Re: Graphics package in v6
- To: mathgroup at smc.vnet.net
- Subject: [mg78411] Re: Graphics package in v6
- From: Bill Rowe <readnewsciv at sbcglobal.net>
- Date: Sun, 1 Jul 2007 03:42:45 -0400 (EDT)
On 6/30/07 at 6:12 AM, Florian.Jaccard at he-arc.ch (Jaccard Florian)
wrote:
>I'm a convinced Mathematica user, I love the new features I discover
>in v6, but I think the Graphics package was very useful in v5 and
>I'm missing it.
>For example, it was very easy to obtain the 3D plot of a surface of
>revolution around the x-axis with SurfaceOfRevolution.
>Example :
><< "Graphics`" ImplicitPlot[x^3*y + x*y^3 == 2, {x, -3, 3}, {y, -3,
>3}]
>Not possible using CountourPlot without cheating, isn't it?
>(I consider the following as cheating : Show[Plot[0, {x, -3, 3},
>PlotRange -> {-3, 3},
>AspectRatio -> Automatic], ContourPlot[x^3*y + x*y^3 == 2, {x, -3,
>3}, {y, -3, 3}, Axes -> True, AxesOrigin -> {0, 0}, PerformanceGoal
>-> "Quality", PlotPoints -> 150]]
>)
On my machine with version 6, I cannot see any difference
between the output of the code above you consider "cheating" and
the much simpler
ContourPlot[x^3*y + x*y^3 == 2, {x, -3, 3}, {y, -3, 3}, Frame -> None,
Axes -> {True, True}]
and except for color of the plotted curve, aliased graphic and
font the results of
ContourPlot[x^3*y + x*y^3 == 2, {x, -3, 3}, {y, -3, 3},
Frame -> {True, True, None, None}]
appear the same as what I see when I do
<< "Graphics`"
ImplicitPlot[x^3*y + x*y^3 == 2, {x, -3, 3}, {y, -3, 3}]
with version 5.2
--
To reply via email subtract one hundred and four