Mar 10
25

2. Maths Pythagoras Theorem

4 Comments



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.

Pythagoras Triangle

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 swiftless@gmail.com

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This entry was posted on March 25, 2010 at 9:36 am. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Posted in Maths by Swiftless 4 Comments

4 Responses to "2. Maths Pythagoras Theorem"

  • Swiftless says:
    April 11, 2010 at 4:11 am
    Reply

    Hi Flekken,

    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.

    Therefore:
    x2 – x1 = A and y2 – y1 = B
    And then:
    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 :)
    Swiftless

  • flekken says:
    April 10, 2010 at 6:35 pm
    Reply

    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 )

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