Mathematics, coordinates with radius of a circle?

sphexion · April 16, 2015

Ok, I did this a fue years ago but I've forgot how I achieved it.

Here are old functions used.

Func _GetAngleDegree($X1, $Y1, $X2, $Y2)
    Local Const $nPI = 4 * ATan(1)
    Local $XDIFF, $YDIFF, $TempAngle
    $YDIFF = Abs($Y2 - $Y1)
    If $X1 = $X2 And $Y1 = $Y2 Then Return 0
    If $YDIFF = 0 And $X1 < $X2 Then
        Return 0
    ElseIf $YDIFF = 0 And $X1 > $X2 Then
        Return $nPI
    EndIf
    $XDIFF = Abs($X2 - $X1)
    $TempAngle = ATan($XDIFF / $YDIFF)
    If $Y2 > $Y1 Then $TempAngle = $nPI - $TempAngle
    If $X2 < $X1 Then $TempAngle = -$TempAngle
    $TempAngle = ($nPI / 2) - $TempAngle
    If $TempAngle < 0 Then $TempAngle = ($nPI * 2) + $TempAngle
    Return $TempAngle * 180 / $nPI
EndFunc   ;==>_GetAngleDegree

Func _Angle($tX, $tY, $cX, $cY)
    $mx = $tX - $cX
    $my = $cY - $tY
    $angle = ATan($my / $mx) * $180_div_pi
    If $mx < 0 Then
        $angle = 180 + $angle
    ElseIf $mx >= 0 And $my < 0 Then
        $angle = 360 + $angle
    EndIf
    Return Int($angle)
EndFunc   ;==>_Angle

Func Angle($X1, $Y1, $Ang, $Length)
    Local $Return[3]
    $Return[1] = $X1 + ($Length * Cos($Ang / 180 * 3.14159265358979))
    $Return[2] = $Y1 - ($Length * Sin($Ang / 180 * 3.14159265358979))
    Return $Return
EndFunc   ;==>Angle

The goal!

Identify what the coordinates are for the radius.

lets just say we are in a 500px by 500px box
The center is 250x250
0 or 360 degrees or north is Radius=160px straight up the x axis. so 90x250

Now my issue, Lets say there is something showing up at 90 degrees.
How do I calculate what the coordinates are for the variable that would be degrees? Witch in the example is 90. and the cords are unknown.

Edited April 16, 2015 by sphexion

MuffettsMan · April 16, 2015

sounds like a game bot in the making... i want in!

SadBunny · April 16, 2015

Let me try and get this straight. You mean, calculate the intersection of a line from the center of the box reaching out at X degrees angle and one of the edges of the square box? Like...

?

Doesn't seem to hard with a bit of geometry...

Please confirm that this is what you mean or offer a better example?

Kyan · April 16, 2015

wow,never done some maths with coding, this may work for the first quadrant properly, for the others my imagination is a bit blurry at this hours

Func _GetIntersect($degrees,$BoxWidth = 300,$BoxHeight = 300) ; [0] = X, [1] = Y
    Local Const $PI = 3.141592653589793, $rads = $degrees*$PI/180
    Local $CircleCenterX = ($BoxWidth * 0.5), $CircleCenterY = ($BoxHeight * 0.5), $coor[2]
    $tan = Tan($rads) * 180 / $PI
    If $degrees < 0 Then $degrees = $degrees * -1
    Switch $degrees
        Case ($degrees/90) < 45
            $coor[0] = $CircleCenterX
            $coor[1] = $CircleCenterX * $tan
        Case ($degrees/90) >= 45
            $coor[0] = $CircleCenterY / $tan
            $coor[1] = $CircleCenterY
        Case Else ; Does this even will happen?
            $coor[0] = "Kim's"
            $coor[1] = "Butt" 
    EndSwitch
    Return $coor
EndFunc

EDIT: I'm trying to solve the 90deg issue, but I'm failing to found a function for the quadrant number (which is stupid, looks like a easy task, but I can't sleep without solving this lol)

