帮助理解函数
下面的函数执行什么操作？
public static double CleanAngle(double angle) {
while (angle < 0)
angle += 2 * System.Math.PI;
while (angle > 2 * System.Math.PI)
angle -= 2 * System.Math.PI;
return angle;
}
这就是它与 ATan2 一起使用的方式。我相信传递给 ATan2 的实际值始终是正值。
static void Main(string[] args) {
int q = 1;
//'x- and y-coordinates will always be positive values
//'therefore, do i need to "clean"?
foreach (Point oPoint in new Point[] { new Point(8,20), new Point(-8,20), new Point(8,-20), new Point(-8,-20)}) {
Debug.WriteLine(Math.Atan2(oPoint.Y, oPoint.X), "unclean " + q.ToString());
Debug.WriteLine(CleanAngle(Math.Atan2(oPoint.Y, oPoint.X)), "cleaned " + q.ToString());
q++;
}
//'output
//'unclean 1: 1.19028994968253
//'cleaned 1: 1.19028994968253
//'unclean 2: 1.95130270390726
//'cleaned 2: 1.95130270390726
//'unclean 3: -1.19028994968253
//'cleaned 3: 5.09289535749705
//'unclean 4: -1.95130270390726
//'cleaned 4: 4.33188260327232
}
谢谢大家的回答。对于每个答案都有另一个问题。
为什么他们要“标准化”角度？这是代码的一部分。
double _theta = Math.ATan2(oEnd.Y - _start.Y, oEnd.X - _start.X);
Point oCenter = new Point();
oCenter.X = (int)(_start.X + _distanceTravelled * Math.Cos(_theta));
oCenter.Y = (int)(_start.Y + _distanceTravelled * Math.Sin(_theta));
//'move barrage
this.Left = oCenter.X - this.Width / 2;
this.Top = oCenter.Y - this.Height / 2;
What does the following function perform?
public static double CleanAngle(double angle) {
while (angle < 0)
angle += 2 * System.Math.PI;
while (angle > 2 * System.Math.PI)
angle -= 2 * System.Math.PI;
return angle;
}
This is how it is used with ATan2. I believe the actually values passed to ATan2 are always positive.
static void Main(string[] args) {
int q = 1;
//'x- and y-coordinates will always be positive values
//'therefore, do i need to "clean"?
foreach (Point oPoint in new Point[] { new Point(8,20), new Point(-8,20), new Point(8,-20), new Point(-8,-20)}) {
Debug.WriteLine(Math.Atan2(oPoint.Y, oPoint.X), "unclean " + q.ToString());
Debug.WriteLine(CleanAngle(Math.Atan2(oPoint.Y, oPoint.X)), "cleaned " + q.ToString());
q++;
}
//'output
//'unclean 1: 1.19028994968253
//'cleaned 1: 1.19028994968253
//'unclean 2: 1.95130270390726
//'cleaned 2: 1.95130270390726
//'unclean 3: -1.19028994968253
//'cleaned 3: 5.09289535749705
//'unclean 4: -1.95130270390726
//'cleaned 4: 4.33188260327232
}
UPDATE
Thank you all for your answers. For every answer there is another question.
Why would they be "normalizing" the angle? Here is the a portion of the code.
double _theta = Math.ATan2(oEnd.Y - _start.Y, oEnd.X - _start.X);
Point oCenter = new Point();
oCenter.X = (int)(_start.X + _distanceTravelled * Math.Cos(_theta));
oCenter.Y = (int)(_start.Y + _distanceTravelled * Math.Sin(_theta));
//'move barrage
this.Left = oCenter.X - this.Width / 2;
this.Top = oCenter.Y - this.Height / 2;
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它以 弧度 为单位规范角度，使其位于区间
[0.. 2pi]。2pi是一个完整的圆，因此角度x等于角度x+2pi、x+4pi,x+6pi...It norms an angle in radians so that it is in the interval
[0..2pi].2piis a full circle and therefore an anglexis equal to the anglesx+2pi,x+4pi,x+6pi...下面的函数执行什么？
这是一种强制将角度（以弧度表示）从 0 写入到 2π 的非常慢的方法
为什么他们要“标准化”角度？ >
因为
ATan()返回 -π 和 π 之间的角度，而他们希望它在 0 到 2π 之间。如果这是唯一使用
CleanAngle的代码，那么由于身份原因不需要它What does the following function perform?
It's a very slow way of forcing an angle (in radians) to be written from 0 to 2π
Why would they be "normalizing" the angle?
Because
ATan()returns an angle between -π and π, and they wanted it between 0 and 2π.If that is the only code that uses
CleanAngle, then it's not needed because of the identities该方法将角度转换为 0..pi*2 范围内的等效角度。
如果您只将正值传递给
Atan2方法，您只能返回第一象限的角度，即 0..pi/2 范围内的角度。因此，您不需要标准化角度。The method converts an angle into the equivalent angle in the range 0..pi*2.
If you only pass positive values into the
Atan2method, you only get back an angle for the first quadrant, i.e. in the range 0..pi/2. Thus, you don't need to normalise the angle.它将以弧度测量的角度归一化到 0 到 2 Pi 的范围内。
It normalizes an angle measured in radians into the range 0 to 2 Pi.