Re: Bug or feature ?
- To: mathgroup at smc.vnet.net
- Subject: [mg72414] Re: Bug or feature ?
- From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
- Date: Wed, 27 Dec 2006 05:15:02 -0500 (EST)
- References: <emggb8$2i1$1@smc.vnet.net> <emj3tj$krc$1@smc.vnet.net>
I should have started my previous reply (see history below) with the following: << Graphics`Graphics` Steve Luttrell West Malvern, UK "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk> wrote in message news:emj3tj$krc$1 at smc.vnet.net... > Evaluate this > > PlotPoints /. Options[PolarPlot] > > to find that the default number of points used to generate the plot is 25. > > You can fix your problem by overriding the default with a larger value. > > For instance: > > PolarPlot[Sin[t]^2 Cos[t]^2, {t, 0, 21},PlotPoints->100] > > Steve Luttrell > West Malvern, UK > > "Giovanni Resta" <g.restaxxx at cutTheXXXiit.cnr.it> wrote in message > news:emggb8$2i1$1 at smc.vnet.net... >> Please, try this with Mathematica (5.2): >> >> << Graphics`Graphics` >> >> PolarPlot[Sin[t]^2 Cos[t]^2, {t, 0, 2Pi}] >> >> PolarPlot[Sin[t]^2 Cos[t]^2, {t, 0, 21}] >> >> compare the two graphics. >> >> In the second (t is between zero and twentyone) >> I got a spurios segment between 3rd and 4th quadrants. >> >> Can you tell me why ? >> >> (Btw, I got this just playing for one minute or two >> with PolarPlot. The "21" was a typo...) >> >> g. >> > >