Error in Stroebel?

[carstenw]

I have been working my way through Leslie Stroebel's "View Camera Technique", 7th Edition, and have now come across some calculations for which I couldn't follow his train of thought, and cannot mathematically reproduce what he has written. The section in question is the Intermediate f-number calculations. He writes (p.71):

"... For 1/2 stops the factor is the square root of 1.5 (1.22). For 1/3 stops, the factor is the square root of 1.33 (1.15), and for 2/3 stops the factor is the square root of 1.67 (1.29). Multiplying f/8, for example, by these factors to determine f-numbers that represent stopping down by 1/3, 1/2, 2/3, and one stop, the f-numbers are f/9.2, f/9.76, f/10.32, and f/11. The factor for 1/10 stop is the square root of 1.1 (1.049)."

I take it that most know that to get one full stop, one multiplies by the sqrt(2) = 1.41, e.g. f/5.6 * 1.41 = f/8. The approximate nature of the old sequence 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45, 64... makes it hard to verify, but it does seem to work out if you go high enough. Each number is a little off, but the sequence works out.

I would think that the factors that he is talking about should be usable to go from any steps to any subsequent step, and thus can also be used repeatedly to go through all steps. The problem is that these numbers do not work like that, except the first one, for obvious reasons. Using full accuracy in my calculator, 1.22*1.22=1.5, not 1.41 (duh), 1.15*1.15*1.15=1.5396, not 1.41. I don't know why he calculates 2/3 stop directly (and wrongly), rather than just multiplying the factor for 1/3 stops by itself, but even so, 1.15*1.15 (with full accuracy) is 1.333333 (of course), not 1.29. Multiplying his calculated value for 1/10th stop by itself 10 times gives 1.61, not 1.41.

I believe that what he should do is to take the nth root of sqrt(2) to get the nth fraction of a stop. For example, to get half stops, he should take the root of sqrt(2) = 1.189. Multiplying this number by itself would give sqrt(2), i.e. 2 half stops equals 1 stop. 1/3 stops would then be the 3rd root of sqrt(2) = 1.122, 1/10th stops would be the 10th root of sqrt(2) = 1.0352, and so on. I won't calculate 2/3 stops His sequence from f/8 to f/11 should be: f/8, f/8.97, f/9.51, f/10.07, f/11. Note that f/11 should actually be f/11.3, if one takes f/8 to be exact, part of the problem with the rounding in the sequence, and verification.

Has someone else come across this, or can someone explain to me where I went wrong?

 

[Jack]

This argument has been around a long time. First off Stroebel is correct, but only as respects convention, not math. Remember that aperture number is diameter of the iris divided by the lens focal length and why the progression is inverse square for the f-number. Then remember too, that lens focal length varies with subject distance, and is only the marked focal at infinity -- so marked apertures are not 100% accurate unless you are focused at infinity. Finally, the true progression should be 1, 1.4(1), 2, 2.8(3), 4, 5.6(6), 8, 11(.32), 16, 22(.64), etc. So in truth we should probably be calling f22 "f23," and then round back at f45. The point here is the existing scale is rounded and not 100% accurate either.

Convention divides the 1/3 (or 1/2 or 1/4) steps between stops equally, and the math used is as Stoebel explains, though using proper math generates a slightly different value based on a progression of fractional inverse squares. However, the diffence in conventional values versus proper values is insignificant as respects exposure, and why convention has held.

For example, the common 1/3 stops between f8 and f11 are stated as f9 and f10. f8 * 1.15 = 9.0 (Stroebel) or f8 * 1.122 = f8.976 (math). The actual difference of those values is three tenths of one percent between the values, and totally insignificant for an exposure. (This is where everybody arguing usually agrees that the point is moot.) Next, to drive the point home, we always use f11 when it should be f11.32 and the actual error here is a full 3% of the value, still insignificant for actual photography but 10 times the error in the previous example using Stroebel's formula...

So the system is flawed due to convention; a convention which induced rounding numbers for convenience; that convenience being the ability to perform most exposure calculations quickly in one's head.

 

[Jack]

EDIT: You'll actually use f-numbes for a lot of calculations in LF photography that have nothing to do with absolute aperture. For example, take the exposure falloff due to bellows extension. Your 150 lens at infinity will have 150mm of bellows extension. How much exposure do you need to add if you focused close and your total extension is 210mm? While you are looking through Stroebel's glossary and grabbing your calculator, I can tell you it's one stop instantly calculating in my head. I just need to think of my 150mm lens as "f15" by dropping the zero, and drop the zero off the total extension of 210 and get my new aperture of f21, which is approximately one stop slower than f15... This is not absolutely accurate either, but it is under 1/3rd stop off, and more than accurate enough for typical photographic purposes.

 

[carstenw]

For example, take the exposure falloff due to bellows extension. Your 150 lens at infinity will have 150mm of bellows extension. How much exposure do you need to add if you focused close and your total extension is 210mm? While you are looking through Stroebel's glossary and grabbing your calculator, I can tell you it's one stop instantly calculating in my head. I just need to think of my 150mm lens as "f15" by dropping the zero, and drop the zero off the total extension of 210 and get my new aperture of f21, which is approximately one stop slower than f15...

:bugeyes: Okay, my thinking hat goes back on again...

 

[Jack]

:bugeyes: Okay, my thinking hat goes back on again...

If you do the math for lens bellows factors, you'll figure it out.

Bottom line is many of the calculations related to exposure follow the aperture scale simply because light falls off at the inverse square of distance, just like the progression of values on an aperture scale, making quick calcs easy.

 

[carstenw]

As a gut feeling, I figure the ratio of bellows extension is tracking the ratio of apertures, and so it works out, but it looks weird-alice the way you wrote it

I was reading up on shutter speed/aperture corrections today, i.e. when you stop down, the shutter efficiency goes up, and so you have to stop down a little to compensate for that. Weird, I had never thought of that. There are a lot of little gotchas in LF. In fact, I was wondering if this one wouldn't also bite MF users of systems with leaf shutters?

 

[Jack]

I was reading up on shutter speed/aperture corrections today, i.e. when you stop down, the shutter efficiency goes up, and so you have to stop down a little to compensate for that. Weird, I had never thought of that. There are a lot of little gotchas in LF. In fact, I was wondering if this one wouldn't also bite MF users of systems with leaf shutters?

It is a fact with ALL leaf shutter systems because of the way the shutter leafs open and close -- small apertures make for less "drag" time as the leaf opens and closes across the aperture, increasing the effective shutter speed by a few percentage points. For most of your applications on film, it is inconsequential. It is also more significant with larger shutters, but again mostly inconsequential for film... Of more importance is the accuracy of the shutter to begin with, and most serious LF shooters keep a small shutter speed checker for just that purpose -- more than half of my lenses had blue tape on the speed ring showing the actual speeds at the detents when they varied by more than 1/4 stop.

 

[carstenw]

So why have I never heard this discussed among Hasselblad users, for example? Do they just get slightly bright photos when they stop down?

 

[Jack]

So why have I never heard this discussed among Hasselblad users, for example? Do they just get slightly bright photos when they stop down?

No, as I said above, for the most part it is inconsequential, especially with smaller shutters, and the MF inter-lens shutters are all small...