为旋转描记器创建椭圆和三角形形状

Question

我一直在修改Computing101.com提供的代码。它可以创建各种图形，但是，我试图将“外圆”修改为圆角三角形或椭圆形以及圆。.我希望它随机选择形状和大小。我还希望笔在每次完整的迭代中更改颜色

import turtle
from math import cos,sin,pi
from time import sleep
from random import randint, uniform, choice

window = turtle.Screen()
window.bgcolor("#FFFFFF")

myColourList = ['#C7980A', '#F4651F', '#82D8A7', '#CC3A05', '#575E76', '#156943', '#0BD055', '#ACD338']

mySpirograph = turtle.Turtle()
mySpirograph.hideturtle()
mySpirograph.speed(0)
mySpirograph.pensize(2)

myPen = turtle.Turtle()
myPen.hideturtle()
myPen.speed(0)
myPen.pensize(3)
myPen.color(choice(myColourList))

R = randint (150, 300) # Radius of outer Circle
r =  randint (10, 100) # Radius of inner Circle
d = randint (5, 100)
s = randint (5, 25)
t = uniform (.01, 1.0)



angle = 0

myPen.penup()
myPen.goto(R-r+d,0)
myPen.pendown()

theta = t
steps = (int(s * pi/theta) * randint (1, 10))

for t in range(0,steps):
    mySpirograph.clear()
    mySpirograph.penup()
    mySpirograph.setheading(0)
    mySpirograph.goto(0,-R)
    mySpirograph.color('#000')
    mySpirograph.pendown()
    mySpirograph.circle(R)
    angle+=theta

    x = (R - r) * cos(angle)
    y = (R - r) * sin(angle)
    mySpirograph.penup()
    mySpirograph.goto(x,y-r)
    mySpirograph.color('#000')
    mySpirograph.pendown()
    mySpirograph.circle(r)
    mySpirograph.penup()
    mySpirograph.goto(x,y)
    mySpirograph.dot(5)

    x = (R - r) * cos(angle) + d * cos(((R-r)/r)*angle)
    y = (R - r) * sin(angle) - d * sin(((R-r)/r)*angle)
    mySpirograph.pendown()
    mySpirograph.goto(x,y)
    mySpirograph.dot(5)
    myPen.goto(mySpirograph.pos())

    mySpirograph.getscreen().update() 
    sleep(0.05)

sleep(0.5)
#Hide Spirograph
mySpirograph.clear()
mySpirograph.getscreen().update()