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?
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. ;^)
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.
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.
Hi Joe,Originally Posted by Joe Smigiel
I have been using my 4x5 for the past 5 years for landscape B&W work- last week I wanted to try a close up self portrait with a 210mm lens and didn't compensate my exposure for the extended bellows. - I was using a flash with manual calculations for exposure. - my exposure was miserable . I searched through the forum and found the solution to my problem. I re shot my portrait today using the table you produced. ( I also checked my exposure against the Quick Disc solution as well). Thank you for the table. I attached a reduced resolution image of a quick digital scan of my negative. ( tomorrow I'll go into my dark room to do a print) My depth of field was a bit shallow - but man, when you fill the 4x5 negative with nothing but "face" you can get some incredible detail-
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