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Re: How to find current ViewAngle, ViewPoint during Rotating graphics?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg98576] Re: [mg98552] How to find current ViewAngle, ViewPoint during Rotating graphics?
  • From: "Nasser Abbasi" <nma at 12000.org>
  • Date: Mon, 13 Apr 2009 03:33:54 -0400 (EDT)
  • References: <7274943.1239523349775.JavaMail.root@n11> <003101c9bb6e$d3c92020$7b5b6060$@net>
  • Reply-to: "Nasser Abbasi" <nma at 12000.org>

From: "David Park" <djmpark at comcast.net>


> Make a Plot, say,
>
> Plot3D[(x - 3) (y - 2), {x, 0, 5}, {y, 0, 5},
> ImageSize -> 300]
>
> Then in the output, position the cursor to the left of the plot and type
> 'AbsoluteOptions['. Next position the cursor to the right of the plot and
> type ',ViewPoint]'. Evaluate and you will obtain the current ViewPoint. 
> You
> can rotate the image and reevaluate to get the new ViewPoint. Then copy 
> the
> ViewPoint into the original plot statement.
>
> Without the ImageSize option it also works but I get warning messages 
> after
> rotating.
>
>
> David Park
> djmpark at comcast.net
> http://home.comcast.net/~djmpark/
>
>

Thanks David, but the above does not seem to work on my plot. I am using 
this 3D plot

ParametricPlot3D[{Sin[u], Cos[u], 0.05*u}, {u, 0, 5*2*Pi}, PlotStyle -> 
{Tube[thick]},
  ViewPoint -> {2.645, 2.057138, 0.468316}, ViewAngle -> Automatic, Axes -> 
False, Boxed -> False,
  ImageSize -> 300]

And the above ViewPoint value was copied as a result of the method you 
described, yet the resulting image orientation does not match that which I 
copied the ViewPoint from.

Here is a screen shot.

http://12000.org/tmp/041209/viewpoint.PNG

Basically, I am trying to make a "spring" which is oriented sideways instead 
of up-down.

This is on Mathematica 7

ps. You are right above getting a warning when not using ImageSize.

--Nasser




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