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

*To*: mathgroup at smc.vnet.net*Subject*: [mg98613] Re: How to find current ViewAngle, ViewPoint during Rotating graphics?*From*: "Alexander Elkins" <alexander_elkins at hotmail.com>*Date*: Tue, 14 Apr 2009 06:17:21 -0400 (EDT)*References*: <7274943.1239523349775.JavaMail.root@n11> <003101c9bb6e$d3c92020$7b5b6060$@net> <grupsu$rp$1@smc.vnet.net>

Here I've incorporated your 3D plot into my example. I've made the tube optional since it slows things down. Note that ViewPoint->{3.38378, 0, 0} with ViewVertical->{0,1,0} gives you a "spring" which is oriented sideways. Manipulate[ ParametricPlot3D[{Sin[u], Cos[u], 0.05*u}, {u, 0, 5*2*Pi}, PlotStyle -> If[tube, Tube[0.05], Automatic], Axes -> Dynamic[axes], Boxed -> Dynamic[boxed], ImageSize -> 300, ViewPoint -> Dynamic[vp], ViewVertical -> Dynamic[vv], ViewAngle -> Dynamic[va], ViewCenter -> Dynamic[vc], SphericalRegion -> Dynamic[sr], Method -> {"RotationControl" -> Dynamic[rc]}], {{vp, ViewPoint /. Options[Graphics3D], "ViewPoint\[Rule]"}, InputField}, {{vv, {0, 0, 1}, "ViewVertical\[Rule]"}, InputField}, {{va, Automatic, "ViewAngle\[Rule]"}, InputField}, {{vc, {{1/2, 1/2, 1/2}, {1/2, 1/2}}, "ViewCenter\[Rule]"}, InputField}, Row[{Labeled[Checkbox[Dynamic[sr]], "SphericalRegion\[Rule]", Left, Spacings -> 0], Labeled[Checkbox[Dynamic[axes]], "Axes\[Rule]", Left, Spacings -> 0], Labeled[Checkbox[Dynamic[boxed]], "Boxed\[Rule]", Left, Spacings -> 0], Labeled[Checkbox[Dynamic[tube]], "Tube\[Rule]", Left, Spacings -> 0]}], {sr, {True, False}, None}, {axes, {False, True}, None}, {boxed, {False, True}, None}, {tube, {False, True}, None}, {{rc, "ArcBall", "RotationControl\[Rule]"}, {"ArcBall", "Globe"}}] Enjoy! Alexander "Nasser Abbasi" <nma at 12000.org> wrote in message news:grupsu$rp$1 at smc.vnet.net... > > 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 > > >