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 [SOLVED] Using distance formula for collision detection?  (Read 1209 times) 0 Members and 1 Guest are viewing this topic.
wreed12345

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http://linebylinecoding.blogspot.com/

 « Posted 2012-11-27 23:38:09 »

WOuld it be effective to use the distance formula to test for collision detection between two moving images (characters)? or is there some other more efficient method. I figured this wouldnt be too hard to implement

Agro
 « Reply #1 - Posted 2012-11-27 23:42:13 »

Assuming your image is a square, no. You could use the AABB method I guess but if its 2 squares you can do it like this:

 1  2  3  4  5  6  7  8 `public boolean collides(Rect a, Rect b) {    if(a.x < b.x + b.width) return true;    if(a.y < b.y + b.height) return true;    if(a.x + a.width > b.x) return true;    if(a.y + a.height > b.y) return true;    return false;}`

However, if your characters are circles, then yes. You can use the distance formula and check if the distance returned is less than the radius of both circles added together. Then there is a collision.

wreed12345

JGO Knight

Medals: 25
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http://linebylinecoding.blogspot.com/

 « Reply #2 - Posted 2012-11-28 00:05:01 »

Well the game i had in mind was a circle colliding with a square image. What do you think would be best for that?

Agro
 « Reply #3 - Posted 2012-11-28 00:06:15 »

Yeah, use AABB: http://www.java-gaming.org/topics/basic-collision-detection/27326/view.html

He explains it in the best possible way ^_^

wreed12345

JGO Knight

Medals: 25
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http://linebylinecoding.blogspot.com/

 « Reply #4 - Posted 2012-11-28 00:12:51 »

thanks for the help!

theagentd

« JGO Bitwise Duke »

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 « Reply #5 - Posted 2012-11-28 14:15:01 »

Assuming your image is a square, no. You could use the AABB method I guess but if its 2 squares you can do it like this:

 1  2  3  4  5  6  7  8 `public boolean collides(Rect a, Rect b) {    if(a.x < b.x + b.width) return true;    if(a.y < b.y + b.height) return true;    if(a.x + a.width > b.x) return true;    if(a.y + a.height > b.y) return true;    return false;}`

However, if your characters are circles, then yes. You can use the distance formula and check if the distance returned is less than the radius of both circles added together. Then there is a collision.
A circle to circle test isn't expensive either...

 1  2  3  4  5 `public boolean collides(Circle a, Circle b) {    float dx = a.x - b.x, dy = a.y - b.y;    float dist = a.radius + b.radius;    return dx*dx + dy*dy < dist*dist;}`

Myomyomyo.
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