Alright, half my brain doesn't work, I am pretty much braindead when it comes to mathematics. So I am asking you mathematically non-challenged people a simple question: To enlarge a 6x6cm negative with a 80mm lens up to 48x48cm (on a 50x60cm -20x24" paper) what should be the distance from the negative plan to the paper? I am working on my darkroom and I need to see if I need a new table as I will be rearanging certain elements such as wall hanging the enlarger.

I suggest that you mount higher and farther from the wall than the required minimum dimensions to allow for cropping.

Z=(m+1)f Z = distance from the lens to the baseboard m = magnification f= focal length So Z=(8+1)8.0 Z=72 cm Roughly. The focal length of your lens is likely a little off the nominal 80mm. You need to measure from the right spot. Still should be pretty close. The negative to the lens is: X=(m+1)/m*f X=(8+1)/8*8 X=.14cm I'm assuming 48cm from 6cm is 8x magnification. You can check the formula with smaller enlargements on your enlarger.

The numbers posted by the previous member are very incorrect. For a 10.4 X magnification the distances are theoretically: negative to lens is 88 mm ( NOT 0.14 cm ??? ) lens to easel is 920 mm Keep in mind that the lens has thickness. As I mentioned previously please consider cropping and allow extra distance from the wall and extra height. I like to use a counter that can be adjusted lower than I typically use.

For a "thin" lens F/f=W/w where: F=Paper-Lens f=neg-lens W=paper width w=neg width For a 10x enlargement W/w = 10 so F/f=10 (approx) This agree's with GMA 90mm x 10 (aprox=) 90 cm. Ian

Let's see. Put the negative in the enlarger. Turn off the room light, turn on the safelight. Raise the enlarger. Focus the image. When the image is the size you want, measure the height from the enlarger to the baseboard. So far, you have two answers and they both disagree. Practicality trumps science every time. This is the difference between scientists and engineers...

Yup did the math wrong for the negative to lens distance. It's 9cm. Which formula are you using to figure lens to paper distance? Or are you using a different magnification factor?

Nick and GMA are using different magnification factors (8x and 10.4x, respectively). I think Nick is judging by the nominal frame size of 6x6 cm, and GMA is assuming a slightly smaller more realistic frame size, plus a little crop space.

Don't forget that the focal length gets shorter as you rack in the bellows for larger print size. Find a rule of thumb. Avoid the calculus.

1/focal length of lens = 1/distance lens to film + 1/ distance lens to easel 1/80mm = 1/ 88mm + 1/ 918mm Magnification = 918/88 = 10.4 X

Ok, I am a bit confused, but I'll give it a shot. The problem with trying it right now is that my current setup doesn't allow me such a magnification. The roof of the darkroom is pretty low, so if I wanted such enlargements I would need to construct a lower or a variable height desk.

A competent scientist or engineer knows when they understand a problem well enough to use theory & equations rather than experiment. Do you think that Boeing and Airbus design airplanes by building full scale models and testing to see whether they break, beefing up parts if they break, or reducing weight if they don't break? Nick, 127 and GMA have all given correct equations. Nick appears to have divided by 8 when his equation called by multiplying by 8. Here is another equation addressing the problem (derivable from the others): The total distance from negative to paper is d_tot = f * (m+1)^2 / m. Here "^2" means squared. This equation neglects the separation of the principal planes of the lens, which is likely to be only a couple of mm and so is reasonable to neglect. If your 6x6 cm negative is actually 56 mm square, then m is 480 mm / 56 mm = 8.6 (rounding up), giving d_tot = 80 mm * (8.6 + 1)^2 / 8.6 = 80 mm * 92.2 / 8.6 = 860 mm. As GMA pointed out, even if you aim to print full frame, you someday will probably wish to crop a negative. If we guess a 20% crop (45x45 mm portion of the negative), then m becomes 10.7. As Nick pointed out, the focal length of the lens might be slightly different from the nominal value. A likely difference is about 1 mm, so we can allow for this with margin by using f = 82 mm. (The focal length is constant and does not vary as the bellows is racked out.) Using these refined values for m and f, d_tot = 82 mm * (10.7 + 1) ^2 / 10.7 = 1050 mm. This is the distance from the negative to the baseboard. The largest uncertainty in these calculations is the largest crop that might be wanted, which determines what value should be used for m. The uncertainty is also present in an experimental determination of the distance -- if you pick some magnifcation for the experiment, you might later find that it doesn't allow for a crop that you want to do. An alternative solution to the problem of getting larger size prints would be to use a wide-angle enlarging lens. These type of lenses cover a given format with a shorter than normal focal length. They cost more and don't show up frequently on the used market. Light falloff in the corners will be worse. For example, the 60 mm f4 Rodagon-WA is rated by Rodenstock to cover 6x6 cm. This lens (for m=10.7) woudl reduce d_tot to 800 mm.

Okay Nick, I've been trying to figure the "wrong math" of .14cm to get 9cm. Is the formula you listed originally wrong or am I misreading sumthin'?

arugran - I suggest that you cut the top of the shelf where the enlarger stands so that it can be lifted off. Then, place rails in a "breadboard" type arrangement so that a board can be slid in on top of the rails to lower the enlarger. You can make a series of these "rails". The ceiling of my darkroom is too low to make 16x20 prints from a 4x5 negative using my 150mm enlargiing lens; I find this arrangement works well. Then, if you have to "crop" the image but still want a 16x20 print, just lower the board to the next lower pair of rails (and so on).