[pure-silver] Re: estimating filter factor: gray card?
- From: "Ralph W. Lambrecht" <info@xxxxxxxxxxxxxxxx>
- To: PureSilverNew <pure-silver@xxxxxxxxxxxxx>
- Date: Tue, 03 Oct 2006 02:45:39 +0200
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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