


Stable neg during wall projection....
Hi,
I am starting to design my mural enlarger and space, pretty much from the ground up and I am wondering about something. I plan on using the bellows stage of a CB7 in wall projection mode on a track system as it takes all my 45MX stuff but has better adjustments in terms of alignment.
For those who are doing wall versus floor projection during mural work, how are you getting the negs to stay put in glass carriers? I have two antinewton 4" x 5" glass carriers I use, one hallowed out a bit larger for 4x5 and one regular with masks for 6x12CM and smaller.
Also, how do you calculate the lens to paper distance using various lenses and formats for given print sizes?

The best carriers for wall projection will sandwich closed with some sort of keeper. The carriers in which the top glass just lays on the other piece are not so good. What makes you think your existing glass carriers won't work? Is there nothing to keep it closed tight when you insert them into the enlarger?
In terms of lens to negative distance. Usually one does not need to calculate that because the enlarger manual will specify which lensboard is needed and then you just focus with the bellows. If you need to eyeball the distance to figure out how much lensboard recess you need, realize that at "Mural" sizes, the lens to negative distance will be close to the lens focal length. And at 1:1 the lens will be at 2x the focal length. If you need something closer than that, use the 'thin lens formula.' (1/p +1/q = 1/f where p = neg to lens and q = lens to wall)
You figured out a way to focus, right? Maybe have your target moveable on rails for fine focus. That way you can use your grain focuser. Otherwise you will need very long arms or an assistant.
Here is a chart, but it is for 8x10 negatives: http://www.durstprousa.com/pdf/Pro...0enlargers.pdf
How big will you be printing? There is a reason there are not many (any?) horizontal 4x5 enlargers and I believe it is because of the light source. If you are printing very big, you need a lot of light. Probably need your condenser head and a brighterthanrecommended bulb with a fan. For example, Bob Carnie's 4x5 setup for big enlargements uses the common Omega D condenser head but he puts a 250W bulb in there (that head is rated for 75w or 150w).
Last edited by icracer; 12062012 at 11:18 PM. Click to view previous post history.

Have you considered putting a frontsurface mirror at 45 degrees below the lens of a vertical enlarger to achieve your wall projections?

My carriers are the ones with no keeper...
I print up to 20x24 on my upright 45MX, so I hope to print between 30x40 to 40x50 on the projection enlarger. The CB7 has a focus motor that I plan to hook a rheostat type of switch onto so I can use a grain focuser as I toggle it.
I had a Aristo D2HI in my 45MX but exposures were way too short...it's damn bright so I think it might work with the mural setup.
IC, I was talking about the lens to paper distance, not neg to lens. I figure I might need up to 8 feet of distance for the bigger prints. Polyglot, I really am after a dedicated mural enlarger so as interesting as the mirror sounds, I am going to have to stick to m planned workflow.

Hi, IC gave you the basic formula, but didn't mention that magnification is proportional to p vs q.
Here's a calc'd example: say that you have a 2 inch wide negative which you want to enlarge to 40 inches wide, a 20x (linear) enlargement. This means the ratio of p/q must also be 20. That is, the lens to paper distance will be 20 times that of the lens to negative distance. The focal length of the lens determines what the exact distances will be.
Rather than try to do this in a proper math method, it's easier for me to simply set up a spreadsheet where I preset the focal length and q (lens to negative distance), then let the spreadsheet calculate p (lens to paper distance). Note: p = 1/(1/f  1/q). I also have the spreadsheet display p/q. Then, I just keep adjusting the value of q until the ratio p/q =~20. Then, p + q = the negative to paper distance.
If we use an 80mm lens (there are 25.4 mm per inch, this is about 3.15 inches), it works out to p = 65.9 inches, and q = 3.31 inches. The neg to paper distance is, therefore, about 69 inches, a bit less than 6 feet. (A 75mm lens would only use 65 inches.)
In other sizes, enlarging a 1" wide neg (35mm film) to 40" width, using a 50mm lens, you'd need about 83 inch total distance, nearly 7 ft.
Or enlarging a 4" width (4x5 film) to 40" width, using a 150mm lens, you'd need about 71.5 inches.
Hope this helps. ps, I didn't doublecheck the calcs, so you might want to verify before building any walls. Note, too, that the measurement points should be from the nodal points of the lens, so it may be an inch or so off, depending on the lens.

