just made an update to the breather surface, with graphing calc file &
Mathematica file.

http://www.xahlee.org/surface/breather_p/breather_p.html

   Xah

   ☄

interesting curve.

i also find it interesting to know its application or how it came up.

   Xah
   xah@...
∑ http://xahlee.org/





On Apr 11, 2005, at 10:58 PM, Narasimham Gudipaty wrote:

An intersting 2D locus, however, I don't know its name. 
 
Perhaps the parameteric form is simpler. It is not difficult to derive
a relation between
 th and an auxiliary angle om(ega). In Mathematica :


   a=1; b=.36; om=ArcTan[Cos[th],Sin[th] (b/a)^2];

cs=Cos[th];sn=Sin[th];CS=Cos[om];SN=Sin[om];

x= a cs; y= b sn; X=a (cs + SN)/2 ; Y= b (sn - CS)/2;

ell= ParametricPlot[{x,y},{th,0,2 Pi}]; anon=
ParametricPlot[{X,Y},{th,0,2Pi}];

Show[ell,anon];
But, what is its application, or , how does it come about ? Depending
on that you could choose its name.
 
Best Regards
 
G.L.Narasimham

garciacapitan <pacoga@...> wrote:

I am new in this group. I am a Spanish math teacher in a secondary
school and I have a question about a curve.

Let P move on the ellipse x^2/a^2+y^2/b^2 = 1 and let Q be one of
points such that QOP is a right angle.

   What is the locus of the midpoint M of chord PQ?

I have found that the equation of this locus is

(a^2 + b^2)*(b^2*X^2 + a^2*Y^2)^2 == a^2*b^2*(b^4*X^2 + a^4*Y^2)

But, what is the mame of this curve?

You can find a picture at

http://garciacapitan.auna.com/problemas/lugar1

Best regards,

Francisco Javier García Capitán
http://garciacapitan.auna.com


G.L.Narasimham, Ex Advisor and Head, Product Design, Composites Group,
Vikram Sarabhai Space Center, Indian Space Research Organization,
Trivandrum 695013

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   ☄

I am new in this group. I am a Spanish math teacher in a secondary school and I have a question about a curve.

Let P move on the ellipse x^2/a^2+y^2/b^2 = 1 and let Q be one of points such that QOP is a right angle.

What is the locus of the midpoint M of chord PQ?

I have found that the equation of this locus is

(a^2 + b^2)*(b^2*X^2 + a^2*Y^2)^2 == a^2*b^2*(b^4*X^2 + a^4*Y^2)

But, what is the mame of this curve?

You can find a picture at

http://garciacapitan.auna.com/problemas/lugar1

Best regards,

Francisco Javier Garc�a Capit�n
http://garciacapitan.auna.com

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Tired of spam? Yahoo! Mail has the best spam protection around
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1) Is there a relationship between the focus of a parabola, its latus rectum, and its evolute?

The 1.5 power curve has a cusp at focus. 

2) The evolute of the spiral  x= cost + tsint,  y= sint - tsint is a
circle.  In general, are the evolutes of most spirals circles?

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On Apr 10, 2005, at 11:52 PM, mcyberliver wrote:



> 1) Is there a relationship between the focus of a parabola, its
> latus rectum, and its evolute?

by definition there is a relation between the focus and the latus
rectum of a parabola. As to focus and its evolute, i suppose so, if not
direct.


> 2) The evolute of the spiral  x= cost + tsint,  y= sint - tsint is a
> circle.  In general, are the evolutes of most spirals circles?

the evolute of spirals are spirals, considering circle a special case
of equiangular spiral with angle π/2 .

   Xah

   ☄

1) Is there a relationship between the focus of a parabola, its
latus rectum, and its evolute?

2) The evolute of the spiral  x= cost + tsint,  y= sint - tsint is a
circle.  In general, are the evolutes of most spirals circles?

I am new in this group. I am a Spanish math teacher in a secondary
school and I have a question about a curve.

Let P mover on the ellipse x^2/a^2+y^2/b^2 = 1 and let Q one of
points such that QOP is a right angle.

