Hi, let’s say I have 2 rectangles on the screen, for example one has center at position xy 100, 100, and second one 100,110. Now I want move 3rd rectangle on top of them and calculate with which object 3rd rectangle is sharing most of the xy. which rectangle is covering the most.
Any xy functions I can look into? some hint?
local function hasCollided(obj1, obj2) if obj1 == nil then return false end if obj2 == nil then return false end local left = obj1.contentBounds.xMin \<= obj2.contentBounds.xMin and obj1.contentBounds.xMax \>= obj2.contentBounds.xMin local right = obj1.contentBounds.xMin \>= obj2.contentBounds.xMin and obj1.contentBounds.xMin \<= obj2.contentBounds.xMax local up = obj1.contentBounds.yMin \<= obj2.contentBounds.yMin and obj1.contentBounds.yMax \>= obj2.contentBounds.yMin local down = obj1.contentBounds.yMin \>= obj2.contentBounds.yMin and obj1.contentBounds.yMin \<= obj2.contentBounds.yMax return (left or right) and (up or down) end -- circle based local function hasCollidedCircle(obj1, obj2) if obj1 == nil then return false end if obj2 == nil then return false end local sqrt = math.sqrt local dx = obj1.x - obj2.x; local dy = obj1.y - obj2.y; local distance = sqrt(dx\*dx + dy\*dy); local objectSize = (obj2.contentWidth/2) + (obj1.contentWidth/2) if distance \< objectSize then return true end return false end
something like this will help me 
Question - Is this question answered?
I ask, because you marked it ‘answered’, but then a the end of you last post you say
something like this will help me
This implies to me you want an answer in that form, but that that isn’t the answer.
?? Confused… :unsure: ??
You need to calculate the area of the intersection of rectangle3 with the other 2 and then test for the bigger one.
Here’s a sample implementation
[lua]
local function intersectRect( x, y, w, h, otherX, otherY, otherW, otherH )
if x < otherX then
w = w - (otherX - x)
x = otherX
end
if (x + w) > (otherX + otherW) then
w = (otherX + otherW) - x
end
if y < otherY then
h = h - (otherY - y)
y = otherY
end
if (y + h) > (otherY + otherH) then
h = (otherY + otherH) - y
end
return x, y, w, h
end
local function intersectArea( r1, r2 )
local bounds1 = r1.contentBounds
local bounds2 = r2.contentBounds
local x, y, w, h = intersectRect(
bounds1.xMin, bounds1.yMin, r1.contentWidth, r1.contentHeight,
bounds2.xMin, bounds2.yMin, r2.contentWidth, r2.contentHeight )
if w <= 0 or h <= 0 then
return 0
else
return w*h
end
end
local function biggerOverlap( r1, r2, r3 )
local area1 = intersectArea( r1, r3 )
local area2 = intersectArea( r2, r3 )
if area1 > area2 then
return “first”
elseif area2 > area1 then
return “second”
elseif area1 == 0 then
return “none”
else
return “same”
end
end
[/lua]
Check out my math lib which includes an ability (with demo) to calculate overlapping areas:
Oh, sorry for not being clear enough. Yes, answered, calculating distance from center works fine enough for my need here.
local function hasCollided(obj1, obj2) if obj1 == nil then return false end if obj2 == nil then return false end local left = obj1.contentBounds.xMin \<= obj2.contentBounds.xMin and obj1.contentBounds.xMax \>= obj2.contentBounds.xMin local right = obj1.contentBounds.xMin \>= obj2.contentBounds.xMin and obj1.contentBounds.xMin \<= obj2.contentBounds.xMax local up = obj1.contentBounds.yMin \<= obj2.contentBounds.yMin and obj1.contentBounds.yMax \>= obj2.contentBounds.yMin local down = obj1.contentBounds.yMin \>= obj2.contentBounds.yMin and obj1.contentBounds.yMin \<= obj2.contentBounds.yMax return (left or right) and (up or down) end -- circle based local function hasCollidedCircle(obj1, obj2) if obj1 == nil then return false end if obj2 == nil then return false end local sqrt = math.sqrt local dx = obj1.x - obj2.x; local dy = obj1.y - obj2.y; local distance = sqrt(dx\*dx + dy\*dy); local objectSize = (obj2.contentWidth/2) + (obj1.contentWidth/2) if distance \< objectSize then return true end return false end
something like this will help me
Question - Is this question answered?
I ask, because you marked it ‘answered’, but then a the end of you last post you say
something like this will help me
This implies to me you want an answer in that form, but that that isn’t the answer.
?? Confused… :unsure: ??
You need to calculate the area of the intersection of rectangle3 with the other 2 and then test for the bigger one.
Here’s a sample implementation
[lua]
local function intersectRect( x, y, w, h, otherX, otherY, otherW, otherH )
if x < otherX then
w = w - (otherX - x)
x = otherX
end
if (x + w) > (otherX + otherW) then
w = (otherX + otherW) - x
end
if y < otherY then
h = h - (otherY - y)
y = otherY
end
if (y + h) > (otherY + otherH) then
h = (otherY + otherH) - y
end
return x, y, w, h
end
local function intersectArea( r1, r2 )
local bounds1 = r1.contentBounds
local bounds2 = r2.contentBounds
local x, y, w, h = intersectRect(
bounds1.xMin, bounds1.yMin, r1.contentWidth, r1.contentHeight,
bounds2.xMin, bounds2.yMin, r2.contentWidth, r2.contentHeight )
if w <= 0 or h <= 0 then
return 0
else
return w*h
end
end
local function biggerOverlap( r1, r2, r3 )
local area1 = intersectArea( r1, r3 )
local area2 = intersectArea( r2, r3 )
if area1 > area2 then
return “first”
elseif area2 > area1 then
return “second”
elseif area1 == 0 then
return “none”
else
return “same”
end
end
[/lua]
Check out my math lib which includes an ability (with demo) to calculate overlapping areas:
Oh, sorry for not being clear enough. Yes, answered, calculating distance from center works fine enough for my need here.