Re: On AspectRatio

*To*: mathgroup at smc.vnet.net*Subject*: [mg72658] Re: [mg72650] On AspectRatio*From*: "Chris Chiasson" <chris at chiasson.name>*Date*: Sun, 14 Jan 2007 04:18:38 -0500 (EST)*References*: <200701131010.FAA25677@smc.vnet.net>

Measure your monitor's active display area with something precise. Write it down in a List. Hypothetical example (in inches): monDim={10,8} Record your current display resolution in a List: Hypothetical example (in pixels): resDim={1280,1024} If resDim/monDim doesn't give a list of two *really* close numbers, (in this case, the horizontal and vertical DPI in each direction) then Mathematica's plots will have the wrong aspect ratio on screen. Similarly, everything else on the screen will be distorted unless a program is aware of the DPI mismatch and adjusts accordingly. Similarly, Divide@@monDim and Divide@@resDim should also be *really* close. On 1/13/07, Virgil Stokes <vs at it.uu.se> wrote: > I have taken the following command from Help on AspectRatio. > > Show[Graphics[Circle[{0, 0}, 1], AspectRatio -> Automatic]]; > > This does not give an aspect ratio of 1 on my monitor. In fact when I > measure the ratio of the height to width of the circle (or the > surrounding box) generated by this command it is greater than 1.0. This > is also true for > > Show[Graphics[Circle[{0, 0}, 1], AspectRatio -> 1.0]]; > > Am I interpreting this option correctly? In any case would you please > tell me how I can get the height to width ratio = 1.0 . > My application is parametric plots (phase-plane analysis) and I need the > aspect ratio to be exactly one. > > Thanks, > --V. Stokes > > -- http://chris.chiasson.name/

**References**:**On AspectRatio***From:*Virgil Stokes <vs@it.uu.se>