BobDude

Test Your Math Skills! Do everyones favorite thing! Math! Instructions-Factor By Grouping (note- ^ means exponent)
x^3 - y^3 + z^3 - x^2y - x^2z + xy^2 - xz^2 - y^2z + yz^2

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jpmarth

Oh Frenchy!!! It's a numbers questions!!

Jukov

the answer is E=mc^2

MrKlaatu

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Jukov

To specify a quadric surface, use quadric {}:

quadric {
< A, B, C >,
< D, E, F >,
< G, H, I >,
J
// pigment and finish
}
where these ten values are coefficients of a quadric polynomial as follows:
Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0
For example, to trace the following quadric surface
10x2 + y2 + 2z2 + 3xy + 4xz + yz + 5x + y + 10z = 0
use the following

quadric {
< 10, 1, 2 >,
< 3, 4, 1 >,
< 5, 1, 10 >,
0
// pigment, finish and transformations
}
Note that in general geometric transformations are required to bring the surface into your camera's coverage. Also note that the above equation is different from what has been discussed in class. The result a flat ellipsoid as shown below:





Cubic Surfaces
POVRAY uses the following cubic equation for tracing a cubic surface:
A1x3 + A2x2y + A3x2z + A4x2 + A5xy2 + A6xyz + A7xy + A8xz2 + A9xz + A10x + A11y3 + A12y2z + A13y2 + A14yz2 + A15yz + A16y + A17z3 + A18z2 + A19z + A20 = 0
There are 20 coefficients organized as follows:

cubic {
< A1, A2, A3, ...., A20 >
sturm // optional
// pigment, finish and transformations
}
All 20 coefficients are in a single vector. Note that a new keyword sturm is introduced. This keyword is used only when the degree of an algebraic surface is greater than 2. The purpose of using this keyword is for more accurate ray and object intersection. As a result, it is more time consuming than without using it. Charles Sturm was a well-known German mathematician of the last century who discovered a method for determining the ranges of the roots of a single-variable polynomial.

To trace the following cubic surface
x3 - 0.11111xz2 + y2 = 0
use the following:

cubic {
// x3 x2y x2z x2 xy2
< 1, 0, 0, 0, 0,
// xyz xy xz2 xz x
0, 0, -0.11111, 0, 0,
// y3 y2z y2 yz2 yz
0, 0, 1, 0, 0,
// y z3 z2 z Const
0, 0, 0, 0, 0 >
sturm
clipped_by {
box { < -2, -2, -2 >, < 2, 2, 2 > }
}
}

Because cubic surfaces extend to infinity, frequently a clipped_by {} is used to cut off unwanted part. It may require several tries to get the result right. Here is the result:




Quartic Surfaces
Quartic surfaces require 35 coefficients as shown below:
A1x4 + A2x3y + A3x3z + A4x3 + A5x2y2 + A6x2yz + A7x2y + A8x2z2 + A9x2z + A10x2 + A11xy3 + A12xy2z + A13xy2 + A14xyz2 + A15xyz + A16xy + A17xz3 + A18xz2 + A19xz + A20x + A21y4 + A22y3z + A23y3 + A24y2z2 + A25y2z + A26y2 + A27yz3 + A28yz2 + A29yz + A30y + A31z4 + A32z3 + A33z2 + A34z + A35 = 0

These coefficients are organized as follows:

quartic {
< A1, A2, ...., A35 >
sturm // optional
// pigment, finish and transformations
}
To trace the following quartic surface
x4 + x2y2 - 3x2y + 2.5x2z2 + 1.5y2z2 - 3yz2 + 1.5z4 = 0
use the following

quartic {
// x^4 x^3y x^3z x^3 x^2y^2
< 1, 0, 0, 0, 1,
// x^2yz x^2y x^2z^2 x^2z x^2
0, -3, 2.5, 0, 0,
// xy^3 xy^2z xy^2 xyz^2 xyz
0, 0, 0, 0, 0,
// xy xz^3 xz^2 xz x
0, 0, 0, 0, 0,
// y^4 y^3z y^3 y^2z^2 y^2z
0, 0, 0, 1.5, 0,
// y^2 yz^3 yz^2 yz y
0, 0, -3, 0, 0
// z^4 z^3 z^2 z Const
1.5, 0, 0, 0, 0 >
sturm
// pigment, finish and transformations
}

The result is the following well-known cross-cup surface

MrKlaatu

Originally Posted by Jukov

To specify a quadric surface, use quadric {}: quadric { < A.....blaaah....blaaah, blaaaaah-b'blah.....-cup surface

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jpmarth

Wow, I think Jukov broke MrK

Jukov

Originally Posted by jpmarth

Wow, I think Jukov broke MrK

JonMisurda

Well, I'm no math major but...

rewriting it like so we get:

x^3 - x^2y - x^2z - y^3 + xy^2 - y^2z + z^3 - xz^2 + yz^2

then we can factor out x^2, y^2, and z^2:

x^2(x - y - z) + y^2(x - y - z) + z^2(z - x + y)

Then we could do:

(x^2 + y^2)(x - y - z) + z^2 (z - x + y)

We could play more with some of the signs, but I'm pretty sure it doesn't reduce any futher as you've written it. If you didn't get all the signs right, it could be:

(x^2 + y^2 + z^2) (x-y-z) or something.

Weird problem. Easy by hand, but Mathematica was zero help.

Jon

JonMisurda

oh and next time do your own homework ;)

CRXGSR

huH? The best I got was (x-y-z)(x²+y²-z²).

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MrKlaatu

Originally Posted by JonMisurda

Well, I'm no math major but... rewriting it like so we get: x^3 - x^2y - x^2z - y^3 + xy^2 - y^2z + z^3 - xz^2 + yz^2 then we can factor out x^2, y^2, and z^2: x^2(x - y - z) + y^2(x - y - z) + z^2(z - x + y) Then we could do: (x^2 + y^2)(x - y - z) + z^2 (z - x + y) We could play more with some of the signs, but I'm pretty sure it doesn't reduce any futher as you've written it. If you didn't get all the signs right, it could be:

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JonMisurda

Originally Posted by CRXGSR

huH? The best I got was (x-y-z)(x²+y²-z²).

Hmm, let me double check my signs... I'm going crosseyed myself like the girl in Mr. K's picture...

Jon

MrKlaatu

No periods, no spaces.



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JonMisurda

Hmm, ok I screwed something up I guess, I'll agree with (x - y - z)(x^2 + y^2 - z^2)

Sorry M R . K .

:hide:

Jon