Calculating the max enlargement

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peoplemerge

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Hi all,

I’ve finally finished the drywall and paint on my new darkroom, and I’m trying to figure the optimal placement for my wall mount Beseler 45V-XL using the 8x10 cold light kit. I plan on using a 240mm lens (I think ... it might be 250). The ceiling is 117”. What’s the maximum projected image?

Key data for the impatient on the 45V-XL:
Column height: 54”
Max height with 810: 61 1/2”. I calculate that means I’m left putting it 55” off the ground. I’m not sure where that leaves the nodal point... I’d have to guess about another foot. For the enlarger head not counting the 810 light box.
 
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I did this sort of math for wall mounting a Durst 1000 a few years ago. It's not as straight forward as you'd think. If you're projecting straight down, at a certain point, the projected image will hit either the wall or the enlarger apparatus. So you've got a practical limit there. This is why very large prints are projected against a wall.

If you can give me the following data, maybe I can help more:

- Film size
- Lens focal length
- The distance from the top of your enlarger (including light source) to the negative plane.
- The height of the your easel or countertop from the floor.
 

gijsbert

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I don't know how precise my method is, but I use the formula: magnification = (height / focal-length) - 1
with height the distance between the lens (nodal point) and the baseboard. Then print size = negative size * magnification

If the max print bumps into the column, you could mount the column away from the wall and make more space below the column for bigger prints.
 

ic-racer

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Subtract the distance from the ceiling to your negative carrier, then you might be able to use a table of values for some other enlarger.

My enlarger won't go all the way to the top, so with a 240mm lens it projects an 8x10 negative only 40 inches across with my baseboard all the way down. To get any bigger I'd have to go horizontal.

Screen Shot 2018-02-17 at 3.41.19 PM.jpg
 
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My math is probably convoluted, but I've found it to be accurate. I find it easiest to start in millimeters, and convert to inches at the end.

The negative to paper distance = lens focal length * ((magnification+1)^2) / magnification

So say you want to make a 32x40 print - a 4x magnification of an 8x10 neg, using a 240mm lens. The math is: 240 * ((4+1)^2) / 4 = 1500mm

So your negative to paper distance for a 32x40 print would 1500mm, or 59.06".

To figure out the counter depth you need, you need to know your nodal point. Let's say on your Beseler set-up, it's 20". You would take half of your print width, and add it to the nodal point.

So in the case of a 32x40" print: (32/2)+ 20 = 36"

That's how much counter space your going to need, at a minimum. The easel gobbles some of that up. My experience was, the ceiling height was less of an issue that the counter depth.

Clear as mud, right?
 

ic-racer

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What’s the maximum projected image?
I just looked and see the 45V-XL has a straight column allowing prints 24" in vertical dimension. I think you might be better with the Non-XL version because you might have to mount the column up high enough so the image can pass below the bottom of the column to make prints larger than 24" high.
 

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I just looked and see the 45V-XL has a straight column allowing prints 24" in vertical dimension. I think you might be better with the Non-XL version because you might have to mount the column up high enough so the image can pass below the bottom of the column to make prints larger than 24" high.

What is a 'non-XL' version? The VXL can be wall mounted... Beseler created the #8562 to mount the 45VXL on a wall or table. It comes with an upper column support bracket that bolts to the wall.

Of course, for really large prints the enlarger needs to be mounted farther away from the wall so the projected image is not striking the wall itself. With standard baseboard and usual FL lens for a given format, the 45VXL can attain 24" print size in the direction to/from the wall. That might assist you in determining how much more enlargement you can get via the 'removable baseboard' wall installation, with added distance to the floor.

Per the Beseler product sheet, use of 240mm FL lens with an 8x10-capable head requires the use of #8037 kensboard.
 
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ic-racer

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No matter how far out from the wall the column is mounted, the head has to be low enough so the image passes below the bottom of the column. With a low ceiling and a tall column, no large prints can be made unless the column is shortened with a saw (Non-XL version). But the OP does not indicate if he even needs anything larger then 24"
 

MattKing

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No matter how far out from the wall the column is mounted, the head has to be low enough so the image passes below the bottom of the column. With a low ceiling and a tall column, no large prints can be made unless the column is shortened with a saw (Non-XL version). But the OP does not indicate if he even needs anything larger then 24"
A drop table can solve this problem by allowing you to do large enlargements with the head not at the top of the column.
 
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A drop table can solve this problem by allowing you to do large enlargements with the head not at the top of the column.
Yes, a drop table is the way to go. All that math I barfed out above was used to design a drop table that would be suited to a particular enlarger. I was trying to figure out the best heights for the different levels.
 

ic-racer

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This picture shows the problem with a low ceiling, long column and short distance between the lens and column. Even with a drop table, the distance from the lens to the column is the issue with making a print larger than 24" vertical. Unless you can shorten the column.

