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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
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