POV-Ray : Newsgroups : povray.newusers : Teardrops Server Time
4 Nov 2024 21:24:26 EST (-0500)
  Teardrops (Message 1 to 10 of 10)  
From: Dancing Bear
Subject: Teardrops
Date: 11 Oct 2001 09:24:11
Message: <3bc59d7b$1@news.povray.org>
I'm not a total newbie but for some reason having a bugger of a time
creating teardrops that look realistic
or don't have cut lines in them... Have tried unions, blob objects and short
of defining every point in a mesh
not having a lot of luck.  Anyone have an idea how to do realistic looking
teardrops?

Thanks
Bear

--
====================================
Mark W. Law, ccp
WWW: http://cabinfever.d2g.com/
Home: http://3entente.d2g.com
"Life is a great big canvas, throw lots of paint at it"
- Danny Kaye
====================================


Post a reply to this message

From: Ken
Subject: Re: Teardrops
Date: 11 Oct 2001 10:10:32
Message: <3BC5A9C5.EC626280@pacbell.net>
Dancing Bear wrote:
> 
> I'm not a total newbie but for some reason having a bugger of a time
> creating teardrops that look realistic
> or don't have cut lines in them... Have tried unions, blob objects and short
> of defining every point in a mesh
> not having a lot of luck.  Anyone have an idea how to do realistic looking
> teardrops?

#declare Teardrop =

  quartic{ <
  0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0,
  -1, 2, 0, -2, 1> }

or maybe

#declare Teardrop =

 merge{
  sphere{0,1.5 scale<1,.65,1>} cone{y*.4,1.37,y*3.1,0}
  translate y*-.6375 scale<.425,.5259,.425>pigment{rgbf .99}}


-- 
Ken Tyler


Post a reply to this message

From: Trevor Quayle
Subject: Re: Teardrops
Date: 11 Oct 2001 10:16:49
Message: <3bc5a9d1@news.povray.org>
However, real teardrops do not have a point, in reality they are spherical
(or rather elliptical or egg shaped as distorted by gravity/friction)

-tgq

"Ken" <tyl### [at] pacbellnet> wrote in message
news:3BC5A9C5.EC626280@pacbell.net...
>
>
> Dancing Bear wrote:
> >
> > I'm not a total newbie but for some reason having a bugger of a time
> > creating teardrops that look realistic
> > or don't have cut lines in them... Have tried unions, blob objects and
short
> > of defining every point in a mesh
> > not having a lot of luck.  Anyone have an idea how to do realistic
looking
> > teardrops?
>
> #declare Teardrop =
>
>   quartic{ <
>   0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0, 0,
>   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0,
>   -1, 2, 0, -2, 1> }
>
> or maybe
>
> #declare Teardrop =
>
>  merge{
>   sphere{0,1.5 scale<1,.65,1>} cone{y*.4,1.37,y*3.1,0}
>   translate y*-.6375 scale<.425,.5259,.425>pigment{rgbf .99}}
>
>
> --
> Ken Tyler


Post a reply to this message

From: Warp
Subject: Re: Teardrops
Date: 11 Oct 2001 11:27:46
Message: <3bc5ba72@news.povray.org>
Trevor Quayle <Tin### [at] hotmailcom> wrote:
: However, real teardrops do not have a point, in reality they are spherical
: (or rather elliptical or egg shaped as distorted by gravity/friction)

  This depends a lot on where the drop is.

  If it's falling freely (ie. a raindrop), then its shape is close to a
lemniscate or like a hamburger-shape (although it depens also a lot in the
size of the drop).
  If it's sliding on a surface (eg. a window) then it gets more the classical
teardrop-shape. The upper part may or may not be sharp, depending on the
physical properties of the surface.
  If it's hanging from a surface (being about to drop), then it's also like
the classical teardrop-shape, but the upper part is more like a hyperbola.

-- 
#macro N(D,I)#if(I<6)cylinder{M()#local D[I]=div(D[I],104);M().5,2pigment{
rgb M()}}N(D,(D[I]>99?I:I+1))#end#end#macro M()<mod(D[I],13)-6,mod(div(D[I
],13),8)-3,10>#end blob{N(array[6]{11117333955,
7382340,3358,3900569407,970,4254934330},0)}//                     - Warp -


Post a reply to this message

From: Nekar Xenos
Subject: Re: Teardrops
Date: 12 Oct 2001 08:27:51
Message: <3bc6e1c7@news.povray.org>
"Warp" <war### [at] tagpovrayorg> wrote in message news:3bc5ba72@news.povray.org...
> Trevor Quayle <Tin### [at] hotmailcom> wrote:
> : However, real teardrops do not have a point, in reality they are spherical
> : (or rather elliptical or egg shaped as distorted by gravity/friction)
>
>   This depends a lot on where the drop is.
>
>   If it's falling freely (ie. a raindrop), then its shape is close to a
> lemniscate or like a hamburger-shape (although it depens also a lot in the
> size of the drop).
>   If it's sliding on a surface (eg. a window) then it gets more the classical
> teardrop-shape. The upper part may or may not be sharp, depending on the
> physical properties of the surface.
>   If it's hanging from a surface (being about to drop), then it's also like
> the classical teardrop-shape, but the upper part is more like a hyperbola.
>

Ok, so what's the isosurface formula for each of these?  =)

- Nekar


---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.282 / Virus Database: 150 - Release Date: 2001/09/25


Post a reply to this message

From: Ken
Subject: Re: Teardrops
Date: 12 Oct 2001 08:47:36
Message: <3BC6E7D8.99C85A@pacbell.net>
Nekar Xenos wrote:

> Ok, so what's the isosurface formula for each of these?  =)

Parametric:

   This short note describes the parametric equations which give
   rise to an approximate model of a drop of water, for example, a
   tear drop.
  
