c33 returns the center of a circle thru 3 points in 3 dimensions. c33[{p1_,p2_,p3_}] := With[{q1 = p1-p3, q2 = p2-p3, q3 = p1-p2}, p3 + ((q2.q2)*(q3.q1)*q1 - (q1.q1)*(q3.q2)*q2)/(2 #.# & @ Cross[q1,q2])] c33 is a closed-form version of c3n, which returns the center of a circle thru 3 points in n dimensions. c3n[{p1_,p2_,p3_}] := Block[{q1 = p1-p3, q2 = p2-p3, c,w1,w2}, c = w1*q1 + w2*q2; p3 + c /. Flatten@Solve[ {c.c == #.#&[q1-c], c.c == #.#&[q2-c]},{w1,w2}]] ----- Ste[hen Gray <stevebg@roadrunner.com> wrote: > I'm looking for a neat formula to find the center of a circle in 3D > through 3 points. I also need a good way to display it, preferably > thickened so I can show several and see whether they are linked, etc. To > my surprise I did not find anything on the Wolfram sites about these > problems. (I have Mathematica 7, if that matters.)

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5/13/2014 2:27:33 AM

Hi Fiz, > > Here is the function > > w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2)) > > r0=1.4 and z0=1.4 The function is Exp, not exp, and it needs square brackets not parentheses. r0 = 1.4; z0 = 1.4; w = Exp[(-2(x^2 + y^2/r0^2))]*Exp[(-2(z^2/z0^2))] // FullSimplify The function you're after is ContourPlot3D, and you use it like Needs["Graphics`ContourPlot3D`"] ContourPlot3D[w, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, Contours -> {.75}, \ PlotRange -> All, Axes -> True, BoxRatios -> {1, 1, 1}]; This produces "something",...

On Fri, 22 May 2009 01:50:51 -0400 (EDT), skkaul wrote: > On May 10, 5:13 am, John Fultz <jfu...@wolfram.com> wrote: > >> <install= >> directory>/SystemFiles/FrontEnd/TextResources/UnicodeFontMapping.tr >> > Besides the comments, is there any documentation on this file? In > particular, what are type V and H entries, and what font is referenced > by -2? > > Thanks, > Shiva It's not documented because it's not intended for user consumption, although being able to edit it very rarely allows working around certain issues...

Hi David, <snippage> Very nice! Begs the question as to why anyone would bother with ContourPlot3D .... Regards, Dave ...

In this particular case we can parametrize the contour surfaces. Looking for the constant values surface, we obtain a simplification to ellipsoids. Exp[-2*(x^2 + y^2/r0^2)]*Exp[-2*(z^2/z0^2)] == c Simplify[%] % /. x^2 + y^2/r0^2 + z^2/z0^2 -> k2 Solve[%, k2][[2,1]] krule = % /. k2 -> k^2 Write the equation for the constant value ellipsoid in terms of contour value c. x^2 + y^2/r0^2 + z^2/z0^2 == k^2 % /. krule (Distribute[#1/(-Log[c]/2)] & ) /@ % Standard parametrization of an ellipsoid (from Alfred Gray, Modern Differential Geometry).... ellipsoid[a_, b_, c_...

Hi David, I suppose because it is not always possible to simply parametrize the contour surfaces. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ From: David Annetts [mailto:davidannetts@aapt.net.au] Hi David, <snippage> Very nice! Begs the question as to why anyone would bother with ContourPlot3D .... Regards, Dave ...

On 3/23/10 at 4:23 AM, hemmecke@gmail.com (hemmecke) wrote: >Does somebody know why I get different behaviour for the following >General::dupsym: The symbol Array with context A` already exists. >message? Why does that message appear at all? If A`Array exists, >then Mathematica should just use it, shouldn't it? >According to >http://reference.wolfram.com/mathematica/tutorial/Contexts.html we >have: >`name a symbol in the current context >So why is Mathematica complaining? The message is not telling you the symbol exists in the current cont...

