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MathGroup Archive 2006

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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.
>>
>
> 



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