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

*To*: mathgroup at smc.vnet.net*Subject*: [mg98612] Re: [mg98576] Re: [mg98552] How to find current ViewAngle, ViewPoint during Rotating graphics?*From*: John Fultz <jfultz at wolfram.com>*Date*: Tue, 14 Apr 2009 06:17:10 -0400 (EDT)*Reply-to*: jfultz at wolfram.com

On Mon, 13 Apr 2009 03:33:54 -0400 (EDT), Nasser Abbasi wrote: > > > 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 You should be getting not only the ViewPoint, but the ViewVertical, as well. Sincerely, John Fultz jfultz at wolfram.com User Interface Group Wolfram Research, Inc.