Xavier Bury
Thu, 21 Sep 2000 14:07:49 -0700
Cartesian equation: x2 + y2 = a2 or parametrically: x = a cos(t), y = a sin(t) Polar equation: r = a but the rest is at... http://www.ussc.alltheweb.com/cgi-bin/search?exec=FAST+Search&type=all&query =code+pixels+circle+points even if you can't plot using the pencil tool, you can estimate the sizes using the circle as a rectangle... scripting left as an excercise! Xavier > > > I've been reading the list heaps and finding out almost all > > that I need to > > > know at this stage but I have one big problem. > > > > > > I am developing a stack that needs to redraw an ellipse and a > > circular arc > > > that intersects with the ellipse at two points. I thought > that the oval > > > graphic was the best tool for the job but how do I find the > > intersection > > > points. The rect of the arc is set and the user sets the rect of the > > > ellipse. I really only need to know the x coordinates so I can > > calculate the > > > arcAngle and the startAngle of the arc. > > > > > > My understanding is that an oval has no real mathematical > > definition so is > > > the oval tool actually an ellipse? > > a little meta geometry should help here... > given that im not sure what you mean with elipses and ovals (pretty > ambiguous here) > you can do a lot of precise enough calculations with geometry... > If you draw a rectangle (rect) over that same oval you have an easier view > to "estimate" your points... > > the thing is algebra level, and it's easy to find code (borrowed from c or > java anywhere on the net... (all the keywords for the search are in that > previous sentence - will take ya right to it). There's also tons of good > programming books on this subject! im a little tempted to do some too > lately... i'll c what I can find! > > http://www.mathcom.com/nafaq/q230.8.html > http://www.mathcom.com/nafaq/q285.html > and > http://www.primenet.com/~grieggs/cg_faq.html > > draw a circle as a Bezier (or B-spline) curve? > http://www.primenet.com/~grieggs/cg_faq.html#Howto_Spline > tell whether a point is within a planar polygon? > http://www.primenet.com/~grieggs/cg_faq.html#Howto_PIP > > in sum without code or more > there I found... > > How do I draw a circle as a Bezier (or B-spline) curve? > The short answer is, "You can't." Unless you use a rational spline you can > only approximate a circle. The approximation may look acceptable, > but it is > sensitive to scale. Magnify the scale and the error of approximation > magnifies. Deviations from circularity that were not visible in the small > can become glaring in the large. If you want to do the job right, consult > the article: > "A Menagerie of Rational B-Spline Circles" by Leslie Piegl and > Wayne Tiller > in IEEE Computer Graphics and Applications, volume 9, number 9, September, > 1989, pages 48-56. > (www.ieee.org or something like that...) > > For rough, non-rational approximations, consult the book: > > Computational Geometry for Design and Manufacture by I. D. Faux and M. J. > Pratt, Ellis Horwood Publishers, Halsted Press, John Wiley 1980. > > For the best known non-rational approximations, consult the article: > > "Good Approximation of Circles by Curvature-continuous Bezier > Curves" by Tor > Dokken, Morten Daehlen, Tom Lyche, and Knut Morken in Computer Aided > Geometric Design, volume 7, numbers 1-4 (combined), June, 1990, > pages 33-41 > [Elsevier Science Publishers (North-Holland)] > > > How do I tell whether a point is within a planar polygon? > Consider a ray originating at the point of interest and continuing to > infinity. If it crosses an odd number of polygon edges along the way, the > point is within the polygon. If the ray crosses an even number of > edges, the > point is either outside the polygon, or within an interior hole > formed from > intersecting polygon edges. This idea is known in the trade as the Jordan > curve theorem; see Eric Haines' article in Glassner's ray tracing book > (above) for more information, including treatment of special cases. > Another method is to sum the absolute angles from the point to all the > vertices on the polygon. If the sum is 2 pi, the point is inside, > if the sum > is 0 the point is outside. However, this method is about an order of > magnitude slower than the previous method because evaluating the > trigonometric functions is usually quite costly. > > www.ddj.com might have some nice algorythms too!!! almost forgot my bible! > > > Archives: http://www.mail-archive.com/metacard%40lists.best.com/ > Info: http://www.xworlds.com/metacard/mailinglist.htm > Please send bug reports to <[EMAIL PROTECTED]>, not this list. > Archives: http://www.mail-archive.com/metacard%40lists.best.com/ Info: http://www.xworlds.com/metacard/mailinglist.htm Please send bug reports to <[EMAIL PROTECTED]>, not this list.