Reciprocity equations

swmcl

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OK. A large format application ...

As stated in the page linked to from the LF home page ... a thing called the 'exposure factor' is = (bellows extension / lens focal length)^2.

In the discussion about exposure factor in the same section on the webpage it discusses how to turn the exposure factor into 'stops' by taking its natural log and dividing by ln(2).

Elsewhere I see a formula for the change in aperture = (bellows extension / lens focal length) (see Kenneth Lee Gallery links).

So ...

Lets give this a go ...

Take a 180mm lens and give it a bellows draw of say 245mm.

The exposure factor is 1.8526. The aperture increase is 1.3611. Yet the 'stops' after the natural logarithm calculation is 0.8896.

I believe the speed would need to increase by the 'exposure factor' multiple of 1.8526.

I think I am tripping over some terminology problems can someone help out ?

Cheers,

Steve
 
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MattKing

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I don't see anything in your scenario that mentions reciprocity (other than the title).
 

snikulin

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[OFF]
APUGuser19 said:
...But certainly, I'd trust Kodak before I'd trust myself...
Click to expand...
And that's why human average IQ peaked at about Ancient Greece' time!
I for one have already lost my ability to navigate by maps and squarely rely on GPS.

[/OFF]
 

Doremus Scudder

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Your thread has nothing to do with reciprocity failure (a characteristic of film at low levels of light), but rather:

Bellows Extension Factors -- i.e., compensating for light loss due to increasing the distance between lens and film. (Maybe a mod can change the thread title?)

So, here's what I use: Bellows extension exposure factors are calculated using the following formula:
Ext² / Fl² = Exposure Factor
Where Ext = Bellows extension measured from film plane to lens nodal point (middle of the lens for most designs) and Fl = The focal length of the lens.

This gives you the Exposure Factor. Multiply the indicated time by the exposure factor for the proper adjusted exposure.

Stops to open the aperture is log2 of the exposure factor.

So, in your example, a 180mm lens and and a bellows draw of 245mm:

The exposure factor is 1.8526. This is the factor to multiply the shutter speed by.

The stops to open is 0.8896. So, you're right on these.

I think the other method you have of calculating aperture change is either faulty or you've missed something copying it.

FWIW, I hate calculating in the field, so I've made a table of bellows extension factors for all my lenses that a carry with me and can easily consult when needed.

Best,

Doremus
 

RalphLambrecht

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I'm not getting it Are you interested in an equation for belows extension or for reciprocity failure?:confused:
 

Sirius Glass

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snikulin said:
[OFF]

And that's why human average IQ peaked at about Ancient Greece' time!
I for one have already lost my ability to navigate by maps and squarely rely on GPS.

[/OFF]
Click to expand...

I have always been able to get lost by myself. A GPS gets me lost faster.
 
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OP

swmcl

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Sorry guys I'm battling two issues at present (reciprocity and bellows extension). I used the wrong phrase in the title. The text is the issue.
 
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swmcl

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I will ask if anyone knows a good bellows extension app for iOS ? The one in Reciprocity Timer is quite limited. I think it has a fixed extension upper limit of 300mm ?

Cheers,

Steve
 
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swmcl

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Also ...

I found a pdf titled, 'The Bellows Extension Exposure Factor: Including Useful Reference Plots for use in the Field' by Robert B. Hallock.

In this document there are several graphs for different focal length lenses. I keep getting different results for the same focal length lens on different graphs.

Is the formula for bellows extension factor the same regardless of lens focal length or is there a wide-angle compensation of some sort ?

I'll accept that there might be differences for tele lenses. I'm just thinking of the standard LF lenses like Sironars and Grandagons etc. (Is it planar lenses ??)

Cheers,
 

Doremus Scudder

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There's a sticky thread on figuring bellows extension factors here: (there was a url link here which no longer exists)

There is more information and methods here than anyone needs! I believe I've posted my tables there as well somewhere. Let's refer the OP to that thread.

Best,

Doremus
 

RobC

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swmcl said:
Is the formula for bellows extension factor the same regardless of lens focal length or is there a wide-angle compensation of some sort ?
Click to expand...

The bellows extension factor is directly related to "effective aperture". When you extend the bellows you are extending the "effective focal length" but the aperture size remains fixed (unless you change it). As you should know, f-no is the relationship between focal length and aperture diameter.

So...

100mm lens @ f4 has an aperture diameter of 25mm. f4 is calculated as (100/25) at infinity. If you extend the effective focal length to 200mm then you get 200/25 = 8. So effective aperture is f8. Thats a two stop bellows extension factor. Wide lens or not is not part of the bellows extension factor calculation.

However, with wide angle lenses you get more light fall off towards the edges of projected image circle which is why some wide angle lenses have centre filters available to correct the falloff. But the closer you focus (the more the lens extension) the less fall off you will get becasue you are using less of the projected image circle as you extend the lens. i.e. you are only using the central portion of the projected image circle so a centre filter may not be required.
If you are metering towards center of image then light falloff will not be a factor for an extended lens (close focus).
 
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