Re: Re: Ellipse equation simplification on Mathematica:
- To: mathgroup at smc.vnet.net
- Subject: [mg76891] Re: [mg76830] Re: Ellipse equation simplification on Mathematica:
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 29 May 2007 05:04:38 -0400 (EDT)
- References: <f2emof$35h$1@smc.vnet.net><200705181006.GAA12812@smc.vnet.net> <200705280500.BAA15766@smc.vnet.net> <A2AFF4CF-D65F-424C-A963-97043E71D6E7@mimuw.edu.pl> <7D2D10D0-B53E-4A9F-A03D-6D1755AA2505@mimuw.edu.pl>
On 29 May 2007, at 12:15, Andrzej Kozlowski wrote: > > On 29 May 2007, at 10:53, Andrzej Kozlowski wrote: > >> *This message was transferred with a trial version of CommuniGate >> (tm) Pro* >> >> On 28 May 2007, at 14:00, Narasimham wrote: >> >>> On May 19, 1:54 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote: >>>> On 18 May 2007, at 19:06, Narasimham wrote: >>>> >>>>> Reference is made to: >>>> >>>>> http://groups.google.co.in/group/geometry.puzzles/browse_thread/ >>>>> threa... >>>> >>>>> Constant[c,d th,ph] ; >>>> >>>>> (* th, ph are spherical cords of tip of tube *) ; >>>> >>>>> cp = Cos[ph] ; sp = Sin[ph] ; cth = Cos[th] ; sth = Sin[th] ; >>>> >>>>> (* earlier typo corrected *) >>>> >>>>> d1 = Sqrt[(x + d cp cth + c )^2 + ( y + d cp sth )^2 + (d sp)^2 ] >>>> >>>>> d2 =Sqrt[(x - d cp cth - c )^2 + ( y - d cp sth )^2 + (d sp)^2 ] >>>> >>>>> FullSimplify[ d1 + d2 + 2 d - 2 a == 0] ; >>>> >>>>> When d = 0, algebraic/trigonometric simplification brings about >>>>> common ellipse form: >>>> >>>>> (x/a)^2 + y^2/(a^2-c^2) = 1 >>>> >>>>> Request help for bringing to standard form involving constants >>>>> a,c >>>>> and the new tube length constant d. >>>> >>>>> Regards, >>>>> Narasimham >>>> >>>> I don't think such a form exists. Consider the following. >>>> >>>> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2), >>>> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2), >>>> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1}; >>>> >>>> id = Prepend[id1, d1 + d2 + 2 d - 2 a]; >>>> >>>> Now consdier first the case of the ellipse: >>>> >>>> d = 0; >>>> >>>> gr = GroebnerBasis[id, {x, y, a, c}, {cp, sp, cth, sth, d1, d2}, >>>> MonomialOrder -> EliminationOrder] >>>> {-a^4 + c^2 a^2 + x^2 a^2 + y^2 a^2 - c^2 x^2} >>>> >>>> This tells us that >>>> >>>> First[%] == 0 >>>> -a^4 + c^2*a^2 + x^2*a^2 + y^2*a^2 - c^2*x^2 == 0 >>>> >>>> is the equation of the ellipse, and this can be easily brought to >>>> standard form by hand. But now consider your "general" case: >>>> >>>> Clear[d] >>>> gr = GroebnerBasis[id, {x, y, a, c, d}, {cp, sp, cth, sth, d1, >>>> d2}, >>>> MonomialOrder -> EliminationOrder] >>>> {} >>>> >>>> This means that elimination cannot be performed and no "standard >>>> form" >>>> of the kind you had in mind exists. Unless of course there is a bug >>>> in GroebnerBasis (v. unlikely) or I have misunderstood what you had >>>> in mind. >>>> >>>> Andrzej Kozlowski >>> >>> I checked for case of tube parallel to x- or y-axis produces >>> ellipses >>> and suspected validity even in 3-D general case. >>> >>> Narasimham >>> >> >> >> OK., now I see that I misundertood you and you wrote that cd,th, >> ph (and presumably a) are supposed to be constants, so you do not >> wish to eliminate them. But now one can easily prove that what you >> get is not, in general, an ellipse. In this situation Groebner >> basis works and you can obtain a rather horrible quartic equation >> of your surface: >> >> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2), >> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2), >> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1}; >> >> id = Prepend[id1, d1 + d2 + 2*d - 2*a]; >> >> >> v = GroebnerBasis[id, {x, y, a, c, cp, cth}, {sth, sp, d1, d2}, >> MonomialOrder -> EliminationOrder][[1]]; >> >> First[v] == 0 >> >> is the equation (I prefer not to include the output here). >> >> Looking at v see that the non zero coefficients are only the free >> coefficient, the coefficients of x^2, y^2, x^2 y^2, x^4 and y^4. >> So only in some cases you will get a quadratic (for example when >> the quartic happens to be a perfect square as in the case d=0, or >> when the free coefficient vanishes, as in the trivial case a=d, >> or when the coefficients of 4-degree terms vanish). One can work >> out all the cases when gets a quadratic but it is also easy to >> find those when one does not. For example, taking both th and ph >> to be 60 degrees (so that cth and cph are both 1/2) we get: >> >> v /. {cp -> 1/2, cth -> 1/2, d -> 4, a -> 2, c -> 1} >> >> y^4 - 48 x^2 y^2 + 120 y^2 + 3600 >> >> This is certianly is not the equation of an ellipse. >> >> Andrzej Kozlowski >> >> > > > Sorry; that last example was a bad one, because with these > parameters (d>a) the original equations do not have solutions. The > point is that the equations that we get after elimination will have > more solutions than the ones we start with, but all the solutions > of the original ones will satisfy the new ones. So to see that we > do not normally get an ellipse we need a different choice of > parameters, with a>d. So take > instead > > w = First[v ]/. {cp -> 1/2, cth -> 1/2, sp -> Sqrt[3]/2, sth -> Sqrt > [3]/2, d -> 1, > a -> 3, c -> 1} > > (1521*x^4)/256 + (2229*y^2*x^2)/128 - (117*x^2)/4 + (3721*y^4)/256 > - (183*y^2)/4 + 36 > > > Now, this is clearly not the equation of an ellipse. We can see now > the relationship between this and your original equation. With the > above values of the parameters your equation takes the form: > > eq = d1 + d2 + 2*d - 2*a == 0 /. {cp -> 1/2, cth -> 1/2, sp -> Sqrt > [3]/2, > sth -> Sqrt[3]/2, d -> 1, a -> 3, > c -> 1} > Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] + > Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0 > > So now look at the graph: > > gr1=ContourPlot[ > Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] + > Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0, {x, -5, > 5}, {y, -5, > 5}] > > > Looks like a nice ellipse, right? Unfortunately it is only a part > of the graph of > > w =First[v] > > gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}] > > The picture is not quite convincing, because the latter contour > plot is not very accurate but when you see them together: > > Show[gr1, gr2] > > the relationship becomes obvious. In any case, what you get a > quartic curve that looks of course like an ellipse but isn't. > > Andrzej Kozlowski Sorry, a small correction is needed. I wrote: > w=First[v] > > gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}] Ignore the first line (w=First[v]) in the above since it will prevent the plot from working. Just evaluate the second line with the value of w defined earlier. Andrzej Kozlowski
- References:
- Re: Ellipse equation simplification on Mathematica:
- From: Narasimham <mathma18@hotmail.com>
- Re: Ellipse equation simplification on Mathematica:
- From: Narasimham <mathma18@hotmail.com>
- Re: Ellipse equation simplification on Mathematica: