Re: tweaking VectorPlot...

*To*: mathgroup at smc.vnet.net*Subject*: [mg111317] Re: tweaking VectorPlot...*From*: Jae Bum Jung <jaebum at wolfram.com>*Date*: Tue, 27 Jul 2010 04:18:20 -0400 (EDT)

There's undocumented option (like "RightArrow", "LeftArrow") you can use in VectorStyle. For example, points={{-1,-1},{-1,1},{1,-1},{1,1}};VectorPlot[{-1-x^2+y,1+x-y^2},{x,-2,2},{y,-2,2},VectorPoints->points,VectorScale->.25,Epilog->{Red,PointSize[Medium],Point[points]}, VectorStyle->"LeftArrow"] or Show[{dp, VectorPlot[f, {x, -3, 3}, {y, -3, 3}, VectorPoints -> pts, VectorScale -> {.15, .2}, VectorStyle -> {"LeftArrow", Black, Arrowheads[0.02]}]}] Just make sure these are undocumented, i.e., it's subject to be changed in the future version of Mathematica. - Jaebum On 7/26/10 5:36 AM, Patrick Scheibe wrote: > Hi, > > this is far away from being *clean*: Take a potential and calculate the > gradient. Combine DensityPlot and ContourPlot (or > LineIntegralConvolution with vectorlength coloring if you like) on the > potential. If it's an artificial student-example use an easy potential > which you solve explicitely in one variable. So you can choose your > points on the wanted level and use VectorPlot to visualize the vectors > only on the curve. > > For the tails of the arrows I have no better idea then replacing: > > << VectorAnalysis` > pot = x^2 + x y - Sin[y^2] > f = -Most@Grad[pot, Cartesian[x, y, z]] > > dp = Show[{DensityPlot[pot, {x, -3, 3}, {y, -3, 3}, > ColorFunction -> "TemperatureMap"], > ContourPlot[pot == 0.2, {x, -3, 3}, {y, -3, 3}, > ContourStyle -> {Thick, Black}]}]; > > ys = Table[y, {y, -4, 4, 6/40.}]; > sol = Solve[pot == 2/10, x]; > pts = Flatten[Table[{x, y} /. sol, {y, ys}], 1]; > > vp = VectorPlot[f, {x, -3, 3}, {y, -3, 3}, VectorPoints -> pts, > VectorScale -> 0.25, VectorStyle -> {Black, Arrowheads[0.02]}] /. > Arrow[{p1_, p2_}] :> Arrow[{p1 + (p2 - p1)/2, p2}]; > > Show[{dp, vp}] > > Hope this gives you a starting point. > > Cheers > Patrick > > On Sun, 2010-07-25 at 01:57 -0400, J Davis wrote: >> Looking at this example in the documentation... >> >> points = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}}; VectorPlot[{-1 - x^2 + >> y, 1 + x - y^2}, {x, -2, 2}, {y, -2, 2}, VectorPoints -> points, >> VectorScale -> .25, >> Epilog -> {Red, PointSize[Medium], Point[points]}] >> >> I would prefer for the *tail* of the vector to be situated at "points" >> rather than the tip (or as Mathematica draws them, the base of the >> arrowhead at the tip). >> >> I was hoping there was an option I could call to VectorPlot to make >> the desired behavior happen but I don't find any. >> >> Suggestions? Or do I need to manually translate all the vectors? I was >> hoping to avoid that since students would be utilizing the code and I >> want to keep it as clean and simple as possible. >> >> Once I accomplish the task above, I was looking for a clean way to >> plot the level curves of a surface and then superimpose the gradient >> vector field---but here's the catch---where the only vectors shown >> were the ones along the level curves (meaning their *tails* are along >> the level curves). I tried RegionFunction but didn't get satisfactory >> results. >> >> Any help is appreciated. >> >> Thanks, >> John >> >> >