Info for Diy 4x5 enlarger

[Ales']

Hi, ehilà I was designing a 7x17 box camera I thouth that it would be usefull to start with somethink smaller and less complicated with cheaper materials to test it.
My idea is to make some sort of magic lantern with color head.
I plan to buy spare color head and adapt it. I prefer OG color head to LED solution thats intrepid is selling. I just realy prefer more natural light sources, I print platinum with sunlight and shoot in studio with redheads instead of strobes.

I plan to enlarge 4x5 color neg to 20 inch wide roll paper. I can’t handle bigger and see no sense to enlarge smaller since the paper is so cheap compared even to uncoated cotton paper.
I never enlarger bigger that 6x6 because i jumped straight to 8x10 for platinum printing and I never had my enlarger so please help me to calculate size to cut the wood.
In my mind since I plan to enlarge x5 with say 135mm lens I need focus travel from 135mm ad infinity to 162 at 1:5(135x1.2), is it right? I found a pair of rails with fine thread focusing, the smallest size is 200mm so Im kinda of safe even with 150mm enlarging lens? My plan to cut a piece of wood 20cm long mount the rail(picture atached) and aroud build the rest, i will probaly also buy a film holder and lens board so as little diy as possibile, I expect good results
Can someone measure film to lens board distance on 4x5 enlarger with 135 or 150 lens?
P.s. Accept advice for somethink to allign it to the projecting plane.

 

[Ian C]

The film-to-lens-board distance in an enlarger (or camera) can be calculated for a given lens and film size. The necessary data is:

Focal length f

Flange distance Fg

Required magnification m

Format dimensions

Most 4” x 5” holders produce exposed dimensions on film of approximately 95 mm x 120 mm. If you want to make a 5X projection to produce a borderless print on 20” wide paper with the 95 mm film dimension mapped onto the 20” width of the paper, you should realize that the projection must be somewhat greater than the width of the paper for full coverage (the projection should overlap the paper).

I’d plan on an approximate 21” projection of the 95 mm dimension of the negative. 21” = 533 mm. Thus, the magnification is m = 533 mm/95 mm = 5.6X.

The equation for the subject distance is s = f(1 + 1/m). Here, the subject is the negative.

I’ll use the 150 mm f/5.6A EL Nikkor whose data is known. This lens has a focal length of 149.6 mm

s = 149.6 mm(1 + 1/533 mm) = 176.3 mm (negative to 2nd nodal point of the lens)

The flange distance of this lens is 135.1 mm. Since the flange distance is less than the focal length, the 2nd nodal point lies forward of the flange surface by the distance f – Fg = 149.6 mm – 135.1 mm = 14.5 mm.

So, at 5.6X, the distance from the negative to the flange (which coincides with the face of the lens board) is s – 14.5 mm = 176.3 mm – 14.5 mm = 161.8 mm.

With a 135 mm f/5.6A EL Nikkor lens, the negative-to-flange distance is 145.1 mm at 5.6X.

Other brands of lenses of the same 6-element 4-group design will require similar dimensions.

https://galerie-photo.com/manuels/el-nikkor-enlarging-lenses.pdf