[cairo] Pen vertex estimation solutions
Keith Packard
keithp at keithp.com
Sat Oct 9 23:16:47 PDT 2004
Thanks to Walter Brisken, we have a solution to computing the number of
vertices needed for our polygonal pen.
He sent a solution for computing the axes lengths for the pen as
transformed to device space (where it becomes an ellipse) along without a
proof; I've asked for a more complete description than 'magic occurs here'
which I'll stick into the code when I receive it.
I then proved that using our equal-angle vertex computation method, the
maximum error in the approximation occurs right at the major axis of the
ellipse (no big surprise there), and that the formula for the error at
that point matches the formula of the error for a circle with radius equal
to the major axis. That the error is independent of the minor axis length
is a bit surprising, but it means that we have one simple formula for the
number of vertices which is independent of whether the pen is an ellipse
or a circle. That's certainly a nice result.
Here's the comment that I've placed in the code.
-------------
The circular pen in user space is transformed into an ellipse in
device space.
We construct the pen by computing points along the circumference
using equally spaced angles.
We show below that this approximation to the ellipse has
maximum error at the major axis of the ellipse.
So, we need to compute the length of the major axis and then
use that to compute the number of sides needed in our pen.
Thanks to Walter Brisken <wbrisken at aoc.nrao.edu> for this
derivation:
1. First some notation:
All capital letters represent vectors in two dimensions. A prime '
represents a transformed coordinate. Matrices are written in underlined
form, ie _R_. Lowercase letters represent scalar real values.
The letter t is used to represent the greek letter theta.
2. The question has been posed: What is the maximum expansion factor
achieved by the linear transformation
X' = _R_ X
where _R_ is a real-valued 2x2 matrix with entries:
_R_ = [a b]
[c d] .
In other words, what is the maximum radius, MAX[ |X'| ], reached for any
X on the unit circle ( |X| = 1 ) ?
3. Some useful formulae
(A) through (C) below are standard double-angle formulae. (D) is a lesser
known result and is derived below:
(A) sin^2(t) = (1 - cos(2*t))/2
(B) cos^2(t) = (1 + cos(2*t))/2
(C) sin(t)*cos(t) = sin(2*t)/2
(D) MAX[a*cos(t) + b*sin(t)] = sqrt(a^2 + b^2)
Proof of (D):
find the maximum of the function by setting the derivative to zero:
-a*sin(t)+b*cos(t) = 0
>From this it follows that
tan(t) = b/a
and hence
sin(t) = b/sqrt(a^2 + b^2)
and
cos(t) = a/sqrt(a^2 + b^2)
Thus the maximum value is
MAX[a*cos(t) + b*sin(t)] = (a*a + b*b)/sqrt(a^2 + b^2)
= sqrt(a^2 + b^2)
4. Derivation of maximum expansion
To find MAX[ |X'| ] we search brute force method using calculus. The unit
circle on which X is constrained is to be parameterized by t:
X(t) = (cos(t), sin(t))
Thus
X'(t) = (a*cos(t) + b*sin(t), c*cos(t) + d*sin(t)) .
Define
r(t) = |X'(t)|
Thus
r^2(t) = (a*cos(t) + b*sin(t))^2 + (c*cos(t) + d*sin(t))^2
= (a^2 + c^2)*cos(t) + (b^2 + d^2)*sin(t)
+ 2*(a*b + c*d)*cos(t)*sin(t)
Now apply the double angle formulae (A) to (C) from above:
r^2(t) = (a^2 + b^2 + c^2 + d^2)/2
+ (a^2 - b^2 + c^2 - d^2)*cos(2*t)/2
+ (a*b + c*d)*sin(2*t)
= f + g*cos(u) + h*sin(u)
Where
f = (a^2 + b^2 + c^2 + d^2)/2
g = (a^2 - b^2 + c^2 - d^2)/2
h = (a*b + c*d)
u = 2*t
It is clear that MAX[ |X'| ] = sqrt(MAX[ r^2 ]). Here we determine MAX[ r^2 ]
using (D) from above:
MAX[ r^2 ] = f + sqrt(g^2 + h^2)
And finally
MAX[ |X'| ] = sqrt( f + sqrt(g^2 + h^2) )
Which is the solution to this problem.
Walter Brisken
2004/10/08
(Note that the minor axis length is at the minimum of the above solution,
which is just sqrt (f - sqrt (g^2 + h^2)) given the symmetry of (D)).
Now to compute how many sides to use for the pen formed by
a regular polygon.
Set
M = major axis length (computed by above formula)
m = minor axis length (computed by above formula)
Align 'M' along the X axis and 'm' along the Y axis and draw
an ellipse parameterized by angle 't':
x = M cos t y = m sin t
Perturb t by ± d and compute two new points (x+,y+), (x-,y-).
The distance from the average of these two points to (x,y) represents
the maximum error in approximating the ellipse with a polygon formed
from vertices 2∆ radians apart.
x+ = M cos (t+∆) y+ = m sin (t+∆)
x- = M cos (t-∆) y- = m sin (t-∆)
Now compute the approximation error, E:
Ex = (x - (x+ + x-) / 2)
Ex = (M cos(t) - (Mcos(t+∆) + Mcos(t-∆))/2)
= M (cos(t) - (cos(t)cos(∆) + sin(t)sin(∆) +
cos(t)cos(∆) - sin(t)sin(∆))/2)
= M(cos(t) - cos(t)cos(∆))
= M cos(t) (1 - cos(∆))
Ey = y - (y+ - y-) / 2
= m sin (t) - (m sin(t+∆) + m sin(t-∆)) / 2
= m (sin(t) - (sin(t)cos(∆) + cos(t)sin(∆) +
sin(t)cos(∆) - cos(t)sin(∆))/2)
= m (sin(t) - sin(t)cos(∆))
= m sin(t) (1 - cos(∆))
E² = Ex² + Ey²
= (M cos(t) (1 - cos (∆)))² + (m sin(t) (1-cos(∆)))²
= (1 - cos(∆))² (M² cos²(t) + m² sin²(t))
= (1 - cos(∆))² ((m² + M² - m²) cos² (t) + m² sin²(t))
= (1 - cos(∆))² (M² - m²) cos² (t) + (1 - cos(∆))² m²
Find the extremum by differentiation wrt t and setting that to zero
∂(E²)/∂(t) = (1-cos(d))² (M² - m²) (-2 cos(t) sin(t))
0 = 2 cos (t) sin (t)
0 = sin (2t)
t = nπ
Which is to say that the maximum and minimum errors occur on the
axes of the ellipse at 0 and π radians:
E²(0) = (1-cos(∆))² (M² - m²) + (1-cos(∆))² m²
= (1-cos(∆))² M²
E²(π) = (1-cos(∆))² m²
maximum error = M (1-cos(∆))
minimum error = m (1-cos(∆))
We must make maximum error ≤ tolerance, so compute the ∆ needed:
tolerance = M (1-cos(∆))
tolerance / M = 1 - cos (∆)
cos(∆) = 1 - tolerance/M
∆ = acos (1 - tolerance / M);
Remembering that ∆ is half of our angle between vertices,
the number of vertices is then
vertices = ceil(2π/2∆).
= ceil(π/∆).
Note that this also equation works for M == m (a circle) as it
doesn't matter where on the circle the error is computed.
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