MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

scaling Plots to millimeters

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38849] scaling Plots to millimeters
  • From: Adalbert Hanssen <hanssen at zeiss.de>
  • Date: Thu, 16 Jan 2003 03:18:49 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hi, MathGroup,

I want to print a plot scaled to a specific size.
According to the online help, ImageSize->72 xi specifies that the image should have a width of xi inches.
In order to enter sizes in mm, I try:

xUmr=yUmr=72/25.4;
Show[Graphics[{Line[{{0,0},{0,1},{1,1},{1,0},{0,0}}]}]
    ,Ticks->None
    ,ImageSize->150*{xUmr,yUmr}
    ,PlotRange->{{0,1},{0,1}}
    ];

and get a printed rectangle about 119 mm wide and 74 mm high.

I am wondering, why the area is non square, since I gave
two PlotRanges and two ImagsSizes.

Only, if I use AspectRatio in addition, I get something closer, to what I want:

xUmr=yUmr=72/25.4;
Show[Graphics[{Line[{{0,0},{0,1},{1,1},{1,0},{0,0}}]}]
    ,Ticks->None
    ,ImageSize->150*{xUmr,yUmr}
    ,PlotRange->{{0,1},{0,1}}
    ,AspectRatio->1
    ];

gives a rectayngle 119.5*119 mm where I expected 150*150.
Ok, the printer is somehow misbehaving, then I set

xUmr=150/119.5*72/25.4;
yUmr=150/119.0*72/25.4;

to obtain almost a square.

Any explanation, why AspectRatio is necessary?

regards

Adalbert




  • Prev by Date: Re: Integrating Abs[Sin[]^2]
  • Next by Date: Re: Modifying arguments of sub-parts of an expression
  • Previous by thread: Re: Mathematica, Windows 2000 and Service Pack 3 (The good, the bad and the ugly ?)
  • Next by thread: Re: scaling Plots to millimeters