How to determine collision in python in right way

Question

Here I have made a code to create random sized bubbles which can be destroyed by collision of another object:

import tkinter
window = tkinter.Tk()
window.title("...")
c = tkinter.Canvas(width=800, height=500, bg="...")
ojct_id1 = c.create_polygon(...)
ojct_id2 = c.create_oval(...)           # A polygon in an oval should constitute the object
def move ojct(event):
    ...
from random import randint
bubbles = list()
bubbles_r = list()                    # radius
bubbles_speed = list()
def create_bub():
    ...
def move_bubbles():
    ...
from time import sleep
while True:
    if randint(1, 10) == 1:
        create_bub()
    move_bubbles()
    window.update()
    sleep(0.01)

The following code determines the position of any bubble:That helps to find out collision.

def hole_coord(id_num):
    pos = c.coords(id_num)
    x = (pos[0] + pos[2])/2
    y = (pos[1] + pos[3])/2
    return x, y

Now I have to make func. for deleting bubbles:

def del_bubbles():
    del bubbles_r[i]
    del bubbles_speed[i]
    c.delete(bubbles[i])
    del bubbles[i]

The following code determines, if the two objects are colliding:

from math import sqrt
def distance(id1, id2):
    x1, y1 = hole_coord(id1)
    x2, y2 = hole_coord(id2)
    return sqrt((x2 - x1)/2 + (y2 - y1)/2)
def collision():
    for bub in range(len(bubbles)-1, -1, -1):
        if distance(ojct_id2, bubbles[bub]) < (15 + bubbles_r[bub]):
            del_bubbles(bub)

Here it is sth. wrong: bubbles get deleted without a hit but if they are hit often they don't get deleted. Can anybody help me? Thanks!

I think your distance function is wrong, it should be return sqrt((x1-x2)**2+(y1-y2)**2) — Thibault D.
– Thibault D., Commented Jul 3, 2018 at 14:35

kutschkem · Accepted Answer · 2018-07-03 14:35:36Z

4

You are not computing the euclidean distance correctly.

def distance(id1, id2):
    x1, y1 = hole_coord(id1)
    x2, y2 = hole_coord(id2)
    return sqrt((x2 - x1)/2 + (y2 - y1)/2)   # <----- this is wrong

should be

def distance(id1, id2):
    x1, y1 = hole_coord(id1)
    x2, y2 = hole_coord(id2)
    return sqrt((x2 - x1)**2 + (y2 - y1)**2)   # <----- square instead of halve

answered Jul 3, 2018 at 14:35

kutschkem

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2 Comments

Timmy Over a year ago

Well, thank you kutschkem, I'm bad at math, you know :-D

kutschkem Over a year ago

@Timmy Don't believe that. ;-)

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