How to determine max enlarging size for my setup?

[Jeff Bannow]

I'm going to build a drop table for my enlarger, and would like to figure out how big I can go to determine how big to make the table.

I'm using a Durst L1200, and will be enlarging mostly 6x6 and 4x5. If I know the maximum distance from the lens to the floor, and the focal length of the lens, can I calculate the biggest print size?

Lens:
6x6 - 80mm or 90mm
4x5 - 135mm or 150mm

 

[Ian C]

Raise the head to maximum height and give us the negative-to-easel distance, focal length, and format.

With this data calculating the maximum print size to within a millimeter or two is simple.

Lens-to-easel distance is imprecise. Negative-to-easel distance is easy to measure accurately.

If you give us the numbers it will take a few seconds to calculate the answer.

 

[Jeff Bannow]

I can do that. I'll measure this afternoon when I get home.

 

[Robert Hall]

Even easier. Don't enlarge more than 3x the negative.

 

[Jeff Bannow]

Even easier. Don't enlarge more than 3x the negative.

That would be no fun! I suppose I could cut up my roll of 42" paper into 8x10s or something ....

 

[Robert Hall]

No, no, no. You've got it all wrong. What you need is a 42" negative!

 

[Jeff Bannow]

No, no, no. You've got it all wrong. What you need is a 42" negative!

Now that would be impressive!

 

[Jeff Bannow]

Raise the head to maximum height and give us the negative-to-easel distance, focal length, and format.

With this data calculating the maximum print size to within a millimeter or two is simple.

Lens-to-easel distance is imprecise. Negative-to-easel distance is easy to measure accurately.

If you give us the numbers it will take a few seconds to calculate the answer.

Maximum from negative carrier to floor is about 65". Thanks.

 

[Ian C]

For any format you’ll always get the greatest magnification with the shortest focal length lens that’s specifically designed for the format.

In order to calculate the largest projection possible you must know the projection distance (negative-to-easel), the dimensions of the window opening in the negative carrier, and calculate the magnification m that the negative-to-easel distance gives for the focal length of the enlarging lens.

The largest projection is m times the dimensions of the carrier’s widow size.

I’ll use the dimensions in my Omega negative carriers as typical. Omega 6 x 6cm carriers have 55.5mm x 55.5mm opening and the 4” x 5” carrier window is 92.5mm x 120.1mm.

If you placed a 1” tall easel on the floor you have a 64” = 1625.6mm negative-to-print distance.

For the stated projecting distance you’ll get the following focal length, magnification, and projection size combinations:


6 x 6cm Negative

80mm, 18.3X, 1013.7mm x 1013.7mm (39.9” x 39.9”)

90mm, 16X, 888mm x 888mm (35” x 35”)


4” x 5” Negative

135mm, 9.9X, 919.5mm x 1193.9mm (36.2” x 47”)

150mm, 8.7X, 806.8mm x 1047.6mm (31.8” x 41.2”)

 

[Jeff Bannow]

Thanks Ian. What's the formula you used here?

 

[Ian C]

Magnification is

m = [d + squareroot(d^2 – 4df)]/2f – 1

d = negative-to-print distance

f = focal length

 

[MattKing]

If you need larger, one of the special purpose wide angle enlarger lenses will give that to you.

 

[Jeff Bannow]

Thanks all.

 

[Martin Aislabie]

You may find that the limiting factor to maximum print size quickly becomes the Enlarger Head to Column/Wall distance

Just something to look out for

Martin

 

[Jeff Bannow]

Definitely something I'll need to watch for. I assume for a 40" print the head needs to be 20" from the wall?

 

[PKM-25]

In order to calculate the largest projection possible you must know the projection distance (negative-to-easel), the dimensions of the window opening in the negative carrier, and calculate the magnification m that the negative-to-easel distance gives for the focal length of the enlarging lens.

The largest projection is m times the dimensions of the carrier’s widow size.

I’ll use the dimensions in my Omega negative carriers as typical. Omega 6 x 6cm carriers have 55.5mm x 55.5mm opening and the 4” x 5” carrier window is 92.5mm x 120.1mm.

If you placed a 1” tall easel on the floor you have a 64” = 1625.6mm negative-to-print distance.

For the stated projecting distance you’ll get the following focal length, magnification, and projection size combinations:


6 x 6cm Negative

80mm, 18.3X, 1013.7mm x 1013.7mm (39.9” x 39.9”)

90mm, 16X, 888mm x 888mm (35” x 35”)


4” x 5” Negative

135mm, 9.9X, 919.5mm x 1193.9mm (36.2” x 47”)

150mm, 8.7X, 806.8mm x 1047.6mm (31.8” x 41.2”)

So I am trying to figure out how to hit the long dimension of paper size 30"x40" and the short of 42" & 56" rolls in terms of neg to wall distance, enlarger will be horizontal, landscape orientation. For example, my mural lenses will be the 50mm Rodagon G for 35mm, 105mm Rodagon G for 6x6 and 150mm Apo N for 4x5. So what distance from the neg stage would I need to attain a monster 56"x70" enlargement from 4x5 via the 150mm lens or 42x42" from 6x6 with the 105mm lens, etc....

