http://cpp.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523?format=rss Tribe.net. Local Connections http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#b9182838-f303-4e25-b636-f5404a0fc776 http://www.gamedev.net/reference/articles/article323.asp This may help or be a start. It sounds like you are trying to do your own Gouraud shading. There are other articles and discussions on the site. Shading a triangle can be quite an enormous topic. Mon, 31 Jul 2006 19:59:35 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#b9182838-f303-4e25-b636-f5404a0fc776 Hypno 2006-07-31T19:59:35Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#0be3afd5-10bd-4fc3-a343-c98f60c47ca5 actually, I haven't considered any different approaches.. :) Graphics is not my area of specialily, but I will be implementing some gradient fill routines in the near future. So, I guess my real question could have been asking for advice in implementing gradient fills, both linear and circular.. my first idea for the non-circular fill was to draw the perpendicular lines all along the fill path.. I've written some code which calculates the colors for each line of the fill, but I don't know how to generate the co-ordinates for the series of lines to draw.. google searches have not provided me with much, except for some great stuff from Chris Lomont regarding circular gradient fills.. anyways, if anyone wants to point me to some other good sources regarding linear fills, etc, that would be great.. peace -cpr Mon, 24 Jul 2006 21:57:20 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#0be3afd5-10bd-4fc3-a343-c98f60c47ca5 cpr 2006-07-24T21:57:20Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#303e9b74-a499-4bbe-84a3-fb3553340d48 That could be done but the gradient wouldn't follow the edge of the triangle in that case without doing a spatial calculation on its distance from the vertices. I'm guessing he just wanted to interpolate between two RGB values following a straight path. Sun, 23 Jul 2006 06:39:42 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#303e9b74-a499-4bbe-84a3-fb3553340d48 Hypno 2006-07-23T06:39:42Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#41b24559-0423-4e28-ba1d-939d1ed8fc40 Have you considered using a scan line approach instead of drawing a bunch of perpendicular lines? Sat, 22 Jul 2006 01:20:24 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#41b24559-0423-4e28-ba1d-939d1ed8fc40 Scott 2006-07-22T01:20:24Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#408a11f7-8b6d-4b6c-877d-a0978759b3d7 Im not sure how you are drawing your lines, but if you use a slope-intercept line equation (y= mx + b) a perpendicular line is any line with a slope of -1/m I believe. taking each point on the initial line, you can project the perpendicular lines out this way since you know each point on the first line is also on the perpendicular line. I dont if this helps for your gradient problem tho :) Fri, 21 Jul 2006 23:42:39 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#408a11f7-8b6d-4b6c-877d-a0978759b3d7 Jason 2006-07-21T23:42:39Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#e0847db5-e18e-419b-8b7b-89e99b0d0db3 Jon, thanks again! This one looks much more promising.. I'll need to wade thru it still.. :) peace -cpr Fri, 21 Jul 2006 22:18:03 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#e0847db5-e18e-419b-8b7b-89e99b0d0db3 cpr 2006-07-21T22:18:03Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#6851f5be-9872-4f81-ace7-6abbd91da57a http://www.cs.fit.edu/~wds/classes/graphics/Rasterize/rasterize/rasterize.html#SECTION00060000000000000000 Fri, 21 Jul 2006 22:13:52 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#6851f5be-9872-4f81-ace7-6abbd91da57a Jon 2006-07-21T22:13:52Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#f051bd5b-d387-4a85-8a0d-15924ff86298 thanks Jon, that's the answer I keep finding 'how to determine the point of intersection of two lines', but that's not the problem I am trying to solve... I have one line, the 'fill path', and I need to create lines that are 90 degrees perpendicular to it... maybe that info is on the page you pointed me to, but I am 'higher level math challenged'.. :) I know my intersection point, ie. one point for each pixel along the fill path, now how do I generate the end points of each of those lines, to pass to a line drawing function? -cpr Fri, 21 Jul 2006 22:05:11 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#f051bd5b-d387-4a85-8a0d-15924ff86298 cpr 2006-07-21T22:05:11Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#c5daa1e3-f47e-4746-b639-fb633b8ff984 http://mathworld.wolfram.com/Line-LineIntersection.html Make MathWorld your friend. Fri, 21 Jul 2006 21:56:30 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#c5daa1e3-f47e-4746-b639-fb633b8ff984 Jon 2006-07-21T21:56:30Z http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#01d86514-2746-49b2-84d1-34722eacc853 this is not c++ specific, and I am working in c to implement it, but, I am starting to work on gradient fill routines, and for the linear version I need to draw lines perpendicular to the fill path. so, given a line from x1,y1 to x2,y2, how to I calculate end points for all of the perpendicular lines (one per pixel along my fill path)? any takers? hehe.. :) Fri, 21 Jul 2006 21:15:48 GMT http://CPP.tribe.net/thread/9894e557-7bd5-4a59-90a9-637d46ab5523#01d86514-2746-49b2-84d1-34722eacc853 cpr 2006-07-21T21:15:48Z