我想绘制sin(θ)*cos(θ)
,但它不起作用。
我可以画 sin 或 cos ,
但我想一起画sin(θ)*cos(θ)
。
这是我的代码
private void button1_Click(object sender, EventArgs e)
{
Graphics drw = this.CreateGraphics();
Pen pen = new Pen(Brushes.Black, 7.0f);
float x1 = 0;
float y1 = 0;
float xoy = 200;
float ef = 20;
for (double i=0;i<40;i+=1)
{
double radi = (float)(i * 180 / Math.PI);
float temp = (float)Math.Cos(radi)*(float)Math.Sin(radi);
drw.DrawLine(pen, x1 * ef, y1 * ef + xoy, ef * (float)i, temp * ef + xoy);
x1 = (float)i;
y1 = temp;
}
}
答案 0 :(得分:1)
实际上,您正在寻找的实际功能有点不同......请参阅示例here。看看关于极地花的this article,我确信它会指向正确的方向,并且它还包含完整的工作源代码。
举一个例子,假设您在表单中使用了一个用于绘制极地花的面板:
panel.OnPaint += Panel_Paint;
private void Panel_Paint(Object sender, PaintEventArgs e)
{
Double scale = ((Panel)sender).Width / 2.0d;
Double repetitions = Math.Round(scale, 0);
Double basis = (2.0d * Math.PI) / scale;
Double petals = 2.0d;
using (Graphics g = e.Graphics)
{
using (Pen pen = new Pen(Brushes.Red, 2.0f))
{
for (Double i = 0.0f; i < (repetitions - 1); ++i)
{
Double t0 = i*basis;
Double t1 = (i + 1)*basis;
Double x0 = Math.Sin(petals * t0) * Math.Cos(t0);
Double x1 = Math.Sin(petals * t1) * Math.Cos(t1);
Double y0 = Math.Sin(petals * t0) * Math.Sin(t0);
Double y1 = Math.Sin(petals * t1) * Math.Sin(t1);
g.DrawLine
(
pen,
(Single) ((scale*x0) + scale),
(Single) ((scale*y0) + scale),
(Single) ((scale*x1) + scale),
(Single) ((scale*y1) + scale)
);
}
}
}
}
基本公式表明,如果petals
变量值为:
因此,如果您定义Double petals = 2.0d;
,您将获得4
个花瓣......如果您定义Double petals = 5.0d;
,您将获得5
个花瓣。
答案 1 :(得分:1)
您可能会发现更容易查看相应的 Parametric Equations 。
private void Form1_Paint(object sender, PaintEventArgs e)
{
var g = e.Graphics;
double pi = Math.PI;
int n = 100;
var t = Enumerable.Range(0, n).Select(p => p * 2 * pi / n).ToArray();
var x = t.Select(p => Math.Sin(2 * p) * Math.Cos(p)).ToArray();
var y = t.Select(p => Math.Sin(2 * p) * Math.Sin(p)).ToArray();
Pen pen = new Pen(Brushes.Black, 3);
int scale = 100;
int shift = 100;
for (int i = 0; i < n - 1; i++)
{
g.DrawLine(pen, scale*(float)x[i] + shift,
scale*(float)y[i] + shift,
scale*(float)x[i + 1] + shift,
scale*(float)y[i + 1] + shift);
}
}