Adjusting Exposure Times With Print Size

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rshepard

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I recall reading a while ago a formula for adjusting print exposure time as the enlargement size changes. But, I cannot recall where I read this. I assume that with an f-stop timer it's relatively simple to make these adjustments.

My reason for asking is that I have several negatives I'd like to print on 5x7", 8x10" and 11x14" paper to test my theory that different images look best at one size; at least, they look better than when larger or smaller. It would be nice to quickly adjust the exposure time (or lens aperture) and make a series of three prints while the negative is still in the enlarger.

Rich
 

Sirius Glass

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Twice the linear dimension on one side means four times the exposure [2 stops more] because you are dealing with area and not the length of one side. You would be better to open the lens two stops rather than increase the time because the longer exposure might take you into the realm of reciprocity failure.

Steve
 

DannL

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Square your lens to paper distance for the first print size.
Square your lens to paper distance for the new enlarged print size.
Divide the second result by the first result.
Take the first good exposure time and multiply it by this result to arrive at your new exposure time.

Example:

10"x10"=100
15"x15"=225
225/100= 2.25
If the first exposure time was 11 seconds then multiply it by 2.25 for the enlargement. ie; 24.75 sec.

My interpretation from "You and Your Prints", by William Hawken ISBN 0-8174-2114-9

I have used this method many times with good results.
 
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Woolliscroft

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It's the same inverse square law used in flash calculations: to double the lens to paper distance you need four times the time. In practice, though, I find that it's easier to re-meter or do a new test strip when you enlarge more (or less).

David.
 

clogz

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You might also find that when enlarging you need a slightly harder grade of paper.

Hans
 

matti

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Or you can look at Dead Link Removed.
by Ralph W. Lambrecht. From his book Way Beyond Monochrome, I think. It should be easy to use if you have a scale on your enlarger.
I mostly use a table in Tim Rudmans "Master printer"-book. I often use it to first make test prints and moving up for the finals.

/matti
 
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dancqu

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Twice the linear dimension on one side means four
times the exposure [2 stops more] because you are
dealing with area and not the length of one side. Steve

The working focal length of the lens must be taken into
account. As the lens is focused for larger prints it's focal
length is lessened and the lens approaches it's rated speed.
The lens becomes faster. The reverse holds true when
increasing the working focal length.

An Ilford EM-10 enlarging meter works well and is just
the thing for what the OP has in mind. Works also as
a ballpark densitometer if calibrated. Under $30. Dan
 

Mike Crawford

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Hello
Another variation of the inverse square law, similar to DannL's is the attached formula. It's easier than my scribble looks. The new length (L2) is divided by the old length. (L1) This figure is then squared and this is the factor to multiply your old exposure (T1) to find out the new one.(T2) In this example, a 5x4 print at 15 second exposure is calculated to 60 at 10x8. That's an easy one as if the paper is four times as big, it will obviously need four times as much light. Bit more usefull if it's something like 17.6 seconds. The formula has certainley helped when I suddenly realise I've only got a few sheets of 20x24 paper left to make one print as you can get a small print correct, then go straight for a big one, though as clogz points out a slight increase in contrast may be neccesary.
Cheers
Mike
 

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rshepard

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Thank you all for validating what seemed reasonable to me. Also, the pointers to Ralph's and Tim's books reminded me where I read about this.

I do have Nick Linden's exposure meter so I can run some tests to calibrate the adjustment to my equipment, lenses, and paper.

Much appreciated,

Rich
 

pentaxuser

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There's also a small programme available at Dead Link Removed where you put in the old and new dimensions and it works out the exposure time for you.

pentaxuser
 

pentaxuser

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Sorry I have this bookmarked under Print Adjustments. I got it from an earlier thread where someone kindly directed posters to this site. I simply copied the address from the site but I have just tried it by clicking on the copied address and it said not found.

I am not computer literate enough to know what to suggest now. Maybe it has simply disappeared

pentaxuser
 

Nicholas Lindan

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I recall reading a while ago a formula for adjusting print exposure time as the enlargement size changes ... I assume that with an f-stop timer it's relatively simple to make these adjustments.

I certainly do hope so.

As the subject of threads seems to be pre-ordained
by the stars and planets, from a recent post on the Pure Silver list:

The easiest solution is to use an enlarging meter and
f-stop timer.

Measure the difference in light with the
meter [the meter has a special function for
doing just this] and add the meter
reading to the f-stop timer setting. You
are done ... accurate to 1/10th of a stop
or better.

And all the dodges and burns made by the timer
automatically track the change.

I do confess I make f-stop timers and enlarging
meters http://www.darkroomautomation.com/da-main.htm

Another solution is to use an enlarging comparator
(Ilford EM-10 or an Analyte) and open/close the
lens so the same amount of light falls on the easel.
However, you are no longer using the lens at optimum
aperture and high-end lenses have fixed apertures
so the method is a non-starter.
 
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Excel formula for scaling prints

Hi,

I made an Excel spreadsheet to do the calculation for me. I have cells labeled, (in the following order):

Original exposure time: Original dimension: New dimension: New exposure time.

These are all on row 5 of my spreadsheet and in columns A,B,C and D respectively.

The formula in the D5 cell is the following (if you use other rows or cells, you must compensate accordingly, of course).

=SUMME((C5/B5)*(C5/B5))*A5 (Note: my software is in German, SUM is probably correct for English versions)

This is based on the formula: To(Ln/Lo)2=Tn
Where:
To is Original exposure time
Ln is New dimension (Length)
Lo is Original dimension (Length)
Tn is New exposure time

As mentioned above, this is not completely accurate, but will get you in the ballpark. There are always small changes in exposure/contrast to be made when scaling a print up or down anyway, since the difference in size changes our perception of detail, contrast, tonality. It does save time, however, to start here instead of making test strips.

