Calculating the max enlargement

NB23 · Jan 17, 2021

ic-racer said:
Read up on optics and magnification. You might be mistaking the inverse square law which relates to EXPOSURE, not MAGNIFICATION.

You have a link to share?

ic-racer · Jan 17, 2021

NB23 · Jan 17, 2021

Yes, linear.

I do not think that the linear model applies to a print size magnification. There is the mention of ratio. Can you explain the meaning of ratio and how it is calculated?

NB23 · Jan 17, 2021

If Linear is the consensus, then fine.

My concern is that an image is X and Y, not only X. A total magnification should be calculated by its volume, not on one side’s length alone.

wiltw · Jan 17, 2021

ic-racer said:
Read up on optics and magnification. You might be mistaking the inverse square law which relates to EXPOSURE, not MAGNIFICATION.

Indeed, geometry says if the 'Adjacent side' distance (lens to paper) is doubled, the 'Opposite side' (paper dimension along one direction) is doubled, per the Tangent relationship. Linear dimensions at simple proportion. Shown in my drawing.
If you double the 'Opposite side' (also doubling the 'Adjacent side' length),...the magnification is 2x.
Tthe area of the image at the longer distance is indeed 4x and the exposure time increases by 4x as well....Inverse Square effect on light intensity by doubling distance.

ic-racer · Jan 17, 2021

Maybe read up on "convention" otherwise we will start calling "Day", "Night"

NB23 · Jan 17, 2021

ic-racer said:
Maybe read up on "convention" otherwise we will start calling "Day", "Night"

I’m fluent in 4 languages with mix-ups between them, here and there. I don’t need to “read-up”. Thanks...

wiltw · Jan 17, 2021

Before there is more added to the apparent p*ssing contest, the issue is that the term 'magnifiication' is CONTEXT DEPENDENT.

If one is talking about 'power' of a loupe or a binocular, megnificaiton indeed is AREA...4x power does NOT make a letter 4x taller, it is only 2x taller than actual.
If one is talking about photographic 'magnification' such as for enlargement, 'magnification' is in LINEAR scale...an '8X' enlargement of a 24x36mm neg results in an 8x12". print...that is what you ordered when photoprocessing services offered in camera stores and drugstores was commonplace. And if you change from 4x6 to 8x12, in relative terms it is a 2x greater enlargment ('4X' print vs '8x' print size)

ic-racer · Jan 18, 2021

NB23 said:
I’m fluent in 4 languages with mix-ups between them, here and there. I don’t need to “read-up”. Thanks...

No offense intended, just kidding a little I wish more people had access to manuals for enlarging equipment, lots of good information on lenses and magnification ratios, etc.

https://www.photrio.com/forum/threa...-hosting-darkroom-equipment-pdf-files.180788/

ic-racer · Jan 18, 2021

wiltw said:
Before there is more added to the apparent p*ssing contest, the issue is that the term 'magnifiication' is CONTEXT DEPENDENT.

If one is talking about 'power' of a loupe or a binocular, megnificaiton indeed is AREA...4x power does NOT make a letter 4x taller, it is only 2x taller than actual.

If one is talking about photographic 'magnification' such as for enlargement, 'magnification' is in LINEAR scale...an '8X' enlargement of a 24x36mm neg results in an 8x12". print...that is what you ordered when photoprocessing services offered in camera stores and drugstores was commonplace. And if you change from 4x6 to 8x12, in relative terms it is a 2x greater enlargment ('4X' print vs '8x' print size)

Good point, those "powered" items produce a virtual image, so almost impossible to measure and challenge advertising claims

Ian C · Jan 18, 2021

To figure the maximum projection size, you need to measure the greatest possible distance, d, from the paper to the negative.

f = focal length of the lens

The greatest image magnification possible is

m = ( [d + Sqrt(d^2 – 4df) ]/2f) – 1

This ignores the nodal distance of the lens. The actual distance, d, is less than the measured value. The true value of d is the measured distance minus the nodal distance. If you disregard the nodal distance, the calculated magnification will be very slightly greater than the actual magnification

Most 8” x 10”-format enlarger lenses have a small nodal distance, so it makes little difference whether you take nodal distance into account or not. For example, the nodal distance of a 240 mm f/5.6A EL Nikkor is given by Nikon as 0.7 mm. Thus, you can ignore the nodal distance and still get a reasonable estimate of the magnification.

Measure the dimensions of the largest part of the illuminated negative visible to the lens. Multiply these by m to obtain the dimensions of the largest possible projection.

You can probably take measurements of your equipment before setting it up to determine the distance d. From that and the focal length, f, of the lens, you should be able to calculate a useful estimate of the dimensions of the projection.

ic-racer · Jan 18, 2021

Actually the most important distance for a Vertical enlarger (like the OP) is the distance from the lens to the column. In this case it limits him to nominal 26" print. For a vertical enlarger the distance from the lens to the ground is the limiting factor.

wiltw · Jan 18, 2021

ic-racer said:
Actually the most important distance for a Vertical enlarger (like the OP) is the distance from the lens to the column. In this case it limits him to nominal 26" print. For a vertical enlarger the distance from the lens to the ground is the limiting factor.

But as I already depicted via illustration, if the ANGLE of the projected cone of light can extend downward (for example, via droptable) the image circle is larger at a longer distance, so the bottom of the enlarger column is NOT NECESSARILY the limiting factor. Per my illustration, the issue is how far from the wall behind the column is mounted, because at a long enough distance, the circle of light will hit the wall behind.

ic-racer · Jan 18, 2021

wiltw said:
But as I already depicted via illustration, if the ANGLE of the projected cone of light can extend downward (for example, via droptable) the image circle is larger at a longer distance, so the bottom of the enlarger column is NOT NECESSARILY the limiting factor. Per my illustration, the issue is how far from the wall behind the column is mounted, because at a long enough distance, the circle of light will hit the wall behind.

Yes, very nice diagram. We need the actual length of the column, though.

wiltw · Jan 18, 2021

ic-racer said:
Yes, very nice diagram. We need the actual length of the column, though.

Per Beseler spec, the column is 54". If it has the 8x10 lightsource, the top of the lightsource is 61.5". But we need to know where the nodal point of the 240mm lens is relative to where the baseboard would be, so that one can compute n (existing distance) vs. Xn (where the floor is relative to where the baseboard would be -- when it is present when not projecting to the floor -- and we can then know the value of the multiplier N, so X * 24" can be computed...the max print size.

Calculating the max enlargement

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