[X3D-Public] X3D-Public Digest, Vol 60, Issue 8

[Luca Martini]

I already saw the lighting equations here
http://www.web3d.org/files/specifications/19775-1/V3.3/Part01/components/lighting.html#Lightsourcesemantics

I explained the problem very badly, sorry, it's my fault, I'll try to be
more specific.
I'm doing my graduation stage in a company, and they have a 3D environment
which allows you to create your own virtual scenes. I've been asked to
implement a solution for an X3D exporter but unfortunately the main
programmer of this environment doesn't work for them anymore and I don't
have all of his Virtual Reality skills, I'm just getting started. I didn't
have any problem in capturing diffuse, specular and emissive color values
of any material object from our environment and exporting them in X3D, but
I'm having some trouble in trying to understand how to do the same with the
material ambient color. Basically we have 3 values for the ambient color
(RGB chromatic space) that the user can set for a material object, but in
X3D the ambientColor attribute for the Material node is missing, unlike of
what happens for diffuseColor, specularColor and emissiveColor. Having a
look at the X3D lighting equations i found that it uses the
ambientIntensity (scalar value) Material attribute to compute the
"attributeColor" value, multiplying it for each component of the
diffuseColor attribute, so if, for example, my ambientIntensity is 0.2 and
my diffuseColor is [1, 1, 0.5] my ambientColor will be [0.2, 0.2, 0.1] *
light[i].ambientIntensity.
I think that our ambient color is instead a vector of 3 ambient intensity,
each corresponding to a color channel: in that case would it be possible to
implement a solution in X3D which allowed to have 3 ambientIntensity, one
for R, another for G and so on, instead of a single ambientIntensity scalar
value? And most importantly could my inference about [our environment's
ambient color] = [ambientIntensityR, ambientIntensityG, ambientIntensityB]
have any sense for somebody who had good VR skills?
Thank you for your patience and help.
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