What is the locus of the midpoint M of chord PQ?

I have found that the equation of this locus is

(a^2 + b^2)*(b^2*X^2 + a^2*Y^2)^2 == a^2*b^2*(b^4*X^2 + a^4*Y^2)

But, what is the mame of this curve?

You can find a picture at

http://garciacapitan.auna.com/problemas/lugar1

Best regards,

Francisco Javier Garc�a Capit�n
http://garciacapitan.auna.com

the one in mma version 2 is here:
http://xahlee.org/SpecialPlaneCurves_dir/Seashell_dir/index.html

formula is:

x = 2*(1 - E^(u/(6*Pi)))*Cos[u]*Cos[v/2]^2,
y = 2*(-1 + E^(u/(6*Pi)))*Cos[v/2]^2*Sin[u],
z = 1 - E^(u/(3*Pi)) - Sin[v] + E^(u/(6*Pi))*Sin[v]}


   Xah
   xah@...
∑ http://xahlee.org/

-----------

On Jan 24, 2005, at 12:23 AM, Narasimham Gudipaty wrote:


Impressive, it was called Cornucopia..

There was one on Version 2 front end images of
Mathematica also.
I also played with such parametrizations before. Shall
send it later (I call it Conch), right now am on
another item.

Regards
Narasimham

--- xah lee <xah@...> wrote:

> Mike Williams sent me some wonderful parametric
> formulas for modeling  seashells.
> -------------------------
> spindle shell
>
> R=1;    // radius of tube
> N=5.6;  // number of turns
> H=4.5;  // height
> P=1.4;  // power
> L=4;    // Controls spike length
> K=9;    // Controls spike sharpness
>
> W = (u/(2*pi)*R)^0.9
>
> Fx = W(u)*cos(N*u)*(1+cos(v))
> Fy = W(u)*sin(N*u)*(1+cos(v))
> Fz = W(u)*(sin(v)+L*(sin(v/2))^K)  + H*(u/(2*pi))^P}
>
> ---------------------------
> top
>
> R=1;    // radius of tube
> N=7.6;  // number of turns
> H=2.5;  // height
> P=1.3;  // power
> T=0.8;  // Triangleness of cross section
> A=-0.3;  // Angle of tilt of cross section (radians)
>
> W = u/(2*pi)*R}
>
> Fx = W(u)*cos(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
> Fy = W(u)*sin(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
> Fz = W(u)*(sin(v+A)+cos(2*v+A)*T/4)  +
> H*(u/(2*pi))^P}
>
> ------------------------
>
>
>   Xah
>   xah@...
>   http://xahlee.org/PageTwo_dir/more.html
>
>


=====
G.L.Narasimham,  Ex Advisor and Head, Product Design, Composites Group,
Vikram Sarabhai Space Center, Indian Space Research Organization,
Trivandrum 695013

   ☄

added a folium of Descartes stamp!

http://xahlee.org/SpecialPlaneCurves_dir/FoliumOfDescartes_dir/
foliumOfDescartes.html

anyone know other stamps featuring curves or surfaces?

   Xah
   xah@...
∑ http://xahlee.org/
   ☄

Thanks for you kind reply. I will have to study this
and perhaps I will have a further question.

