如何在python中使用乌龟画一个圆圈？

Question

I wanted ask how can I draw a circle using turtle module in python just using turtle.forward and turtle.left? I use the code below:

 for i in range(30):
     turtle.forward(i)
     turtle.left(i)  
 turtle.done()

What I get is that the line does not stop once I get the full cirle. How can I code so that I have a circle of specific radius and that I have a full stop once the circle is drawn (without using turtle.circle).

Answer 1

If you want to draw a circle the best thing to do is to simplyfy the problem, if we consider moving 1 space for each degree of the circle then we can simply write this as

def draw_circle1():
    for _ in range(360):
        turtle.forward(1)
        turtle.left(1)

Now what do we know about this basic circle that we drew? well we know it took 360 steps and each step was 1. so the circle has a circumference of 360. we can use a bit of math to calculate the radius.

circumference = 2 * 3.14... * radius
360 = 2 * 3.14... * radius
360 / 2 / 3.14... = radius
radius = 57.29...

So now we can reverse this, if we want to specify a circle of a given radius, we can calculate what circumference that circle should have. divide that by the 360 degrees and we know what size step to take before each turn of 1 degree.

def draw_circle(radis):
    circumfrence = 2 * math.pi * radis
    step_size = circumfrence / 360
    for _ in range(360):
        turtle.forward(step_size)
        turtle.left(1)

if we run this for 3 separate circles each increasing in size you see it gives us a consistent result

draw_circle(20)
draw_circle(40)
draw_circle(60)
turtle.hideturtle()
turtle.done()

So now we have a function which can accept a radius and draw a circle based on that radius

Answer 2

I made this image as a reference,

Essentially you need to draw the inscribed polygon with n sides.

The initial left turn will be ϴ/2.

Then forward by a = 2rsin(ϴ/2).

Each forward is followed by a left turn of the full ϴ, except that after the last forward we want only a left turn of ϴ/2 so that the heading will be correctly updated to be tangential to the circle (or arc).

Something like this,

import turtle
import math

def circle2(radius,extent=360,steps=360):
    if extent<360 and steps==360:
        steps=extent
    
    theta=extent/steps
    step_size=2*radius*math.sin(math.radians(theta/2))
    turtle.left(theta/2)
    turtle.forward(step_size)
    for i in range(1,steps):
        turtle.left(theta)
        turtle.forward(step_size)
    
    turtle.left(theta/2)
    

turtle.hideturtle()
turtle.speed(0)
turtle.getscreen().tracer(False)

circle2(50)
circle2(100,180)
turtle.up()
turtle.home()
turtle.down()
circle2(130)
circle2(130,360,10)

turtle.update()
turtle.mainloop()

Answer 3

Create a SPIROGRAPH using turtle

[Final output][1] [1]: https://i.stack.imgur.com/unlAl.png

import random

import turtle

from turtle import Turtle, Screen

tim = Turtle()

tim.shape('arrow')

turtle.colormode(255)

def random_colour( ):

r = random.randint(0, 255)
g = random.randint(0, 255)
b = random.randint(0, 255)
return (r, g, b)

tim.speed('fastest')

def draw_spirograph(size_of_gap):

for _ in range(int(360/size_of_gap)):
    tim.color(random_colour())
    tim.circle(100)
    tim.setheading(tim.heading()+size_of_gap)

draw_spirograph(5)

screen = Screen() screen.exitonclick()

Answer 4

A spirograph code

from turtle import Turtle
import random

josh = Turtle()
josh.color('DarkRed')

def random_color()->tuple:
    r = random.randint(0,255)
    g = random.randint(0,255)
    b = random.randint(0,255)
    return (r,g,b)

josh.speed('fastest')
josh.pensize(2)
for i in range(72):
    josh.circle(100)
    josh.right(5)
    colormode(255)
    josh.pencolor(random_color())
        
screen = Screen()
screen.setup(800,800)
screen.exitonclick()

Answer 5

Example here,

import turtle

def circle(distance, sections):
    angle = 360/sections
    for i in range(1, sections+1):
        turtle.forward(distance)
        turtle.left(angle)

circle(20, 30)
turtle.done()