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Message: <blockquote data-quote="elkcub" data-source="post: 1391592" data-attributes="member: 14473">Hi Ron,I think you're right on track.  You expressed the observed R, G, and B values from the binocular-plus-camera as a proportion of the R, G, and B values obtained from the camera alone.  When these proportions are taken of pure white (255, 255, 255) it models the binoculars' (pure) color shift.  Of course, this assumes constant proportionality, which makes sense.    To get an idea of what this tint (241, 239, 237) looks like one can use the RGB calculator at: <a href="http://webdeveloper.earthweb.com/repository/javascripts/2001/06/48571/hex.html" target="_blank">http://webdeveloper.earthweb.com/repository/javascripts/2001/06/48571/hex.html</a>.  In HSV terms this comes out to (21, 4, 241).  (Unfortunately, the calculator only produces .tiff files, which BF doesn't accept.)But, our calculations don't quite agree.  Using your RBG numbers I would calculate the binoculars' transmission as 215/229.45 = 93.7%, where 229.45 is Sum(RGB)/3.  Also, the calculator's HSV is 21 deg. not 30.  What am I missing?  The basis for my RGB preference is that it is essentially the Young-Hemholtz theory.  A good summary of this and the many normalizations and transformations later developed by psychologists and CIE can be found in "Optics" by Freeman and Hull.  <a href="http://books.google.com/books?hl=en&id=lXNFnybj9wwC&dq=optics+freeman+and+hull&printsec=frontcover&source=web&ots=P6ica2IRcT&sig=HwiLJ1xSgankWiMgADnGY-MV1CI&sa=X&oi=book_result&resnum=1&ct=result#PPA368,M1" target="_blank">http://books.google.com/books?hl=en&id=lXNFnybj9wwC&dq=optics+freeman+and+hull&printsec=frontcover&source=web&ots=P6ica2IRcT&sig=HwiLJ1xSgankWiMgADnGY-MV1CI&sa=X&oi=book_result&resnum=1&ct=result#PPA368,M1</a>  Unfortunately, the whole section on colorimetry isn't available to read on the Internet, but at least some of it is.  (I bought the book some time ago.)Ed</blockquote>

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