Pooper200000 at 15:15, 20 February 2010

2010-02-20T15:15:19Z

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			[[Category:Scripting Reference]]
	== Definitions ==		== Definitions ==
	''θ'': greek letter ''theta''<br />		''θ'': greek letter ''theta''<br />

Stefan: Reverted edits by Anti-Up (Talk); changed back to last version by Tolnaftate2004

2007-07-01T13:19:29Z

Reverted edits by Anti-Up (Talk); changed back to last version by Tolnaftate2004

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Anti-Up at 02:31, 1 July 2007

2007-07-01T02:31:40Z

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Tolnaftate2004: This is Polar Rotation. Euler Rotation is a bit more basic. My mistake. I will fix it soon.

2007-04-21T22:59:27Z

This is Polar Rotation. Euler Rotation is a bit more basic. My mistake. I will fix it soon.

New page

== Definitions ==
''θ'': greek letter ''theta'' 
''φ'': greek letter ''phi'' 
''theta'' and ''phi'' are commonly used in trigonometry for angles (but phi is most commonly used as a constant: 1.618...). 
''α'': 1st greek letter ''alpha''. For our purposes, it will represent rotation about the y-axis. 
''β'': 2nd greek letter ''beta''. For our purposes, it will represent rotation about the x-axis. 
''γ'': 3rd greek letter ''gamma''. For our purposes, it will represent rotation about the z-axis. 

sin(''θ'') = y / ''r'' (Δy of a point from the center on a circle with radius ''r'')
cos(''θ'') = x / ''r'' (Δx of a point from the center on a circle with radius ''r'')
sin(''θ'' ± ''φ'') = sin(''θ'') cos(''φ'') ± cos(''θ'') sin(''φ'')
-cos(''θ'' ± ''φ'') = -cos(''θ'') cos(''φ'') ± sin(''θ'') sin(''φ'')

In addition (not really useful):
tan(''θ'') = sin(''θ'') / cos(''θ'')
cot(''θ'') = cos(''θ'') / sin(''θ'')
sec(''θ'') = 1 / cos(''θ'')
csc(''θ'') = 1 / sin(''θ'')

== Deriving the 2D equations ==

Simply: 
x² + y² = ''r''²; the basic equation of a circle with center (0,0)
x² + y² + z² = ''r''²; the basic equation of a sphere with center (0,0,0)
sin²(''θ'') + cos²(''θ'') = 1

arctan(y/x) = ''θ''; arctan() being the inverse of tangent
tan(arctan(y/x)) = tan(''θ'')
y/x = tan(''θ'')
y/x = ''r'' sin(''θ'') / ''r'' cos(''θ'')

Thus, 
''r'' sin(''θ'') = y
''r'' cos(''θ'') = x

Likewise,
sin(''θ'') = y / ''r''
cos(''θ'') = x / ''r''

Plugging into the circle equation, we get
''r''² = ''r''² cos²(''θ'') + ''r''² sin²(''θ'')

Compare:
√(a² + b²) = c (the distance formula)
x² + y² = ''r''² (established as the equation of a circle)

√(a² + b²) = c
( √(a² + b²) )² = c²
a² + b² = c²

== Rotation on a single plane ==

Think of a circle with some radius, ''r'', and a point somewhere on it. This circle is a two-dimensional figure on the xy [or y(x)] plane. The angle between 0º and the point is ''θ''. 
Trigonometric functions sine and cosine are specific to determine x and y values according the the angle passed as its parameter. 

To keep things simple, we will rotate about the (invisible) z-axis. In doing so, x and y are changing. Any z values that a point may have remain constant. ''φ'' is a second angle, representing the amount of rotation. 

y = ''r'' sin(''θ''), so,
y = ''r'' sin(''θ'' + ''φ'')
y = ''r'' ( sin(''θ'') cos(''φ'') + cos(''θ'') sin(''φ'') )
y = ''r'' sin(''θ'') cos(''φ'') + ''r'' cos(''θ'') sin(''φ'')
Using substitutions form above, we can simplify to 
y = y cos(''φ'') + x sin(''φ'')

Alternatively, 
x = ''r'' cos(''θ'')
x = ''r'' cos(''θ'' + ''φ'')
x = ''r'' ( cos(''θ'') cos(''φ'') - sin(''θ'') sin(''φ'') )
x = ''r'' cos(''θ'') cos(''φ'') - ''r'' sin(''θ'') sin(''φ'')
x = x cos(''φ'') - y sin(''φ'')

== Deriving the 3D equations ==
[[Image:Pointonsphere.png|thumb|right|This image shows how to determine the cartesian coordinates of a point on a sphere in terms of sine and cosine. The point is at the upper tip of the green triangle.]]
As above, we can change the sphere's equation into a series of distance formulas
x² + y² + z² = ''r''²
( √(x² + y²) )² + z² = ''r''²
( √( ( √(x² + y²) )² + z² ) )² = ''r''²
√( x² + y² + z² ) = ''r''
''r''√( ( sin²(''θ'') + cos²(''θ'') ) sin²(''φ'') + cos²(''φ'') ) = ''r''
√( ( √( sin²(''θ'') + cos²(''θ'') ) )² sin²(''φ'') + cos²(''φ'') ) = 1
( sin²(''θ'') + cos²(''θ'') )² sin²(''φ'') + cos²(''φ'') = 1
sin²(''θ'') sin²(''φ'') + cos²(''θ'') sin²(''φ'') + cos²(''φ'') = 1