Joseph Ferrara


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> There is 1 message in this issue.
>
> Topics in this digest:
>
>       1. Re: Question about a curve constructrion
>            From: Narasimham Gudipaty
> <glnarasimham@...>
>
>
>
________________________________________________________________________
>
________________________________________________________________________
>
> Message: 1
>    Date: Sun, 20 Mar 2005 08:25:32 -0800 (PST)
>    From: Narasimham Gudipaty
> <glnarasimham@...>
> Subject: Re: Question about a curve constructrion
>
> Let a ( = 10 ) be intercept on x- or y- axis .
>
> Any straight line is x / c + y / (a -c) = 1 is
> obtained by taking c as an arbitrary parameter. By
> C- discriminant method one can get the envelope.
> Difffernetiate the above equation with respect to c
> ;   - x /c^2 + y/ ( a -c)^2 = 0 ;
>
> Eliminate c between the the above equations to get
> sqrt( x/a) + sqrt( y/a) = 1 which is a parabola
> whose axis is 45 degrees to either axis. Also the
> x-, y- axes are tangents to the parabola.
>
> ... then someone told me it could not be a parabola
> because two tangents were at right angles.... But
> tangents can be at any angle !
>
>  HTH       Narasimham
>
> jpferrara06379 <jpferrara06379@...> wrote:
>
> This is just to describe the construct of the curve
> and then I want
> to ask what kind of curve is it, is it a parabola?
> Secondly how can I
> get into the math of the curve?
>
> Consider the x axis and the y axis on graph paper (1
> cm ruled lines)
>
> On the y axis number from the origin at 1 cm
> distances
> 0,1,2,3,4,5,6,7,8,9,10
>
> On the x axis note the numbering is reversed but the
> distances are
> the same
> 0, 10,9,8,7,6,5,4,3,2,1
>
> Simply connect the 10's, 9's, 8's etc with straight
> lines and the
> result are tangents that describe a curve. All my
> life I was under
> the impression it was 90 degree circle arc, then
> someone pointed out
> it is a parabola, then someone told me it could not
> be a parabola
> becaue two tangents were at right angles. So finally
> I want to
> investigate further but am weak in math.
>
> Thank you in advance.
>
> Joe
>
>
> G.L.Narasimham,  Ex Advisor and Head, Product
> Design, Composites Group, Vikram Sarabhai Space
> Center, Indian Space Research Organization,
> Trivandrum 695013
>
> ---------------------------------
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>  Yahoo! Small Business - Try our new resources site!
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This is just to describe the construct of the curve and then I want
to ask what kind of curve is it, is it a parabola? Secondly how can I
get into the math of the curve?

Consider the x axis and the y axis on graph paper (1 cm ruled lines)

On the y axis number from the origin at 1 cm distances
0,1,2,3,4,5,6,7,8,9,10

On the x axis note the numbering is reversed but the distances are
the same
0, 10,9,8,7,6,5,4,3,2,1

Simply connect the 10's, 9's, 8's etc with straight lines and the
result are tangents that describe a curve. All my life I was under
the impression it was 90 degree circle arc, then someone pointed out
it is a parabola, then someone told me it could not be a parabola
becaue two tangents were at right angles. So finally I want to
investigate further but am weak in math.

Thank you in advance.

Joe

This is just to describe the construct of the curve and then I want
to ask what kind of curve is it, is it a parabola? Secondly how can I
get into the math of the curve?

Consider the x axis and the y axis on graph paper (1 cm ruled lines)

On the y axis number from the origin at 1 cm distances
0,1,2,3,4,5,6,7,8,9,10

On the x axis note the numbering is reversed but the distances are
the same
0, 10,9,8,7,6,5,4,3,2,1

Simply connect the 10's, 9's, 8's etc with straight lines and the
result are tangents that describe a curve. All my life I was under
the impression it was 90 degree circle arc, then someone pointed out
it is a parabola, then someone told me it could not be a parabola
becaue two tangents were at right angles. So finally I want to
investigate further but am weak in math.

Thank you in advance.

Joe

recently i learned about:
http://en.wikipedia.org/wiki/Ogee

is there a geometric definition?

  Xah Lee

I added a photo page on Curlicue. See
http://xahlee.org/SpecialPlaneCurves_dir/Curlicue_dir/curlicue.html

has anyone seen interesting examples?

i know that Michael Trott of Wolfram Research has done curlicue
graphics. Also i've seen mac software that generates it.

  Xah

Nice connection of involute, evolute, and parallels.

I've added some illustrations using sine curve as example.
http://xahlee.org/SpecialPlaneCurves_dir/Parallel_dir/parallel.html

will supply proofs in the coming days.

how's Leibnitz viewed these concepts?
do you have a source?

