# A way to show all the 3D curve, with view still centered on {0, 0, 0}

I frequently have this graphics output problem :

I draw a 3D curve using NDSolve, and depending on the initial data, there's no way I could predict the shape and extent of the curve in 3D space. In the ParametricPlot3D code, I could use PlotRange -> All to show the whole curve, but then the 3D box isn't centered on the origin : {0, 0, 0}. Rotating the view is annoying in this case. I could use something like

PlotRange -> {{-100, 100}, {-100, 100}, {-100, 100}},

but most of the time the box is too small or too large to display the whole curve. The scale isn't adequate and there's no way I could predict in advance which scale should be best to show all the curve (and no more useless space, since it could be a pain to zoom in).

Using FindMax, I could find the fartest distance to the curve, from the origin, and use that value in PlotRange, but this may have a strong impact on performances in a Manipulate box (depending on the curve/solution complexity). I much prefer to use another way to show the whole curve.

So my question is this :

Using PlotRange (or another option ?), how can we tell Mathematica to show the whole curve, while maintaining the central view position on the origin ({0, 0, 0}) ?

EDIT : An example of a curve not centered on the origin :

ParametricPlot3D[
{Sin[3 t] - 2, Cos[5 t], 0.03 t},
{t, 0, 2Pi},
PlotRange -> All, (* Not good in this case *)
Boxed -> True,
ViewVector -> {0, 0, 0} (* Ugly option *)
]


I need to show all that curve, while fixing the view centered on the origin, and have a symetrical box all around the origin.

• Have you tried ViewVector? This maintainces the view Position on the origin. – Kay Mar 25 '16 at 18:27
• How ? Can you give a simple example of its use ? The documentation isn't very helpfull for these things ; it always give useless and fancy examples ! – Cham Mar 25 '16 at 18:28
• Could you give an example of you curve? Btw, the 2. option you might Need is ViewAngle. This changes the visible range, without need of adjusting PlotRange etc. – Kay Mar 25 '16 at 18:30
• Yes, I know the ViewAngle option, but then the problem is still the same : how to define a proper scale ? – Cham Mar 25 '16 at 18:32
• Just a quick example : ParametricPlot3D[ {Sin[3 t] - 2, Cos[5 t], 0.03 t}, {t, 0, 10 Pi}, PlotRange -> All, Boxed -> True, ViewVector -> {0, 0, 0} ]. – Cham Mar 25 '16 at 18:32

first make the plot and use AbsoluteOptions to get the range:

p=ParametricPlot3D[{Sin[3 t] - 2, Cos[5 t], 0.03 t}, {t, 0, 2 Pi},
PlotRange -> All,(*Not good in this case*)Boxed -> True,
ViewVector -> {0, 0, 0} (*Ugly option*)];(*<-this semicolon will be red, is ok*)
rad = Max@Abs@Flatten[ (PlotRange /. AbsoluteOptions[p])]


now combine the plot with a sphere ( or a cube or whatever ) that is centered and encompasses the figure.

Show[{Graphics3D[{Opacity[.1], Sphere[{0, 0, 0}, rad]}], p}]


You can set Opacity here if you don't want to see the sphere at all..but I think it sort of helps anyway. drawing axes is another option:

Show[{Graphics3D[{Arrowheads -> {-.02, .02},
Arrow[{-#, #}] & /@ Permutations[{0, 0, rad}]}], p}, Boxed -> False] • This solution is hard to apply to a numerical curve found with NDSolve, that may change inside a Manipulatebox. And is has a cost on performance. – Cham Mar 25 '16 at 19:09
• You can put all that in Mainpulate. I think it doesn't add much overhead to generating the plot in the first place.. give it a try. Or put up a real example. – george2079 Mar 25 '16 at 19:20
• just noticed I forgot to assign the plot to p.. fixed. Hope that didn't confuse. – george2079 Mar 25 '16 at 19:47
• It's working. I'll study this. – Cham Mar 25 '16 at 20:47
• Since I don't want the sphere, this appears to work : rad = Max[Abs[Flatten[(PlotRange/.AbsoluteOptions[graphicsName])]]]; Show[graphicsName, PlotRange -> rad{{-1, 1}, {-1, 1}, {-1, 1}}]. But I don't understand the AbsoluteOptions. Could you add some comments in your answer about what your code is doing ? – Cham Mar 25 '16 at 20:50

If I got that right I suggest the following solution for your example:

ParametricPlot3D[{Sin[3 t], Cos[5 t], 0.03 t}, {t, 0, 1 Pi},
PlotRange -> Full, Boxed -> True, ViewAngle -> 20 Degree,
ViewVector -> {{0, -5, 5}, {0, 0, 0}}]


Now, play with the ViewAngle to adjust the angle of view... • You changed the function, which is now centered, so you have eliminated the source of the problem ("throwing the baby with the bath water"). Add "-2" to the "x" component. Your solution doesn't work. – Cham Mar 25 '16 at 19:00
• Sry, my mistake. Didn't realized, that I throw something important away... – Kay Mar 25 '16 at 19:18