Enlarger Bellows Extension & Height for Size?

[rknewcomb]

Hi,
Is there a formula with which I can calculate the bellows extension and distance between negative and easel for a given size of enlargement desired?

Say, if I want to enlarge an 8x10 negative to 16x20 when using a 300mm enlarger lens, what would be the distance between the neg and easel and how much bellows extension would be needed?

I'm trying to figure out about converting a 5x7 enlarger to an 8x10 and what the geometry of all that would be.

thanks to any smart person who understands all this.

[willwilson]

D/F - 2 = M

D = Distance from negative to baseboard
F = focal length
M = Magnification


This formula does not work too well for smaller reproduction ratios. It looks like it has a error of 10% at 3:1 and 100% at 1:1.

I would suggest m = f/(d-f) where d is the distance from the lens to the negative.

The only problem is that the effective location of the lens (its rear nodal point) is unknown for any practical enlarging lens. If precision is required then set up the enlarger to project a negative at 1:1 on the baseboard (some trial & error will be required) and measure down two focal lengths from the negative and make a mark on some part of the enlarger that is rigidly attached to the lens. That will be the effective location of the lens for measuring the "d" needed to calculate an unknown magnification in the future.

I have searched around for additional details on this as well, so I am eager to hear the responses from some of our members that are more fluent in math and optics.

[rknewcomb]

My math skills are a little rusty so I also invite help.

One more thing - A 16x20 inch print is four times the area of an 8x10, but is it actually a 2x magnification? - 16x20 from an 8x10?

[Paul Ewins]

I finally found a comprehensive answer for this in Langford's Advanced Photography. It gives a multipart formula with lots of choices for inputs:

m = v/u = I/O = (d-u)/u = v/(d-v) = f/(u-f) where

m = Magnification
v = distance from Image (paper) to lens
u = distance from Object (neg) to lens
I = Image (paper) height
O = Object (neg) height
d = distance from Object to Image
f = focal length of lens

8x10 to 16x20: we can use
m = I/O
m = 640/320
m = 2 -> magnification = 2x

using m = f/(u-f)
2 = 300/(u-300)
2(u-300) = 300
u-300 = 300/2
u-300 = 150
u = 450 -> 450mm from the neg to the lens

using m = v/u
2 = v/450
900 = v -> 900mm from lens to paper

so total distance from neg to paper (d) = v + u = 450 + 900 = 1350mm

Alternately, once we have worked out U we could also use
m = (d - u)/u
2 = (d - 450)/450
900 = (d - 450)
1350 = d

[ic-racer]

I checked my 300mm lens and to get a 20" wide projection of the 10" side of an 8x10 negative, the negative carrier is 53" from the baseboard with about 17" of bellows extension.

[aduncanson]

Neg to Paper Distance, D, is given by

D = f * ( M + 2 + 1/M) + P

f = focal length

M = Magnification

P = Principal point separation, posted by Schneider on their older lenses here,
http://www.schneideroptics.com/info/vintage_lens_data/enlarging_lenses/enlarging.pdf but is a constant and is small in most cases.

I have verified that this equation is accurate over a range of magnifications and focal lengths using a spreadsheet that confirms that the image and object distances satisfy the lens equation and produce the selected magnification.

This verification does not take into account the principal point separation because it used the thin lens assumption, i.e. P = 0.

By the way, this equation agrees with ic's 53" measured value for M=2 with a 300mm lens.

[Paul Ewins]

For the metricly challenged, 53" = 1346mm so either formula works. My method was a little more convoluted but along the way also gave the bellows extension. A better formula for that would be:

u=f+f/m

So with a 300mm lens the minimum extension will be 300mm. At 2x it is 450mm, at 1:1 it is 600mm and if somebody ever asks you to make a 5x7 print from an 8x10 neg you will need around 730mm of extension.

[Ray Astley]

What Height Would We Need To Make A 24 X 20 Print Or 30" Wide Print?

[Paul Ewins]

Ray, for 8x10 it would be 1445 mm (about 57") from neg to paper with a 300mm lens and 1156mm (45.5") with a 240mm lens. If you were enlarging 4x5 to 20x24 the distance would be 1050mm with a 150mm lens and 945mm with a 135mm lens.

For 30" wide that is 1600mm with a 300mm lens and 1280mm with a 240mm lens.

[Ray Astley]

Cheers Paul,that Is All I Wanted To Know