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

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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