Question

我正在尝试编写一个代码来在XNA中绘制一个椭圆。假设我们有以下参数：

半长轴（a）。
半短轴（b）。
椭圆中心（h，k）。
和2D旋转（theta）。

如何使用上述参数绘制椭圆。

这是函数

public VertexPositionColor Set2dEllipse(int x, int y, int z, Color color)
{
    VertexPositionColor vertices= new VertexPositionColor[100];
    for(int i= 0;i<100;i++)
    {
        double angle = (i / 100 * Math.PI * 2);
        vertices[i].Position = New Vector3((x + (Math.Cos(angle)) * size.Width), (y + Math.Sin(angle) * size.Height), z);
        vertices[i].Color = color;
    }
}

Answer 1

我修改了从here获取的代码以满足您的要求。

public void CreateEllipse(int a, int b, int h, int k, float theta)
{
    VertexPositionColor[] vertices = new VertexPositionColor[vertexCount];
    //Drawing an Ellipse with its major axis parallel to the x-axis. Rotation can be applied to change this.
    Vector3 position;
    const float max = MathHelper.Pi;
    //2 * max since we're moving from -Pi to +Pi in the loop.
    float step = 2 * max / (float)vertexCount;
    int i = 0;
    //Optional Axis and angle rotation for the ellipse (See later notes):
    //Vector3 axis = new Vector3(0, 0, -1);
    float angle = MathHelper.ToRadians(theta);

    for (float t = -max; t <= max; t += step)
    {
        //Formula shamelessly taken from wikipedia
        position = new Vector3(h + a * (float)Math.Cos((double)t), k + b * (float)Math.Sin((double)t), 0f);
        //Optional Rotation for the Ellipse:
        //position = Vector3.Transform(position, Matrix.CreateFromAxisAngle(axis, angle));
        vertices[i] = new VertexPositionColor(position, Color.DarkOrange);
        i++;
    }
    //Optional Rotation for the Ellipse:
    position = Vector3.Transform(position, Matrix.CreateFromAxisAngle(axis, angle));
    //then add the first vector again so it's a complete loop (sounds familiar)
    position = new Vector3(h + a * (float)Math.Cos((double)-max), k + b * (float)Math.Sin((double)-max), 0f);
    vertices[vertexCount - 1] = new VertexPositionColor(position, Color.DarkOrange);

    vertexBuffer = new VertexBuffer(device, vertexCount * VertexPositionColor.SizeInBytes,
        BufferUsage.WriteOnly);
    vertexBuffer.SetData<VertexPositionColor>(vertices);
}

Answer 2

您可以单独根据x或y找出椭圆的方程。然后递增/递减x或y以基于方向获得另一个坐标。这就是我为圆圈所做的事情

 this.x+=1.5*Gdx.graphics.getDeltaTime() * direction;

 if(direction==1)
 {

     if(this.x>=(centerx+radius))
    {


        direction=-1;
    }

 }

 else
 {

     if(this.x<=(centerx-radius))
        {


            direction=1;
        } 
 }

    this.y=(  (float) Math.sqrt(Math.abs(((radius*radius)- ( (this.x -centerx )*    (this.x -centerx )) )) )*direction  +  centery )  ;

PS：代码在java中用于libgdx。你可以相应调整

在XNA中绘制一个椭圆

2 个答案: