Factor for enlarger head height adjustment?

[RalphLambrecht]

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.

CHECK PAGE 511 FOR A GRAPHICAL METHOD!

 

[RalphLambrecht]

Exposure Height and Exposure Correction Whenever the enlarger head is raised or lowered, and the negative magnification is changed, print exposure must be corrected. In ‘Tables and Templates’, you will find a chart to determine the magnification of your enlargement and another to estimate the exposure compensation required to accommodate a change in enlarger height. Strictly speaking, projected print exposures fail to follow the inverse-square law, but they follow the inverse square of the lens-to-paper distance if the paper reciprocity failure is ignored, in which case, a new theoretical exposure time (t2) is given by:

 

[RalphLambrecht]

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?

EnlargerHeightExpTimeEqn.eps

 

[RalphLambrecht]

There 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.

it's allhere

 

[kapro]

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.

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.

 

[RalphLambrecht]

I cannot attach the file with the info you desire because it's too big for this forum.send me a private email to rlambrec@ymail.com and you shall receive a copy.

 

[BHuij]

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.

I was going to recommend the same thing. I picked up my EM-10 hoping to eliminate test strips forever. Turns out it's much good for that pipe dream. But it is great for keeping print density the same across different print sizes

 

[pentaxuser]

This thread come up again as a result of today's post by BHuij so I re-read all the posts. Something about Bruce Osgood's calculation on page 1 #3 looked strange to me and later I noticed that kapro #55 has said the same thing. kapro's result is the same as the formula that hoffy gives although hoffy's formula is expressed differently

In Bruce Osgood's example on page 1 his old elevation was 24 and his old time was 16 secs, his new elevation is 30. I make the new time to be 25 secs as kapro does not the 39.4 secs that Bruce arrives at.

Are kapro and I correct?

Thanks

pentaxuser

 

[BHuij]

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.

 

[RalphLambrecht]

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.

You'll find it in the attached file

 

[MattKing]

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.

Actually, it is appreciated when someone revives and enhances an existing thread rather than starting a new almost duplicate thread.

 

[wiltw]

The Kodak Master Darkroom Dataguide has an enlarging computer in it. All of the folks who submitted formulae should try their solution vs. Kodak's dial computer.

• If original is 4X magnification 15 sec., an 8X magnification enlargement is suggested at 60 seconds.
• If original is 3X magnification 15 sec., a 9X magnificaiton enlargement is suggested at 130 sec.
• If original is 5X magnification 15 sec., an 10X magnification enlargement is suggested at 60 seconds.

 

[darkroommike]

The Kodak Darkroom Data Guide (most editions) have a nifty calculator wheel for time changes vs. magnification, no need to reinvent the (ahem) wheel.

 

[faberryman]

Seriously, just make a test strip. You might find that a different - not an equivalent - exposure is appropriate for the re-cropped image.

 

[Kilgallb]

Measure the diagonal of the original and divide into the diagonal size of the new exposure. The ratio is almost perfect.

The ratio of of the image area does not work as it is actually an inscribed rectangle inside a circle. The diagonal size is similar to the radius of the circle. I have found this works well. Taking the square root of the ratio of the image size gives the same result.

Using a formula based on one side only fails as aspect ratio is different from 35mm to MF and LF.

That said, the same Dmax and Dmin at two different sizes might not be esthetically pleasing.

 

[cornflower2]

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.

Sorry 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.

 

[pentaxuser]

Sorry 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.

Could you explain this in more detail with reasons? Thanks

pentaxuser

 

[cornflower2]

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?

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.

 

[cornflower2]

Could you explain this in more detail with reasons? Thanks

pentaxuser

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

 

[RalphLambrecht]

A

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

drew, you have a unique talent to make things complicated.

 

[pentaxuser]

Thanks for your explanation cornflower2. A brief example of how each of these unique factors affect the calculation would have helped. I wonder how each of these factors are taken care of in your App as it sounds as if the App needs to know what these are. That being the case what does each owner of your App have to "input" into the App to get the right answer for his set-up?

It sounds as if the i-phone etc does instantly a complicated calculation that any manually operated calculator takes 15 minutes to do and is so many-stepped as to risk mental strain. What is it about a i-phone that gives it this kind of calculating power?

Can I just finally ask briefly how you came to decide that your App was needed and assuming your App is the only way to get it right, what did you do before you devised the App. For instance how far out were the results given by other methods such as the inverse square law i.e. if a 5x7 print is enlarged to say 8x10 by, for want of a better phrase, traditional methods how many seconds/fractions of a second will this be wrong.

