Reply To: Trapezoidal ring

[Kirill]

Hello Rahul,

You’ll need to adjust your px and py coordinates. Check this example:

# Parameters
start_angle = 0.0;       # Start angle in degrees
end_angle = 45.0;        # End angle in degrees
radius = 10.0;           # Radius of the arc
m = 0.55191502449;       # "Magic number" for a 90° arc approximation

# Convert angles to radians
start_angle_rad = start_angle * pi / 180;
end_angle_rad = end_angle * pi / 180;

# Calculate arc angle
arc_angle = end_angle_rad - start_angle_rad;

# Adjust the "magic number" for the specified arc angle
m_adj = m * arc_angle / (pi / 2);

# Calculate control points for Bezier curve approximation
# Starting point (p0)
p0_x = cos(start_angle_rad) * radius;
p0_y = sin(start_angle_rad) * radius;

# First control point (p1)
p1_x = p0_x - sin(start_angle_rad) * m_adj * radius;
p1_y = p0_y + cos(start_angle_rad) * m_adj * radius;

# Ending point (p3)
p3_x = cos(end_angle_rad) * radius;
p3_y = sin(end_angle_rad) * radius;

# Second control point (p2)
p2_x = p3_x + sin(end_angle_rad) * m_adj * radius;
p2_y = p3_y - cos(end_angle_rad) * m_adj * radius;

# Create array of control points
px = [p0_x; p1_x; p2_x; p3_x];
py = [p0_y; p1_y; p2_y; p3_y];
p = [px, py];

# Output
?("Control points:");
?num2str(p, "%.3f");

So for:

px = [10.000; 10.000; 9.022; 7.071];
py = [ 0.000;  2.760; 5.120; 7.071];

you should get an approximation of a 45° arc.

I also suggest reviewing the following resources:

1. Approximation of a cubic Bézier curve by circular arcs and vice versa – This provides a solid mathematical discussion on the topic.
2. Approximate a circle with cubic Bézier curves – This source presents a clear discussion on approximation accuracy.
3. Bézier curves - Desmos calculator – A nice interactive tool to speed up your understanding of Bézier curves.

I hope you find this helpful.

Best regards,
Kirill