Re: Re: Mapping Data

*To*: mathgroup at smc.vnet.net*Subject*: [mg57341] Re: [mg57321] Re: Mapping Data*From*: "David Park" <djmp at earthlink.net>*Date*: Wed, 25 May 2005 06:02:46 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

You can obtain a nice circular contour plot, without warning messages by using... T = Table[Exp[-x], {x, 0, 1, 0.1}]; R = Table[x, {x, 0, 1, 0.1}]; fun = ListInterpolation[T, {R}]; Needs["DrawGraphics`DrawingMaster`"] Draw2D[ {(ContourDraw[fun[r], {r, 0, 1}, {\[Theta], 0, 2Pi}, ColorFunction -> Hue] // FineGrainPolygons[0.1, 4]) /. DrawingTransform[#1Cos[#2] &, #1Sin[#2] &]}, AspectRatio -> Automatic, Frame -> True, Background -> Linen]; That first draws the contour plot in r-theta coordinates and then uses DrawingTransform to convert all the Lines and Polygons to Cartesian coordinates. FineGrainPolygons is needed to breakup the long straight lines at the edges so they will transform smoothly. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: dh [mailto:dh at metrohm.ch] To: mathgroup at smc.vnet.net Hi Othman, If I understand your question, you want to change the temperature radially and keep the same value for all angles. Here is an example: T = Table[Exp[-x], {x, 0, 1, 0.1}]; R = Table[x, {x, 0, 1, 0.1}]; fun = ListInterpolation[T, {R}]; ContourPlot[fun[Norm[{x, y}]], {x, -1, 1}, {y, -1, 1}] This will give some warnings because the PlotRange is recangular and we extrapolate the function in the corners. Sincerely, Daniel othman wrote: > Hi All > I'm a new user of Mathematica.The problem is that I started using Mathematica from the middle not the beginning. > Anyway, My problem is: > I have two Matrices > Temperature= T = [Ax8] > Radial position = RR = [1x8] > as an example, let A=1, > instead of Plotting T vs. RR at xy-axes, I want to map the data in a cylinderical Coordinates as a contour mapping, whic means I need to "copy" my data to N times. > My proposed procedure is to do the following: > > Generate a position matrix, [x,y], as follows: > > Do [x,y] RR=0,1,.145, Theta = 0,360,M ( M is the step 360/N) > x = RR*Cos(theta) > y = RR*Sin(theta) > and > T=T @ RR > > I would like to know: > 1. if this is the easy way to do it ? > 2. How can translate it to Mathematica Code? > > Hopefully it is clear! > > Thank you in advance! > > Othman >