   Xah

On Mar 12, 2005, at 3:50 AM, Narasimham Gudipaty wrote:


Yes,Xah, I also found it intriguing... the base circle
is evolute of all parallel involutes, in perhaps the
simplest 2D plane example.
In fact,this is how Liebnitz viewed parallel curves,
involutes /evolutes together.The involutes form a
family of parallel curves, parallel distance between
any two curves being an arbitrary constant.
In formulation, si=0 furnishes the cusp locus,where si
is tangent rotation to principal coordinate direction.

Regards,
Narasimham


--- Xah Lee <xah@...> wrote:
>
> i just learned that the locus of cusps of parallel
> curves is the
> evolute of the given curve.
>
> this is interesting, because it is a alternative
> definition of
> evolute. (the locus of centers of tangent circles)
>
> and from the locus of cusps perspective, one gains
> insight of the
> original curve...
>
>  Xah Lee
>
>
>
>

G.L.Narasimham,  Ex Advisor and Head, Product Design, Composites Group,
Vikram Sarabhai Space Center, Indian Space Research Organization,
Trivandrum 695013



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Some notes on “Scalar Field”, “Vector Field”, “Gradient”,
“Divergent”

A function that returns one real number is often called a Scalar Field.

A function that returns a vector of two values is often called a Vector
Field.

A Scalar Field, is defined by a function of the form f:R^n→R^1.

a Vector Field, is defined by a fuction of the form f:R^n→R^m.

The Gradient of a Scalar Field f:R^n→R^1 is a vector field g:R^n→R^n
that indicates the direction and change at each point. It is given by
taking the derivative with respect to each parameter. For example, if a
scalar field is given by f[x,y], then its gradient fuction is
{D[f[x,y],x],D[f[x,y],y]}.

The Divergent of a vector field f:R^n→R^n is scalar field g:R^n→R^1. It
is defined as the sum of the derivative of each component. For example,
if the vector field is f[x,y]:={a[x,y],b[x,y]}, then its divergent is
D[a[x,y],x]+D[b[x,y],y].

Given a curve f[x,y]==0, the normal at {x,y} can be computed as the
gradient of f at {x,y}. Similarly for a surface f[x,y,z]==0.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

Yes,Xah, I also found it intriguing... the base circle
is evolute of all parallel involutes, in perhaps the
simplest 2D plane example.
In fact,this is how Liebnitz viewed parallel curves,
involutes /evolutes together.The involutes form a
family of parallel curves, parallel distance between
any two curves being an arbitrary constant.
In formulation, si=0 furnishes the cusp locus,where si
is tangent rotation to principal coordinate direction.

Regards,
Narasimham


--- Xah Lee <xah@...> wrote:
>
> i just learned that the locus of cusps of parallel
> curves is the
> evolute of the given curve.
>
> this is interesting, because it is a alternative
> definition of
> evolute. (the locus of centers of tangent circles)
>
> and from the locus of cusps perspective, one gains
> insight of the
> original curve...
>
>  Xah Lee
>
>
>
>

G.L.Narasimham,  Ex Advisor and Head, Product Design, Composites Group, Vikram
Sarabhai Space Center, Indian Space Research Organization, Trivandrum 695013



__________________________________
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i just learned that the locus of cusps of parallel curves is the
evolute of the given curve.

this is interesting, because it is a alternative definition of
evolute. (the locus of centers of tangent circles)

and from the locus of cusps perspective, one gains insight of the
original curve...

  Xah Lee

Just created a page on curvature and tangent circle. See

http://xahlee.org/SpecialPlaneCurves_dir/Curvature_dir/curvature.html

the interesting idea is the development of the definition of curvature.
One is by the limit of two approaching normal lines's intersection.
Another, is as the rate of turning of the tangent line.

I have yet to add illustrations, and also work out these details.

i think another way the definition of curvature might have developed,
is to simply begin by defining curvature as the 1/r of a circle. Then,
find the fittest circle on the curve. The criterion for which circle is
fittest i haven't thought in detail yet, but i'm sure it can go by
measuring the distance from the curve to the circle in the vicinity of
the contact point.

i like this idea the most, as it is most geometric and static; does not
require the concept of moving things.

does any one know the historical detail of curvature? or any tech info
on the above?
Most math texts simply throw out the curvature formula one way or
another, totally forego any explanation how the curvature concept and
definition came about.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

new page on cusps, with illustrations:
http://xahlee.org/SpecialPlaneCurves_dir/Cusp_dir/cusp.html

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

here are 3 curves illustrating 3 cusps (at t=0) that has different
curvatures.