The angled column Omegas are better suited to big enlargements via drop table, but don't have the 8x10 head option the OP wants.

tumblr_omtowixh301snmasho1_1280.jpg
 
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wyofilm

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This picture shows the problem with a low ceiling, long column and short distance between the lens and column. Even with a drop table, the distance from the lens to the column is the issue with making a print larger than 24" vertical. Unless you can shorten the column.

The angled column Omegas are better suited to big enlargements via drop table, but don't have the 8x10 head option the OP wants.

tumblr_omtowixh301snmasho1_1280.jpg


NO! Cameras as paperweights!?
 

MattKing

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Actually, I think ic-racer's problem is that his floor isn't low enough! :smile:
I too noticed the cameras.
 
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peoplemerge

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So! I’ll need to digest some of these posts, but a few comments on the first few:
* I plan to project on the floor. The Beseler only has wall mount, no table.
* My book Build Your Own Home Darkroom has plans for wall mount any enlarger and constructing drop tables all out of wood.
* My challenge is where along the wall to put this beast. I’ll have a pair of Omega D5s on either side, each with their own drop table.

So to wrap up, I’m stuck planning what goes where in the room.
 
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peoplemerge

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- Film size
- Lens focal length
- The distance from the top of your enlarger (including light source) to the negative plane.
- The height of the your easel or countertop from the floor.

So:
Film: 8x10
Lens: 240mm
Distance from top of enlarger to lens stage: about 12”. I can’t say for sure because I haven’t unpacked the conversion yet. It is kind of a delicate cold light setup I don’t want to touch until the enlarger is mounted to it’s permanent position.
Height of easel from floor: 0mm
 
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peoplemerge

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I know about the wall projection idea, I really don’t have room space for a permanent setup like this.
 
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peoplemerge

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I have an 117” ceiling, which is luxuriously high. I buy paper in rolls to do mural sized prints. Isn’t that a good use of 8x10 film?

I’ll build the adjustable table so the highest position will accommodate the smallest print I expect to make from 8x10, I think 11x14.
 

wiltw

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No matter how far out from the wall the column is mounted, the head has to be low enough so the image passes below the bottom of the column. With a low ceiling and a tall column, no large prints can be made unless the column is shortened with a saw (Non-XL version). But the OP does not indicate if he even needs anything larger then 24"

I have made a drawing depicting the situation the OP describes, where 24" max on the baseboard COULD indeed result in a much larrger projection to the floor...the key is the distance of the column from the wall. The cone of projected light does not have to change....the distance n is merely increased to 2n in order to 2x the size of the print (Double the distance = 4x the exposure because the area covered is 4x...or 2x in a single dimension)



So if n is increased by only 1.4x, the print on the floor is 1.4*24
Simple geometric relationship.
 
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NB23

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That is not a 4x magnification!

A 4x magnification of a 8x10 (80 sq inches) is a print containing 320sq inches (80x4=320), as opposed to your 1280 sq inches.

The 32x40 print you are refering to is actually a 16X magnification, not 4x.

A 4x magnification of a 8x10 is a 16x20 print.

Edit: The method of calculation is extremely important because it will determine your fixer’s efficiency. If you are into archival printing, the fixer capacity goes hand in hand with the print size, or more exactly with the square inches.

My math is probably convoluted, but I've found it to be accurate. I find it easiest to start in millimeters, and convert to inches at the end.

The negative to paper distance = lens focal length * ((magnification+1)^2) / magnification

So say you want to make a 32x40 print - a 4x magnification of an 8x10 neg, using a 240mm lens. The math is: 240 * ((4+1)^2) / 4 = 1500mm

So your negative to paper distance for a 32x40 print would 1500mm, or 59.06".

To figure out the counter depth you need, you need to know your nodal point. Let's say on your Beseler set-up, it's 20". You would take half of your print width, and add it to the nodal point.

So in the case of a 32x40" print: (32/2)+ 20 = 36"

That's how much counter space your going to need, at a minimum. The easel gobbles some of that up. My experience was, the ceiling height was less of an issue that the counter depth.

Clear as mud, right?
 
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ic-racer

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4x mag of 8x10 is a 40" print. The convention is based on the diameter of the projected image circle, or the linear size relation of objects in the 'image' or 'subject' fields.
 
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NB23

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4x mag of 8x10 is a 40" print. The convention is based on the diameter of the projected image circle.

There may be a standard of which I am not aware, but as far as I know, there are only Four 8x10 prints fitting into One 16x20, which makes 16x20 a 4x enlargement. Not a 2x enlargement, as per your calculation.
 

ic-racer

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There may be a standard of which I am not aware, but as far as I know, there are only Four 8x10 prints fitting into One 16x20, which makes 16x20 a 4x enlargement. Not a 2x enlargement, as per your calculation.
Read up on optics and magnification. You might be mistaking the inverse square law which relates to EXPOSURE, not MAGNIFICATION.
 
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