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!
>
>
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