On Mon, 04 Jun 2001 16:58:11 -0700, Ken wrote:
>See Warp's reply. My bad.

Your bad what?

-- 
#macro R(L P)sphere{L F}cylinder{L P F}#end#macro P(V)merge{R(z+a z)R(-z a-z)R(a
-z-z-z a+z)torus{1F clipped_by{plane{a 0}}}translate V}#end#macro Z(a F T)merge{
P(z+a)P(z-a)R(-z-z-x a)pigment{rgbt 1}hollow interior{media{emission T}}finish{
reflection.1}}#end Z(-x-x.2y)Z(-x-x.4x)camera{location z*-10rotate x*90}

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"Ron Parker" <ron### [at] povrayorg> wrote in message
news:slr### [at] fwicom...
> On Mon, 04 Jun 2001 16:58:11 -0700, Ken wrote:
> >See Warp's reply. My bad.
>
> Your bad what?
>
It would mean "mistake."

--
Me (Daniel Lin (dli### [at] yahoocom))

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On Mon, 4 Jun 2001 22:16:48 -0400, Daniel Lin wrote:
>"Ron Parker" <ron### [at] povrayorg> wrote in message
>news:slr### [at] fwicom...
>> On Mon, 04 Jun 2001 16:58:11 -0700, Ken wrote:
>> >See Warp's reply. My bad.
>>
>> Your bad what?
>>
>It would mean "mistake."

It would, if it did.  But it doesn't.  "Bad" is still an adjective.

-- 
#macro R(L P)sphere{L F}cylinder{L P F}#end#macro P(V)merge{R(z+a z)R(-z a-z)R(a
-z-z-z a+z)torus{1F clipped_by{plane{a 0}}}translate V}#end#macro Z(a F T)merge{
P(z+a)P(z-a)R(-z-z-x a)pigment{rgbt 1}hollow interior{media{emission T}}finish{
reflection.1}}#end Z(-x-x.2y)Z(-x-x.4x)camera{location z*-10rotate x*90}

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Ron Parker wrote:
> 
> On Mon, 04 Jun 2001 16:58:11 -0700, Ken wrote:
> >See Warp's reply. My bad.
> 
> Your bad what?

I know it was a bad thing to say but the Devil made me do it.

-- 
Ken Tyler

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Ken <tyl### [at] pacbellnet> wrote:
: I know it was a bad thing to say but the Devil made me do it.

  Ah, so it was your evil brother? ;)

-- 
#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 -

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Rune wrote:
> 
> I know a little about what nurbs are, but I'd like to know more about the
> details.
> 
> First: what are the differences between nurbs and bicubic patches?
> 
> Rune
> --

NURBS in PovRay - my dream or 150 MB of memory necessary for mesh of my
last plane's model exported from Rhino (please see
www.angelfire.com/sk/gatial)

Jozef

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James Tonkin wrote:

> In article <3b17c237@news.povray.org>, Warp  <war### [at] tagpovrayorg> wrote:
> 
>>Ron Parker <ron### [at] povrayorg> wrote:
>>: Not really.  Bezier patches/bicubic patches are neither non-uniform nor
>>: rational.
>>
> 
> Ok, there's about 4 types of curves being discussed here, from (mathematically)
> simplest to most complex, they are:
> 
> 	Bezier curves
> 	Bicubic patches
> 	B-Splines (uniform, non-rational)
> 	NURBS (which stands for non-uniform, rational, b-spline)
> 
> All of them are based on the idea of a set of control points, and a set of
> blending functions.
> 
> Starting in 2 dimensions, with Bezier curves:
> 
> A curve can be defined in terms of a single parameter, (typically 's' or 't'
> are used, depending on your mathematical background) which represents the
> normalized distance along the curve, from it's starting point (control point 0) 
> towards it's ending point (for a Bezier curve, this is control point 3, 
> although the algorithim could be generalized to any number of points).  By
> 'normalized distance' I mean that the parameter (I'll use 't') is 0 at the 
> starting point, and 1 and the ending point.
> 
> Each control point will have an x and y co-ordinate (<x0,y0> to <x3,y3>), and
> an associated blending function.  
> 
> For Bezier curves, the blending functions are just the standard expansion
> of ( t * (1-t) ) ^3, seperated according to the power of the 't' term.
> 
> (Historically, I don't think this was the motivation for development of
>  this set of blending functions, but this is how it works out)
> 
> So blending function 3 is t^3.
> 	b.f. 2 = 3 * (t ^2)*(1-t)
> 	b.f. 1 = 3 * t * ( (1-t)^2)
> 	b.f. 0 = (1-t)^3
> 



Could you please tell me what separation of a formule is? i.e., how do 
you get the blending functions from ( t*(1-t))^3?

