Re: [Nuke-users] Gamma and Alpha

2011-11-15 Thread Ivan Busquets
Hi Gavin, As you said yourself, the equation cannot be solved UNLESS you know both variables on one of the sides. In other words, you'd need to have the BG image in order to prep a FG image so it can be comped in sRGB space and match the results of a linear comp. So is there no way to output a

Re: [Nuke-users] Gamma and Alpha

2011-11-15 Thread Ron Ganbar
I knew I was right. (You guys just proved an old argument I had with someone). Oh, the joys of self gratification. Ron Ganbar email: ron...@gmail.com tel: +44 (0)7968 007 309 [UK] +972 (0)54 255 9765 [Israel] url: http://ronganbar.wordpress.com/ On 15 November 2011 10:28, Ivan Busquets

Re: [Nuke-users] Gamma and Alpha

2011-11-15 Thread Ivan Busquets
I hope there was a bet involved... :) If you need further proof, you could use this script: set cut_paste_input [stack 0] version 6.2 v4 BackdropNode { inputs 0 name BackdropNode2 tile_color 0x7171c600 label BG note_font_size 42 selected true xpos 2434 ypos 13246 bdheight 156 }

Re: [Nuke-users] Gamma and Alpha

2011-11-15 Thread Ron Ganbar
Truth is, I don't even remember who the argument was with - I just remember being annoyed. Thanks Ivan. Ron Ganbar email: ron...@gmail.com tel: +44 (0)7968 007 309 [UK] +972 (0)54 255 9765 [Israel] url: http://ronganbar.wordpress.com/ On 15 November 2011 19:57, Ivan Busquets

Re: [Nuke-users] Gamma and Alpha

2011-11-15 Thread Julik Tarkhanov
On 14 nov 2011, at 21:13, Gavin Greenwalt wrote: How are Nuke users handling workflows in which they need to deliver images with alpha that will be composited in sRGB space not linear space? It might be mathematically impossible but we do just that when going from nuke to flame with

[Nuke-users] Gamma and Alpha

2011-11-14 Thread Gavin Greenwalt
How are Nuke users handling workflows in which they need to deliver images with alpha that will be composited in sRGB space not linear space? Essentially we have a situation where you would need to find equations for u and v such that (xy + z(1-y))^(1-2.2) = (uv + z^(1-2.2)(1-v)). My initial