[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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