Any calculation for exposure as enlargement changes?

[Matt5791]

What agreat forum - have only just joined!

I have recenly st up my own darkroom and have been doing quite a bit of enlarging.

One question I have is:

Say I decide that 30 seconds exposure is about right for my 5X5 print I am about to make.

Is there any calculation to enable me to decide on comparable exposure if I increase the size of the enlargement to say 8X8?

I have discovered that as the enlargement is bigger I need to expose for longer, but sometimes I don't want to have to go through the busines of making test strips.

Mabey there is a rough estimation or an actual calculation?

Thanks for any help,

Matt

 

[Thomassauerwein]

Other than instinct and testing, I've found this Ilford tool really handy. It is an enlrger light meter of sorts called an EM10. really cheap and very accurate. You can measure a highlight then enlarge then re-set to the same spot, and finaly adjust the f-stop till you have the same exposure.

 

[gordrob]

Try and find an old copy of the Kodak Darkroom Guide for B&W. They have dial calculators built into them for just this purpose as well as a lot of other useful darkroom information
Gord

 

[Gerald Koch]

Kodak publishes a book called the Black & White Darkroom Guide. Among some very useful information it contains a calculator (sort of a circular sliderule). Once you have made a good print you put the information into the calculator and it will tell you the time for any other size print. Most good photostores will have it or order it online. No one just starting out in the darkroom should be without it.

 

[Pastiche]

What I do is look at the height of the enlarging head on the chasis. Many chasi's have a scale on one of the posts, or the post, somewhere.. you dont need to be acurate as far as what height the film actually is, just consistent (precise) on how you read the height. On the Beseler MX45, I always read the height of the motor switch (but any part of the machine that goes up or down with the film will do).

If say, I do an enlargment and the (enter your choice of "marker" here) switch is at 8inches, and I then decide to make things larger, say I raise the head so that my marker is now reading 11" on the post, I double the exposure. It's not dead acurate, but it gets me whithin one print of where I want to be.

What it relies on is that the way light falls off is consistent whether it's going into the camera, or comming out of it. As the distance increases, the light falls off at a predictable rate. I just choose to use aproximate f/stop height readings to judge more or less how much to change exposure times. If I started with the enlarger head at 44", and dropped it down to 6", I'd cut exposure by the same # of stops between f/5.6 and f/45 . . . It's harder to write about than to actually do... Just think about the way light is behaving, and what it's doing as you raise and lower the head. If the times get out of hand b/c of raising or lowering the head, I finish the adjustments with the lens aperture control. It's something I've been trying out to cut down on wated paper myself.. and it seems to be giving me workable approximations. YMMV.

 

[johnnywalker]

Is there any reason why you can't simply calculate it using the square inches of the print? If a 4x5 (20 sq. in) print takes 20 seconds, then an 8x10 (80 sq in) will take 80/20 = 4 times as long at the same f stop.

 

[Pastiche]

JW, I might be wrong - but it seems to me that the relation between sizes is linear, while the behaviour of light is log... (arithmetic? Im no mathematician.)

without knowing if I'm right, I'd guess that each time you double the size of the enlargement, in terms of total square inches, you have to double the exposure (EV). BUT - I think the trouble comes in where you begin to have to account for the kind of lens, and the angle of view (projection in this case) in each case... obviously a longer lens will have a narrower angle of view/projection... hm... I'm just out of my depth here... I'd better leave that last thought unfinished... better to get a solid answer.

 

[Pastiche]

pure speculation

just to throw more sand in your eye...

I have put my sekonic 358 w/ incident bulb on the easel a couple of times... nothing worth feeling confident about.. but I THINK (stress THINK) that it should work. All you really care about is total vol. of light after all...

so -

after you have made one sucessful print at size X -
pull the neg. from the carrier but dont change the head height -
set your incident/spot meter on the easel -
take a reading, and adjust the ISO untill the time matches your exposure time for the last sucessful print -
raise lower the head as desired -
take another reading -

I'd guess it will give you the right exposure for the print.
dial in your time, put the neg back in the carrier-
and expose the paper-
if it comes out totaly screwed up -
say my name in vain along with a good many explicative/descriptive terms

 

[Donald Miller]

This is the formula for changing exposure time for different enlarger heights for the same print:

TO*(LN/LO)^2=TN (the "*" is a multiplication sign, "^2" is "squared")

Where: TO is the original time

LN is the new length

LO is the original length

TN is the new time (at the same f-stop).

Exposure calculations are calculated using the law of inverse square.

 

[Pastiche]

what DM said , which is exactly the same as bellows factor calcs + time variables (aha! a little light goes on)

 

[Lee L]

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.

Lee

 

[Pastiche]

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)

 

[Matt5791]

Other than instinct and testing, I've found this Ilford tool really handy. It is an enlrger light meter of sorts called an EM10. really cheap and very accurate. You can measure a highlight then enlarge then re-set to the same spot, and finaly adjust the f-stop till you have the same exposure.

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.

Mabey this is the thing!!

 

[Ole]

That's it, Matt!

If you have the manual for it, throw it away. Then search APUG for advice on how to use it

 

[Lee L]

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.

Mabey this is the thing!!

Sounds like the Ilford EM-10 (EM =enlarging meter?). See this thread:
(there was a url link here which no longer exists)
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.

Lee

 

[Konical]

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.

Konical

 

[mmmichel]

Noah,

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.

