mirror of
https://github.com/ppy/osu.git
synced 2024-11-06 09:07:25 +08:00
100 lines
3.7 KiB
C#
100 lines
3.7 KiB
C#
// Copyright (c) 2007-2018 ppy Pty Ltd <contact@ppy.sh>.
|
|
// Licensed under the MIT Licence - https://raw.githubusercontent.com/ppy/osu/master/LICENCE
|
|
|
|
using System;
|
|
using System.Collections.Generic;
|
|
using osu.Framework.MathUtils;
|
|
using OpenTK;
|
|
|
|
namespace osu.Game.Rulesets.Objects
|
|
{
|
|
public class CircularArcApproximator
|
|
{
|
|
private readonly Vector2 a;
|
|
private readonly Vector2 b;
|
|
private readonly Vector2 c;
|
|
|
|
private int amountPoints;
|
|
|
|
private const float tolerance = 0.1f;
|
|
|
|
public CircularArcApproximator(Vector2 a, Vector2 b, Vector2 c)
|
|
{
|
|
this.a = a;
|
|
this.b = b;
|
|
this.c = c;
|
|
}
|
|
|
|
/// <summary>
|
|
/// Creates a piecewise-linear approximation of a circular arc curve.
|
|
/// </summary>
|
|
/// <returns>A list of vectors representing the piecewise-linear approximation.</returns>
|
|
public List<Vector2> CreateArc()
|
|
{
|
|
float aSq = (b - c).LengthSquared;
|
|
float bSq = (a - c).LengthSquared;
|
|
float cSq = (a - b).LengthSquared;
|
|
|
|
// If we have a degenerate triangle where a side-length is almost zero, then give up and fall
|
|
// back to a more numerically stable method.
|
|
if (Precision.AlmostEquals(aSq, 0) || Precision.AlmostEquals(bSq, 0) || Precision.AlmostEquals(cSq, 0))
|
|
return new List<Vector2>();
|
|
|
|
float s = aSq * (bSq + cSq - aSq);
|
|
float t = bSq * (aSq + cSq - bSq);
|
|
float u = cSq * (aSq + bSq - cSq);
|
|
|
|
float sum = s + t + u;
|
|
|
|
// If we have a degenerate triangle with an almost-zero size, then give up and fall
|
|
// back to a more numerically stable method.
|
|
if (Precision.AlmostEquals(sum, 0))
|
|
return new List<Vector2>();
|
|
|
|
Vector2 centre = (s * a + t * b + u * c) / sum;
|
|
Vector2 dA = a - centre;
|
|
Vector2 dC = c - centre;
|
|
|
|
float r = dA.Length;
|
|
|
|
double thetaStart = Math.Atan2(dA.Y, dA.X);
|
|
double thetaEnd = Math.Atan2(dC.Y, dC.X);
|
|
|
|
while (thetaEnd < thetaStart)
|
|
thetaEnd += 2 * Math.PI;
|
|
|
|
double dir = 1;
|
|
double thetaRange = thetaEnd - thetaStart;
|
|
|
|
// Decide in which direction to draw the circle, depending on which side of
|
|
// AC B lies.
|
|
Vector2 orthoAtoC = c - a;
|
|
orthoAtoC = new Vector2(orthoAtoC.Y, -orthoAtoC.X);
|
|
if (Vector2.Dot(orthoAtoC, b - a) < 0)
|
|
{
|
|
dir = -dir;
|
|
thetaRange = 2 * Math.PI - thetaRange;
|
|
}
|
|
|
|
// We select the amount of points for the approximation by requiring the discrete curvature
|
|
// to be smaller than the provided tolerance. The exact angle required to meet the tolerance
|
|
// is: 2 * Math.Acos(1 - TOLERANCE / r)
|
|
// The special case is required for extremely short sliders where the radius is smaller than
|
|
// the tolerance. This is a pathological rather than a realistic case.
|
|
amountPoints = 2 * r <= tolerance ? 2 : Math.Max(2, (int)Math.Ceiling(thetaRange / (2 * Math.Acos(1 - tolerance / r))));
|
|
|
|
List<Vector2> output = new List<Vector2>(amountPoints);
|
|
|
|
for (int i = 0; i < amountPoints; ++i)
|
|
{
|
|
double fract = (double)i / (amountPoints - 1);
|
|
double theta = thetaStart + dir * fract * thetaRange;
|
|
Vector2 o = new Vector2((float)Math.Cos(theta), (float)Math.Sin(theta)) * r;
|
|
output.Add(centre + o);
|
|
}
|
|
|
|
return output;
|
|
}
|
|
}
|
|
}
|