On Tue, 28 Nov 2000 23:54:45 -0000, "Alf Peake"
<alf### [at] peake42freeservecouk> wrote:

>I don't know how to do a smooth colour change with s_sweeps, I don't
>see any way to apply a pigment to each vector.

In this particular case you might get away with a radial pigment, I
think.


Peter Popov ICQ : 15002700
Personal e-mail : pet### [at] vipbg
TAG      e-mail : pet### [at] tagpovrayorg

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Tor Olav Kristensen wrote:

> > Thanks Tor. Glass takes too long though :-\
> > BTW, wish I had a maths brain like yours.
>
> :)
>
> It seems like you too know some math.

Speaking of which, I just figured out that to use complex bases and
exponents you can convert the rectangular vector (a+bi) to circular
(ai^b) and then you only have one term and it becomes (somewhat) doable!
Mind you that I have never had this sort of material in school, I
conjecture it playing with my TI-89. :) So now I know that n^i =
cos(ln(n))+i*sin(ln(n)), but I have no idea why. :)

--
David Fontaine  <dav### [at] faricynet>  ICQ 55354965
My raytracing gallery:  http://davidf.faricy.net/

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Sorry David

- I didn't see your reply before now.


David Fontaine wrote:
> 
> ...
> Speaking of which, I just figured out that to use complex bases and
> exponents you can convert the rectangular vector (a+bi) to circular
> (ai^b) 

Hmmm... I think it would be more correct to write it like this:

a + b*i  =  radius*e^(angle*i)

(The latter is called polar form)


> and then you only have one term and it becomes (somewhat) doable!

What becomes doable ?


> Mind you that I have never had this sort of material in school, I
> conjecture it playing with my TI-89. :) So now I know that n^i =
> cos(ln(n))+i*sin(ln(n)), but I have no idea why. :)

Does this mean that also the TI calculators are 
capable of doing symbolic calculations now ?


Below is my try at an explanation of why you get 
these results.

If you know Euler's formula:

e^(theta*i) = cos(theta) + i*sin(theta)


And that n can be written as e^ln(n)
and also that (a^b)^c = a^(b*c)

then you'll see that

n^i = (e^ln(n))^i = e^(ln(n)*i) = cos(ln(n)) + i*sin(ln(n))


Euler's formula shows one interesting relation
between these 3 constants: 1, e and i

-e^(pi*i) = 1

(try to feed it to your TI-89 or to a HP-48)


If you want to know more about complex numbers, 
then here's a link to a page about the basics:

http://www.math.fsc.qut.edu.au/~gustafso/mab112/week4/

Look at the bottom of the page for info about the 
exponential notation of complex numbers.


Also note the 9. property of the exponential form 
of a complex number:

That e^(theta*i) is periodic of period 2*pi.


The 10. properties are also noteworthy and often used:

sin(theta) = (e^(theta*i) - e^(-theta*i))/2/i

cos(theta) = (e^(theta*i) + e^(-theta*i))/2



Best regards,

Tor Olav
-- 
mailto:tor### [at] hotmailcom
http://www.crosswinds.net/~tok/tokrays.html

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Poor IB math program, just when it thought it was catching up to me... :)

--
David Fontaine  <dav### [at] faricynet>  ICQ 55354965
My raytracing gallery:  http://davidf.faricy.net/

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