Re: ContourPlot and lines vrs. 8.0
- To: mathgroup at smc.vnet.net
- Subject: [mg116292] Re: ContourPlot and lines vrs. 8.0
- From: kristoph <kristophs.post at web.de>
- Date: Thu, 10 Feb 2011 05:21:12 -0500 (EST)
- References: <iir4ih$6lk$1@smc.vnet.net> <iitepi$jmu$1@smc.vnet.net>
On 9 Feb., 08:12, "Sjoerd C. de Vries" <sjoerd.c.devr... at gmail.com>
wrote:
> Hi Kris,
>
> The syntax desciption for ContourPlot does not include the syntax you
> are using (one function with a specific contour level and no
> specification of levels for the other). If you're providing multiple
> functions it should be done by providing a contour level for all. To
> achieve this you could generate a set of copies of the first function
> each at a different contour level, like this:
>
> \[Alpha]h1 = 16;
> \[Alpha]h2 = 16;
> cl = {Table[(\[Alpha]h1 xh - xh^2) + (\[Alpha]h2 xf - xf^2) == i, {=
i,
> 0, 100, 10}] // Evaluate, xf - (\[Alpha]h1 - xh) == 0}=
//
> Flatten;
> ContourPlot[cl // Evaluate, {xh, 0, 10}, {xf, 0, 10},
> ContourStyle -> Flatten[{Table[Red, {Length[cl] - 1}], Green}]]
>
> Cheers -- Sjoerd
>
> On Feb 8, 11:06 am, kristoph <kristophs.p... at web.de> wrote:
>
>
>
> > Hi,
>
> > I would like to plot a contour plot of two different functions. One of
> > them is a line in the contour plot. The problem I have is that the
> > plot only shows the line and only ONE curve of the other function.
> > What I would like to have is a usual contour plot of the function and
> > the line of the other function. Below you find the code of the contour
> > plot:
>
> > \[Alpha]h1 = 16;
> > \[Alpha]h2 = 16;
> > ContourPlot[{(\[Alpha]h1 xh - xh^2) + (\[Alpha]h2 xf - xf^2),
> > xf - (\[Alpha]h1 - xh) == 0}, {xh, 0, 10}, {xf, 0, 10},
> > Contours -> 250]
>
> > Here is the contour plot of the first function only. It would be great
> > to have this plot with only one line expressed by the function xf - =
(\
> > [Alpha]h1 - xh) == 0.
>
> > ContourPlot[{(\[Alpha]h1 xh - xh^2) + (\[Alpha]h2 xf - xf^2)}, {xh, 0,
> > 10}, {xf, 0, 10}, Contours -> 250]
>
> > Thanks for help,
> > Kris
Thanks to both of you. Always very helpful!