Paul Kinzer

19-May-2017, 21:12

I recently bought the cells for a 90mm f/5.6 Super Angulon lens, and I've put them into a Seiko 0 shutter. (The Seiko 0 and the Copal 0 have the same specs, so the cells fit just fine). The Seiko shutter once held a different lens, and the aperture scale goes from f/6.3-64. I have to close the aperture past that f/6.3 setting before the blades show on the SA. I want to make my own aperture scale for this lens, and have some questions that I hope folks here can advise me on.

I've done a bunch of reading on this, and want to make a few points before getting to my questions. First, I will not send the lens in to have a scale professionally made. I got very good deals on both the glass and the shutter, and after checking on the cost of pro-made scales, I found that I would add nearly 50% to the cost of the lens by buying one. I'm cheap, so that's right out. Second, I know that shutters are not perfectly accurate, and that imaging is at least somewhat forgiving, so a perfectly accurate scale is not needed (one more reason to forego a pro-made scale). Finally, the only light meter I own is in my DSLR, so the descriptions of making scales based on the actual light coming through is not easy for me to understand, and may be impossible to do.

In my reading, several folks expressed the view, based on some of my points above, that simply using a ruler to measure the opening as seen through the front of the lens is good enough to use as a tool for marking a DIY aperture scale. Dividing the focal length by the f/stop gives what the diameter of the opening should be. This is what I've been messing about with so far, and my questions concern what I've found.

First, when the lens is wide open, the aperture should be 16.07 mm across (90/5.6 = 16.07). But when I use a ruler to measure the wide open lens, it looks to be 15mm across. Does this mean that the glass in the lens bends the light in such a way that the diameter I'm seeing is off of the 'true' aperture? (I imagine this is so; either that, or Schneider just gave their f/6 lens [90/15 = 6] a generous rounding off to f/5.6.) If this is true, should I, when making my calculations, subtract a proportional amount from each calculation? What I mean is, 15 is 0.93 x 16.07, so should I multiply each calculation by that amount before taking me measurements? 90/8 = 11.25; 11.25 x 0.93 = 10.46. So I would look for about 10.5 mm as the diameter of my f/8 setting?

Next, how do I measure the 'diameter' of the aperture when the opening is a pentagon? That 15mm measurement is the wide open circular view through the glass in the cells, but when I stop down even to about f/8, the sides of the five-bladed aperture get pretty flat. I'm not a math guy, so I looked online to try to find a formula for figuring this out. This confirmed that, yes, I'm not a math guy. Still, I found a calculator that showed me the area of pentagons drawn on virtual graph paper. I found that, if I measure from a point of the pentagon straight across to the middle of the opposite side, that 'diameter' is very close. A circle with the same diameter has 1.07 times the area. Is this close enough?

And here's an odd thing: if the first thing I described -- the 16.07 supposed aperture measuring a visual 15mm -- means I should shrink my measurements a bit (x 0.93), then the second thing -- a measurement across a pentagon is somewhat smaller than the diameter of a circle that has the same area -- means I should make my measurements a bit bigger (x 1.07). Again, I'm not a math guy, but it seems to me that, for this particular glass in this shutter, I should multiply those two numbers together to tell me how to measure the apparent aperture. And it turns out that 0.93 x 1.07 = 0.9951! So I should just use the ruler and go by the distance across the pentagon and mark based on that.

I hope someone can check my thinking for me.

I also hope that what I've written, if it makes sense and is close enough, might help others who want to make their own aperture scales.

I've done a bunch of reading on this, and want to make a few points before getting to my questions. First, I will not send the lens in to have a scale professionally made. I got very good deals on both the glass and the shutter, and after checking on the cost of pro-made scales, I found that I would add nearly 50% to the cost of the lens by buying one. I'm cheap, so that's right out. Second, I know that shutters are not perfectly accurate, and that imaging is at least somewhat forgiving, so a perfectly accurate scale is not needed (one more reason to forego a pro-made scale). Finally, the only light meter I own is in my DSLR, so the descriptions of making scales based on the actual light coming through is not easy for me to understand, and may be impossible to do.

In my reading, several folks expressed the view, based on some of my points above, that simply using a ruler to measure the opening as seen through the front of the lens is good enough to use as a tool for marking a DIY aperture scale. Dividing the focal length by the f/stop gives what the diameter of the opening should be. This is what I've been messing about with so far, and my questions concern what I've found.

First, when the lens is wide open, the aperture should be 16.07 mm across (90/5.6 = 16.07). But when I use a ruler to measure the wide open lens, it looks to be 15mm across. Does this mean that the glass in the lens bends the light in such a way that the diameter I'm seeing is off of the 'true' aperture? (I imagine this is so; either that, or Schneider just gave their f/6 lens [90/15 = 6] a generous rounding off to f/5.6.) If this is true, should I, when making my calculations, subtract a proportional amount from each calculation? What I mean is, 15 is 0.93 x 16.07, so should I multiply each calculation by that amount before taking me measurements? 90/8 = 11.25; 11.25 x 0.93 = 10.46. So I would look for about 10.5 mm as the diameter of my f/8 setting?

Next, how do I measure the 'diameter' of the aperture when the opening is a pentagon? That 15mm measurement is the wide open circular view through the glass in the cells, but when I stop down even to about f/8, the sides of the five-bladed aperture get pretty flat. I'm not a math guy, so I looked online to try to find a formula for figuring this out. This confirmed that, yes, I'm not a math guy. Still, I found a calculator that showed me the area of pentagons drawn on virtual graph paper. I found that, if I measure from a point of the pentagon straight across to the middle of the opposite side, that 'diameter' is very close. A circle with the same diameter has 1.07 times the area. Is this close enough?

And here's an odd thing: if the first thing I described -- the 16.07 supposed aperture measuring a visual 15mm -- means I should shrink my measurements a bit (x 0.93), then the second thing -- a measurement across a pentagon is somewhat smaller than the diameter of a circle that has the same area -- means I should make my measurements a bit bigger (x 1.07). Again, I'm not a math guy, but it seems to me that, for this particular glass in this shutter, I should multiply those two numbers together to tell me how to measure the apparent aperture. And it turns out that 0.93 x 1.07 = 0.9951! So I should just use the ruler and go by the distance across the pentagon and mark based on that.

I hope someone can check my thinking for me.

I also hope that what I've written, if it makes sense and is close enough, might help others who want to make their own aperture scales.