Func _GetIntersect($degrees,$BoxWidth = 300,$BoxHeight = 300) ; [0] = X, [1] = Y
    Local Const $PI = 3.141592653589793, $rads = $degrees*$PI/180
    Local $CircleCenterX = ($BoxWidth * 0.5), $CircleCenterY = ($BoxHeight * 0.5), $inDeg = $degrees, $coor[2], $chang = 1
    If IsInt($degrees/90) Then
        $tan = 0
    Else
        $tan = Tan($rads)
    EndIf
    If $degrees < 0 Then
        $degrees = $degrees * -1
        $chang = -1
    EndIf
    Switch $degrees
        Case ($degrees/90) < 45 And $tan <> 0
            $coor[0] = $CircleCenterX * $chang
            $coor[1] = $CircleCenterX * $tan
        Case ($degrees/90) >= 45 And $tan <> 0
            $coor[0] = $CircleCenterY / $tan
            $coor[1] = $CircleCenterY * $chang
        Case $tan = 0
            If $chang = 1 Then ; Positive angle
                ; #### AND I'M STUCK HERE
                $port = $degrees / 90
                Select
                    Case Mod($degrees,
                    Case
                $coor[0] = 
                $coor[1] = 
            Else
                $coor[0] = 
                $coor[1] =
            EndIf
        Case Else ; Does this even will happen?
            $coor[0] = "Kim's"
            $coor[1] = "Butt" 
    EndSwitch
    Return $coor
EndFunc

EDIT2: Negative angles need to be looked at...just made for the positive ones in case of Mod(degrees,90) = 0

Func _GetIntersect($degrees,$BoxWidth = 300,$BoxHeight = 300) ; [0] = X, [1] = Y
    Local Const $PI = 3.141592653589793, $rads = $degrees*$PI/180
    Local $CircleCenterX = ($BoxWidth * 0.5), $CircleCenterY = ($BoxHeight * 0.5), $inDeg = $degrees, $coor[2], $chang = 1
    If IsInt($degrees/90) Then
        $tan = 0
    Else
        $tan = Tan($rads)
    EndIf
    If $degrees < 0 Then
        $degrees = $degrees * -1
        $chang = -1
    EndIf
    Switch $degrees
        Case ($degrees/90) < 45 And $tan <> 0
            $coor[0] = $CircleCenterX * $chang
            $coor[1] = $CircleCenterX * $tan
        Case ($degrees/90) >= 45 And $tan <> 0
            $coor[0] = $CircleCenterY / $tan
            $coor[1] = $CircleCenterY * $chang
        Case $tan = 0
            If $chang = 1 Then ; Positive angle
                $quad = Mod($degrees/90,4)
                Switch $quad
                    Case 0
                        $coor[0] = $CircleCenterX
                        $coor[1] = 0
                    Case 1
                        $coor[0] = 0
                        $coor[1] = $CircleCenterY
                    Case 2
                        $coor[0] = - $CircleCenterX
                        $coor[1] = 0
                    Case 3
                        $coor[0] = 0
                        $coor[1] = - $CircleCenterY
                EndSwitch
            Else
                $coor[0] = 
                $coor[1] = 
            EndIf
        Case Else ; Does this even will happen?
            $coor[0] = "Kim's"
            $coor[1] = "Butt" 
    EndSwitch
    Return $coor
EndFunc

My sketch pad

0*90 ->x=W/2 y = 0
1*90 ->x=0  y=H/2
2*90 ->x=-W/2 y = 0
3*90 ->x = 0 y =-H/2

4*90 ->x = W/2 y = 0

EDIT3: This might work for all, not sure if I got straight the negative quadrants

