我需要帮助来发展一个循环模式

Question

我正在尝试生成这个模式，而不是添加for loops和近似角度（我认为朝向末尾的角度可能是关闭的）。我想知道是否有人能提供一些关于用更少的代码生成这种模式的见解。你知道吗

最终目标是简单地说明半径以及我希望生成多少层圆

import numpy as np
import matplotlib.pyplot as plt

__author__ = 'George Pamfilis'

def rotate_around_origin(r,angle):
    # http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx
    x = r * np.cos(angle)
    y = r * np.sin(angle)
    return x, y

if __name__ == '__main__':
    r = 1
    r_cylinder = 1
    p1 = (0, 0)
    ps = [p1]

angle = 60

# first layer
for i in range(6):
    p_new = rotate_around_origin(2*r + (r/2) * 0, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60

# second layer


angle = 30

for i in range(6):
    p_new = rotate_around_origin(3*r+(r/2) * 1, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60
for i in range(6):
    p_new = rotate_around_origin(4*r+(r/2)*0, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60 + 30/2 + 4

for i in range(6):
    p_new = rotate_around_origin(5*r+(r/3), np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60 +30/2 + 4 -8 + 30

for i in range(6):
    p_new = rotate_around_origin(5*r+(r/3), np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60
for i in range(6):
    p_new = rotate_around_origin(5*r+(r/1)*1, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60

angle = 30
for i in range(6):
    p_new = rotate_around_origin(6 * r + (r / 1) * 1, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60 +13.75
for i in range(6):
    p_new = rotate_around_origin(6*r+(r/1)*1+r/4, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 46.25
for i in range(6):
    p_new = rotate_around_origin(6*r+(r/1)*1+r/4, np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


angle = 60
for i in range(6):
    p_new = rotate_around_origin(6*r+(r/1)*1+r/4 + (r/1.25), np.deg2rad(angle))
    ps.append(p_new)
    angle += 60


plt.figure(figsize=(20, 20 * (1/1)))

fig = plt.gcf()
# circle1 = plt.Circle((0, 0), 6*r+r+r+r/2+r, color="r")
# this is added just to enclose the pattern.
circle1 = plt.Circle((0, 0), 6*r+r+r+r, color="r")

fig.gca().add_artist(circle1)
for p in ps:
    circle1 = plt.Circle((p[0], p[1]), r)
    fig.gca().add_artist(circle1)


plt.xlim(-10, 10)
plt.ylim(-10, 10)
plt.show()

Answer 1

求给定大R和层数的小圆半径（环包括中心小圆）

查找水平层数

对于每个水平层，计算第一个和最后一个小圆的位置

工作示例：

var
  cx, cy, R, NL, irs: Integer;
  rs, yshift, dx: Double;
  iymax, ix, iy, mx, my, il, ir, Nl4: Integer;
begin
  R := 180;
  NL := 8;
  cx := 200;
  cy := 200;
  Canvas.Brush.Color := clRed;
  Canvas.Ellipse(cx - R, cy - R, cx + R + 1, cy + R + 1);
  Canvas.Brush.Color := clBlue;

  rs := R / (2 * NL - 1); // small radius
  irs := Round(rs); // integer radius for drawing
  Nl4 := 4 * (NL - 1) * (NL - 1);
  iymax := Floor((2 * NL - 2) / Sqrt(3)); // horizontal layers from center to top

  for iy := -iymax to iymax do
  begin
    yshift := iy * Sqrt(3) * rs; // relative to center
    my := cy + Round(yshift); // y-coordinate

    // find left small circle inside big one
    if Odd(iy) then
    begin
      il := Floor(0.5 * Sqrt(Nl4 - 3 * iy * iy) + 0.5);
      dx := rs;
      ir := 1;
    end
    else
    begin
      il := Floor(0.5 * Sqrt(Nl4 - 3 * iy * iy));
      dx := 0;
      ir := 0;
    end;

    for ix := -il to il - ir do
    begin
      mx := cx + Round(ix * 2 * rs + dx);
      Canvas.Ellipse(mx - irs, my - irs, mx + irs + 1, my + irs + 1);
    end;
  end;