Comparing with the original equation we can split it up and get individual equations for x,y, and z.
y = sin(''θ'') sin(''φ'') (1)
x = cos(''θ'') sin(''φ'') (1)
z = cos(''φ'')

The equation is
( sin²(''θ'') + cos²(''θ'') ) ( sin²(''φ'') + cos²(''φ'') ) = 1

So, why doesn't the ( sin²(''θ'') + cos²(''θ'') ) distribute to the cos²(''φ'')? If you'll recall, ( sin²(''θ'') + cos²(''θ'') ) = 1, so it ''could be'' that it was. But also, it was left out so that there were only 3 terms as opposed to 4 (We only want three: x, y, and z!).

<hr width="33%" NOSHADE>
(1) The portion used contained the original x or y equation, so was split up to match, but it doesn't matter in the end.

== Rotation on three planes ==

Now in three dimensions: 
Imagine three planes that intersect: xy [y(x)], yz [z(y)], zx [x(z)] (using these names helps remember the order of calculations if re-writing the script... you'll see). 
For this, instead of merely one rotation angle, we will need three (''α'', ''β'', and ''γ''). Also, we will need to replace the simple x and y with their three-dimensional counterparts and add a z (To begin, ''f'' = ''f''0, where ''f'' is a function, x, y, or z). 
x0 = ''r'' sin(''α'') cos(''β'')
y0 = ''r'' sin(''α'') sin(''β'')
z0 = ''r'' cos(''α'')

About the x-axis: 
Finally, rotate our three planes to replace z with x, y with z, and x with y. 
nz = z cos(''β'') + y sin(''β'')
ny = y cos(''β'') - z sin(''β'')
Note how as the axes are switched, the x, y, and z or replaced respectively from the 2D rotation equations!

We then replace y with ny and z with nz. So in a sense, 
z = nz
y = ny

About the y-axis: 
Now imagine the 3 planes rotate forward 90º and 90º to the left. The z has taken the place of the x, x in the stead of y, and y in that of z. The idea of this rotation is now just like that above, but y is constant. 
nx = x cos(''α'') + z sin(''α'')
nz = z cos(''α'') - x sin(''α'')

x = nx

The two-dimensions calculations work for rotations about the z-axis. 
About the z-axis: 
ny = y cos(''γ'') + x sin(''γ'')
nx = x cos(''γ'') - y sin(''γ'')

We expand to simplify: 

x = x cos(''γ'') - y sin(''γ'')
x = ( x0 cos(''α'') + ( z0 cos(''β'') + y0 sin(''β'') ) sin(''α'') ) cos(''γ'') - ( y0 cos(''β'') - z0 sin(''β'') ) sin(''γ'')
x = x0 cos(''α'') cos(''γ'') + z0 cos(''β'') sin(''α'') cos(''γ'') + y0 sin(''β'') sin(''α'') cos(''γ'') - y0 cos(''β'') sin(''γ'') + z0 sin(''β'') sin(''γ'')
x = x0 cos(''α'') cos(''γ'') + y0 ( sin(''β'') sin(''α'') cos(''γ'') - cos(''β'') sin(''γ'') ) + z0 ( cos(''β'') sin(''α'') cos(''γ'') + sin(''β'') sin(''γ'') )

y = y cos(''γ'') + x sin(''γ'')
y = ( y0 cos(''β'') - z0 sin(''β'') ) cos(''γ'') + ( x0 cos(''α'') + ( z0 cos(''β'') + y0 sin(''β'') ) sin(''α'') ) sin(''γ'')
y = y0 cos(''β'') cos(''γ'') - z0 sin(''β'') cos(''γ'') + x0 cos(''α'') sin(''γ'') + z0 cos(''β'') sin(''α'') sin(''γ'') + y0 sin(''β'') sin(''α'') sin(''γ'')
y = x0 cos(''α'') sin(''γ'') + y0 ( cos(''β'') cos(''γ'') + sin(''β'') sin(''α'') sin(''γ'') ) + z0 ( -sin(''β'') cos(''γ'') + cos(''β'') sin(''α'') sin(''γ'') )

z = z cos(''α'') - x sin(''α'')
z = ( z0 cos(''β'') + y0 sin(''β'') ) cos(''α'') - x0 sin(''α'')
z = z0 cos(''β'') cos(''α'') + y0 sin(''β'') cos(''α'') - x0 sin(''α'')

As is, the object will display in an orthographic projection, which means that z has no bearing on the perspective. To give a perspective projection, the x and y can be distorted by some equation like: 
x = centerx + h / (h-z)
y = centery - h / (h-z)

== Links ==
[http://forums.graalonline.com/forums/showthread.php?t=54167 Amon-ra's explanation and models (continues on p. 2)]

Polar Rotation - Revision history

Pooper200000 at 15:15, 20 February 2010

Stefan: Reverted edits by Anti-Up (Talk); changed back to last version by Tolnaftate2004

Anti-Up at 02:31, 1 July 2007

Tolnaftate2004: This is Polar Rotation. Euler Rotation is a bit more basic. My mistake. I will fix it soon.