and Deke... thank you
________________________________
From: Torax Unga <tungau...@yahoo.com>
To: Ivan Busquets <ivanbusqu...@gmail.com>; Nuke user discussion
<nuke-users@support.thefoundry.co.uk>
Sent: Thursday, June 23, 2011 1:43 PM
Subject: Re: [Nuke-users] linear to AlexaV3LogC math
Thanks Ivan and Donat... this is great help.
________________________________
From: Ivan Busquets <ivanbusqu...@gmail.com>
To: Torax Unga <tungau...@yahoo.com>; Nuke user discussion
<nuke-users@support.thefoundry.co.uk>
Sent: Thursday, June 23, 2011 1:37 PM
Subject: Re: [Nuke-users] linear to AlexaV3LogC math
Based on that formula, the reverse should be:
x>0.0106232?(log10((x +
0.00937677) / 0.18)*0.2471896) + 0.385537:(((x+ 0.00937677) / 0.18) +
0.04378604) * 0.9661776
Where 0.0106232 comes from solving the second part of the equation using the
threshold value (0.1496582)
(0.149658 / 0.9661776 - 0.04378604) * 0.18 - 0.00937677 = 0.0106232
On Wed, Jun 22, 2011 at 11:28 AM, Torax Unga <tungau...@yahoo.com> wrote:
Keeping it going with the math questions.
>
>If this is the math for going from AlexaV3LogC to Linear:
>
>(x > 0.1496582 ? pow(10.0, (x - 0.385537) / 0.2471896) : x / 0.9661776 -
>0.04378604) * 0.18 - 0.00937677
>
>What's the math to revers it? linear to AlexaV3LogC.
>
>Gracias!
>
>
>
>
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