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Author Topic: Another idea for math to give the "true 3D" Mandy  (Read 1297 times)
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David Makin
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« on: December 01, 2012, 02:22:22 PM »

Hi all,

Am just trying a new theory - not finished implementing the test version yet but then again my programming is somewhat better than my math so it's taking a while for me to be sure I've got the math correct, however I know there are some more acute math brains out there so I though I'd just mention where I'm going - to see if someone else gets there first to give my brain a rest !!

First of all I wondered if we're missing a trick in that to go beyond 2 dimensions we need an equivalent to trigonometry but in 3D.

Now for a triangle Pythagoras' theorem relates the squares of the sides to the hypoteneuse and I figured that in 3D the equivalent should actually be relating the squares of the *areas* of the 3 sides of a right-tetrahedron to the square of the area of the base and found a remarkable thing which I then discovered (via Wiki) is already known as De Gua's theorem (see http://en.wikipedia.org/wiki/Trirectangular_tetrahedron ) i.e. that this is exactly the relationship of the areas of the sides and base.

Specifically (0.5*x*y)^2 + (0.5*x*z)^2 + (0.5*y*z)^2 = area of base^2

Now this gives us 3 ratios equivalent to sin and cos in 2D - the problem is that they still don't fit nicely in a rotation matrix and anyway it's been proved that there's no true "field" in 3D - but of course go to 4D and you can get a nice matrix.

Now (also from wiki) the *volume* of a right-tetrahedron of sides from origin (x,y,z) is x*y*z/6 so if we go to the 4D equivalent of the tetrahedron I assume that:

(x*y*z/6)^2 + (x*y*w/6)^2 + (x*z*w/6)^2 + (y*z*w)^2 = area (edit:read volume!) of "base" tetrahedron of 4D shape^2

Which given a 4D number (x,y,z,w) led me to the following:

a = x*z*w
b = y*z*w
c = z*x*y
d = w*x*y

With a rotation matrix:

 
Code:
 |  a   b   c   d |
  | -b   a   d  -c |
  | -c   d   a  -b |
  |  d   c   b   a |

Also one can perfectly convert back from volume form to vector form:

a = x*z*w
b = y*z*w
c = z*x*y
d = w*x*y

So:

x = cubert(a*c*d/b^2)
y = cubert(b*c*d/a^2)
z = cubert(a*b*c/d^2)
w = cubert(a*b*d/c^2)

There are complications of course when one or more of x,y,z,w is zero - but we just need to consider the same rotation matrix but wth appropriate rows/columns removed e.g. if z and w are 0 we get standard complex - same if y and w are 0 but obviously in the x/z rather than the x/y.

« Last Edit: December 03, 2012, 02:26:54 AM by David Makin » Logged

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David Makin
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« Reply #1 on: December 01, 2012, 08:03:51 PM »

Interesting point is that for 1 dimension we could say that:

x^2/1 = r^2/1 (where r is the only dimension anyway)

For 2:

x^2/1 + y^2/1 = r^2/1 (where r is the hypoteneuse)

For 3:

(xy/2)^2 + (xz/2)^2 + (yz/2)^2 = r^2 (where r is the base area)

For 4 (possibly):

(xyz/6)^2 + (xyw/6)^2 + (xzw/6)^2 + (yzw/6)^2 = r^2 (where r is the "base" volume)

So for n dimensions (possibly):

Sum of all (a0*a1*...*an-2*an-1 /n!)^2 with one element a0 to an-1 missing in each term = r^2

Gives a solution for fractional dimensions ? Maybe even points to a purely algebraic way to find n! (or gamma(n)) for real n.
But I'm not really a mathematician wink
« Last Edit: December 01, 2012, 08:06:23 PM by David Makin » Logged

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M Benesi
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« Reply #2 on: December 03, 2012, 06:38:59 PM »

Now for a triangle Pythagoras' theorem relates the squares of the sides to the hypoteneuse and I figured that in 3D the equivalent should actually be relating the squares of the *areas* of the 3 sides of a right-tetrahedron to the square of the area of the base and found a remarkable thing which I then discovered (via Wiki) is already known as De Gua's theorem (see http://en.wikipedia.org/wiki/Trirectangular_tetrahedron ) i.e. that this is exactly the relationship of the areas of the sides and base.

Specifically (0.5*x*y)^2 + (0.5*x*z)^2 + (0.5*y*z)^2 = area of base^2
 Interesting idea.

  Imagine how they arrived at that (De Gua's theorem) by looking at the image on the wikipedia page.  Should be able to pick out all of the sin and cosine relationships between the 3 variables.  I'll look at it more later, want coffee now.


http://en.wikipedia.org/wiki/Simplex#Simplexes_with_an_.22orthogonal_corner.22
« Last Edit: December 03, 2012, 10:45:00 PM by M Benesi » Logged

David Makin
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« Reply #3 on: December 04, 2012, 10:32:55 PM »

smiley Let me know if you have a play with the idea - the problem I'm having is what to do in the case where only one of the 4 dimensions is zero - here we would go back to using 3D "trig" but there isn't one as far as a nice matrix is concerned, so I'm having trouble deciding exactly what to do in that situation wink

Also the base 4*4 to use isn't cut and dried, if the 4 volume based trig values are a.b.c.d then I've been using:

Code:
|  a   b   c   d |
| -b   a   d  -c |
| -c   d   a  -b |
|  d   c   b   a |

But since IMO our definite base is only:

Code:
|    a   b   c   d |
|   -b   a         |
| +/-c             |
| +/-d             |

There are considerable other possibilities !!
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The meaning and purpose of life is to give life purpose and meaning.

http://www.fractalgallery.co.uk/
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M Benesi
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« Reply #4 on: December 06, 2012, 08:19:05 AM »

  If you check a little above where my link takes you, you will see that it confirms what you stated in your second post.

