[pure-silver] Re: estimating filter factor: gray card?

Here's my formula;

After a little preliminary testing, if you are still in doubt, overexpose by another half-stop. You will *rarely* ever berate yourself later for overexposing. But I have cursed myself many times for underexposing. Film has way more overexposure latitude than it has underexposure latitude.



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


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