> -----Original Message----- > From: Steven T. Hatton [mailto:hattons@globalsymmetry.com] > Sent: 15 December 2005 10:30 > Subject: Re: Mathematica Programmer vs. Programming > in Mathematica > ....... > > I wonder what value there woudl be in trying to explain what makes > > Mathematica "functions" different from functions in > languages such as > > C in a book addressed to readers most of whom have no > knowledge of C > > and are not particualry interested in getting it? > > I suspect you will not fi...

I think most people this entire discussion does not have any practical importance. Obviously something like this: 2*Unevaluated[1+1] 2 Unevaluated[1+1] is extremly unlikely to have any practical use. After all, we use Unevaluated when we want something to remain unevaluated, whereas here only two things can happen: either one will be left with Unevaluated[something] or the "something" will evaluate. Both of these outcomes are obsiously undesirable. Why should one ever use anything like this in a program? In fact if for some unimaginable reason somone neede...

Berthold, I would suggest spending an hour per day following threads that seem interesting to you on this MathGroup's postings. Each example of problem or issue related to the use of Mathematica will expose you to practical solutions and very often deep insight that is hard to replicate with books or documentation read in isolation. Yours truly ... Syd Geraghty Syd Geraghty B.Sc., M.Sc. sydgeraghty@me.com San Jose, CA (408) 532 6852 Mathematica 8.0 for Mac OS X x86 (64-bit) (November 6, 2010) Licenses: L2983-5890, L3028-2592 MacOS X V 10.6.5 Snow Leopard MacBook Pro ...

I must agree about the debugger. I was very excited by the release of Workbench 1.0 because of the promise of a good debugger. I even took a course on it. The reality is that Workbench is so hard to use ( I can't bring in my old code and debug changes- it just doesn't work) that I never use it. So I'm back to using Print statements again. I love Mathematica but would love to have an easy to use debugger with break points, etc. Oh well. Cliff Nasser Abbasi <nma@12000.org> wrote: "Murray Eisenberg" wrote in message news:fdf236$20u$1@smc.vnet.net... &...

So far I could resist the temptation to participate in this discussion. However, in his mail Maxim Rytin presents some examples of which he thinks the result is unpredictable. Maybe there is some interest in how I predict the results of simple commands in which Unevaluated occurs. Of course these examples are of no practical interest. Unevaluated is meant to pass unevaluated arguments to a function body and as such it works perfectly. No one in practice is interested in (1+1)*Unevaluated[2+2]. The basic principle has been clearly explained by Andrzej Kozlowsky. Suppose we have a ...

Sorry again, but your previous message said >=, not <=. It's still posted on Google Groups, and I checked to make sure. DrBob www.eclecticdreams.net -----Original Message----- From: AGUIRRE ESTIBALEZ Julian [mailto:mtpagesj@lg.ehu.es] Subject: RE: Re: Mandelbrot Set & Mathematica On Tue, 11 May 2004, DrBob wrote: > Sorry, but that just doesn't work, even after changing =BE to >=. There are > only two colors (even using your rainbow function), and no fractal > "antennae". As noted in a previous message, it should be "<=...

I know this option. But say, if I can't convert my data (list of points in 3D) as array of height values, then can I plot my data of list of points in 3D directly as we plot in case of a function? Ananya. On Tue, 13 Apr 2004, Tomas Garza wrote: > There are some fine points regarding the nature of the points you want to > plot in 3D. First, have a close look at the definition of ListPlot3D (in the > on-line help browser, e.g.). Notice that the argument you must provide is an > array of height values, not a list of points in 3D. So, for example > > ...

On Thu, 14 Feb 2008 01:00:12 -0500 (EST), J=E1nos wrote: > > > On Feb 13, 2008, at 4:04 AM, David Park wrote: > >> But be aware that if you are buying a new computer, from Dell at >> least, and >> you specify a 64 bit microprocessor, you will not necessarily get a >> 64 bit >> operating system, and may not even be able to install a 64-bit >> operating >> system. So if you are looking to use 64-bit Mathematica check out very >> carefully before purchase that you will indeed have a 64-bit operating >> system. >>...

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