Re: Elasticity functions for Bezier and rectangular forms
- To: mathgroup at smc.vnet.net
- Subject: [mg132181] Re: Elasticity functions for Bezier and rectangular forms
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Sat, 11 Jan 2014 02:36:11 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
- Delivered-to: mathgroup-newsendx@smc.vnet.net
- References: <20140109065010.D66676A08@smc.vnet.net>
Either it is a version issue or you need to start with a fresh kernel.
$Version
"9.0 for Mac OS X x86 (64-bit) (January 24, 2013)"
pts = {p[0], p[1], p[2], p[3]} =
Rationalize[{{0, .65}, {1.8, 2.6}, {4, 2.5}, {4, 5}}];
n = Length[pts] - 1;
bcurve = Sum[Binomial[n, k]*(1 - t)^(n - k)*t^k p[k], {k, 0, n}] //
Simplify;
Solve[{x == bcurve[[1]], 0 <= x <= 4}, t]
{{t -> ConditionalExpression[
Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
1], 0 < x < 4]},
{t -> ConditionalExpression[
Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
2], 0 < x < 4]},
{t -> ConditionalExpression[
Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
3], 0 < x < 4]}}
First /@ (bcurve[[2]] /. %)
{(1/20)*(13 + 117*
Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
1] - 123*
Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
1]^2 +
93*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
1]^3), (1/20)*
(13 + 117*Root[
5*x - 27*#1 - 6*#1^2 +
13*#1^3 & , 2] -
123*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
2]^2 +
93*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
2]^3), (1/20)*
(13 + 117*Root[
5*x - 27*#1 - 6*#1^2 +
13*#1^3 & , 3] -
123*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
3]^2 +
93*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
3]^3)}
y1[x_] = Select[%,
Chop[N[# /. x -> p[0][[1]]]] == p[0][[2]] &&
Chop[N[# /. x -> p[3][[1]]]] == p[3][[2]] &][[1]]
(1/20)*(13 +
117*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & ,
2] - 123*
Root[5*x - 27*#1 - 6*#1^2 +
13*#1^3 & , 2]^2 +
93*Root[5*x - 27*#1 -
6*#1^2 + 13*#1^3 & , 2]^
3)
Bob Hanlon
On Fri, Jan 10, 2014 at 6:19 PM, E. Martin-Serrano <
eMartinSerrano at telefonica.net> wrote:
> Hi Bob,
>
> Thank you very much for your answer, but I am afraid, I am unable to see
> how
> from:
>
> Select[
> First /@ (bcurve[[2]] /.
> Solve[{x == bcurve[[1]], 0 <= x <= 4}, t]),
> Chop[N[# /. x -> p[0][[1]]]] == p[0][[2]] &&
> Chop[N[# /. x -> p[3][[1]]]] == p[3][[2]] &][[1]]
>
> We get the roots expression:
>
> (1/20)*(13 + 117*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & ,
> 2] - 123*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & , 2]^2 +
> 93*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & , 2]^3)
>
> Since
>
> Solve[{x == bcurve[[1]], 0 <= x <= 4}, t]
>
> Yields just the empty list '{}' (no possible solutions), and therefore
> nothing can be selected.
>
> In fact from
>
> y1[x_] = Select[
> First /@ (bcurve[[2]] /.
> Solve[{x == bcurve[[1]], 0 <= x <= 4}, t]),
> Chop[N[# /. x -> p[0][[1]]]] == p[0][[2]] &&
> Chop[N[# /. x -> p[3][[1]]]] == p[3][[2]] &][[1]]
>
> the result that I obtain is
>
> 1[[1]]
>
> The same with the alternative solution expressed with radicals.
>
> From this, it is clear that I am missing something fundamental.
>
> E. Martin-Serrano
>
>
>
> -----Mensaje original-----
> De: Bob Hanlon [mailto:hanlonr357 at gmail.com]
> Enviado el: viernes, 10 de enero de 2014 8:49
> Para: mathgroup at smc.vnet.net
> Asunto: Re: Elasticity functions for Bezier and rectangular
> forms
>
>
> pts = {p[0], p[1], p[2], p[3]} =
> Rationalize[{{0, .65}, {1.8, 2.6}, {4, 2.5}, {4, 5}}];
>
>
> n = Length[pts] - 1;
>
>
> bcurve = Sum[Binomial[n, k]*(1 - t)^(n - k)*t^k p[k],
> {k, 0, n}] // Simplify;
>
>
> Solving for solution expressed as Root objects
>
>
> y1[x_] = Select[
> First /@ (bcurve[[2]] /.