   The equations as functions of longitute phi and lattitude theta are:
  
  x = 0.5 *(1-cos(8)) sin(8) cos(circle with verticle line through it)
  y = 0.5 *(1-cos(8)) sin(8) sin(circle with verticle line through it)
  z = cos(8)
  
    where 0 <= circ w/line tru/it  <= 2pi
    and 0   <= 8 <= pi


   When theta is 0 there is a discontinuity at the apex where 
   x = 0  y = 0  z = 1

  An implicit equation for the aforementioned tear drop is:

  1 - 4x^2 - 4y^2 - 2z + 2z^3 - z^4 = 0,

  or, it simplifies a bit as 4(x^2+y^2)=(1+z)(1-z)^3, which is a surface
  of revolution bounded by -1 <= z <= 1.  The POV command is:

  quartic{ <
  0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0,
  -1, 2, 0, -2, 1> }

  Sometimes, a parametric map involving trig functions can be converted
  to an algebraic implicit equation by using new variables:

  u=cos(t),
  v=sin(t),

  and introducing a new implicit relation:

  u^2+v^2=1.

  The idea is to make the parametric equations into polynomials, at the
  cost of increasing the number of variables and equations.  Then u and
  v can (sometimes) be eliminated from the system of polynomial
  equations to get an implicit equation in x,y,z.

-- 
Ken Tyler


Post a reply to this message

From: Marc-Hendrik Bremer
Subject: Re: Teardrops
Date: 12 Oct 2001 13:30:16
Message: <3bc728a8@news.povray.org>
I don't know if it helps you, but the "piriform" and "piriform_2d" functions
looks like a classic teardrop. Maybe it's not very realistic, but perhaps
you can manipulate it in a way you like it. Look at the Smellenbergh's
IsoTutorial to see some working parameters.

Hth,

Marc-Hendrik


Post a reply to this message

From: Marc-Hendrik Bremer
Subject: Re: Teardrops
Date: 12 Oct 2001 13:31:45
Message: <3bc72901$1@news.povray.org>
The "glob"-function might help you, too.


Marc-Hendrik Bremer schrieb in Nachricht <3bc728a8@news.povray.org>...
>I don't know if it helps you, but the "piriform" and "piriform_2d"
functions
>looks like a classic teardrop. Maybe it's not very realistic, but perhaps
>you can manipulate it in a way you like it. Look at the Smellenbergh's
>IsoTutorial to see some working parameters.
>
>Hth,
>
>Marc-Hendrik
>
>
>


Post a reply to this message

From: Mike Williams
Subject: Re: Teardrops
Date: 12 Oct 2001 15:11:25
Message: <wx3XBEAd9zx7EwIC@econym.demon.co.uk>
Wasn't it Ken who wrote:
>
>
>Nekar Xenos wrote:
>
>> Ok, so what's the isosurface formula for each of these?  =)
>
>Parametric:
>
>   This short note describes the parametric equations which give
>   rise to an approximate model of a drop of water, for example, a
>   tear drop.
>  
>   The equations as functions of longitute phi and lattitude theta are:
>  
>  x = 0.5 *(1-cos(8)) sin(8) cos(circle with verticle line through it)
>  y = 0.5 *(1-cos(8)) sin(8) sin(circle with verticle line through it)
>  z = cos(8)
>  
>    where 0 <= circ w/line tru/it  <= 2pi
>    and 0   <= 8 <= pi

For those of you who prefer stuff in POV language:-

#version 3.5;

camera { location  <0, 0, -2.4> look_at 0}

light_source {<100,100,-100> colour rgb 1}                              

#declare Fx = function(u,v) {0.5 *(1-cos(u)) *sin(u) *cos(v)}
#declare Fz = function(u,v) {0.5 *(1-cos(u)) *sin(u) *sin(v)}
#declare Fy = function(u,v) {cos(u)}
                        
#declare Umin = 0;
#declare Umax = pi;
#declare Vmin = 0;                       
#declare Vmax = 2*pi;

#declare Iter_U = 50;
#declare Iter_V = 50;

#include "param.inc"
Parametric()

object {Surface
  pigment {rgb 0.9}
}

I've used Ingo's "param.inc" because when I tried it as a real
parametric isosurface the "pixels per hour" figure looked like it was
heading for zero. I've switched the z and y values so that the pointy
bit is at the top.


-- 
Mike Williams
Gentleman of Leisure


Post a reply to this message

From: Ken
Subject: Re: Teardrops
Date: 12 Oct 2001 21:07:27
Message: <3BC79542.D0E62858@pacbell.net>
Mike Williams wrote:

> For those of you who prefer stuff in POV language:-

Thanks Mike. I'll add it to my private collection and hopefully can
share it with others again in the future.

-- 
Ken Tyler


Post a reply to this message

Copyright 2003-2023 Persistence of Vision Raytracer Pty. Ltd.