Thanks in advance,

Remco Poelstra

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Remco Poelstra  <rjp### [at] homenl> wrote:
>James Tonkin wrote:

[a bunch of stuff about the math behind bicubic patches]

>Could you please tell me what separation of a formule is? i.e., how do 
>you get the blending functions from ( t*(1-t))^3?

Well, it would help if I had of put the correct formula down in the first
place... sorry bout that.  Here's the full derivation

(t + (1-t) ) ^3

= (t + (1-t)) * (t + (1-t)) * (t + (1-t))

= [ t*t + t*(1-t) + (1-t)*t + (1-t)*(1-t)] * [ t + (1-t)]

= [ t^2 + 2 * t * (1-t) + (1-t)^2] * [t + (1-t)]

= ( t^2 * t) + (2 * t * (1-t) * t) + ( (1-t)^2 * t) + (t^2 * (1-t)) 
	+ (2 * t * (1-t) * (1-t) ) + ( (1-t)^2 * (1-t))

= t^3 + ( 2 * t^2 * (1-t) ) + (t * (1-t)^2) + (t^2 * (1-t)) 
	+ ( 2 * t * (1-t)^2) + (1-t)^3

= t^3 + (3 * t^2 * (1-t)) + (3 * t * (1-t)^2) + (1-t)^3

So the 4 terms in the last line correspond to the 4 blending functions.

Hope that helps,
Jamie

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James Tonkin wrote:

> 
> Well, it would help if I had of put the correct formula down in the first
> place... sorry bout that.  Here's the full derivation
> 
> (t + (1-t) ) ^3
> 
> = (t + (1-t)) * (t + (1-t)) * (t + (1-t))
> 
> = [ t*t + t*(1-t) + (1-t)*t + (1-t)*(1-t)] * [ t + (1-t)]
> 
> = [ t^2 + 2 * t * (1-t) + (1-t)^2] * [t + (1-t)]
> 
> = ( t^2 * t) + (2 * t * (1-t) * t) + ( (1-t)^2 * t) + (t^2 * (1-t)) 
> 	+ (2 * t * (1-t) * (1-t) ) + ( (1-t)^2 * (1-t))
> 
> = t^3 + ( 2 * t^2 * (1-t) ) + (t * (1-t)^2) + (t^2 * (1-t)) 
> 	+ ( 2 * t * (1-t)^2) + (1-t)^3
> 
> = t^3 + (3 * t^2 * (1-t)) + (3 * t * (1-t)^2) + (1-t)^3
> 
> So the 4 terms in the last line correspond to the 4 blending functions.
> 


Thanks, exactly what I wanted to know.

Remco Poelstra

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In article <3b40f28b$1@news.povray.org>, James Tonkin says...
> Remco Poelstra  <rjp### [at] homenl> wrote:
> >James Tonkin wrote:
> 
> [a bunch of stuff about the math behind bicubic patches]
> 
> >Could you please tell me what separation of a formule is? i.e., how do 
> >you get the blending functions from ( t*(1-t))^3?
> 
> Well, it would help if I had of put the correct formula down in the first
> place... sorry bout that.  Here's the full derivation
> 
> (t + (1-t) ) ^3
> 
> = (t + (1-t)) * (t + (1-t)) * (t + (1-t))
> 
> = [ t*t + t*(1-t) + (1-t)*t + (1-t)*(1-t)] * [ t + (1-t)]
> 
> = [ t^2 + 2 * t * (1-t) + (1-t)^2] * [t + (1-t)]
> 
> = ( t^2 * t) + (2 * t * (1-t) * t) + ( (1-t)^2 * t) + (t^2 * (1-t)) 
> 	+ (2 * t * (1-t) * (1-t) ) + ( (1-t)^2 * (1-t))
> 
> = t^3 + ( 2 * t^2 * (1-t) ) + (t * (1-t)^2) + (t^2 * (1-t)) 
> 	+ ( 2 * t * (1-t)^2) + (1-t)^3
> 
> = t^3 + (3 * t^2 * (1-t)) + (3 * t * (1-t)^2) + (1-t)^3
> 
> So the 4 terms in the last line correspond to the 4 blending functions.
> 
> Hope that helps,
> Jamie
> 
Not to criticize the way you calculate the result, starting from 
(t + (1-t) ) ^3, but in school we went from (a+b)^3 straight to
a^3 + 3*a^2*b + 3*a*b^2 + b^3.... :)

-- 
Regards,  Sander

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