Using your example and Donald's formula (which is the correct way to calculate this):
TO = 30 seconds
LO = 5 inches
LN = 8 inches
LN/LO = 1.6
(LN/LO)^2 = 2.56
TN = 30 seconds *2.56 = 76.8 seconds
With some papers and large jumps in exposure length, you may need to add more time for reciprocity failure. Or you can also compensate with aperture if you have some room to breathe there.
Thanks Lee and Donald. It makes perfect sense once I recognized the format of the equasion.. it IS bellows extention... only we are IN the "bellows" (i.e. darkroom)
What does this ilford tool look like? I bought all my darkroom kit on ebay as a job lot and there is this Ilford thing with a dial and little lights that light up as you turn it. It is white and about 4" long by 1.5" wide.
Originally Posted by Thomassauerwein
Mabey this is the thing!!
That's it, Matt!
If you have the manual for it, throw it away. Then search APUG for advice on how to use it
-- Ole Tjugen, Luddite Elitist
Sounds like the Ilford EM-10 (EM =enlarging meter?). See this thread:
Originally Posted by Matt5791
on how some people use it.
Search Ilford EM-10 in google to see some photos. I think there were a couple of versions over the years. I use a Beseler Analite 500 occasionally, which is essentially the same in function.
Sponsored Ad. (Subscribers to APUG have the option to remove this ad.)
Good Afternoon, Matt,
Assuming that your enlarger has a scale along one standard, along with an indicator giving the distance of the head above the baseboard, the following formula should work well in most cases.
New Distance/Original Distance (Squared), Multiplied by Original Printing Time = New Printing Time.
Example: Original Print Exposure--5 seconds with Enlarger head at 13 inches.
Next Print with the enlarger head at 26 inches.
26/13 = 2; 2 squared = 4; 4 x 5 seconds(OT) = 20 seconds for the new larger print, assuming, of course, no change in f-stop. It's just an illustration that light falls off as the square of the distance.
This is, if I understand correctly, essentially the same idea as Donald and Lee L have described; it just uses enlarger head height instead of print length as the measurement. Since my Beseler MCR-X has a scale on the right-hand standard, it's the most convenient way for me.
Note that JohnnyWalker was correct and that his method is equivalent. In his example, TO = 20sec, LO = 4in, LN = 8in therefore TN = 20*2^2 or 80 seconds, which is 4 times the original time. This is, of course, because (LN/LO)^2 = LN^2/LO^2.
That's it exactly. The change in height of the enlarger is directly proportional to the change in width of the print, so percentage change in enlarger height (new height/old height) would result in the same number as LN/LO, so you square that and multiply the original exposure time by that number.
Originally Posted by Konical
Just to be clear on precedence of operations for the sake of those like me, who are 30 years past algebra classes or otherwise out of practice:
Originally Posted by mmmichel
TN = 20*(2^2) or 80 seconds, which is correct
(as opposed to TN = (20*2)^2 or 1600 seconds, which is incorrect.)
Exposure follows the "square law" vs distance. (i.e. Divide the new distance from lens to paper by the old distance, then square the result, then multiply the old exposure time by that to get the new time.) If you double the distance between the lens and the paper, you must quadruple the exposure. You will have to make a test exposure at the new time and distance, since papers suffer noticeably from reciprocity failure.
The easy way to handle this, as noted above, is to use a meter. Take a reading in a key area (preferably about zone VI in the print), then change the enlargement size, read the same area again, and adjust the aperture until you get the same light level as before.