effective aperture help

Hi everyone, I've joined this forum because I have a class in large format this quarter and I'm struggling to

comprehend and apply the formulas given to me to get a proper exposure.

 

I was first given the formula to get the Effective Aperture, which is;

 

Indicated Aperture multiplied by Bellows Draw divided by Focal Length.

 

I guess my problem is that I'm not sure what to do with the answer. Is that the aperture I shoot at or am I

doing something else with that number?

 

I have another equation that's for the Bellows Extension Factor;

 

Bellows Draw squared divided by Focal Length squared

 

I wasn't sure if I was supposed to apply my Bellows Extension Factor to my answer for the Effective Aperture or

not. I can do the math, I just don't know what to do with my answers or how these equations relate to each

other, if at all. I appreciate any help anyone can give me here. I usually don't struggle with stuff like this,

so this class is especially frustrating for me.

The easiest formula to remember for this is M+1 squared--M is magnification. An example: at 1:1 magnification, 1 + 1=2, 2 squared=4.

4x is your exposure factor, which equals 2 f/stops to correct for exposure. If you don't want to do math, Calumet used to sell a two part

plastic target/ruler that would allow you to read out the exposure correction directly. You could make your own target, and use the normal

centimeter markings that are present on Toyo and other ground glasses. For 35 mm, I took some time and made markings on a metal ruler

that allow me to directly read out the correction by comparing the 6 mm squares on my ground glass with the markings on the ruler.

As a college photography instructor, however, I realize that this may not help you on your final exam...

 

Andrew Gillis

Bellows extension is a function of the focal length of the lens and the total bellows extension. Example: A 210 mm lens is approximately 8 inches. If you're doing macro and you're in focus when you have a total of 12 inches of bellows, you use the rule of thumb "a 50 percent increase in bellows extension requires a 100 percent increase in exposure." Hence, if you have that 12 inches of bellows extension on your 8 inch lens and your meter told you to expose at f11 for 1/4 second, your bellows factor corrected exposure would be f11 at 1/2 second. Greater or lesser bellows extension for a given lens is adjusted accordingly. The Calumet chip does work well on a large format groundglass. There are also bellows extension adjustment formulas that get you an exact number, but I've never had a problem doing the above. Even if you do use the math, it's handy to have a rule of thumb to check your answers.

thanks for the input. I'll try to take some time to see if I can get it to sink in.

The purpose of bellows calculations is to correct your exposure when your lens is significantly farther from the film than the infinity position. This spreads the light out so that extra exposure is needed. Since LF cameras rarely have through the lens metering, compensation of the meter reading is necessary in these cases.

 

The first and simplest rule: if the subject that you are focused on is more than 10 focal lengths away, you don't need to apply any exposure correction.

 

In your first formula, "indicated aperture" is the aperture mark at which you have set the control lever on your lens. If you were focused on infinity, this would be the "effective aperture". What if you are focused on a much closer subject, which means longer bellows draw, the "effective aperture" is a smaller aperture / larger f-number. You should use the effective aperture f-number value to calculate your exposure time.

 

If you focus so that the image is the same size as the subject, the bellows extension will be twice the focal length. Suppose you have the lens set at f16. The factor "Bellows Draw divided by Focal Length" will be 2, so you should calculate your exposure time based in the effective f-number of f32.

 

If you want you can calculate the exposure correction in stops in advance for your lenses and make a table. You can use google to do the calculations. The correction in stops for a lens of focal length F with bellows extension B is 6.64 * log (B/F), with log base 10. Use the same units for B and F. For example, for a 18 cm lens at 26 cm extension: 6.64 * log (26 / 18) = 1.06 stops. You could round this to 1 stop. If you don't have a calculator with the log function, you can enter "6.64 * log (26 / 18)" into the search field of google and google will do the calculation. I figured out the corrections for my lenses in advance so that I don't have to think about it when taking a photo -- I just measure the extension and look up the number of extra stops needed.

 

There are many ways to figure the correction. Also, if the exposure becomes long, typically over one second, you may need to add even more exposure to correct for reciprocity failure.

