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