Bullet collision??

[micha]

If you just want to detect, if there is a hit, or not, you can use this code inside the box.lua.

Code: Select all

function update( dt )
  local p1 = line.point1
  local p2 = line.point2
  hit = true
  if p1.x < x and p2.x < x then
    hit = false
    return
  end
  
  if p1.x > x + w and p2.x > x + w then
    hit = false
    return
  end
  
  if p1.y < y and p2.y < y then
    hit = false
    return
  end
  
  if p1.y > y+h and p2.y > y+h then
    hit = false
    return
  end  
  
  -- last case: first build normal vector
  local nx,ny = p2.y-p1.y, -(p2.x-p1.x)
  local vx,vy = x-p1.x, y-p1.y
  
  local min = nx*vx+ny*vy
  local max = min
  
  vx,vy = x+w-p1.x, y-p1.y
  min = math.min(min,nx*vx+ny*vy)
  max = math.max(max,nx*vx+ny*vy)
  
  vx,vy = x+w-p1.x, y+h-p1.y
  min = math.min(min,nx*vx+ny*vy)
  max = math.max(max,nx*vx+ny*vy)
  
  vx,vy = x-p1.x, y+h-p1.y
  min = math.min(min,nx*vx+ny*vy)
  max = math.max(max,nx*vx+ny*vy)    
  
  if min*max>0 then
    hit = false
  end 
end

It is based on the separating axis theorem. The code is not commented and it does not allow you to find the intersection point. It will only determine if it is a hit or not. If you want to understand this code, it is probably best to look into vector analysis. But if you only want to use it, feel free to use this code.

[jag_e_nummer_ett]

--snip--

Your code works as expected and will probably be used. I will though want to know how to effectively getting the intersection point for maybe a more precise project. But I think I will leave that to another day.

[davisdude]

I would show you the actual code, but it's really quite long and lengthy, so instead, I'll show you how I've done it. Feel free to steal the code if you'd like (but credit wouldn't hurt). Like I said, lengthy and complicated. Lots of math stuff. Enjoy!

[kikito]

I must point out that this particular case is neatly solved without vectors with an algorithm called the Liang-Barsky intersection.

Here's one implementation in Lua (warning, untested):

Code: Select all

local function getSegmentIntersectionIndices(l,t,w,h, x1,y1,x2,y2)
  local t1, t2 = 0, 1
  local dx, dy = x2-x1, y2-y1
  local p, q, r

  for side = 1,4 do
    if     side == 1 then p,q = -dx, x1 - l     -- left
    elseif side == 2 then p,q =  dx, l + w - x1 -- right
    elseif side == 3 then p,q = -dy, y1 - t     -- top
    else                  p,q =  dy, t + h - y1 -- bottom
    end

    if p == 0 then
      if q <= 0 then return nil end
    else
      r = q / p
      if p < 0 then
        if     r > t2 then return nil
        elseif r > t1 then t1 = r
        end
      else -- p > 0
        if     r < t1 then return nil
        elseif r < t2 then t2 = r
        end
      end
    end
  end

  local xi1, yi1 = x1 + dx*t1, y1 + dy*t1
  local xi2, yi2 = x1 + dx*t2, y1 + dx*t2

  return xi1, yi1, xi2, yi2
end

It will return nil if the segment does not intersect with the box, or the coordinates of the two points of intersection if the segment intersects with the box.

[davisdude]

I took a look at the Wikipedia you linked, but I couldn't figure out what I was calling. Mind explaining?

[kikito]

I took a look at the Wikipedia you linked, but I couldn't figure out what I was calling. Mind explaining?

Sure. The wikipedia article is a bit hard if you are not big on math.

kinda-elvish.jpg (39 KiB) Viewed 2987 times

I actually got the initial code from this other place, and then adapted it to Lua.

The parameters are:

• l,t : "left" and "top" corners of the box
• w,h: "width" and "height" of the box
• x1,y1: The start of the segment (the point "from which the bullet is shot")
• x2,y2: The end of the segment (the "bullet target")

[Azhukar]

The parameters are:

• l,t : "left" and "top" corners of the box
• w,h: "width" and "height" of the box
• x1,y1: The start of the segment (the point "from which the bullet is shot")
• x2,y2: The end of the segment (the "bullet target")

I suspect that doesn't work for rotated rectangles.

[jag_e_nummer_ett]

I suspect that doesn't work for rotated rectangles.

Well I bet not concidering it doesn't got an angle as parameter

[kikito]

I suspect that doesn't work for rotated rectangles.

Well of course it doesn't. That's what AABB means - Axis-Aligned Bounding Box. If it's rotated, then it's not axis-aligned any more