2. Maths Pythagoras Theorem
Most people who have done maths in high school, would have heard of Pythagoras theorem.
This is used to work out the unknown side of a triangle, if you know the other 2 sides.
Here you can see the sides, A, B and C.
Now Pythagoras theorem states that A^2 + B^2 = C^2
So if you know any two sides, you add them in, rearrange the formula if needed, and then you can solve for the unkown sides.
So how would you use this in a game? What about collision detection? Using this you can find quite easily in a 2 dimensional plane when 2 points will meet.
They will meet when A^2 + B^2 = 0
Quite simple really, you take the position of A and B on the X axis and do the equation, then if it comes up with an answer of 0, you then take the position of A and B on the Y axis, then if that also comes to 0, they must be colliding.
So to put that in other words, we are finding the distance 🙂 If the distance is nothing, then they are located in the same position.
Not sure if that is right, sounds right from my position, although it is 1:18 am.
If you have any questions, please email me at firstname.lastname@example.org
That is exactly true, A^2 + B^2 will only equal 0 if A and B is 0.
While the formula you have given is the true formula for working out the distance between two points, look at what it is actually doing. I will look at 2D space:
Now if we have a collision when the result is 0, if we apply some basic maths to the distance formula, we can remove the sqrt:
0 = sqrt((x2-x1)^2 + (y2-y1)^2)
0^2 = sqrt((x2 – x1)^2 + (y2 – y1)^2)^2
0 = (x2 – x1)^2 + (y2 – y1)^2
And now look at what x2-x1 and y2-y1 do. They give us the length of the line segments A and B.
x2 – x1 = A and y2 – y1 = B
0 = (A)^2 + (B)^2
0 = A^2 + B^2
Therefore we have a collision. Both this shorthand version and the full version of the distance formula are the same thing.
Hope this helps clarify 🙂
I don’t understand how can you use this for calculating distance. The only time A^2 + B^2 would be 0 is when A and B is 0. The proper formula for calculating distance between 2 point is:
sqrt( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 )
What if A = 1 and B = i ???
Then I would have to ask why are you using imaginary numbers as lengths?