{t^2, t^3}
{t^2, t^5}
{t^2 + t^3, t^4}

first case has curvature Infinity. Second has curvature 0. Third has
curvature 2.

This example is from C G Gibson's book.
Does anyone has other examples?

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

ha ha, here's the answer:

In[3]:=
(1/f'[x])//InputForm

Out[3]//InputForm=
Derivative[1][f][x]^(-1)

In[4]:=
Format[f'[z],OutputForm]

Out[4]=
\!\(\*
    InterpretationBox[
      FrameBox["\<\"f'[z]\"\>",
        BoxFrame->False,
        BoxMargins->False],
      Format[
        Derivative[ 1][ f][ z], OutputForm],
      Editable->False]\)

In[5]:=
Format[Derivative[n_][f_][t_],InputForm]:=
    StringJoin[ToString[f],
      StringReplace[ToString@Table["'",{i,1,n}],{"{"->"",",
"->"","}"->""}],"[",
      ToString[t],"]"]

In[6]:=
(1/f'[x])//InputForm

Out[6]//InputForm=
"f'[x]"^(-1)


Not a good solution but a work around.

anyone know the right solutions?

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html


---------------------------------
On Feb 18, 2005, at 12:25 AM, xah lee wrote:


here's a problem that took me a while to find an answer.

yesterday i've been updating the evolute page.
http://xahlee.org/SpecialPlaneCurves_dir/Evolute_dir/evolute.html

Given a curve in parametric form {x[t], y[t]}, its evolute is
{x + (y'*(x'^2 + y'^2)) / (  y'*x''  - x'*y''),
    y + (x'*(x'^2 + y'^2)) / (-(y'*x'') + x'*y'')}

(The traling [t] is ommited for easy reading.)

What i had on that page is a image that displayed the formula. However,
i wanted a flat ascii representation. But, if i use Mathematica
InputForm, what i get is this:

In[15]:=
Evolute[{xf[t],yf[t]},t]//InputForm

Out[15]//InputForm=
{xf[t] + (Derivative[1][yf][t]*(Derivative[1][xf][t]^2 +
        Derivative[1][yf][t]^2))/
      (Derivative[1][yf][t]*Derivative[2][xf][t] -
       Derivative[1][xf][t]*Derivative[2][yf][t]),
    yf[t] + (Derivative[1][xf][t]*(Derivative[1][xf][t]^2 +
        Derivative[1][yf][t]^2))/
      (-(Derivative[1][yf][t]*Derivative[2][xf][t]) +
       Derivative[1][xf][t]*Derivative[2][yf][t])}

very unsavory.
Here, in Mathematica, the Derivative[n][f][t] represents the nth
derivative of f, with parameter t.

Now the problem is that I want the ascii output from InputForm (i.e.
expression in flat ascii text), but i want derivatives written as
f'[t], f''[t] etc.

How to do that?

after some 30 minutes i've figured out. (the last period i'm immersed
in Mathematica is about 1998.)

See if you can figure it out as a fun exercise. I'll post answer
tomorrow.

    Xah
    xah@...
    http://xahlee.org/PageTwo_dir/more.html





Yahoo! Groups Links

hi Hop,

you might find this wikipedia article useful:
http://en.wikipedia.org/wiki/Vector_calculus

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html


------------------
On Feb 18, 2005, at 10:13 AM, Hop David wrote:

I want to learn some orbital mechanics and electrodynamics. So I'm
interested in vector fields.

Don't have a text handy but IIRC there are level surfaces; points in
space where vectors all have the same magnitude. For example the points
where the acceleration vector is 9.8 meters/sec forms a nearly spherical
surface nearly corresponding with earth's surface.

