Volute aperture scale

[Jim C.]

I'm trying to figure out the aperture scale on my Volute shutter, from what I gleaned
here and elsewhere f16 is the same on the Unified Scale ( US ) as on a modern lens
and I need to double it or divide ( ? ) to get the modern equivalent.
The lens cells are marked 12 and 18 in ( Gundlach - Manhattan Optical ) so I'm guessing its
a Turner Reich design and it's a convertible and there is probably a missing cell since the
aperture scale has three scales.

The aperture numbering has me scratching my head, since there is no f16, how would I convert
them to modern f-stops ? any guidance is appreciated.

[BrianShaw]

Are you sure that is US? Everything I've ever seen in US goes from 8 to 128.

What an odd assortment of numbers. I suppose interpolation is the way to go: 16 is somewhere between 13.6 and 19.

[Peter Gomena]

I can answer part of the question. There are three scales because one is for both lens elements together, one for the rear element used singly behind the shutter, and the third for the front element used singly behind the shutter. The first is the outside scale, the third the inside scale.

I can find no reference to this particular f/stop scale in my files, but this can't be the only shutter ever manufactured with these numbers.

[Jim C.]

BrianShaw - Actually, I'm not sure anymore if it is US or not. The numbers are certainly a riddle ....

Peter - Thanks ! I never occurred to me to think of the scales the way you explained it, I've seen some old catalogs that
mention three sets of lenses as a kit for this type shutter, so I assumed that there was an missing cell and one of the scales was
for that.

[bentleyR]

It looks like what they did was start from the wide-open f/# and mark out the stops from there. For example, for the one that starts at 6.8 the next stop would be 6.8 * sqrt(2) ~ 9.6 and the next 6.8X2= 13.6. On a modern version I suppose there would still be a mark for f/6.8 (wide open) but then they would mark it in the usual sequence for f/8, f/11, etc.

[Mark Sawyer]

What bentlyR said. Each scale doubles every other number, indicating the f/stop progression. F/6.8 is a fairly common starting point, (Gundlach's Turner Reich was an f/7; this might be an early version of that lens), and the other scales start where their single elements convert to.

[Jim C.]

The doubling of every other number I understand for the scale that is on the shutter, but how does that convert to modern f stops ?
Do assume that f6.8 is in between f 5.6 and f8 and proceed from there ?
I'm not understanding bentlyR's formula - 6.8 times the square root of what ?

I'm a bit advanced math challenged

[BrianShaw]

This might be a "good enough" approximation of your f-stop scale to a modern f-stop scale:
6.8 = 8
9.6 = 11
13.6 = 16
19 = 32
72 = 64

[Jim C.]

This might be a "good enough" approximation of your f-stop scale to a modern f-stop scale:
6.8 = 8
9.6 = 11
13.6 = 16
19 = 32
72 = 64

How did you get those f stop values ?

[Peter Gomena]

Basically, they're saying that f/6.8 is half-way between f/5.6 and f/8.

F/numbers double by the square root of two. 1, 1.4, 2, 2.8, etc. each larger number is the preceding number multiplied by the square root of two (and rounded a bit). Your shutter is marked with f/numbers that multiply each previous number by the square root of two, but it starts with a different number. 6.8xsquare root of two is 9.6, so that's a full stop.

Most shutters aren't marked on the half-stop because you can't fit all the numbers on the darned scale. I'd just read them as 5.6 1/2, 8 1/2, 11 1/2, etc. so they work a little more easily with your light meter.