Normally the vectors in VectorPlot3D are attached the middle. How to get them attached at the beginning (what is typical conventions in most textbooks) by use of VectorPlot3D?
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1 Answer
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2 Update V.11.3: In version 11.3+ the new option VectorMarkers can be used with Placed to control the position of vectors:
points = Tuples[{-1, 1}, {2}]; Row[VectorPlot[{-1 - x^2 + y, 1 + x - y^2}, {x, -2, 2}, {y, -2, 2}, VectorPoints -> points, VectorMarkers -> Placed["Arrow" , #], VectorScale -> {.5, .4}, ImageSize -> 300, Prolog -> {Yellow, Opacity[.5], Rectangle[{-1, -1}, {1, 1}], Opacity[1], Red, PointSize[Large], Point[points]}] & /@ {"Start", "End"}] 
Original answer:
You can post-process the graphics output to shift the arrows:
points = Tuples[{-1, 1}, {2}]; vp1 = VectorPlot[{-1 - x^2 + y, 1 + x - y^2}, {x, -2, 2}, {y, -2, 2}, VectorPoints -> points, VectorScale -> {.5, .4}, ImageSize -> 400, Prolog -> {Yellow, Opacity[.5], Rectangle[{-1, -1}, {1, 1}], Opacity[1], Red, PointSize[Large], Point[points]}]; vp1b = vp1 /. Arrow[x_] :> Arrow[{Mean[x], Mean[x] + Last[x] - First[x]}]; (* or vp1 /. Arrow[x_] :> Translate[Arrow[x],Mean[x]-First[x]] *) Row[{vp1, vp1b}, Spacer[10]] 
Similarly, for VectorPlot3D:
points2 =Tuples[{-1, 1}, {3}]; vp2 = VectorPlot3D[{x, y, z}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, VectorPoints -> points2, VectorScale -> .25, ImageSize -> 400]; vp2 = Show[vp2, Graphics3D@{Yellow, Opacity[.5], Cuboid[{-1, -1, -1}, {1, 1, 1}], Opacity[1], Red, PointSize[.03], Sphere[points2, .2]}]; vp2b = vp2 /. Arrow[x_] :> Arrow[{Mean[x], Mean[x] + Last[x] - First[x]}]; Row[{vp2, vp2b}, Spacer[10]] 
Update: A function that shifts the arrows to start from the designated points:
trF = MapAt[# /. Arrow[x_] :> Arrow[{Mean[x], Mean[x] + Last[x] - First[x]}] &, #, {1}] &; (* or trF = MapAt[#/.Arrow[x_] :> Translate[Arrow[x],Mean[x]-First[x]]&,#,{1}]&; *) Row[trF /@ {vp1, vp2}, Spacer[15]] 
Update 2: For 3D arrow glyphs, we need to modify the replacement rule:
vp3 = VectorPlot3D[{x, y, z}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, VectorPoints->points2, VectorStyle -> "Arrow3D", VectorScale -> .25, ImageSize -> 400]; vp3 = Show[vp3, Graphics3D@{Yellow, Opacity[.5], Cuboid[{-1, -1, -1}, {1, 1, 1}], Opacity[1], Red, PointSize[.03], Sphere[points2, .2]}]; vp3b =vp3/. Arrow[Tube[x_, r__]]:>Arrow[Tube[{Mean[x], Mean[x] + Last[x] - First[x]}, r]]; Row[{vp3, vp3b}, Spacer[10]] 
- $\begingroup$ An exhaustive answer with nice figures, thanks. In my tests the function with Translate[..] appeared to be 40% faster, is that right ? $\endgroup$sebqas– sebqas2015-01-12 20:04:02 +00:00Commented Jan 12, 2015 at 20:04
- $\begingroup$ @sebqas, i haven't done any timing test, but it makes sense that
Translateis faster. $\endgroup$kglr– kglr2015-01-13 06:54:15 +00:00Commented Jan 13, 2015 at 6:54