 

[Lee L]

Good Afternoon, Matt,
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.

Konical

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.

Lee

 

[Lee L]

Noah,

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.

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:

TN = 20*(2^2) or 80 seconds, which is correct

(as opposed to TN = (20*2)^2 or 1600 seconds, which is incorrect.)

Lee

 

[nworth]

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.

 

[MichaelBriggs]

The statement of nworth that the exposure can be scaled as the square of the lens distance is correct. (To be really precise, the image distance).

Some of the other statements are either imprecise or not accurate.

If you want to figure out the exposure time compensation by measuring linear dimensions of the prints, i.e., by the two magnifications, the correct formula is (M2 + 1) ^ 2 / (M1 + 1)^2. That is, the square of the ratios of the factors (magnification plus one). It seems obvious that the inverse square law should use the just the magnifications, but this neglects refocusing the lens and isn't accurate. The refocusing changes the effective f-stop.

For a derviation, see Lenses in Photography by Rudolf Kingslake or Applied Photographic Optics by Sidney Ray.

For example, going from M1 = 2 (8x10 from a 4x5 neg) to M2 = 4 (16 x 20 from a 4x5 neg), the change in exposure time should be x2.78 longer, not x4 longer. If you are doing enlargements from a small negative, the "+1" tends not to matter, for example, an 8x11 from 35 mm is M=8 and M and M+1 are about the same, and simplifying the equation gives an answer that is close.

Of course, reciprocity effects or artistic choices depending on print size might cause departures from this equation, which only accounts for light.

It also works to square the ratio of image distances because d_image = focal length * (magnifcation + 1). Maybe this is another way to see that there should be a "+1" in the magnification version of the equation.

For the numeric example above, the 8x10 print from 4x5 neg has d_image = 450 mm and the 16x20 has d_image = 750 mm. The ratio squared is 2.78.

In this example, the two distances between head and the paper (i.e., the the negatives and paper) are 675 mm and 937.5 mm. It doesn't work to take the ratio these distances and square.

 

[Pastiche]

summing up... or is that... squaring up . ..

OK -
Bottom line...
slip a square, double the height of the height,
punch your lights out,
Meter for the low life,
movements and jiggle aren't dance steps, but good derivations,
if you inverse the law of doubling the print you get twice as many prints,
and pack that into a pint real tight and it just may,
turn out O... K...

... ah... and don't forget to pull the filter, or your meter will come off a foot,
and you'll have to wobble on home without leaving a print...

are we all on the same f* . . . STOP!?
when you get there, you'll know, you'll be exausted.
get off the bus.
In the morning.. all you'll remember are the highlights baby, the highlights...

Thanks all for the info. I think I've been overexposed. Matt- how you dealing?

 

[morpheuse]

hmm, i think somone needs to create a software for this - i have a pc in my darkroom, it would be so cool to just punch in the numbers and have the new exposure out. =P

...and i dont mean the calculator either...

 

[Konical]

Good Morning, Morpheuse,

I think a computer would be overkill. A calulator does the job in just a few seconds, and, with experience, you soon reach the point where a guesstimate is usually about right.

Konical

 

[Troy Hamon]

As a mathematical biologist, I find that very often people choose the most difficult line between two points when they are trying to develop a mathematical description. I'd say the same thing applies here. With the exception of Michael Briggs, who has given us a new result that is different than all the others, all of these equations are functionally identical, though not all identical in application (enlarger height versus print dimension).

In terms of Michael's equation, there is no reason that you should have to add one to the two terms. As far as the physics of the situation, there is no basis for it. If it helps to obtain better prints, then sure, go ahead, and since I've never tried it that way perhaps it works reasonably well. As Michael said, for larger prints, the difference between his equation and the others in terms of the results provided becomes smaller and smaller. For small prints, though, it could provide pretty different prints at the different sizes.

The basic physics of the problem is, as others have said, that light falls off at an exponential rate with distance. This does not mean that the problem is particularly difficult, though. A simple way to look at it is to think of it solely in terms of the light put out by the enlarger. There is a certain quantity of light that is coming through the enlarger lens. That same amount is going to come out of the lens whether the image on the print is covering 4x5 or 8x10 or 16x20. So, by simply taking the ratio of the areas in the prints (or the ratio of the squares of the enlarger heights) you can accurately measure the ratio of exposure times. So...a 4x5 is 20 inches square, an 8x10 is 80 inches square, an 8x10 takes 4 times the exposure of a 4x5. The mathematical version of this is

Tn = To*(An/Ao)

where Tn = new exposure time, To = old exposure time, An = new print area, and Ao = old print area. To alter this equation for enlarger distance, it simply becomes

Tn = To*(Dn^2/Do^2)

where Dn = new enlarger distance and Do = old enlarger distance. Neither of these are new, others above have already provided these results in different forms, but this presentation makes more sense to me, so I hope it is helpful to somebody else.

The missing link in the above description is the light scattered during transmission through air. The greater the distance to the print, the greater the potential light loss, affecting our assumption above that there is a uniform amount of light falling on the prints of various sizes. In practice, the only way this would be important is if you were printing over exceptional distances or the air in the darkroom was particularly dirty...

The overall line of reasoning laid out above is related to the reason that teleconverters reduce the effective aperture of your lenses, and why the physical aperture of a lens is larger to achieve the same F-stop (which is a ratio) for a longer focal length...but if people are interested in that I'll post it separately...and otherwise I'll happily keep it to myself.