   I'm thinking tangentially to you (more about the trig):  it's possible to calculate the actual sin and cosine combinations if we want to use them, but I really like the idea of defining 3 dimensional volumetric sine/ cosine/ ???ine.  I don't know how to orient them however.

  Which one is \frac{.5\times x\times y} {\sqrt{0.5 \times x \times y + 0.5 \times x \times z + 0.5 \times y \times z}}?

  Is there a taylor series expansion that gives us the 3 ratios (.5xy/... , .5xz/....  and .5yz/...)?  This would allow us to double the "angle", and square  (or tetrate??) the area.  Probably have to follow your 4d method, unless a combined exponential function method works...

  
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fivex
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« Reply #5 on: September 17, 2013, 11:52:00 AM »

Hm.
On the image in the wiki article for De Gua's theorem, Aabo could be defined as the primary adjacent, Aaco could be defined as the secondary adjacent, Abco as the opposite, and Aabc being the hypotenuse or base.
The 3d trigonometric functions would need to take two angles: the angle between Aab and Aao(θ) and the angle between Aac and Aao(ϕ?)
Cosine, tangent, cotangent, and secant  would operate using the primary adjacent. There would also be a secondary cosine (scos), secondary tangent (stan), secondary cotangent (scot), and secondary secant (ssec) which would operate on the secondary adjacent instead of the primary. Thus cos would be:
\frac{primary\ adjacent} {base}
And scos would be:
\frac{secondary\ adjacent} {base}
There would also be a function that is:
\frac{primary\ adjacent} {secondary\ adjacent}
Which I'm going to call the adversant (adv)
And another that is:
\frac{secondary\ adjacent} {primary\ adjacent}
Which I'm going to call the coadversant (cad)

I'm going to use sin2, cos2, etc. to refer to the 2d trigonometric function and sin3, cos3, etc. to refer to the 3d ones.
You can construct the result of the 3d functions easily using the 2d ones. To do so for (12░,16░):
Let's say that the length of Aao is 1. Since Aao is the adjacent of Aabo, Abo is simply cot212░ and Aab is sec212░. The area of Aabo (the primary adjacent) is thus \frac{1}{2}A_{ao}A_{bo}=2.3523150547392273...

Likewise, Aco is cot216░ and Aac is sec216░. The area of Aaco (the secondary adjacent) is thus \frac{1}{2}A_{ao}A_{co}=1.7437072219204546...

Since we know the two legs of Abco (the opposite), it's area is \frac{1}{2}A_{bo}A_{co}=8.203497498362001..

And thus by De Gua's theorem, the area of Aabc (the base) is \sqrt{A_{abo}^2 + A_{aco}^2 + A_{bco}^2} = 8.710411712322328...
From here, we can calculate the functions:
sin3(12░,16░) = 0.941803644798648...
cos3(12░,16░) = 0.2700578494368396...
scos3(12░,16░) = 0.2001865445066954...
tan3(12░,16░) = 3.4874144438409096...
stan3(12░,16░) = 4.7046301094784555
cot3(12░,16░) = 0.28674538575880787
scot3(12░,16░)=0.2125565616700221
sec3(12░,16░)=3.7029103286030476...
ssec3(12░,16░)=4.995340733135809...
csc3(12░,16░)=1.061792450605555...
adv3(12░,16░)=1.3490309755949021...
cad3(12░,16░)=0.741272823301196...
It should be possible to easily create a 4d trigonometry using a similar method.
Now to figure out atan3(and/or astan3?) and it should be possible to create a 3d complex plane analog.
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fivex
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« Reply #6 on: September 28, 2013, 04:37:59 AM »

tan3(θ, ϕ) = cot2(ϕ)
Thus:
atan3(x) = acot2(x)
And since
stan3(θ, ϕ) = tan3(ϕ,θ)
astan3 is the same.
From there, it's trivial to correct create a atan2 equivalent.

To convert (re, im, jm) into polar coordinates:
r = \sqrt{re^2 + im^2 + jm^2}
θ = atan3(im, re)
ϕ = atan3(im, jm)
To convert back is simply:
re = r*cos3(θ,ϕ)
im = r*sin3(θ,ϕ)
jm = r*scos3(θ,ϕ)
Which means that j is a second real axis?
Also, now that I think about it, that doesn't work at all if im = 0, because then θ = ϕ and thus re = jm
Edit: Actually, re and jm are complex axis? Because cos(0,0) = 0, scos(0,0) = 0 and sin(0,0) = 1, which is the opposite of 2d trig.
Edit: atan2 isn't nearly as simple as I thought.
« Last Edit: September 28, 2013, 06:04:50 AM by fivex » Logged
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