# A Spot of Maths?? Help with new bellows factor formula!

Discussion in 'Large Format Cameras and Accessories' started by holmburgers, Oct 19, 2011.

1. ### holmburgersMember

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Spot of maths anyone? That's inspired by the "Americanisms" thread... but down to brass tacks here.

So I proposed this idea years ago and got scoffed and belittled out the door, but I've revisited the idea and I stand resolved that it's useful and good!!

The idea is to calculate bellows factor from subject distance. Everyone told me, use the magnification formula, use the bellows draw formula! Well bah humbug to that!

Just kidding... but I think it's undeniable that subject distance would be a useful chart to make. Subject distance is much easier to estimate by eye than bellows draw or magnification, and if you had a chart made up for each lens, well it'd be an easy thing to say "for this portrait I need to add 1 stop".

eXposure compensation = (bellows draw)²/(focal length)²
Magnification = (image height)/(subject height)
eXpsoure = (Magnification + 1)²

eXposure compensation as a function of subject distance = ??

Your expertise in algebra is much appreciated!

2. ### Ian CMember

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This can be done with the understanding that s = lens-to-subject distance.

The calculation might be tedious, but once you’ve compiled a table of values for a particular lens and format, no further calculation is needed.

Let f = focal length, then

Time factor = [s/(s – f)]^2

Δf = 2*ln[s/(s – f)]/ln2

Magnification = s/(s – f) -1

Example: s = 3 meters = 3000mm

f = 200mm

Time factor = [3000/(3000 – 200)]^2 = 1.15

Δf = 2*ln[3000/(3000 – 200)]/ln2 = 0.20 stops (not enough to need compensation in most cases)

Magnification = 3000/(3000 – 200) – 1 = 0.07X

As you can see in the above example, you don’t need compensation unless the subject is quite close. At typical portrait distances compensation is generally not needed.

When compensation is needed, most of us prefer measuring the bellows draw as it’s quick and accurate.

3. ### holmburgersMember

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Thanks Ian,

I have to run for the day, but next time I log in I'm going to mull this over.

But I would know that at 3 meters (about 9 feet) I wouldn't need to compensate. I'm thinking more in the 4 foot to 1 foot range; those are distances that I can easily "eye-ball". I also just find it odd that there isn't a well respected formula for measuring bellows factor from this way of thinking. To someone who hasn't been entrenched in the conventions, this way seems perfectly logical to me. This might be quicker and accurate.

Anyways, thanks, and questions to come.

4. ### polyglotMember

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The reason people say to go with magnification is that it is lens-independent and directly related to bellows extension. You can certainly generate a bellows-factor table but it will be specific to each focal length if the table is based on subject distance.

If you're doing portraits, an easy approach is "head widths per frame" - heads are about 5" across and if you shoot 4x5" then the number of heads you can fit on the long side of the frame is 1/magnification and therefore tells you compensation required. And of course it doesn't matter what lens you have on!

If you want a no-compensation rule-of-thumb, start with Ian's time-compensation formula as a function of s and f:

Time factor = [s/(s – f)]^2

Therefore at 10x the focal length, you're about a third of a stop out and need to start compensating if shooting chromes and and at 6x the focal length, you need half a stop, which is the point where you want to start compensating even for B&W. Any distances further than that, you can ignore it. Those rules go out the window if your lens is telephoto or retrofocus; at that point you MUST use bellows draw.

Say you shoot with a 150mm, you can ignore bellows compensation outside 2m, add 1/3 stop at 1.5m, 1/2 stop at 900mm and do calculations for any point closer. For a 300mm, double those distances, etc...

(hopefully I haven't buggered up the maths)

5. ### holmburgersMember

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Ok, so Ian, you've given me 3 formulas. I see now... It's the middle one that threw me off. Natural logarithms... who comes up with this stuff, seriously?

And the purpose of those is to have the output in f/stops, not an exposure factor I guess.

Time factor = exposure compensation, right? So if you're time factor is 2, you need to add 1 stop, 4 - 2 stops, 8 - 3 stops and so on. If that's the case, then I'm going to just write down this formula:

eXposure compensation = [subject distance/(subject distance - focal length)]²

If that's the case, I can dig it!

So for example, if I'm using my 180mm Symmar and the subject is 1 foot from the lens, then I get roughly an exposure/time factor of 6. So about 2-1/2 stops.

And on a somewhat related note, exposure factors, whether they're filter factors, time factors, etc., do they follow the "f/stop" sequence? That is, I know a factor of 2, 4 and 8 corresponds to 1, 2 and 3 stops. So does 2.8, 5.6 and 11 correspond to 1.5, 2.5 and 3.5 f/stops? I think so, but I just need some reassurance.

Really appreciate the help guys!

6. ### Ian CMember

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Whenever we have an equation with the unknown as the exponent, the use of logarithms is the only tool that can isolate the unknown.

Most calculators priced at \$15 or more have the common base 10 log and natural log functions conveniently on the keyboard. You need only press a single key in most cases to calculate the value.

Δf, the unknown difference in f-stops, is just such a case. It occurs as the unknown exponent in an equation.

Filter factors are simply time multiplication factors just like the time factor above.

If you first calculate the Time Factor, you can then find Δf as

Δf = ln(Time Factor)/ln(2)

For example, if f = 200 and s = 400, then

Time Factor = [400/(400  200)]^2 = 4

so

Δf = ln(4)/ln(2) = 2 stops.

In the example you cited of a 180mm lens at 1 foot (304.8mm),

Time Factor = 5.96, so

Δf = 2.58

Your estimated value of 2 1/2 stops is close enough for most purposes.

Regarding,

I know a factor of 2, 4 and 8 corresponds to 1, 2 and 3 stops. So does 2.8, 5.6 and 11 correspond to 1.5, 2.5 and 3.5 f/stops?

If you mean that 2.8, 5.6, and 11 are time factors, then yes, they do correspond as (factor, Δf) as:

(2.8, 1.5 stops), (5.6, 2.5 stops), and (11, 3.5 stops).

Note: the conventions adopted by camera, lens, and meter makers is to mark the f-stops in convenient values that are not quite accurate in the case of odd integer powers of the square root of 2.

The true values are:

(sqrt 2)^3 = 2.8284

(sqrt 2)^5 = 5.6585

(sqrt 2)^7 = 11.3137

Similar simplifying conventions apply to the marking of shutter speeds. What are marked as:

1s, 1/2s, 1/4s, 1/8s, 1/15s, 1/30s, 1/60s, are actually

1s, 1/2s, 1/4s, 1/8s, 1/16s, 1/32s, 1/64s,, 1/2^n, for integer values of n.

7. ### holmburgersMember

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Thanks Ian, that answers all my questions I think! I now have a new bellows-factor formula that I will find very useful.

And connecting the dots regarding filter factors and f/stops is helpful for a host of things (namely... filter factors). I was struggling with say, a filter factor of 10, or 6. But knowing the logarithmic scale (and further knowing the 1/2 and 1/3 f/stop designations) will allow me to easily relate the FF to an f/stop change within a reasonable degree of accuracy.