Exactly! On 2006-10-03 15:44, "BOB KISS" <bobkiss@xxxxxxxxxxxxx> wrote: > The relationship between f stops and factor is NOT a square root. As > demonstrated in my previous e-mail: " the square root of a factor of 8 is > 2.8 but we all know that a factor of 8 equals exactly 3 stops (2X2X2)." > The relationship is exponential factors of 2 NOT the square root which is > ONE SPECIFIC exponential factor of 2. > Therefore it DOES happen to correspond to the square root in some SPECIFIC > cases but not all. > CHEERS! > BOB > > Please check my website: http://www.bobkiss.com/ > > "Live as if you are going to die tomorrow. Learn as if you are going to > live forever". Mahatma Gandhi > > -----Original Message----- > From: pure-silver-bounce@xxxxxxxxxxxxx > [mailto:pure-silver-bounce@xxxxxxxxxxxxx]On Behalf Of Ralph W. Lambrecht > Sent: Monday, October 02, 2006 8:46 PM > To: PureSilverNew > Subject: [pure-silver] Re: estimating filter factor: gray card? > > I still don't see how you want to use square roots to convert filter factors > into stops or visa versa. I believe the equations given below do that much > simpler. > > > > > > Regards > > > > Ralph W. Lambrecht > > http://www.darkroomagic.com > > > > > > > > On 2006-10-03 02:07, "Richard Knoppow" <dickburk@xxxxxxxxxxxxx> wrote: > >> >> ----- Original Message ----- >> From: "Ralph W. Lambrecht" <info@xxxxxxxxxxxxxxxx> >> To: "PureSilverNew" <pure-silver@xxxxxxxxxxxxx> >> Sent: Monday, October 02, 2006 2:57 AM >> Subject: [pure-silver] Re: estimating filter factor: gray >> card? >> >> >>> The square root function is not used to turn filter >>> factors into f/stops or >>> visa versa? I think this has caused some confusion before. >>> >>> The conversion can be done using the following: >>> >>> x = 2^N >>> log(x) = log(2)*N >>> N = log(x)/log(2) >>> >>> where x is the filter factor and N is the amount of stops. >>> >>> >>> >>> >>> >>> Regards >>> >>> >>> >>> Ralph W. Lambrecht >>> >>> http://www.darkroomagic.com >> >> Stops are definitely square and square root functions. >> One can do roots in log form by multiplying or dviding by >> the root required, that is 2^4 can be found by taking >> anti-log of (Log 2 *4) and roots by deviding the log by the >> root required. The (2) factor in your equations above are >> square and square-root functions. Most calculators do >> exponentials by means of log look up tables. >> >> --- >> Richard Knoppow >> Los Angeles, CA, USA >> dickburk@xxxxxxxxxxxxx >> >> > ============================================================================ > == >> =============================== >> To unsubscribe from this list, go to www.freelists.org and logon to your >> account (the same e-mail address and password you set-up when you > subscribed,) >> and unsubscribe from there. > > > ============================================================================ > ================================= > To unsubscribe from this list, go to www.freelists.org and logon to your > account (the same e-mail address and password you set-up when you > subscribed,) and unsubscribe from there. > > > ============================================================================== > =============================== > To unsubscribe from this list, go to www.freelists.org and logon to your > account (the same e-mail address and password you set-up when you subscribed,) > and unsubscribe from there. ============================================================================================================= To unsubscribe from this list, go to www.freelists.org and logon to your account (the same e-mail address and password you set-up when you subscribed,) and unsubscribe from there.