[pure-silver] Re: estimating filter factor: gray card?
- From: "BOB KISS" <bobkiss@xxxxxxxxxxxxx>
- To: <pure-silver@xxxxxxxxxxxxx>
- Date: Tue, 3 Oct 2006 09:44:55 -0400
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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