View Full Version : Linear EXR to cineon-log lut?
Gunleik Groven
11-16-2011, 07:27 AM
Hi
Need to have a linear to log LUT for a bunch of EXRs that come in...
Thought that was built into Resolve, but: Oh Boy was I wrong... :)
How do you guys resolve this?
Cheers
Gunleik
Nick Shaw
11-16-2011, 10:41 AM
If the EXRs contain float data above 1.0 (which is generally the reason for using EXR) it may not be possible to do the conversion with a LUT. If the data is in the 0-1 range the attached LUT should do it. Otherwise you may want to try batch converting the EXRs to Log DPXs (preferably 16 bit if Resolve supports those) using e.g. Nuke.
Gunleik Groven
11-16-2011, 10:44 AM
Thanks a bunch!
Will try!
Cheers G
jake blackstone
11-16-2011, 12:06 PM
try batch converting the EXRs to Log DPXs (preferably 16 bit if Resolve supports those) using e.g. Nuke.
Resolve supports 16 bit DPX.
Nick Shaw
11-16-2011, 12:28 PM
Resolve supports 16 bit DPX.
Thanks Jake. I can't run Resolve on my system to test, as there is a conflict with my Baselight.
Graeme Nattress
11-16-2011, 12:43 PM
I'd probably use the standard lin2log node in Nuke - I know that math is correct. You may also be able to do it in shake with it's float support and it's lin2log.
Graeme
Nick Shaw
11-16-2011, 01:41 PM
I'd probably use the standard lin2log node in Nuke
Yes, or the the Colorspace node, with in=linear and out=Cineon. The LUT I posted is just that transform converted to a cube. But as I said, it is limited to the 0-1 range of the linear input. But then I have seen an awful lot of EXRs which were clamped to 0-1 due to people not understanding float workflows, and using inappropriate tools.
Graeme Nattress
11-16-2011, 01:42 PM
Lut's that clamp are a pain.... But they're not hard to code for full float range if you know what you're doing, but I don't know much that works that way. Maybe in the future....
Graeme