Func _GetIntersect($degrees,$BoxWidth = 300,$BoxHeight = 300) ; [0] = X, [1] = Y
    Local Const $PI = 3.141592653589793, $rads = $degrees*$PI/180
    Local $CircleCenterX = ($BoxWidth * 0.5), $CircleCenterY = ($BoxHeight * 0.5), $inDeg = $degrees, $coor[2], $chang = 1
    If IsInt($degrees/90) Then
        $tan = 0
    Else
        $tan = Tan($rads)
    EndIf
    If $degrees < 0 Then
        $degrees = $degrees * -1
        $chang = -1
    EndIf
    Switch $degrees
        Case ($degrees/90) < 45 And $tan <> 0
            $coor[0] = $CircleCenterX * $chang
            $coor[1] = $CircleCenterX * $tan
        Case ($degrees/90) >= 45 And $tan <> 0
            $coor[0] = $CircleCenterY / $tan
            $coor[1] = $CircleCenterY * $chang
        Case $tan = 0
            If $chang = 1 Then ; Positive angle
                $quad = Mod($degrees/90,4)
                Switch $quad
                    Case 0
                        $coor[0] = $CircleCenterX
                        $coor[1] = 0
                    Case 1
                        $coor[0] = 0
                        $coor[1] = $CircleCenterY
                    Case 2
                        $coor[0] = - $CircleCenterX
                        $coor[1] = 0
                    Case 3
                        $coor[0] = 0
                        $coor[1] = - $CircleCenterY
                EndSwitch
            Else
                $quad = Mod(-$degrees/90,4)+4
                Switch $quad
                    Case 1
                        $coor[0] = 0
                        $coor[1] = $CircleCenterY
                    Case 2
                        $coor[0] = - $CircleCenterX
                        $coor[1] = 0
                    Case 3
                        $coor[0] = 0
                        $coor[1] = - $CircleCenterY
                    Case 4
                        $coor[0] = $CircleCenterX
                        $coor[1] = 0
                EndSwitch
            EndIf
        Case Else ; Does this even will happen?
            $coor[0] = "Kim's"
            $coor[1] = "Butt" 
    EndSwitch
    Return $coor
EndFunc

EDIT4: Looking good, didn't found any flaw on the EDIT3 code

Test:

Global $TestAngles[11] = [-450,-360,-270,-180,-90,0,90,180,270,360,450]
For $a = 0 To UBound($TestAngles)-1
    $intersect = _GetIntersect($TestAngles[$a])
    ConsoleWrite("Angle:"&$TestAngles[$a]&@LF&"X:"&$intersect[0]&@LF&"Y:"&$intersect[1]&@LF&"=============="&@LF)
Next
Exit

Output

Angle:-450
X:0
Y:-150
==============
Angle:-360
X:150
Y:0
==============
Angle:-270
X:0
Y:150
==============
Angle:-180
X:-150
Y:0
==============
Angle:-90
X:0
Y:-150
==============
Angle:0
X:150
Y:0
==============
Angle:90
X:0
Y:150
==============
Angle:180
X:-150
Y:0
==============
Angle:270
X:0
Y:-150
==============
Angle:360
X:150
Y:0
==============
Angle:450
X:0
Y:150
==============

BTW, I'm using coordinates relative to circle's center, and the same angle origin as the unit circle.

EDIT5: Code with lesser lines (don't know why, but can't make inline if statements work)

Func _GetIntersect($degrees,$BoxWidth = 300,$BoxHeight = 300) ; [0] = X, [1] = Y
    Local Const $PI = 3.141592653589793, $rads = $degrees*$PI/180
    Local $CircleCenterX = ($BoxWidth * 0.5), $CircleCenterY = ($BoxHeight * 0.5), $coor[2], $ChangeSignal= 1, $tan = Tan($rads), $add = 1
    If IsInt($degrees/90) Then $tan = 0
    If $degrees < 0 Then
        $degrees = $degrees * -1
        $ChangeSignal= -1
    EndIf
    Switch $degrees
        Case ($degrees/90) < 45 And $tan <> 0
            $coor[0] = $CircleCenterX * $ChangeSignal
            $coor[1] = $CircleCenterX * $tan
        Case ($degrees/90) >= 45 And $tan <> 0
            $coor[0] = $CircleCenterY / $tan
            $coor[1] = $CircleCenterY * $ChangeSignal
        Case $tan = 0
            If $ChangeSignal = -1 Then $add = 4 ; If the number is negative
            $quad = Mod($degrees/90,4)+$add
            Switch $quad
                Case 4
                    $coor[0] = $CircleCenterX
                    $coor[1] = 0
                Case 1
                    $coor[0] = 0
                    $coor[1] = $CircleCenterY
                Case 2
                    $coor[0] = - $CircleCenterX
                    $coor[1] = 0
                Case 3
                    $coor[0] = 0
                    $coor[1] = - $CircleCenterY
            EndSwitch
    EndSwitch
    Return $coor
EndFunc

Edited April 16, 2015 by Kyan

sphexion · April 16, 2015

OK, thanks for the support.
I have figured it out with my previous code after a nights rest, it was more simple then I thought.
The Angle($centerX,$centerY2, $_ang, $radius) is not dependent on the other function.
It will return the coordinates of center axis with designated angel and radius.

