CHECK PAGE 511 FOR A GRAPHICAL METHOD!
I use the area of the print: double the area, double the light (twice as long an exposure, or 1 f-stop); quadruple the area = 4 X the exposure or two f-stops.
Following the lens to paper measurement ratio method (for those of us without metering devices), wouldn't it be more accurate to measure the difference in the lamp to paper difference, which is really the material issue, and would be a different ratio, given the re-focusing necessary?
it's allhereThere was a thread about this a while back. Now that this new thread is a 'sticky' it needs to come to the correct conclusion.
The equation I use is:
new_time = old_time x (new_M +1)^2 / (old_M+1)^2
where M = new magnification (print/neg) and m = old magnification (print/neg)
The exposure time factor would be:
Factor = (M + 1)^2 / (m + 1)^2
I wonder if Nicholas could supply the magnification in the above examples. I would like to see how this equation (converted from 'factor' to stops) compares to the metered values measured by Nicholas.
You're dealing with the Inverse Square Law. Moving a light source farther from the subject it diminishes proportionately to the distance.
I've worked out a simple Excel program that I simply plug in three variables and come up with New Elevated Exposure
New elevated exposure equals:
((New elevation / Old Elevation) ^2) X Original Exposure
Supposing your 8x10 elevation from lens to paper was 24 inches @ 16 seconds and your New elevation is 30
inches, then: ((24/30)^2) = 2.44
2.44 X Original Exposure (16 seconds) = 39.4 New Elevated Exposure.
I have always used an Ilford EM-10 for things like this, it's a nice handy baseboard light meter..... just remember to use it with your safe light off for greater accuracy.
You'll find it in the attached fileI use the area of the print: double the area, double the light (twice as long an exposure, or 1 f-stop); quadruple the area = 4 X the exposure or two f-stops.
Actually, it is appreciated when someone revives and enhances an existing thread rather than starting a new almost duplicate thread.My apologies for necroing this thread BTW. It was on the first page in the subforum so I didn't bother checking the date. Shouldn't have assumed it was a relatively new topic... haha.
Either my calculator doesn't work or I should return university diploma:
((New elevation / Old Elevation) ^2) X Original Exposure
Supposing your 8x10 elevation from lens to paper was 24 inches @ 16 seconds and your New elevation is 30
inches, then: ((30/24)^2) = (1,25)^2)=1,5625
1,56 X Original Exposure (16 seconds) = 25s New Elevated Exposure.
Could you explain this in more detail with reasons? ThanksSorry but you’re both wrong! The real answer is that it depends exactly upon which enlarger, lamphouse and lens you’re using. With my Leitz Focomat IIc fitted with a 60mm lens, the new exposure would be 26.6 secs, but if the lens is changed to a 100mm then the exposure would be 30.8 secs.
Hi, yes it depends upon the particular optical setup of your enlarger including the construction of the lamphouse and the focal length of the enlarging lens. There is a new app on the Apple App Store called enLARGE, made for iPhone, iPod touch and iPad, that lets you work it out exactly with NO error. Otherwise you’ll never work it out just right, ‘inverse square law’ doesn’t do it.I was printing from one negative last night and had the 8x10 landscape print down pretty well using split filtration, then decided I wanted to crop in on it a bit more and orient it in portrait mode. I ended up raising the enlarger head probably from 10" to 15" or so. It was getting late and I didn't want to bother with another set of test strips, so I just ballparked it a bit and increased the time about 50%. The print came out OK, but definitely too light. I probably should have taken better notes before posting, but I do recall from a failed experiment in the past where I doubled the height of the head and I increased the time by 4x figuring the inverse-square law would work, but it just toasted the print. Has anyone done this experiment before and have a good factor for this sort of thing?
Could you explain this in more detail with reasons? Thanks
pentaxuser
drew, you have a unique talent to make things complicated.
Hi! There is a lot of discussion going on here about how to (and how not to) compute a new enlarging exposure time for a change in print magnification. Basing the computation on, say, the height of the lens, negative or lamp from the print paper, or the new size or area of the print, does NOT work accurately and is, in fact, a waste of time and materials and will NEVER give you a perfectly matching print. The reason for this has to do with the physical make-up of the interior of the enlarger's lamphouse, specifically the fact that (in every enlarger that I know of) the components inside the lamphouse - being its lamp/s, reflecting surfaces and condenser/s (or diffusor/s) do not change their relative positions to each other and to the negative with a change in image magnification. They SHOULD but they DON'T. As a result of this, the brightness of illumination at the negative gate effectively changes as the enlarger is set for different print magnifications, and this introduces an error - call it an 'exposure error curve' which is different and unique for every enlarger, and is also compounded by the focal length of enlarging lens. So when you change the magnification of the enlarger you have two forces at work to contend with: one of them is the simple change in size (or print area or magnification, describe it how you like) which some people (erroneously) relate to the Inverse Square Law (and which cannot be computed by it due to the fact that the enlarger is not a simple point light source), and then you also have the enlarger's own unique 'exposure error curve' caused by its fixed lamphouse components. So how do you make the computation? The answer is to use an app available on the Apple App Store called enLARGE, which I have produced and have posted there. It answers this problem with utter precision and absolutely no error. It answers the question, "If I make a perfect looking print at size X using an exposure time of Y seconds, then what should be the equivalent exposure times for subsequent enlargements of all other sizes?' It does what a computer (your iPhone, iPod touch or iPad) does best, and what would take you at least fifteen minutes to do on a calculator and would probably put you in the nut house first. Every time I talk about it I get my head kicked in by other experts, but that's the bald truth. Best regards - Andrew Wittner, Melbourne, Australia
Could you explain this in more detail with reasons? Thanks
... use the inverse square law to get a reasonable new time, and use a test strip to fine tune the exposure and contrast. ...
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