Time calculations when switching print sizes
I am very curious how you folks calculate the exposure time change when switching print size in the darkroom. In my case, I take the longer size of the prints, square them up and then divide. So let’s say I am making my test print in 5x7 and switching to 10x8. I square up 10 and 7 and divide each other 10² : 7² = 2.04
So my new time would double, or I open one stop the lens.
This pretty much worked well for me. But is there a simpler way? What about those height markings on the enlarger column? Are they helpfull when rising or lowering the enlarger? Any ideas would be appreciated.
I'm lazy, and use lots of different sizes including 24x30cm, which isn't "double" anything at all.
So I cheat, and use a EM-10 meter: Highlights are just almost-white when exposed for 20 seconds at an aperture which corresponds to "80" on the meter.
Varies with the paper, of course...
-- Ole Tjugen, Luddite Elitist
Assuming you are using the same negative, your maths seems o.k. The time is proportional to the area of the image therefore it is proportional to a linear dimension squared.
As for the height markings, I tend to make notes so I can get a starting point next time. E.g. I know that with a 6x6 or 6x4.5 negative with the enlarger head 16" high I get an exposure time of about 20 seconds with the lens stopped down two stops. In fact this height is the key to exposure rather than the paper size (assuming a normal negative).
In this case, if I drop the enlarger to 8" I would expect four times as much light (in a quarter of the area) so the exposure should drop to 5 seconds.
Some paper sizes are designed to have 1 stop difference to each other (5x7, 8x10, 11x14, 16x20). You can also use the column height to apply the distance-squared law to calculate an exposure factor, but you need to take the lens-to-paper distances (not the negative-to-paper distances). In either case, don't forget to add some compensation for reciprocity failure. Rule-of-thumb is 1/12 stop for every doubling of exposure.
I use the height markings, and I use them the same way you use the paper dimensions. Suppose my enlarger is at a height of 60 cm. I move it to 90 cm. Suppose I have the first print at f/8 and 22 seconds. My new exposure time is 22 seconds * (90^2/60^2) which is 22*(8100/3600) which is 49.5 seconds.
You can use inches if you like. I am more used to metric measurements. You can use mm if you prefer, too, as long as your units are consistent.
This ignores reciprocity failure, which can be an issue with very long or very short exposures. It will result in underexposure, so if you notice that density is lower after making an adjustment, that might be why.
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Originally Posted by metod
Your maths is correct. If you double the area of a print the exposure must increase by one stop and vice versa, because one stop equals a factor of two. If you double one dimension (as per 5x7 to 10x8) then the area increases four times, so you need an additional two stops.
That formula is probably as easy as it's going to get.
For those who use f-stop printing (and if not why not? ) use
T = 2log(L2/L1) / 0.3
where L2 is the new length of the print, L1 is the original length, and T is the adjustment in f-stops. E.g., if the new size is 14x11 and the old was 10x8, the adjustment required is 1 stop. If the new size is 16x12 and the old was 10x8, the adjustment is 1.4 stops, and so on.
The height scale on an enlarger column isn't a lot of help unless it's marked in magnification rather than inches or mm. However, using the f-stop formula above you could (at least in theory, I haven't tried it) derive a scale for your enlarger column marked directly in f-stops, and then when you changed the height you could read off the difference directly in stops and use that as the correction factor.
This math is correct, but only if you use the lens-to-paper distance (after focusing) and not the negative-to-paper distance. You can use the enlarger column if it is marked in magnification (mine is for several focal lengths), but you can't use the height of the negative above the paper. The distance-squared-law only works for the lens-to-paper distance. Anything else is likely to produce significant errors in exposure, and reciprocity is only of secondary influence.
... the lens-to-paper distance...[QUOTE]
The lens to negative distance is a factor. Just as
with a camera lens the speed an enlarging lens is a
function of it's distance from the film plane.
Camera or enlarger, lens speed is maximum at
infinity. Enlarger lenses are, in use, close-up or
even macro lenses. So, as with LF, adjust
exposure for bellows extension.
Or do as I and Ole do. Buy an Ilford EM-10.
See the EM-10 thread this page. I've calibrated
my EM-10 to use as a ball-park densitometer.
A step wedge is needed to calibrate. Dan
I use a Metrolux II about half the time and I have a gismo that plugs into the timer and reads densitys at the orignial size and then you move the enlarger head and re read the neg and it makes the correction for you.
Lee, it must be a joy to use that gizmo of yours….maybe one day I should get it.
Originally Posted by lee
Anyway, thanks to all for the great suggestions. Richard your formula for f-stop printing is interesting, but isn’t kind of hard to adjust f-stops in decimals numbers?
Jim, your calculation taking into account the height of enlarger is more precise than mine with the paper. I think I’ll stick to that.
Reading up some stuff about the light I discovered another interesting formula. I found it actually in the book about plants and calculating the intensity of light for the plants to grow.
It is I = L⁄D² where I = intensity, L=light output and D is distance.
So I would calculate the Intensity 1 for the first print, then taking into account the distance change for the second print, calculate the Intensity 2. Then divide each other to get the needed ratio. A few different ways to get to the same point.