本文整理汇总了C#中System.Vector2.LSProjectOn方法的典型用法代码示例。如果您正苦于以下问题：C# Vector2.LSProjectOn方法的具体用法？C# Vector2.LSProjectOn怎么用？C# Vector2.LSProjectOn使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类System.Vector2的用法示例。
在下文中一共展示了Vector2.LSProjectOn方法的3个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: GetClosestOutsidePoint
public static Vector2 GetClosestOutsidePoint(Vector2 from, List<Geometry.Polygon> polygons)
{
var result = new List<Vector2>();
foreach (var poly in polygons)
{
for (var i = 0; i <= poly.Points.Count - 1; i++)
{
var sideStart = poly.Points[i];
var sideEnd = poly.Points[(i == poly.Points.Count - 1) ? 0 : i + 1];
result.Add(from.LSProjectOn(sideStart, sideEnd).SegmentPoint);
}
}
return result.MinOrDefault(vector2 => Vector2.Distance(from));
}示例2: GetSpellProjection
public static Vector2 GetSpellProjection(this Spell spell, Vector2 pos, bool predictPos = false)
{
if (spell.spellType == SpellType.Line
|| spell.spellType == SpellType.Arc)
{
if (predictPos)
{
var spellPos = spell.currentSpellPosition;
var spellEndPos = spell.GetSpellEndPosition();
return pos.LSProjectOn(spellPos, spellEndPos).SegmentPoint;
}
else
{
return pos.LSProjectOn(spell.startPos, spell.endPos).SegmentPoint;
}
}
else if (spell.spellType == SpellType.Circular)
{
return spell.endPos;
}
return Vector2.Zero;
}示例3: FindLineCircleIntersections
//http://csharphelper.com/blog/2014/09/determine-where-a-line-intersects-a-circle-in-c/
// Find the points of intersection.
public static int FindLineCircleIntersections(
Vector2 center, float radius,
Vector2 from, Vector2 to,
out Vector2 intersection1, out Vector2 intersection2)
{
float cx = center.X;
float cy = center.Y;
float dx, dy, A, B, C, det, t;
dx = to.X - from.X;
dy = to.Y - from.Y;
A = dx * dx + dy * dy;
B = 2 * (dx * (from.X - cx) + dy * (from.Y - cy));
C = (from.X - cx) * (from.X - cx) +
(from.Y - cy) * (from.Y - cy) -
radius * radius;
det = B * B - 4 * A * C;
if ((A <= 0.0000001) || (det < 0))
{
// No real solutions.
intersection1 = new Vector2(float.NaN, float.NaN);
intersection2 = new Vector2(float.NaN, float.NaN);
return 0;
}
else if (det == 0)
{
// One solution.
t = -B / (2 * A);
intersection1 =
new Vector2(from.X + t * dx, from.Y + t * dy);
intersection2 = new Vector2(float.NaN, float.NaN);
var projection1 = intersection1.LSProjectOn(from, to);
if (projection1.IsOnSegment)
{
return 1;
}
else
{
return 0;
}
}
else
{
// Two solutions.
t = (float)((-B + Math.Sqrt(det)) / (2 * A));
intersection1 =
new Vector2(from.X + t * dx, from.Y + t * dy);
t = (float)((-B - Math.Sqrt(det)) / (2 * A));
intersection2 =
new Vector2(from.X + t * dx, from.Y + t * dy);
var projection1 = intersection1.LSProjectOn(from, to);
var projection2 = intersection2.LSProjectOn(from, to);
if (projection1.IsOnSegment && projection2.IsOnSegment)
{
return 2;
}
else if (projection1.IsOnSegment && !projection2.IsOnSegment)
{
return 1;
}
else if (!projection1.IsOnSegment && projection2.IsOnSegment)
{
intersection1 = intersection2;
return 1;
}
return 0;
}
}