PDA

View Full Version : Bellows Extension/Focusing Distance Formula



Vlad Soare
16-Sep-2010, 06:45
Hi guys,

If D is the distance between the lens and the subject (focused), F is the focal length of the lens, and E is the distance between the lens and the ground glass, is there a mathematical formula that calculates D as a function of F and E?

Thank you.

aduncanson
16-Sep-2010, 08:08
The lens equation:

1/F = 1/D + 1/E

Works for "thin lenses". For real lenses D is measured from subject to Front Nodal Point and E is measured from ground glass to Rear Nodal Point. The difficulty comes in knowing the locations of these Nodal Points. For telephoto and retrofocus lenses they can be quite far away from the center of the lens.

memorris
16-Sep-2010, 08:16
I was taught a simple way to calculate exposure compensation for bellows extension.

Convert the focal length of the lens to inches and think of that as an aperture setting... For example, a 210mm lens id about f/8 1/3. So then look at the distance between the standards in inches as another aperture setting. The difference in stops is the compensation.

If you have a 210mm lens and have the bellows extended to 16 inches you should adjust the exposure by 1 2/3 stops.

Jerry Bodine
16-Sep-2010, 08:26
Vlad,

Using your nomenclature, then according to my Leica Manual:

D = (E x F) / (E - F), where D and E are measured to the lens diaphragm

Vlad Soare
16-Sep-2010, 11:05
Thanks, guys.
I understand that it's difficult to know where the nodal point is, but I don't need extreme precision. A rough approximation will do.

I'm trying to figure out what the longest lens usable on my camera is, and under what circumstances (at what distance from the subject, that is). The camera is a Chamonix 045n-1. The maximum extension of the camera is 395mm.
Let's say I wanted to use a 360mm lens. With the bellows extended out to 395mm, the lens would be focused at 360*395/(395-360) = 4063mm.
So, can I assume that a 360mm lens can be used on my camera with subjects from about 4 meters to infinity?

Dan Fromm
16-Sep-2010, 11:10
Vlad, I just checked y'r arithmetic. You're right.

aduncanson
16-Sep-2010, 13:06
Yes, but with a 360/5.5 Tele-Xenar with back focal length of 184mm (https://www.schneideroptics.com/info/vintage_lens_data/large_format_lenses/tele-xenar/data/5,5-360mm.htm) you could focus to just about 1 meter.

Take maximum image distance = focal length + max extension - back focal length.

Jerry Bodine
16-Sep-2010, 16:18
I've not done much close-up work, but it seems to me that a WA lens (e.g., 90mm Super Angulon) would not require very much bellows draw. Any drawbacks to this?

cowanw
16-Sep-2010, 18:22
Here's another one from Deardorff's (Kellsey) Corrective Photography.
if r=(times of reduction;i.e. linear object size divided by linear image size)
D=(r+1) x F
e=D/r
F=D/r+1
Regards
Bill

Dan Fromm
17-Sep-2010, 02:01
I've not done much close-up work, but it seems to me that a WA lens (e.g., 90mm Super Angulon) would not require very much bellows draw. Any drawbacks to this?Short working distance. Possibly, great stress possibly, so-so image quality.

Jerry Bodine
17-Sep-2010, 11:55
Short working distance. Possibly, great stress possibly, so-so image quality.

Dan,
Would the "so-so image quality" be due more to the lens design or to the focal length?
Jerry

Dan Fromm
17-Sep-2010, 12:34
design

mircea nicolae
9-Feb-2013, 16:48
Hi guys,

I am looking at a similar problem as the one in this thread, but from a different point of view. My questions sounds like this.

If D is the distance between the lens and the subject (focused), F is the focal length of the lens, and E is the distance between the lens and the ground glass, is there a mathematical formula that calculates E as a function of D and F?


I am trying to figure out the lens to flange distance for a 135 mm apo sironar s lens placed at 12 meters from the subject, when focused.


Thanks.

Dan Fromm
9-Feb-2013, 17:51
mircea, I find all this easiest to think of in terms of magnification.

The magic formulas are, in your notation:

D = F*(m + 1)/m where D = front node to subject distance, F = focal length and m = magnification

whence m = F/(D - F)

E = F * (m + 1) where E = rear node to subject distance

whence E = F + (F^2/(D - F))

For most of the lenses used on LF cameras, the nodes, front and rear, are approximately at the diaphragm. This is far from true for telephoto lenses.

The rear node to flange distance for lenses is very variable, depends on the lens' mechanical design. Measure!

mircea nicolae
10-Feb-2013, 10:02
Thanks a lot :)