围绕另一个点旋转一个点 [英] Rotate a point around another point

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本文介绍了围绕另一个点旋转一个点的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的任务是绘制一个特定的图形.作为这项任务的一部分,我需要将一些点旋转 45 度.

I have a task to draw a specific graphic. As part of this task I need to rotate some dot's on 45 degrees.

我已经花了 2 天时间试图计算一个公式,但就是无法正确计算.我一直在到处搜索,包括这个特定的网站,我已经非常接近了,但我仍然不在那里.

I've spent already 2 days trying to calculate a formula, but just couldn't get it right. I've been searching all over the place including this particular website, I'm getting very close, but I'm still not there.

这里是:我需要画4个不同的点

Here it is: I need to draw 4 different points

我有一个特定的公式来计算那里的位置,这超出了问题的范围,但这是我得到的结果:

I have a specific formula to calculate there position, which is out of scope of the question, but here is what I'm getting as a result of it:

int radius = 576;
int diameter = radius * 2;
Point blueA = new Point(561, 273);
Point greenB = new Point(273, 561);
Point yellowC = new Point (849, 561);
Point redD = new Point (561, 849);

现在我需要将这些点旋转 45 度.我使用下面的代码来实现它:

Now I need to rotate this dots on 45 degrees. I use the following code to achieve it:

double rotationAngle = 45;
double rotationRadians = rotationAngle * (Math.PI / 180);
int center = radius;    
result.X = (int)(Math.Cos(rotationRadians) * ((double)result.X - (double)center) - (double)Math.Sin(rotationRadians) * ((double)result.Y - center) + (double)center);
result.Y = (int)(Math.Sin(rotationRadians) * ((double)result.X - (double)center) + (double)Math.Cos(rotationRadians) * ((double)result.Y - center) + (double)center);

但这就是我得到的:

任何帮助将不胜感激

推荐答案

问题是你设置的 int center = radius int radius = 576.这没有任何意义,因为您肯定正在围绕应该具有 x 和 y 位置的点旋转.

The problem is int center = radius which you are setting int radius = 576. This doesn't make sense as surely you are rotating about a point that should have an x and y location.

假设您围绕原点旋转,中心 xy 都应该是 0 而不是 576.

Given you are rotating around the origin the center x and y should both be 0 not 576.

所以,既然如此,试试这个.

So, given that, try this.

/// <summary>
/// Rotates one point around another
/// </summary>
/// <param name="pointToRotate">The point to rotate.</param>
/// <param name="centerPoint">The center point of rotation.</param>
/// <param name="angleInDegrees">The rotation angle in degrees.</param>
/// <returns>Rotated point</returns>
static Point RotatePoint(Point pointToRotate, Point centerPoint, double angleInDegrees)
{
    double angleInRadians = angleInDegrees * (Math.PI / 180);
    double cosTheta = Math.Cos(angleInRadians);
    double sinTheta = Math.Sin(angleInRadians);
    return new Point
    {
        X =
            (int)
            (cosTheta * (pointToRotate.X - centerPoint.X) -
            sinTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.X),
        Y =
            (int)
            (sinTheta * (pointToRotate.X - centerPoint.X) +
            cosTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.Y)
    };
}

这样使用.

Point center = new Point(0, 0); 
Point newPoint = RotatePoint(blueA, center, 45);

显然,如果中心点始终为 0,0,那么您可以相应地简化函数,或者通过默认参数或重载方法使中心点成为可选.您可能还希望将一些可重用的数学封装到其他静态方法中.

Obviously if the center point is always 0,0 then you can simplify the function accordingly, or else make the center point optional via a default parameter, or by overloading the method. You would also probably want to encapsulate some of the reusable math into other static methods too.

例如

/// <summary>
/// Converts an angle in decimal degress to radians.
/// </summary>
/// <param name="angleInDegrees">The angle in degrees to convert.</param>
/// <returns>Angle in radians</returns>
static double DegreesToRadians(double angleInDegrees)
{
   return angleInDegrees * (Math.PI / 180);
}

/// <summary>
/// Rotates a point around the origin
/// </summary>
/// <param name="pointToRotate">The point to rotate.</param>
/// <param name="angleInDegrees">The rotation angle in degrees.</param>
/// <returns>Rotated point</returns>
static Point RotatePoint(Point pointToRotate, double angleInDegrees)
{
   return RotatePoint(pointToRotate, new Point(0, 0), angleInDegrees);
}

这样使用.

Point newPoint = RotatePoint(blueA, 45);

最后,如果您使用 GDI,您也可以简单地执行 RotateTransform.请参阅:http://msdn.microsoft.com/en-us/library/a0z3f662.aspx

Finally, if you are using the GDI you can also simply do a RotateTransform. See: http://msdn.microsoft.com/en-us/library/a0z3f662.aspx

Graphics g = this.CreateGraphics();
g.TranslateTransform(blueA);
g.RotateTransform(45);

这篇关于围绕另一个点旋转一个点的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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