The Gradient points in the direction of greatest change between level
surfaces? and the Gradient is perpendicular to the level curve's tangent
plane at that point?

I hope this will generate conversations on vector fields, div, curl and
such as my understanding is still not what I'd like it to be.

--
Hop David
http://clowder.net/hop/index.html

xah lee wrote:
> recently i started a mailing lists on Python and Java programing. I
> started it about a month ago, in which i give daily tips or commentary
> about them as i learn these languages. The result has been extremely
> positive, as i've wanted to master these languages but never had the
> incentive. But with these daily writings, quickly in the past month
> i've learned tremendously of these languages with community's feedback.
>
> I thought i'd do the same with the subject of curves and surfaces.
> Recently i've been trying to study the formal mathematics of curves &
> surfaces. Starting with differential aspects. (and will be followed
> with Algebraic aspects, or the study of mappings with complex
> functions, and or projective geometry) The last time i'm fluent in the
> ins and outs of derivative and integration is some decade ago. :)

I want to learn some orbital mechanics and electrodynamics. So I'm
interested in vector fields.

Don't have a text handy but IIRC there are level surfaces; points in
space where vectors all have the same magnitude. For example the points
where the acceleration vector is 9.8 meters/sec forms a nearly spherical
surface nearly corresponding with earth's surface.

The Gradient points in the direction of greatest change between level
surfaces? and the Gradient is perpendicular to the level curve's tangent
plane at that point?

I hope this will generate conversations on vector fields, div, curl and
such as my understanding is still not what I'd like it to be.

>
> ... there are several interesting questions and tips posted here in the
> past that is also worthy of further exploration. I hope we can have a
> more fruitful and busier forum with daily dosages.
>
>   Xah
>   xah@...
>   http://xahlee.org/PageTwo_dir/more.html
>
>
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--
Hop David
http://clowder.net/hop/index.html

here's a problem that took me a while to find an answer.

yesterday i've been updating the evolute page.
http://xahlee.org/SpecialPlaneCurves_dir/Evolute_dir/evolute.html

Given a curve in parametric form {x[t], y[t]}, its evolute is
{x + (y'*(x'^2 + y'^2)) / (  y'*x''  - x'*y''),
   y + (x'*(x'^2 + y'^2)) / (-(y'*x'') + x'*y'')}

(The traling [t] is ommited for easy reading.)

What i had on that page is a image that displayed the formula. However,
i wanted a flat ascii representation. But, if i use Mathematica
InputForm, what i get is this:

In[15]:=
Evolute[{xf[t],yf[t]},t]//InputForm

Out[15]//InputForm=
{xf[t] + (Derivative[1][yf][t]*(Derivative[1][xf][t]^2 +
       Derivative[1][yf][t]^2))/
     (Derivative[1][yf][t]*Derivative[2][xf][t] -
      Derivative[1][xf][t]*Derivative[2][yf][t]),
   yf[t] + (Derivative[1][xf][t]*(Derivative[1][xf][t]^2 +
       Derivative[1][yf][t]^2))/
     (-(Derivative[1][yf][t]*Derivative[2][xf][t]) +
      Derivative[1][xf][t]*Derivative[2][yf][t])}

very unsavory.
Here, in Mathematica, the Derivative[n][f][t] represents the nth
derivative of f, with parameter t.

Now the problem is that I want the ascii output from InputForm (i.e.
expression in flat ascii text), but i want derivatives written as
f'[t], f''[t] etc.

How to do that?

after some 30 minutes i've figured out. (the last period i'm immersed
in Mathematica is about 1998.)

See if you can figure it out as a fun exercise. I'll post answer
tomorrow.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

recently i started a mailing lists on Python and Java programing. I
started it about a month ago, in which i give daily tips or commentary
about them as i learn these languages. The result has been extremely
positive, as i've wanted to master these languages but never had the
incentive. But with these daily writings, quickly in the past month
i've learned tremendously of these languages with community's feedback.