Without going on at great length, the indicated aperture is only precisely accurate when focused at infinity. The effective aperture formula tells you what the new aperture is when focused closer than infinity. If the F were 200 and the BE 250 and the indicated aperture were f16, the effective aperture would be f 20 using that formula. The hard part of this is that you often have effective apertures that are different from the actual indicated aperture and perhaps you can't tell exactly were to set the f. therefore I prefer the Bellows Extension Factor (BEF) system which changes the film speed and lets you set the pricise indicated aperture.

 

The BEF formula is used primarily for changing the ISO/ASA so that all the other aspects can be used as indicated. If you use the BE divided by the F the result will be more than 1.0 so you then divide the ASA by that factor. On the other hand, if you divide the F by the BE, the factor will be less than 1.0 so you multiply the ASA by that factor. They come out exactly the same way.

 

BE 250/F200 squared = 1.56, ASA of 100 divided by 1.56 = 64

 

F 200/BE 250 squared = .64 times ASA 100 + 64

 

Lynn

I cheat. What if the lens and scale said nothing at all? Aperture is simply a function of dividing the size of the hole the light travels through into the length from the film plane to the node of the lens. Last weekend I was using an antique Pinkham lens on a big studio camera. The lens says it's a 15 inch but my bellows were typically 22 to 27 inches for my still life shots. So I measured the aperture, 55mm. Then I measured the film plane to node distance. 605mm. 605 / 55 = 11. I measured the light and shot for f11. Perfect. But learn all the equations too because that's what will be on the test.

Jim,

 

Your method works pretty well in the situation you describe You have a relatively long lens. Also, you

apparently initially set the aperture at about f/7, presumably because your exposure meter told you that was

right without including the bellows extension. That means the diameter of the aperture is pretty large and can

be measured reasonably accurately.

 

But consider a more typical case. Suppose I am using my 90 mm lens, and my exposure meter tells me that f/22 is

right if bellows extension is ignored. And suppose the reproduction ratio is close to 1:2, which seems to be

the case in your example. That means the bellows extension would be about 135 mm. On the other hand, 90/22 ~

4 mm would be the diameter of the aperture. I have little confidence that I can measure a 4 mm opening in the

diaphragm by holding a scale against the front of the lens at all accurately. Suppose I measured it at closer to

5 mm. I would then decide that the effective f-number were 135/5 or 27, when in fact it would be 1.5 x 22 = 33.

That would put me off by over half a stop. Suppose on the other hand, I decided it were 3 mm. That would

give me 135/3 = 45 for the effective f-number. That would put me off by almost a full stop.

 

So maybe there is some use for those formulas anyway.

Leonard, in the case you just described I would do things slightly differently but equally simple. 90mm / 135mm is roughly the equivalent of f9 / f13.5 round this back to f8 / f12.5. I would compensate for 1 1/3 stops. That's all the math that's needed.

Jim,

 

I'm not sure I understand your method. I think I understand why you rounded f/9 down to f/8, but why did you round f/13.5 to f/12.5? Are you rounding down to the nearest 1/3 stop and you happen to know that f/12.5 is about 1/3 stop down from f/11. Does this method require knowing f-numbers in 1/3 stop increments?

 

I use a tape, on which I've indicated the corrections in f-stops for each of my lenses. I just stretch the tape along the rail and read the correction. Of course, I had to do the calculations once in preparing the tape. But, in practice, it requires no calculations at all.

Leonard,

 

I rounded down because it's easier for my brain to get a handle on an f8 to something comparison than having 2 non standard stops to work with. That's the only reason. The answer is the same either way. f9 to f13.5 is like all black keys on the piano. Hurts my brain.......which doesn't take much.

<p>I think I have a simple and correct answer to your question. Lets take for examlpe that once focused and your indicated aperature is 16 and your total bellows extension is 32 inches and you are shooting a 16 inch lens. Take 16 x 32 ( indicated aperature times total bellows extension) and divide that by the focal length of the lens in inches ( divide mm of lens by 25.4mm equals inches or you can approximate). This will give you an effective aperature of 32. Divide the effecive aperature by the indcated aperature and that will give you 2 or how many stops to open up.<br>

Steve S</p>