got the sf20, but I am a little confused about one thing. some of you, in previous posts, suggested to reduce the flash exposure to -2 or so. but when I set the flash to 'minus' something, it increases the meter mesure, indicating larger space cover. when I set it to 'plus', it decreeses the space covered. so I guess to reduce the flash power I have to set it to '+', right?

-- rami (rg272@columbia.edu), February 07, 2002

Answers

No. You set the exposure compensation to "-". First off, make sure the flash is on your TTL camera, the flash is turned on, AND the camera is activated by slightly pressing the shutter release. I don't have my set-up in front of me now, but if the SF20 is still denoting additional range, it is because the flash is telling you that you now have the same relative amount of light available at a longer distance, because you have reduced the amount of light you expect out of it.

:-),

-- Jack Flesher (jbflesher@msn.com), February 07, 2002.

Jack, thanks for the answer. it makes some sense, which is much more than it did before. but still, for example, with -1, it changes the meters mesure from 9.9 to 14. as you put it, I expect less flash light, but I reduced it for the 9.9 meters, there is where I expect less light (I knew from the begging that I would get less light on 14 meters, I didn't have to change anything for that?) so do I get it all wrong? I am not just trying to argue, I spend 4 hours last night trying to figure out what could it mean. thanks again.

-- rami (rg272@columbia.edu), February 07, 2002.

Rami:

Let's try this explanation:

1)Keep in mind that light falls off according to the inverse square law just like f-stops -- hence we lose 1/2 the effective output with our flash every time we increase the distance by a factor of the square root of 2, or 1.4.

2)Furthermore, keep in mind that the flash has a maximum output, and that with TTL we will get the exact amount of light we need if it is equal to or less than the maximum output we asked for.

So, if we tell the flash we want to reduce its EFFECTIVE output by 1/2 or 1 stop (by dialing in -1 on the exposure compensation) we now can gain extra distance by the factor of 1.4. BUT, anything exposed at the longer distance will be underexposed by 1 stop since that is what we asked for. Hence, your 9.9M going to 14M makes perfect sense: At full power with normal exposure, it can get out to 10M; but since you are asking for it to give you the EFFECT of half-power (1 full stop less) it can produce that same effect by using its full power at 14M (10M X 1.4).

:-),

-- Jack Flesher (jbflesher@msn.com), February 07, 2002.

Jack, exellent explanation. I sware I already guessed it has to do with squares... but to tell you the truth, I would never have been able to clarify to myself exactly how. so now it does make full sense. thanks.

-- rami (rg272@columbia.edu), February 07, 2002.

Rami:

Always happy to be of assistance to a fellow Leica shooter!

;-),

-- Jack Flesher (jbflesher@msn.com), February 07, 2002.