Hope this helps,

Doremus Scudder

www.DoremusScudder.com
 

Dave Miller

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Square your lens to paper distance for the first print size.
Square your lens to paper distance for the new enlarged print size.
Divide the second result by the first result.
Take the first good exposure time and multiply it by this result to arrive at your new exposure time.

Example:

10"x10"=100
15"x15"=225
225/100= 2.25
If the first exposure time was 11 seconds then multiply it by 2.25 for the enlargement. ie; 24.75 sec.

My interpretation from "You and Your Prints", by William Hawken ISBN 0-8174-2114-9

I have used this method many times with good results.

I always use this method because any resizing / cropping of the image does not effect the calculation. By the way, stating the obvious, but it also works in reverse.
 

Bruce Osgood

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My Excel program works thus:

NEW TIME = ((NEW DISTANCE / OLD DISTANCE)^2) X OLD TIME

This works for either enlarging or reducing size by distance and is is based on the NASA explanation of the Inverse Square Law.

http://quest.arc.nasa.gov/pioneer10/education/temp/

What is lacking in this is the fact that contrast is affected and we do not believe it should be, or at least I don't. But it is and I don't know why.

When enlarging from 8X10 to 11X14 there is an ever-so slight reduction in contrast. I thought there was a problem with the math but the math is correct.

I'm not too concerned with this mystery as the results are generally quite pleasing.
 

Monophoto

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Many years ago, Unicolor offered a series of workshops on the use of their products. One of the handouts you received was a card that showed the approximate adjustment in printing time as you moved from one print size to another - 4x5 to 5x7 to 8x10 to 11x14 to 16x20, etc.. While I haven't printed in color for many years, I still keep this card hanging in my darkroom.

It doesn't guarantee a perfect print, but it gets me into the ballpark so that the first workprint is close.

I would offer to scan it, but there are those copyright issue. But you can generate a similar card/table from the formulas that others have suggested.
 

Jean Noire

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Dancu writes:-

"The working focal length of the lens must be taken into
account. As the lens is focused for larger prints it's focal
length is lessened and the lens approaches it's rated speed.
The lens becomes faster. The reverse holds true when
increasing the working focal length."

This may well explain why the inverse square law calculations quoted here have never worked that well for me, I have always had to make adjustments to time but have not been able to incorporate the differences into a new model.

Thanks Dan,
Regards,
John.
 

dancqu

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Dancu writes:-

"The working focal length of the lens must be taken into
account. As the lens is focused for larger prints it's focal
length is lessened and the lens approaches it's rated speed.
The lens becomes faster. The reverse holds true when
increasing the working focal length."

This may well explain why the inverse square law calculations
quoted here have never worked that well for me, I have always
had to make adjustments to time but have not been able to
incorporate the differences into a new model. Thanks Dan,
Regards, John.

The exact formula is somewhat more complex. I've never
made a point of having it at hand and do not use it. Likely
it is some where on the Web. The exact exposure will be
some little shy of that indicated by the inverse square
law on greater magnification and the reverse when
down sizing. Dan
 

Dave Miller

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My Excel program works thus:

NEW TIME = ((NEW DISTANCE / OLD DISTANCE)^2) X OLD TIME

This works for either enlarging or reducing size by distance and is is based on the NASA explanation of the Inverse Square Law.

http://quest.arc.nasa.gov/pioneer10/education/temp/

What is lacking in this is the fact that contrast is affected and we do not believe it should be, or at least I don't. But it is and I don't know why.

When enlarging from 8X10 to 11X14 there is an ever-so slight reduction in contrast. I thought there was a problem with the math but the math is correct.

I'm not too concerned with this mystery as the results are generally quite pleasing.

You are correct, I believe the effect is an optical illusion
 

Sirius Glass

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When enlarging from 8X10 to 11X14 there is an ever-so slight reduction in contrast. I thought there was a problem with the math but the math is correct.

I'm not too concerned with this mystery as the results are generally quite pleasing.

That is because the physics of light is linear [predictable], while the chemistry of photography is non-linear [not totally predictable].

Steve
 

brian steinberger

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This evening I re-printed a photograph at a larger size. The previous print was 7x7" on 8x10 Ilford MGIV FB glossy. The exposure was for 9.5 sec. at f/16 at grade 3 1/2.

Tonight I printed the image up to 10x10" on 11x14 Ilford MGIV FB glossy. I had read this thread previously and instead of doubling the exposure, I just opened the lens up one stop (essentially the same thing). After a few test strips my final exposure was at 9.5 sec. at f/11 at grade 3 1/2... exactly one stop more exposure than the smaller print. Its funny how simple things are sometimes. But then I thought about it, the 7x7" print has 42 inches of area, and the 10x10" has 100 inches of area.. Approx. double.

I just wanted to add this for anyone, because this was an eye opener for me and will certainly give me a great starting point when re-printing from 8x10 to 11x14 (on MGIV anyway). No calculations or densitometer needed.
 

Vincent Brady

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Spoilsport Brian, you know how we love our secret ways and calculations.

-TEX
 

ic-racer

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Just curious, why do we have three page thread here when there is a "sticky" thread on the same topic?
 

dancqu

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But then I thought about it, the 7x7" print has 42 inches of area,
and the 10x10" has 100 inches of area.. Approx. double.

Although the larger print is close to 140% larger you found
that a 100% increase in exposure gave same results. How
do you account for that? In my experience that difference
of 40% is easily discernable. Dan
 
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