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I'd like to plot a vector field with a legend for the magnitude of a reference vector. I have read all the document of VectorPlot and found only PlotLegends is possible for this purpose. But I cannot come up with a solution. I also found a similar question here without the desired solution. So I post a minimal question here. Note that I do not want to visualize the magnitudes with colors, so please do not waste your time.

Question: How can I plot a vector field with a reference legend showing the reference magnitude? For example, in this figure, the arrow length of the legend should be related to that of the vector field to indicate a reference length.

VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, PlotLegends -> {"Ref. vector"}]

enter image description here

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  • $\begingroup$ What do you mean by reference length? if you run InputForm[VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, PlotLegends -> {"Ref. vector"}]] you will see that PlotLegend is formed from an arrow of unit length Arrow[{{0, 0.5}, {1, 0.5}}] placed After. So, this unit arrow has reference length, in some sense at least. $\endgroup$
    – Alx
    Sep 1, 2019 at 13:37
  • $\begingroup$ @Alx reference length means a 'ruler' with a given length, which can be used to determine/estimate the magnitudes of other vectors in the field. Considering point (x=-3, y=3) gives vector (3,3) which has a length of about 4.243. So it is expected that the length of this arrow there has a length of 4.243 times of the reference length, i.e. the length of the legend. $\endgroup$
    – user55777
    Sep 1, 2019 at 14:30
  • $\begingroup$ What about VectorColorFunction -> "Rainbow", PlotLegends -> Automatic? $\endgroup$
    – Michael E2
    Sep 1, 2019 at 15:02
  • $\begingroup$ Maximal arrow length is Maximize[{Norm[{y, -x}], -3 <= x <= 3 && -3 <= y <= 3}, {x, y}] // First = 3 Sqrt[2]. Lengths of vectors are scaled to the diagonal of the plot range, in your case this is 6*Sqrt[2]. So, using VectorScale option and Legended: Legended[VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, VectorScale -> {0.5, 0.1}], Graphics[{Arrowheads[0.3], Arrow[{{0, 0}, {1/(3 Sqrt[2]), 0}}]}]]. $\endgroup$
    – Alx
    Sep 1, 2019 at 16:09
  • $\begingroup$ @Alx so the absolute length of the legend is $1/(3 Sqrt[2])$ and the relative length in respect to the vector field is 1, right? $\endgroup$
    – user55777
    Sep 1, 2019 at 16:15

1 Answer 1

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In help page on VectorPlot (and VectorScale as well) we can read: "By default, the vector length is scaled by the norm of the vector field". And from the documentation on VectorScale it has the following settings: {unitlen, aratio, sfun}, where "unitlen is given as a fraction of the diagonal of the overall bounding box", i.e. diagonal of plot range.

For the example in question, maximal arrow length is given by Maximize[{Norm[{y, -x}], -3 <= x <= 3 && -3 <= y <= 3}, {x, y}] // First = 3 Sqrt[2], and the diagonal is 6 Sqrt[2]. Hence if we want to see arrows having their "native" length (not scaled as Mathematica does by default) we need to apply appropriate scaling:

VectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, VectorScale -> {0.5, 0.15},
PlotLegends -> {"Ref. vector"}]

enter image description here

If needed, number of arrows can be adjusted by VectorPoints option.

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  • $\begingroup$ it is almost there! while it is still necessary to specify the arrow length of the legend to be unity (relative length), which is about $29\text{cm}/4.243=6.8$cm in absolute length. $\endgroup$
    – user55777
    Sep 2, 2019 at 3:40

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