I thought i'd do the same with the subject of curves and surfaces.
Recently i've been trying to study the formal mathematics of curves &
surfaces. Starting with differential aspects. (and will be followed
with Algebraic aspects, or the study of mappings with complex
functions, and or projective geometry) The last time i'm fluent in the
ins and outs of derivative and integration is some decade ago. :)

... there are several interesting questions and tips posted here in the
past that is also worthy of further exploration. I hope we can have a
more fruitful and busier forum with daily dosages.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html

Mike has sent others, one latest is the wentletrap:

--------------------
Here's a Precious Wentletrap.

N=6.6;  // number of turns
H=4.0;  // height
P=2;    // power
A=0.12; // Ridge Amplitude
F=15;   // Ridge Frequency

W = (u/(2*pi))^P

Fx = W(u)*(cos(N*u)-A*cos(N*F*u))*(1+cos(v))}
Fy = W(u)*(sin(N*u)+A*sin(N*F*u))*(1+cos(v))}
Fz = W(u)*sin(v)  + H*(u/(2*pi))^(1+P)

Approximating this surface requires a very large number of polygons,
particularly in the "u" direction. The attached image uses a mesh of
300,000 triangles. Use too few polygons in the approximation and the
details of the ridge structure gets lost.
-----------

i haven't been able to plot in mma yet, because it crashes for perhaps
too high a resolution...
here's the code:
-----------
Clear[cW, seashell, cN, cH, cP, cA, cF]
cW[u_] := (u/(2*\[Pi]))^cP
seashell[u_, v_] := {cW@u*(Cos[cN*u] - cA*Cos[cN*cF*u])*(1 + Cos@v),
        cW@u*(Sin[cN*u] + cA*Sin[cN*cF*u])*(1 + Cos@v),
        cW@u*Sin@v + cH*(u/(2*\[Pi]))^(1 + cP)};

cN = 6.6;(*number of turns*)
cH = 4.0;(*height*)
cP = 2;(*power*)

cA = 0.12;(*Ridge Amplitude*)
cF = 15;(*Ridge Frequency*)

ParametricPlot3D[Evaluate@seashell[u, v], {u, 4, 5}, {v, 2, 6},
    PlotPoints -> {200, 40}]

----------------
now, try to increase the u range to 0 to 6. And the plot points should
be also be increased.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html




---------------------------------------------------------


On Jan 24, 2005, at 12:23 AM, Narasimham Gudipaty wrote:


Impressive, it was called Cornucopia..

There was one on Version 2 front end images of
Mathematica also.
I also played with such parametrizations before. Shall
send it later (I call it Conch), right now am on
another item.

Regards
Narasimham

Impressive, it was called Cornucopia..

There was one on Version 2 front end images of
Mathematica also.
I also played with such parametrizations before. Shall
send it later (I call it Conch), right now am on
another item.

Regards
Narasimham

--- xah lee <xah@...> wrote:

> Mike Williams sent me some wonderful parametric
> formulas for modeling  seashells.
> -------------------------
> spindle shell
>
> R=1;    // radius of tube
> N=5.6;  // number of turns
> H=4.5;  // height
> P=1.4;  // power
> L=4;    // Controls spike length
> K=9;    // Controls spike sharpness
>
> W = (u/(2*pi)*R)^0.9
>
> Fx = W(u)*cos(N*u)*(1+cos(v))
> Fy = W(u)*sin(N*u)*(1+cos(v))
> Fz = W(u)*(sin(v)+L*(sin(v/2))^K)  + H*(u/(2*pi))^P}
>
> ---------------------------
> top
>
> R=1;    // radius of tube
> N=7.6;  // number of turns
> H=2.5;  // height
> P=1.3;  // power
> T=0.8;  // Triangleness of cross section
> A=-0.3;  // Angle of tilt of cross section (radians)
>
> W = u/(2*pi)*R}
>
> Fx = W(u)*cos(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
> Fy = W(u)*sin(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
> Fz = W(u)*(sin(v+A)+cos(2*v+A)*T/4)  +
> H*(u/(2*pi))^P}
>
> ------------------------
>
>
>   Xah
>   xah@...
>   http://xahlee.org/PageTwo_dir/more.html
>
>


=====
G.L.Narasimham,  Ex Advisor and Head, Product Design, Composites Group, Vikram
Sarabhai Space Center, Indian Space Research Organization, Trivandrum 695013

__________________________________________________
Do You Yahoo!?
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the following are some wonderful seashell formulas, of Mike's original
message.

