Originally Posted by ChristopherCoy
So if I understand this correctly... Instead of 2-4-6-8-10 second intervals, I would do 2.8-4-5.6-8-11-16 second exposures?

And this refers only to time, correct? My lens setting on the enlarger would stay on f11 or f16 the entire time?
Exactly right. "f/stop printing" is a really bad misnomer, it should be called "logarithmic printing" and it has absolutely nothing (sorry ME Super) to do with lens apertures at all*. Paper is a negative just like film, though the curve is funkier and S-shaped. To get a specific change in tone, you want to multiply the exposure by some factor, not add on some constant. Leave your enlarger lens at its optimum setting (generally 2 stops from open) unless that causes problems with stupidly long (>60s) or short (<4s) print exposures and control your density with printing time.

So the 2.8-4-5.6-8-11-16 second sequence gives you a factor of 1.4=sqrt(2), which is half-stop spacing when used as times. When you're printing at lower grades you need bigger steps and when printing at higher grades, you need smaller steps to achieve the same shift in tone; the reason is the changed slope of the paper's response.

I think I get the gist, but with linear intervals, I set my timer on the time, then slide a sheet from one side of the print to the other. How do you do that with fstop printing? If I added 5.6 seconds on top of the 8 second, that would give me 13.6 seconds. I would have to cover that 8 second portion right?
You want the total exposure for each bit of paper to be the number from the sequence, so (assuming you're doing the covering-up thing) each exposure is the difference between respective steps: 2.8, 4-2.8=1.2, 5.6-4=1.6, 8-5.6=2.4, 11-8=3, 16-11=5, 23-16=7, 32-23=9.

* the numerical sequence looks familiar only because it is a geometric sequence with factor sqrt(2). The numbers for full stops on aperture match the numbers for half stops on time because aperture area and therefore exposure varies quadratically with respect to f-number.