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Glass carriers used in horizontal projection might require temporarily anchoring the negative in place with some lowtack drafting tape such as made by 3M. I secure the upper corners of the negtive to the lower glass with the minimum tape overlap needed. The adhesive touches the base side only of the negative and comes away cleanly afterwards. The areas contacted by the adhesive of the glass and the top surface of the negative can be lightly cleaned with naphtha if any adhesive gets transferred.
To estimate the easel position, first determine the required magnification m. Then for a lens of focal length f,
Lenstoprojection distance = (m + 1)f
For example, if you have a 6 x 7cm negative whose 55mm format width you wish to project at 54” wide to cover a 50” strip of roll paper, you need a magnification of 25X.
Lets suppose that you’re using a 105mm Rodagon G. Then
Lenstoprojection distance = (25 + 1)*105mm = 2730mm
This is the thinlens approximation and will be a bit low since we’re using a compoundelement thick lens. You can get a more accurate estimate by first calculating the focal length of the equivalent thin lens that conforms to your optical system (can be done, but somewhat tedious). For a 6element symmetric lens at enlarger distances this is usually about 2.6% greater than the actual focal length. So if you substitute the approximate equivalent thin lens focal length of 107.7mm you’ll get a more accurate estimate.
Then
Lenstoprojection distance = (25 + 1)*107.7mm = 2801mm
This is only an estimate, but is usually reasonably accurate within practical needs.

Thanks Ian, that is exactly what I am looking for. And yes, I will be using both a 105mm Rodagon G and hope to soon have 150 Apo N.

Originally Posted by Ian C
For a 6element symmetric lens at enlarger distances this is usually about 2.6% greater than the actual focal length. So if you substitute the approximate equivalent thin lens focal length of 107.7mm you’ll get a more accurate estimate.
Hi Ian, how do you arrive at 2.6% greater? That seems to be a proper lens "rack out" at about 38X magnification. I don't see how the lens design (symmetric) would affect this?

When we use a realworld lens: multiple thick elements and cemented compound elements having different indexes of refraction, the lens simply won’t conform exactly to the thin lens equation. This is annoying insofar as the calculated values differ from the actual values produced by the lens. If our needs aren’t fussy, the approximations the TLE provides are useful and reasonably accurate. The calculated values for typical enlarger lenses are somewhat less than the actual measured values. The enlarger lens behaves as though it had a somewhat longer focal length than specified by the maker.
By measuring the negativetoprint distance and magnification of the projection, we can calculate the focal length of the imaginary equivalent thin lens that corresponds to these measured values. By substituting the equivalent focal length into subsequent equations we can get better approximations to the actual values of our system.
I’ve found that for the 6element enlarging lenses I’ve tested and for which I have the maker’s specified focal length, the ratio (equivalent thin lens focal length)/(actual focal length) is about 1.026, i.e. the equivalent focal length exceeds the actual focal length by about 2.6%.

Thanks Ian; ok, I think I see what's going on. It turns out that we are both doing exactly the same thing (before you "correct"). My method finds the the lens extension (ie., how far the lens needs to be "racked out" from infinity focus) by trial and error. This insures that the distances are accurate.
I first thought perhaps you were adding on 2.6% to roughly correct for this. But it appears the "+1" in your formula actually does the correction. Both of our methods give the exact same result. So the only error should be due to the location of the lens nodal points (same as principal points here).
That said, I think that using the 2.6% factor is not the best way to correct for this. A single distance adjustment would be better. I once had some Rodenstock data sheets showing locations of the principal points, but not to be found. (The current website data also does not show principal points.) Just for fun, I thought I would check a Rodenstock lens  it's a 75mm Apo Rodagon D, for 1:1 reproduction.
It appears that my Rodagon D has overlapping principal points  that is, it should make images in LESS distance than the calcs indicate. Since you had found the opposite, I thought I'd test a bit further. I set up for 1:1 reproduction, which requires that the subject and object be 4 focal lengths apart, plus the correction for principal point offsets. Actual results were that my lens did it in about 3/4 inch LESS than 4 focal lengths, which is the same as the estimated overlap of principal points. So it is not a given that more distance is always required.
I know that few people care about this sort of thing, but just wanted to point out that the two calculation methods give the same result, and that the only error it might have is due to the (generally) unknown location of the lens' principle points.
ps: the linked diagram shows standard notation for the principal points, H and H_prime. https://www.schneideroptics.com/nomenclature.htm