Local Const $180_div_pi = 180 / ACos(-1)
$centerX = 19
$centerY = 762
$radius = 160
$targetX =194
$targetY = 922

$_ang = _Angle($targetX,$targetY,$centerX,$centerY) ; test 90deg
$rod = Angle($centerX,$centerY2, $_ang, $radius)

ClipPut($_ang&"o:"&$rod[1] &":"& $rod[2])
ToolTip($_ang&"o:"&$rod[1] &":"& $rod[2])
Sleep(5000)

Func _Angle($tX, $tY, $cX, $cY)
    $mx = $tX - $cX
    $my = $cY - $tY
    $angle = ATan($my / $mx) * $180_div_pi
    If $mx < 0 Then
        $angle = 180 + $angle
    ElseIf $mx >= 0 And $my < 0 Then
        $angle = 360 + $angle
    EndIf
    Return Int($angle)
EndFunc   ;==>_Angle

Func Angle($X1, $Y1, $Ang, $Length)
    Local $Return[3]
    $Return[1] = $X1 + ($Length * Cos($Ang / 180 * 3.14159265358979))
    $Return[2] = $Y1 - ($Length * Sin($Ang / 180 * 3.14159265358979))
    Return $Return
EndFunc   ;==>Angle

sphexion · April 16, 2015

Full test functionality.

For $i = 0 To 360 Step +1
$rod = Angle($centerX,$centerY, $i, $radius)
ToolTip($i&"o:"&$rod[1] &":"& $rod[2])
MouseMove($rod[1] , $rod[2],0)
Sleep(10)
Next

+90 degree offset check

For $i = 0 To 360 Step +1
$angle = $i + 90
$rod = Angle($centerX,$centerY, $angle, $radius)
ToolTip($angel&"o:"&$rod[1] &":"& $rod[2])
MouseMove($rod[1] , $rod[2],0)
Sleep(10)
Next

Edited April 16, 2015 by sphexion

sphexion · April 16, 2015

Now I have given birth to the _PixelSearchCircumference()
I wanted to design a way of searching a radar, witch is outer Circumference for targets to keep from poinless searches and useless scans out of the region.
Now it works perfectly. I am sure it can be twked to search inner Circumference also.

Local Const $180_div_pi = 180 / ACos(-1)

$it = _PixelSearchCircumference(194, 762, 160, 0xFA3C32)
MouseMove($it[1],$it[2],0)
Sleep(5000)

Func _Angle($tX, $tY, $cX, $cY)
    $mx = $tX - $cX
    $my = $cY - $tY
    $angle = ATan($my / $mx) * $180_div_pi
    If $mx < 0 Then
        $angle = 180 + $angle
    ElseIf $mx >= 0 And $my < 0 Then
        $angle = 360 + $angle
    EndIf
    Return Int($angle)
EndFunc   ;==>_Angle

Func Angle($X1, $Y1, $Ang, $Length)
    Local $Return[3]
    $Return[1] = $X1 + ($Length * Cos($Ang / 180 * 3.14159265358979))
    $Return[2] = $Y1 - ($Length * Sin($Ang / 180 * 3.14159265358979))
    Return $Return
EndFunc   ;==>Angle

Func _PixelSearchCircumference($centerX, $centerY, $radius, $color)
For $i = 0 To 360 Step +1
$angle = $i + 90
$rod = Angle($centerX,$centerY, $angle, $radius)
$iColor = PixelGetColor($rod[1] , $rod[2])
If $iColor == $color Then
Return $rod
ExitLoop
EndIf
Next
EndFunc

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Mathematics, coordinates with radius of a circle?

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