所以我一直试图让一些点不仅走向一个圆圈，而且还让它们围绕它旋转。要做到这一点，我使用余弦和正弦，但是我遇到的问题是让点向前移动以及设置它们的距离。使用下面的代码，点可以围绕较大的点形成一个圆圈，也可以跟随它，但是当它们与t1的距离缩放时，它们不会接近点也不会到达那个位置，来到那个位置，而是做一些时髦的东西。这是指专线
t2.goto(2 * (t1.xcor() + math.degrees(math.cos(math.radians(t1.towards(t2)))) // 1), 2 * (t1.ycor() + math.degrees(math.sin(math.radians(t1.towards(t2)))) // 1))
我用以下代替：
t2.goto(dist * (t1.xcor() + math.degrees(math.cos(math.radians(t1.towards(t2)))) // 1), dist * (t1.ycor() + math.degrees(math.sin(math.radians(t1.towards(t2)))) // 1))
这让我看到了试图跟随更大点的点的零星观点。
这一行可以在follow（）函数中找到。 Create（）使得较小的点，move（）移动较大的点并且grow（）在与较小的点碰撞时增大较大的点。 Produce（）和redraw（）应该是程序的第2阶段，但这些函数与问题无关。最后，quit（）退出Screen（）并退出程序。
感谢cdlane提供有关组织数据和更有效地更新屏幕的帮助。
截至目前的代码：
from turtle import Turtle, Screen
import sys
import math
CURSOR_SIZE = 20
def move(x, y):
""" has it follow cursor """
t1.ondrag(None)
t1.goto(x, y)
screen.update()
t1.ondrag(move)
def grow():
""" grows t1 shape """
global t1_size, g
t1_size += 0.1
t1.shapesize(t1_size / CURSOR_SIZE)
g -= .1
t1.color((r/255, g/255, b/255))
screen.update()
def follow():
""" has create()'d dots follow t1 """
global circles, dist
new_circles = []
for (x, y), stamp in circles:
t2.clearstamp(stamp)
t2.goto(x, y)
dist = t2.distance(t1) / 57.29577951308232 // 1
t2.goto(2 * (t1.xcor() + math.degrees(math.cos(math.radians(t1.towards(t2)))) // 1), 2 * (t1.ycor() + math.degrees(math.sin(math.radians(t1.towards(t2)))) // 1))
t2.setheading(t2.towards(t1))
if t2.distance(t1) < t1_size // 1:
if t2.distance(t1) > t1_size * 1.2:
t2.forward(500/t2.distance(t1)//1)
else:
t2.forward(3)
if t2.distance(t1) > t1_size // 2:
new_circles.append((t2.position(), t2.stamp()))
else:
grow() # we ate one, make t1 fatter
screen.update()
circles = new_circles
if circles:
screen.ontimer(follow, 10)
else:
phase = 1
produce()
def create():
""" create()'s dots with t2 """
count = 0
nux, nuy = -400, 300
while nuy > -400:
t2.goto(nux, nuy)
if t2.distance(t1) > t1_size // 2:
circles.append((t2.position(), t2.stamp()))
nux += 20
count += 1
if count == 40:
nuy -= 50
nux = -400
count = 0
screen.update()
def quit():
screen.bye()
sys.exit(0)
def redraw():
t2.color("black")
t2.shapesize((t2_size + 4) / CURSOR_SIZE)
t2.stamp()
t2.shapesize((t2_size + 2) / CURSOR_SIZE)
t2.color("white")
t2.stamp()
def produce():
#create boundary of star
global t2_size, ironmax
t1.ondrag(None)
t1.ht()
t2.goto(t1.xcor(), t1.ycor())
t2.color("black")
t2.shapesize((t1_size + 4) / CURSOR_SIZE)
t2.stamp()
t2.shapesize((t1_size + 2) / CURSOR_SIZE)
t2.color("white")
t2.stamp()
#start producing helium
while t2_size < t1_size:
t2.color("#ffff00")
t2.shapesize(t2_size / 20)
t2.stamp()
t2_size += .1
redraw()
screen.update()
ironmax = t2_size
t2_size = 4
while t2_size < ironmax:
t2.shapesize(t2_size / 20)
t2.color("grey")
t2.stamp()
t2_size += .1
screen.update()