> Solve[{x == bcurve[[1]], 0 <= x <= 4}, t]),
> Chop[N[# /. x -> p[0][[1]]]] == p[0][[2]] &&
> Chop[N[# /. x -> p[3][[1]]]] == p[3][[2]] &][[1]]
>
>
> (1/20)*(13 + 117*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & ,
> 2] - 123*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & , 2]^2 +
> 93*Root[5*x - 27*#1 - 6*#1^2 + 13*#1^3 & , 2]^3)
>
>
>
> Solving for solution expressed with radicals
>
>
> y2[x_] = Select[(First /@ (bcurve[[2]] /.
> Solve[{x == bcurve[[1]], 0 <= x <= 4}, t])) //
> ToRadicals //
> FullSimplify,
> Chop[N[# /. x -> p[0][[1]]]] == p[0][[2]] &&
> Chop[N[# /. x -> p[3][[1]]]] == p[3][[2]] &][[1]]
>
>
> (1/20)*(-((465*x)/13) + (1/4394)*(-245386 +
> (30482562*(-1)^(1/3)*2^(2/3))/
> (718 - 845*x + 13*Sqrt[5]*Sqrt[(-4 + x)*
> (1944 + 845*x)])^(2/3) +
> (11676984*(-1)^(2/3)*2^(1/3))/
> (718 - 845*x + 13*Sqrt[5]*Sqrt[(-4 + x)*
> (1944 + 845*x)])^(1/3) - 48252*(-1)^(1/3)*2^(2/3)*
> (718 - 845*x + 13*Sqrt[5]*Sqrt[(-4 + x)*
> (1944 + 845*x)])^(1/3) - 1041*(-1)^(2/3)*2^(1/3)*
> (718 - 845*x + 13*Sqrt[5]*Sqrt[(-4 + x)*
> (1944 + 845*x)])^(2/3)))
>
>
>
> Note that y2 is a different branch than that obtained by converting y1 to
> radicals (see also plot below)
>
>
> y1r[x_] = y1[x] // ToRadicals // FullSimplify
>
>
> (1/20)*(-((465*x)/13) + (1/4394)*(-245386 -
> (30482562*(-2)^(2/3))/(718 - 845*x +
> 13*Sqrt[5]*Sqrt[(-4 + x)*(1944 + 845*x)])^(2/3) -
> (11676984*(-2)^(1/3))/(718 - 845*x +
> 13*Sqrt[5]*Sqrt[(-4 + x)*(1944 + 845*x)])^(1/3) +
> 48252*(-2)^(2/3)*(718 - 845*x + 13*Sqrt[5]*
> Sqrt[(-4 + x)*(1944 + 845*x)])^(1/3) +
> 1041*(-2)^(1/3)*(718 - 845*x + 13*Sqrt[5]*
> Sqrt[(-4 + x)*(1944 + 845*x)])^(2/3)))
>
>
>
> Expressed as an interpolation function
>
>
> y3 = Interpolation[Table[bcurve, {t, 0, 1, .01}]];
>
>
>
> Note that y1r is not the proper branch:
>
>
> Grid[
> {{Graphics[{BezierCurve[pts], Red,
> AbsolutePointSize[4], Point[pts]},
> AspectRatio -> 1, Frame -> True, Axes -> False,
> PlotLabel -> "Bezier Curve"],
> pp = ParametricPlot[BezierFunction[pts][x],
> {x, 0, 4}, AspectRatio -> 1, Frame -> True, Axes -> False,
> PlotLabel -> "BezierFunction",
> Epilog -> {Red, AbsolutePointSize[4], Point[pts]}],
> ParametricPlot[bcurve, {t, 0, 1},
> AspectRatio -> 1, Frame -> True, Axes -> False,
> PlotLabel -> "bcurve",
> Epilog -> {Red, AbsolutePointSize[4], Point[pts]}]},
> {Plot[y1[x], {x, 0, 4}, AspectRatio -> 1,
> Frame -> True, Axes -> False,
> PlotLabel -> "Root Object (y1)",
> Epilog -> {Red, AbsolutePointSize[4], Point[pts]}],
> Plot[y2[x], {x, 0, 4}, AspectRatio -> 1,
> Frame -> True, Axes -> False,
> PlotLabel -> "Radicals (y2)",
> Epilog -> {Red, AbsolutePointSize[4], Point[pts]}],
> Plot[y3[x], {x, 0, 4}, AspectRatio -> 1,
> Frame -> True, Axes -> False,
> PlotLabel -> "Interpolation of bcurve (y3)",
> Epilog -> {Red, AbsolutePointSize[4], Point[pts]}]},
> {Plot[y1r[x], {x, 0, 4}, AspectRatio -> 1,
> PlotStyle -> Red, Frame -> True, Axes -> False,
> PlotLabel -> "Root converted to Radicals (y1r)"],
> "", ""}}]
>
>
>
> Bob Hanlon
>
>
>
>
> On Thu, Jan 9, 2014 at 1:50 AM, E. Martin-Serrano <
> eMartinSerrano at telefonica.net> wrote:
>
> >
> > Hi,
> >
> >
> >
> > (*I have the Bezier curve*)
> >
> >
> >
> > pts = {p[0], p[1], p[2], p[3]} = {{0, .65}, {1.8, 2.6}, {4, 2.5}, {4,
> > 5}};
> >
> >
> >
> > n = Length[pts] - 1;
> >
> >
> >
> > bcurve = Sum[Binomial[n, i]* (1 - t)^(n - i) *t ^i p[i] , {i, 0, n}];
> >
> >
> >
> > (* the plot range for the resulting Bezier curve is obviously
> > {p[0][[1]], p[3][[1]} -> {0,4} *)
> >
> >
> >
> > (*Then, I want to transform the above parametric Bezier curve into a
