Found cordinates on circumference

07/05/2014 14:56 fuso98#1

Hi guys, i've a problem:
i want to find a specific point on a circumference (that there is not on the screen but i can only calculate every point).
My script need to find a picture on the screen, so it get direction line that pass by the center of my circumference and by the center of the picture, so now I have to find the intersection point originated from the line and the circle.
This image will help you:

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Until now i made this, but it not find the point well :(

PHP Code:


			
  $Midx=@DesktopWidth/2
  $Midy=@DesktopHeight/2
  $x=0
  $y=0
$tag =_IMAGESEARCH(@ScriptDir & "\Images\pt.bmp", 1,$x,$y, 18)
If $tag=1 Then
  $dir=point_direction($Midx,$Midy,$x,$y)
ClickCircleFromDir($dir,200)
EndIf

Func ClickCircleFromDir($dir,$radius)
  $Midx=@DesktopWidth/2
  $Midy=@DesktopHeight/2
 Mouseclick ("left",$Midx + ($radius * Cos($dir)),$Midy + ($radius * Sin($dir)),1,0)
EndFunc

Func point_direction($x1,$y1,$x2,$y2)
  Local $dir1,$dir2,$dist
  $dist=point_distance($x1,$y1,$x2,$y2);get the radius
  $dir1=ACos(($x2-$x1)/$dist)*(360/3.14);X-derrived angle
  $dir2=ASin(($y1-$y2)/$dist)*(360/3.14);Y-derrived angle
   if ($dir2<0) Then 
  return $dir1-(2*$dir2);
   else 
  return $dir1
  EndIf
 return -1; //Error!
EndFunc

Func point_distance($x1, $y1, $x2, $y2)
    Local $a, $b, $c
    If $x2 = $x1 And $y2 = $y1 Then
        Return 0
    Else
        $a = $y2 - $y1
        $b = $x2 - $x1
        $c = Sqrt($a * $a + $b * $b)
        Return $c
    EndIf
EndFunc   ;==>Pixel_Distance

07/05/2014 15:14 butter123#2

seems like a basic math problem. though i dont really understand wich information are given. you want to get the x,y of the blue point?
and you have the x,y of the center, the radius and...?

07/05/2014 16:03 fuso98#3

the angle of the direction line

07/05/2014 17:04 butter123#4

so this is the interesting part:

PHP Code:


			
Func ClickCircleFromDir($dir,$radius)
  $Midx=@DesktopWidth/2
  $Midy=@DesktopHeight/2
 Mouseclick ("left",$Midx + ($radius * Cos($dir)),$Midy + ($radius * Sin($dir)),1,0)
EndFunc

math is correct, if angle is 0 at at x axis and is 90� if at top. chek this in ur point_direction func (i didnt check the math here). also check if it has to be � or radian.

alternative way: create a function f=a+b*x from pic and center position and use the function value with binary search untill you have the point with needed radius.

07/06/2014 13:12 fuso98#5

I need to do it fast, so binary search will take much time maybe

07/06/2014 14:29 butter123#6

thats how i would have made the angle calculation:
angle =atan(y/x) ;y = ydistance between center and pic...
if x<0 and y >0 then angle = Pi - angle
if x<0 and y< 0 then angle = Pi + angle
if x>0 and y< 0 then angle = 2*Pi - angle
(didnt test it, jsut from mind. if... could be mixed) (result in radian)

07/10/2014 00:05 FacePalmMan#7

Use this:

Code:

Func CalcAngle($x1,$y1,$x2,$y2)
	Local $radAdd=0
	Local $X=($x2-$x1)
	Local $Y=($y2-$y1)
	Select
		Case $Y<0 and $X<0
			$radAdd=$pi
			Return ATan((-$Y)/(-$X)) + $radAdd
		Case $Y<0
			$radAdd=2*$pi
			Return $radAdd - ATan((-$Y)/$X)
		Case $X<0
			$radAdd=$pi
			Return $radAdd - ATan($Y/(-$X))
		Case Else
			Return ATan($Y/$X)
	EndSelect
EndFunc

Func MovXYbyAngle($x,$y,$angle,$Distance,$IsDeg=1)
	If $IsDeg=1 then $Angle/=180/3.14159
	Local $aReturn[2]=[$x+(Sin($Angle)),$y+(Cos($Angle))]
	Return $aReturn
Endfunc

and with the second function you can even make games.