tag:blogger.com,1999:blog-1840985738235902482.post6806566167312875049..comments2024-03-27T07:14:48.488-04:00Comments on John the Math Guy: How do you define a color?John Seymourhttp://www.blogger.com/profile/11350487038873935295noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-1840985738235902482.post-80858929157934502792017-08-03T01:49:00.498-04:002017-08-03T01:49:00.498-04:00I cant get past "the ace of hearts is white&q...I cant get past "the ace of hearts is white"¿Hanshttp://www.colourisintheeyeoftheoneviewingit.comnoreply@blogger.comtag:blogger.com,1999:blog-1840985738235902482.post-56378200977857413292017-08-02T21:06:48.064-04:002017-08-02T21:06:48.064-04:00Ah, excellent! Thanks!Ah, excellent! Thanks!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1840985738235902482.post-79643067434030952162017-08-02T12:20:49.559-04:002017-08-02T12:20:49.559-04:00Three years ago, I predicted that you would ask th...Three years ago, I predicted that you would ask that question, so I blogged on it:<br />http://johnthemathguy.blogspot.com/2014/11/measuring-fluorescent-inks.html<br /><br />I agree, the practical answer is that if you have a well defined light source (which is defined down into the UV) that you can have a meaningful definition of the color of something that fluoresces. Such a standard light source was defined for the print industry in ISO 13655.<br />http://johnthemathguy.blogspot.com/search?q=m3<br /><br />For more information on fluorescence:<br />http://johnthemathguy.blogspot.com/2017/04/on-nature-of-emitted-light-part-3.html<br />John Seymourhttps://www.blogger.com/profile/11350487038873935295noreply@blogger.comtag:blogger.com,1999:blog-1840985738235902482.post-5878080703913277742017-08-02T11:38:12.911-04:002017-08-02T11:38:12.911-04:00Does this discussion also apply to fluorescent col...Does this discussion also apply to fluorescent colors (e.g. DayGlo)?<br /><br />My initial guess is yes: You can illuminate a fluorescent color with a defined light source (e.g. D50) and measure the resulting reflectance spectrum, which can then be converted to L*a*b*. Unlike a non-fluorescent color, there will be more energy in certain parts of the spectrum than was present in the light source, but I don't think that should affect the calculations. The color will probably just have a brightness and/or saturation beyond what would be expected given the white point. (Maybe L* could end up greater than 100?)Anonymousnoreply@blogger.com