PDA

View Full Version : Linear Burn/Dodge = Subtractive/Additive?


EricChadwick
01-05-2004, 05:41 PM
Does anyone know if these two layer blending modes are the same as the traditional non-PS Additive and Subtractive blending?

I think Linear Burn is basically Subtractive:
one = Color(1,1,1);
colorout = Layer1 + BaseMap - one;
But I might be wrong.
Here's the description from the Photoshop Help file :
Looks at the color information in each channel and darkens the base color to reflect the blend color by decreasing the brightness. Blending with white produces no change.

And I think Linear Dodge is basically Additive:
colorout = Layer1 + BaseMap;
Here's the description from the Photoshop Help file :
Looks at the color information in each channel and brightens the base color to reflect the blend color by increasing the brightness. Blending with black produces no change.

The equations are from Michael Spaw's site (http://www.morphographic.com/Diversions/MaxPlugins/Diversions_3DMAXPluginsCompositeMode.htm).

Also curious if anyone has the equations for Vivid Light, Linear Light, and Pin Light...

Thanks!

halo
01-05-2004, 08:47 PM
hmm...i always thought Multiply was close additve, adding black produces a result still...but adobe are probably the ones to ask.

but i cant find those forumlae on that site...

Stroker
01-05-2004, 10:02 PM
I did some playing this a few weeks ago.

Linear Dodge = Add
Linear Burn = Subtract, but is "upside-down"
Linear Light = Add/Substract using 128 as the mid-point with either x2 or /2 in there somewhere

Something like that.

Stroker
01-06-2004, 03:49 AM
Dude, that material plug seriously rocks.
There is something that I've been trying to figure out how to do in Max for several weeks, and Composite Mode had it done in no time at all (2 minutes of fiddling around).

Thanks.
Woohoo!

CGTalk Moderation
01-17-2006, 02:00 AM
This thread has been automatically closed as it remained inactive for 12 months. If you wish to continue the discussion, please create a new thread in the appropriate forum.