# variables
t1_size = 6
circles = []
phase = 0
screen = Screen()
screen.screensize(900, 900)
#screen.mode("standard")
t2 = Turtle('circle', visible=False)
t2.shapesize(4 / CURSOR_SIZE)
t2.speed('fastest')
t2.color('purple')
t2.penup()
t2_size = 4
t1 = Turtle('circle')
t1.shapesize(t1_size / CURSOR_SIZE)
t1.speed('fastest')
r = 190
g = 100
b = 190
t1.color((r/255, g/255, b/255))
t1.penup()
t1.ondrag(move)
screen.tracer(False)
screen.listen()
screen.onkeypress(quit, "Escape")
create()
follow()
#print(phase)
screen.mainloop()
答案 0 :(得分：1)
我对此采取了另一个裂缝，只是看着在行星周围蜂拥而至的流星问题。或者在这种情况下，月亮我选择Deimos作为我的模型。我试图按比例工作，使坐标系1像素= 1公里。起初，Deimos坐落在一个流星的领域，每个流星都有一个随机的标题，但它们都具有相同的大小和速度：
from turtle import Turtle, Screen
from random import random
METEOR_VELOCITY = 0.011 # kilometers per second
METEOR_RADIUS = 0.5 # kilometers
SECONDS_PER_FRAME = 1000 # each updates represents this many seconds passed
UPDATES_PER_SECOND = 100
DEIMOS_RADIUS = 6.2 # kilometers
G = 0.000003 # Deimos gravitational constant in kilometers per second squared
CURSOR_SIZE = 20
def follow():
global meteors
new_meteors = []
t = SECONDS_PER_FRAME
for (x, y), velocity, heading, stamp in meteors:
meteor.clearstamp(stamp)
meteor.goto(x, y)
meteor.setheading(heading)
meteor.forward(velocity * t)
meteor.setheading(meteor.towards(deimos))
meteor.forward(G * t * t)
meteor.setheading(180 + meteor.towards(x, y))
if meteor.distance(deimos) > DEIMOS_RADIUS * 2:
new_meteors.append((meteor.position(), velocity, meteor.heading(), meteor.stamp()))
screen.update()
meteors = new_meteors
if meteors:
screen.ontimer(follow, 1000 // UPDATES_PER_SECOND)
def create():
""" create()'s dots with meteor """
count = 0
nux, nuy = -400, 300
while nuy > -400:
meteor.goto(nux, nuy)
if meteor.distance(deimos) > DEIMOS_RADIUS * 2:
heading = random() * 360
meteor.setheading(heading) # all meteors have random heading but fixed velocity
meteors.append((meteor.position(), METEOR_VELOCITY, meteor.heading(), meteor.stamp()))
nux += 20
count += 1
if count % 40 == 0:
nuy -= 50
nux = -400
screen.update()
meteors = []
screen = Screen()
screen.screensize(1000, 1000)
screen.setworldcoordinates(-500, -500, 499, 499) # 1 pixel = 1 kilometer
meteor = Turtle('circle', visible=False)
meteor.shapesize(2 * METEOR_RADIUS / CURSOR_SIZE)
meteor.speed('fastest')
meteor.color('purple')
meteor.penup()
deimos = Turtle('circle')
deimos.shapesize(2 * DEIMOS_RADIUS / CURSOR_SIZE)
deimos.color("orange")
deimos.penup()
screen.tracer(False)
create()
follow()
screen.mainloop()
要调查的第一个变量是METEOR_VELOCITY。在提供的设置下，大多数流星会撞到月球，但有一些流星会获得轨道速度。如果你减半它的值，所有的流星都将坠入月球。如果你加倍它的值，一些流星获得逃逸速度，离开窗口;少数人可能会撞上月球;大多数将形成一个变得越来越小的轨道云。
我扔了三角函数并恢复到度数而不是弧度。我使用向量加法逻辑来计算运动。
最后，它只是一个粗糙的模型。
答案 1 :(得分：1)
在cdlane's代码的def follow()中将180更改为其他一些偏移，例如195，
meteor.setheading(195 + meteor.towards(x, y))
提供了很好的例子！