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fralexis
16-Nov-2016, 14:26
I just bought a beautiful convertible lens from Central Camera in Chicago. It is an Ilex-Calumet Series 8 Caltar. On the aperture ring it has the settings for 4.8 through 45 and a scale marking 8 1/2". Just above this ring is another scale going from 10 to 45. That scale says 14". So I assume the 14" scale is for the lens when used as a 14" lens and the other when used as a 8 1/2" lens.

Do I simply convert inches to millimeters? So the 14" lens would be 355.6 mm? And the 8 1/2" would be a 215.9 mm lens?

According to the scale, the 14" scale shows a maximum aperture of 4.8 and the 8 1/2" lens has a maximum aperture of 10. Does this sound like I am understanding this?

Thanks!

Alexis

Dan Fromm
16-Nov-2016, 14:54
The lens has two cells, front and rear. When both cells are in the shutter the complete lens' focal length is 8 1/2" = 215 mm and its maximum aperture is f/4.8. In this configuration, use the aperture scale the starts at 4.8.

The lens is convertible. To convert the lens, remove the front cell. The rear cell's focal length is 14" = 355 mm. Its maximum aperture is f/10. In this configuration use the aperture scale that starts at 10.

Mark Sawyer
16-Nov-2016, 15:00
I just bought a beautiful convertible lens from Central Camera in Chicago. It is an Ilex-Calumet Series 8 Caltar.

Technically, it's a Series S, not a Series 8. A re-branded Ilex Acuton, (single-coated Plasmat design), and a very fine lens!

Do I simply convert inches to millimeters? So the 14" lens would be 355.6 mm? And the 8 1/2" would be a 215.9 mm lens?

Yes, that's right.

According to the scale, the 14" scale shows a maximum aperture of 4.8 and the 8 1/2" lens has a maximum aperture of 10. Does this sound like I am understanding this?

You have it backwards; the 8 1/2 inch configuration has a maximum aperture of f/4.8, and the 14 inch is a maximum of f/10.