]

You need to measure, don’t be lazy

[@fotolookconde]

@@TheNakedPhotographer what I meant was after take measurements then using an app to work out the aperture

[@jimpix8019]

I used to work with large format cameras for many years. Yes, the Bellows factor system was always used but some people had problems with it. So I found that anything less than 25% reproduction size. i.e. the size of the image on the focus screen, compared with its actual size, would be ok to shoot at the settings on the light meter. Anything bigger, i.e. using a 10x8 camera: with an image focused to 8” high on screen. But the real life size of the item is 2” high, the ‘Repro size’ is 400% (aka: a factor of 2) 100-200 is 1 stop extra light. Then 200-400 is another 1 stop extra light. Ergo, 2 stops extra light to add to the original light meter reading. This method is quick and easy to apply for all sizes in between or higher. i.e. let’s say reproduction size 7 times bigger than the original subject. This would equate to 700% repro size. So 200% = +1stop 400%= +2 stops 800%= +3 stops (& so on) -------- So 700%= + 2.75 stops Whether you want to work at 1/2 or 1/3 stop increments, This will be easy to apply by just taking the 2 measurements. 1. The actual subject. And 2. The Reproduction size on the focus screen. Use a calculator to calculate exact percentage which can then indicate how much extra light is needed. Remember, anything less than 25% , just go by the light meter reading. Sorry to be long winded, but I just wanted to avoid confusion. May the Force be with you.🌀 Jim🌀 (-: :-)

[@unbroken1010]

That more confusing then what he was explaining

[@heatonize]

i don't know how i just realized how similar your voice is to howard stern

[@andresmills7915]

The new intro caught me off guard 😆

[@alanhuntley55]

OMG, your method is certainly comprehensive, but unnecessarily difficult IMO. Nowadays, I would simply pull out my phone and plug the extension into an app for the correction. For 30 years prior to having a smartphone, I worked like this: let's use a 210mm (8 1/4") lens as an example: convert the lens to an f/number (f/8 in this case), measure the extension and convert to the closest f/number (say, 12" extension would be f/11), and the difference is your extension factor (1 stop more exposure for 12" extension.) Is it exact? No, but as you said, close enough is good enough.

[@stealthvanlife6867]

I agree with you. Certain things like this particular subject can get off the rails and keep people from exploring large format because it seems complicated or intimidating.

[@jonjon1217]

Or, you can try doing what I do and that's simply smashing my camera on the ground and stomping my feet as I mumble to myself about hating school, hating math and now, hating to use a bellows system! 😳

[@carlosenriquez2092]

I freaking hate math.