Is there a different formula?

I am just trying to figure out the size of room needed to do max enlargements on Ilford 30"x40", 42" & 56"roll paper, not counting the enormous room to soup the prints in. Seems like a good mural room should be at least 10'x10 feet, soup room *much* larger.

 

[ac12]

Without doing the math transformation to calculate distance, just put the formula into Excel.
You can calculate the magnification you need to get to the print size. Print size / negative frame size = magnification.
You have the focal length of the lens.
Just start putting in different d (distance) values till the magnification value comes out to the number you need.
It is a trial and error process, but you can narrow in on the distance pretty fast.

 

[ac12]

I did a google search and found this

easel_to_carrier_distance = focal_length * (2 + 1/M + M)

where M is the negative to print magnification.

For a 7.2 X enlargement, the distance with a 300 mm lens would be 300 * 9.34 = 2801 mm = 110.3 inches.
For a 240, the distance would be 240 * 9.34 = 2241 mm = 88.25 inches.

Validate the formula by testing it on your enlarger as you have it setup.

 

[Ian C]

So what distance from the neg stage would I need to attain a monster 56"x70" enlargement from 4x5 via the 150mm lens

The 4” x 5” film image is about 95mm x 120mm as limited by the window in the film holder when the photo was made in the camera. You’d need 15X to do this. Realistically, you’ll likely have to project an image slightly larger than the intended print for proper coverage if you intend to make a borderless print. I’d plan on making the small dimension of the projection at least 57”. That requires 15.24X.

The negative-to-print distance in this situation with a 150mm lens is 2596mm = 102.2”. The formula of post #18 gives the same value and is easy to use.

 

[Chan Tran]

Magnification is

m = [d + squareroot(d^2 – 4df)]/2f – 1

d = negative-to-print distance

f = focal length

The formula works assuming that front and rear nodal planes of the lens are the same. But I think for enlarging lenses they are close enough for the purpose. Camera lenses especially those for 35mm would not be close.

 

[Ian C]

I don’t know the nodal distance of the 150/4 APO Rodagon N. However, I have the nodal distance of the 150/5.6A EL Nikkor (1.2mm). That would inflate the true negative-to-print distance by 1.2 mm compared to the calculated value for the EL Nikkor. The error is small enough that the calculation is still sufficiently accurate for a reasonable estimate.

The nodal distance of the 150/4 APO Rodgon N is likely not greatly different than that of the 150/5.6A EL Nikkor.

A greater potential calculation error is the true focal length of the lens. A good example of this is the 50/2.8N EL Nikkor. Nikon gives its true focal length as 52.0 mm—4% greater than marked. Some makers supply this information. Others simply state the nominal focal length only.

 

[PKM-25]

Good info, I am pretty much trying to decide on the amount of travel needed to go from roughly 24"x36" to 50"x70"-ish for all formats. I am likely going to be using an 80" length of 7-1/2" lawn track that is commonly used for model railroading purposes, it should keep that axis *roughly* aligned and the fine tuning will be done with adjustments on the carriage and the CB7 head it self. So it sounds like 80" of travel should indeed cover the range, room will need to be at last 10' feet deep in order to hit the largest size, not that I am aiming for those huge sizes any time soon.....I still don't know how the hell I would pull off huge prints other than a Clyde Butcher or Bob Carnie style long tray room, but I will worry about that later, 30"x40" at the biggest for now...


This ought to be fun to make too, I am ordering a model 3b LED Modern Enlarging Lamp for the CB7 head this Spring, should work great for this purpose. It's a really slick setup, the b version is not listed yet but his video covers it here:

http://www.youtube.com/watch?v=Ein89audefA

 

[MattKing]

This ought to be fun to make too, I am ordering a model 3b LED Modern Enlarging Lamp for the CB7 head this Spring, should work great for this purpose. It's a really slick setup, the b version is not listed yet but his video covers it here:

http://www.youtube.com/watch?v=Ein89audefA

Dan:

While you are at it, see if you can convince the light source manufacturer to:

1) provide an Android App for it; and
2) develop an f-stop timer App as well .

 

[Erik L]

..

 

[Doremus Scudder]

Okay,

I'm trying to write the Excel formula for the equation for magnification given above, which is:

Magnification is

m = [d + squareroot(d^2 – 4df)]/2f – 1

d = negative-to-print distance

f = focal length

However, I don't seem to have it right for some reason, and I can't figure out what I've done wrong.

Here's my Excel formula (with the above variables instead of cell coordinates):

=((SQRT((d^2)-(4*d*f)))+d)/((2*f)-1)

Any help from the Excel experts out there?

TIA

Doremus