base board to negative distance

[Ralph Miyashiro]

Putting together my first darkroom and have want to know how much distance I need from the negative to my paper for a 16x20 print (largest I can do) using 5x7 format and 180mm lens, and 4x5 format with 135mm lens. Is there a formula to make this calculation? (I may eventually get longer lenses if fall off becomes an issue). Thanks for any advise.

[Luis-F-S]

On my Durst, 5x7 with a 180mm Rodagon 37" from the negative to the baseboard gets about 14"x20" which is about as big I can go without lowering the baseboard. On 4x5 with a 135 Rodagon I get 35" for a 16x20; a 150 Rodagon gets about a 14x17. Hope that helps. L

[AtlantaTerry]

Even if you knew the distances involved you need to consider that most likely you will want to crop or enlarge images. When you do that you will, of course, need to put a greater distance between the negative carrier and paper.

If your darkroom ceiling height is limited consider building an enlarging table where you can drop the working surface down from the normal height of about waist level. Doing that will give you another 3 feet or so of distance between the enlarging paper and negative.

[Luis-F-S]

I can drop the table on the Durst, just didn't want to go through it since I'd have to move some "stuff" that's on the floor below the table. Since I don't print above 11x14; I don't need to lower the table very often. Was just in Atlanta this past weekend; after the "mess". Bought the LPL 4550XLG that was on Craigslist. Nice enlarger. L

[Bob Salomon]

And a 120mm Rodagon-WA will give you the same print size from a 45 at 30% less column height.

[ic-racer]

1/p + 1/q = 1/180
Mag= 3x
So q= p*3
So 1/p + 1/(3*p) = 1/180
So 4/p = 1/180
So p= 720 = lens to print
So lens to neg= 720/3=240
So neg to print= 720+ 240=960mm

[Ralph Miyashiro]

1/p + 1/q = 1/180
Mag= 3x
So q= p*3
So 1/p + 1/(3*p) = 1/180
So 4/p = 1/180
So p= 720 = lens to print
So lens to neg= 720/3=240
So neg to print= 720+ 240=960mm

Thanks for this. Algebra was a long time ago. If I plug in numbers I get results that seem reasonable but I don't really know what I am doing. what does p and q represent? Would it be asking too much to write all of the steps in your equation? Thanks again.

[ic-racer]

That is the "thin lens" equation. It relates subject-to-lens (p) and lens-to-film(q) distances to lens focal length (f). We also know the ratio of subject and film distances = the ratio of subject height to image height. Lets say you are not cropping so magnification will be the ratio of film-to-print linear dimension. To get 16x20 print from 5x7 I rounded to 3x magnification.
We know magnification = p/q and not only will the image be 3 times bigger, it will be 3 times farther from the lens.
So we can indicate p = 3 times q and put that into the thin lens equation. Solve it for p then divide by 3 to get q and add them together to get total negative to print distance (then check against empiric data , like the post showing a measurement of 37" which is close to a yard or meter and close to our calculated 0.96 meters)
http://www.physicsclassroom.com/class/refrn/u14l5f.cfm

[Ralph Miyashiro]

Got it. Just what I was looking for.

[Tin Can]

Missed this until today, just what i have been looking for.

Thanks!