Euler Rotation

Definitions

Euler - Pronounced Oiler; discovered that as 1 is continuously compounded, it reaches a number, e. However, Euler rotation is taking a 3-dimensional object and rotating the axes on which it lies, in effect rotating the object as well. For objects that inherit the property 'eulerrotation', the format is supposedly "Δβ Δα Δγ". Read further for a better explanation.

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: 1^st greek letter alpha. For our purposes, it will represent rotation about the y-axis.
β: 2^nd greek letter beta. For our purposes, it will represent rotation about the x-axis.
γ: 3^rd greek letter gamma. For our purposes, it will represent rotation about the z-axis.

Therefore, images and text could be rotated, something that could not be done before, so that said image or text would have a three-dimensional effect. If all said is true, eulerrotation = "-pi/3 0 0"; for text would project it in a Star Wars-like fashion.

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 = r²; the basic equation of a circle with center (0,0)
x² + y² + z² = r²; 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

r² = r² cos²(θ) + r² sin²(θ)

Compare:

√(a² + b²) = c (the distance formula)
x² + y² = r² (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

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 = r²
( √(x² + y²) )² + z² = r²	 
( √( ( √(x² + y²) )² + z² ) )² = r²
√( 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!).

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⁽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 = f₀, 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

Amon-ra's explanation and models (continues on p. 2)