Are you the sole inventor of this App or are there others out there who have similar methods?

I should say that some of your anticipated answers to my questions on the technicalities may be over my head but please take this risk as on this forum we have many whose knowledge of such matters exceed mine many times and their collective comments will get us all on APUG to a collective understanding of what appears to be a a complicated matter

Thanks

pentaxuser

 

[mmerig]

Regarding the enLARGE APP:

The explanation on the Apple App store mentions a calibration procedure, to obtain a 'Unique Exposure Curve" for a particular enlarger and lens focal length. So, apparently, the approach is empirical, where the enlarger is raised and actual light fall-off is measured at various increments of enlarger heights, with the iPhone, I assume, and this curve is used within the iPhone to calculate exposures for enlargement sizes along this curve. At least, that's the way it looks to me, but I could be wrong.

The curve would roughly follow the inverse-square law, but would incorporate the effects of bellows extension and the characteristics of the lamp house. As cornflower2 mentions, the inverse square law assumes a point source, so unless the lamp is very small compared to the negative-to-paper distance, there will be some noticeable error using the inverse square law, even if bellows adjustments are included.

Strictly speaking a point has no dimension* -- there is no length or size to it. In other words, it's "the small end of nothing whittled to a point". But a light source very far away from an illuminated surface, like a star, is essentially a point source for most practical purposes. An enlarger may be fantastic, but its lamp is hardly a star.

I don't have an iPhone, or even a cellphone, so the enLARGE App is not an option for me. When making bigger enlargements based on a known, good exposure, I use the inverse square law to get a reasonable new time**, and use a test strip to fine tune the exposure and contrast. But thanks to cornflower2's comments, I may make my own calibration curve for each set-up I have (one enlarger, three formats and their associated lenses). My guess is that a quadratic or power function that models the height/exposure observations could either be put in a spreadsheet, or used in a calculator. The latter would be my preference -- my calculator has red LED display (a Texas Instruments model 30***), so I can use it in the dark with a safe light. I am printing today, so I may try and make a curve. If it makes sense and works, I'll post what I have done.

* There are lots of ways to think about dimensions -- a zero dimension for a point is a Hausdorff dimension
**of course the f-stop can be changed too, but I am trying to keep this simple.
*** as the original owner of a 41-year old calculator, it gives an idea of how apt I am to buy new gadgets. But you already knew that (no cellphone).

 

[Mr Bill]

Could you explain this in more detail with reasons? Thanks

If I may, I can probably improve the explanations, at least for the case of a condenser system.

We can see it as two overlapping optical systems - the first is the condenser/light-collection system. It basically collects a sizeable amount of the light energy coming from the lamp, then tries to direct that light through the enlarging lens. The second system is the enlarging lens focusing an image of the negative onto the paper.

Regarding the condensing system, imagine all of the light it collects formed as a "cone of light" which ideally "fills" the pupil of the enlarging lens with an "image" of the source. But there is a problem with the second system - it is necessary to move the enlarging lens in order to focus as the print size changes. So we can see that if the lens is moved too far "up," the outer parts of the cone of light don't make it into the lens, and are therefore "lost" light energy. Things are more complicated than this, but it's about as far as I can explain without a lot of thinking things through. Everything should ultimately be calculable, but an empirical method is probably a more "cost effective" way to do things. Anyway, it's not as simple as it initially seems.

 

[pentaxuser]

Thanks both. We will need to wait until cornflower2 responds and it is now about 6:00am Sunday in OZ in terms of U.K. time so we can expect to wait a few hours minimum. I think it best to say no more until we hear from him and give him a chance to answer. Given your interest you may want to look at his contribution in his thread entitled "How to calculate a new print exposure time for a change in enlarger head" dated Dec 29 2012.

All that requires you to do is to measure the two easel to negative distances and the App which must contain the complex calculation does the rest. No mention is made of empirically establishing an exposure curve. However and perhaps unfortunately no-one seemed inclined to ask him many if any questions. He had been using the method and the maths for 20 years then so must have had the calculator tools since 1992 but presumably this was a scientific hand-held calculator which made things complex until he devised the App

I'd attach the thread to this post if I knew how but I don't. Hopefully I have given enough info for you to find it

pentaxuser

 

[David Brown]

... use the inverse square law to get a reasonable new time, and use a test strip to fine tune the exposure and contrast. ...

Seriously! Just make a new test strip. It solves all of the problems.