   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html
---------------------------------------------

Begin forwarded message:

From: Mike Williams <mike@...>
Date: January 19, 2005 12:35:17 AM PST
To: xah@...
Subject: Paramnetric equations of seashells

I hear that you're interested in parametric equations that approximate
spiral seashells. Here are a few that I've made.

They're all based on this parameterisation of the torus

    Fx = R*cos(u)*(R2+cos(v))
    Fy = R*sin(u)*(R2+cos(v))
    Fz = R*sin(v)

I then replaced R with R*u/(2*pi) so that the radius of the tube
increases linearly as u increases, and replaced sin(u) with sin(N*u) so
that the tube spirals round the origin N times. Then I added an offset
factor H*(u/2*pi)^P so that the turns of the spiral are offset in the z
direction  by a distance that varies from 0 to H.


This gives a periwinkle:

R=1;    // radius of tube
N=4.6;  // number of turns
H=2;    // height
P=2;    // power

W(u) = u/(2*pi)*R
Fx = W(u)*cos(N*u)*(1+cos(v))
Fy = W(u)*sin(N*u)*(1+cos(v))
Fz = W(u)*sin(v) + H*(u/(2*pi))^P


And a top

R=1;    // radius of tube
N=7.6;  // number of turns
H=2.5;  // height
P=1.3;  // power

W = u/(2*pi)*R
Fx = W(u)*cos(N*u)*(1+cos(v))
Fy = W(u)*sin(N*u)*(1+cos(v))
Fz = W(u)*sin(v) + H*(u/(2*pi))^P


A cone shell is something like this:

R=1;    // radius of tube
N=4.6;  // number of turns
H=0.5;  // height
P=2;    // power

W = u/(2*pi)*R
Fx = W(u)*cos(N*u)*(1+cos(v))
Fy = W(u)*sin(N*u)*(1+cos(v))
Fz = W(u)*sin(v)*1.25 + H*(u/(2*pi))^P +W(u)*cos(v)*1.25


And here's a wrinkled periwinkle

R=1;    // radius of tube
N=4.6;  // number of turns
H=2.5;  // height
F=80;   // wave frequency
A=0.2;  // wave amplitude
P=1.9;  // power

W = u/(2*pi)*R
Fx = W(u)*cos(N*u)*(1+cos(v)+cos(F*u)*A)
Fy = W(u)*sin(N*u)*(1+cos(v)+cos(F*u)*A)
Fz = W(u)*sin(v) + H*(u/(2*pi))^P

--
Mike Williams
Gentleman of Leisure

Mike Williams send me some wonderful parametric formulas for modeling
seashells.

-------------------------
spindle shell

R=1;    // radius of tube
N=5.6;  // number of turns
H=4.5;  // height
P=1.4;  // power
L=4;    // Controls spike length
K=9;    // Controls spike sharpness

W = (u/(2*pi)*R)^0.9

Fx = W(u)*cos(N*u)*(1+cos(v))
Fy = W(u)*sin(N*u)*(1+cos(v))
Fz = W(u)*(sin(v)+L*(sin(v/2))^K)  + H*(u/(2*pi))^P}

---------------------------
top

R=1;    // radius of tube
N=7.6;  // number of turns
H=2.5;  // height
P=1.3;  // power
T=0.8;  // Triangleness of cross section
A=-0.3;  // Angle of tilt of cross section (radians)

W = u/(2*pi)*R}

Fx = W(u)*cos(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
Fy = W(u)*sin(N*u)*(1+cos(v+A)+sin(2*v+A)*T/4)}
Fz = W(u)*(sin(v+A)+cos(2*v+A)*T/4)  + H*(u/(2*pi))^P}

------------------------


   Xah
   xah@...
   http://xahlee.org/PageTwo_dir/more.html