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Re: Ellipse equation simplification on Mathematica:
*To*: mathgroup at smc.vnet.net
*Subject*: [mg76946] Re: Ellipse equation simplification on Mathematica:
*From*: Narasimham <mathma18 at hotmail.com>
*Date*: Wed, 30 May 2007 05:29:28 -0400 (EDT)
*References*: <f2emof$35h$1@smc.vnet.net><200705280500.BAA15766@smc.vnet.net>
On May 29, 2:19 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> On 29 May 2007, at 12:15, Andrzej Kozlowski wrote:
>
>
>
>
>
> > On 29 May 2007, at 10:53, Andrzej Kozlowski wrote:
>
>
> >> 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
Thanks Andrzej so much for all the cumbersome & involved
verifications.
Ok,are we now in a situation to find out what exhaustively all
situations/conditions give ellipses?
Should coefficients of third and fourth order term vanish,or should it
have repeated roots etc.?
At least don't we get ellipses for tube center when th = 0,?
(inclusive when the tube is parallel or perpendicular to x- axis, th =
0,ph = 0,Pi/2? For this case new a,b need to be defined in terms of
c,a and d).
Regards,
Narasimham
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