> > rectangular {x,y} curve one by solving curve[[1]] x= x(t) for t and
> > plugging the result t-> x(t) into curve[[2]] or y = y(t).*)
> >
> >
> >
> > sols= Solve[x == bcurve[[1]], t, Reals],
> >
> >
> >
> > bcurve[[2]]/.%
> >
> >
> >
> > yvalues = (% /. x -> #) & /@ Range[0, 4]
> >
> >
> >
> > (*For testing purposes, by visual inspection I discard directly the
> > resulting yvalues which overflow the plot range interval {0,4} and
> > select only those values of y which are members of the interval.*)
> >
> >
> >
> > #[[2]]&/@yvalues
> >
> >
> >
> > (*But that done, none of these selected {x,y} points belong to the
> > actual Bezier curve bcurve yet.*)
> >
> >
> >
> > (*In other words the apparent direct procedure to convert a parametric
> > curve into rectangular one does not seem to work here.*)
> >
> >
> >
> > (*Remark: I need the basic Bezier curve bcurve to be reshapable
> > (dynamic) and be able to calculate the elasticity point function of
> > each new resulting function as the Bezier curve bcurve gets reshaped,
> > and for some reason I am not being able of getting directly the
> > elasticity point function of bcurve = Sum[Binomial[n, i]* (1 - t)^(n -
> > i) *t ^i p[i] , {i, 0, n}]; no problem for any other similar (or
> > approximate) function given in rectangular form*) ;
> >
> >
> >
> > Any help will be welcome.
> >
> >
> >
> > E. Martin-Serrano
> >
> >
> >
> > P.S. For the sake of clarity I have enclosed the text in as in
> > (*text*) and symbols or experssions within text as in symbol (when
> > necessary).
> >
> >
> >
> > __________________________________________
> >
> > This e-mail and the documents attached are confidential and intended
> > solely for the addressee; it may also be privileged. If you receive
> > this e-mail in error, please notify the sender immediately and destroy
> > it. As its integrity cannot be secured on the Internet, no sender's
> > liability can be triggered for the message content. Although the
> > sender endeavors to maintain a computer virus-free network, he/she
> > does not warrant that this transmission is virus-free and will not be
> > liable for any damages resulting from any virus transmitted.
> >
> > Este mensaje y los ficheros adjuntos pueden contener informaci=F3n
> > confidencial destinada solamente a la(s) persona(s) mencionadas
> > anteriormente y su contenido puede estar protegido por secreto
> > profesional y en cualquier caso el mensaje en su totalidad est=E1
> > amparado y protegido por la legislaci=F3n vigente que preserva el
> > secreto de las comunicaciones, y por la legislaci=F3n de protecci=F3n
> > de datos de car=E1cter personal. Si usted recibe este correo
> > electr=F3nico por error, gracias por informar inmediatamente al
> > remitente y destruir el mensaje. Al no estar asegurada la integridad
> > de este mensaje sobre la red, el remitente no se hace responsable por
> > su contenido.
> > Su contenido no constituye ning=FAn compromiso para el remitente,
> > salvo ratificaci=F3n escrita por ambas partes. Aunque se esfuerza al
> > m=E1ximo por mantener su red libre de virus, el emisor no puede
> > garantizar nada al respecto y no ser=E1 responsable de cualesquiera
> > da=F1os que puedan resultar de una transmisi=F3n de virus.
> >
> >
>
>
>
>
- Follow-Ups:
- Re: Elasticity functions for Bezier and rectangular forms
- From: "E. Martin-Serrano" <eMartinSerrano@telefonica.net>
- Re: Elasticity functions for Bezier and rectangular forms
- References:
- Elasticity functions for Bezier and rectangular forms
- From: "E. Martin-Serrano" <eMartinSerrano@telefonica.net>
- Elasticity functions for Bezier and rectangular forms