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Math Check: Distance Between 2 Obbjects


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#1 Shippou   Members   -  Reputation: 1278

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Posted 14 January 2012 - 01:01 PM

I would like to knoow if my math is correct in calculating the distance between 2 objects on a 2D plain:
def distance(x1,x2,y1,y2):
	# A squared + B squared = C squared
	return ( ( (x2 - x1)**2) + ( (y2-y1)**2) )**.5
# Object 1 Location
coordinate1 = 2,8	 # x,y
# Object 2 Location
coordinate2 = 5,10	 # x,y
print (distance(coordinate1[0],coordinate2[0],coordinate1[1],coordinate2[1]))


Thanks.

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#2 jjd   Crossbones+   -  Reputation: 2062

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Posted 14 January 2012 - 01:15 PM

I would like to knoow if my math is correct in calculating the distance between 2 objects on a 2D plain:

def distance(x1,x2,y1,y2):
	# A squared + B squared = C squared
	return ( ( (x2 - x1)**2) + ( (y2-y1)**2) )**.5
# Object 1 Location
coordinate1 = 2,8	 # x,y
# Object 2 Location
coordinate2 = 5,10	 # x,y
print (distance(coordinate1[0],coordinate2[0],coordinate1[1],coordinate2[1]))


Thanks.


Yes, that is correct.

-Josh

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--linkedin
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#3 HappyCoder   Members   -  Reputation: 2194

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Posted 14 January 2012 - 01:16 PM

yup. That distance equation will give you the distance between two points.

#4 Cornstalks   Crossbones+   -  Reputation: 6966

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Posted 14 January 2012 - 01:16 PM

Yes, your math is correct.

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#5 Shippou   Members   -  Reputation: 1278

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Posted 14 January 2012 - 01:42 PM

Thank you.

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#6 Narf the Mouse   Members   -  Reputation: 318

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Posted 16 January 2012 - 03:37 PM

As a side note, it's also useful to include a "DistanceSquared" function, which omits the square root - If you're simply comparing which is farther, and don't need to know how far, omitting a costly square root can save significant time in time-sensitive applications.




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