How to normilize rotation to zero

up vote
-2
down vote

favorite

I have a rotated rectangle on a 2d surface

enter image description here

Known values are:

x,y (upper left corner)

width/height of rectangle

rotation

How can I rotate the rectangle back to zero with the point of origin being the center of the rectangle...

... and then get the new x,y values of the upper left rectangle corner?

enter image description here

example php function:

    function getPositionWithoutRotation(

        float $rotation,

        float $x,

        float $y,

        float $width,

        float $height

    ) {

        $angleRadian = ($rotation * pi()) / 180;



        $xRelativeToCenter = ($width / 2) * cos($angleRadian) - ($height / 2) * sin($angleRadian);

        $yRelativeToCenter = ($width / 2) * sin($angleRadian) + ($height / 2) * cos($angleRadian);



        $cx = $x - $xRelativeToCenter;

        $cy = $y - $yRelativeToCenter;



        $x0 = $cx + ($width / 2);

        $y0 = $cy + ($height / 2);



        return [

            'x0' => $x0,

            'y0' => $y0,

        ];

    }

result:

enter image description here

data:

x = 453

y = 244

w = 139

h = 139

rotation = 16



angleRadian = 0.27925268031909



xRelativeToCenter = 47.650891638432

yRelativeToCenter = 85.964484096995



cx = 405.34910836157

cy = 158.03551590301



x0 = 474.84910836157

y0 = 227.53551590301

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

add a comment |

up vote
-2
down vote

favorite

I have a rotated rectangle on a 2d surface

enter image description here

Known values are:

x,y (upper left corner)

width/height of rectangle

rotation

How can I rotate the rectangle back to zero with the point of origin being the center of the rectangle...

... and then get the new x,y values of the upper left rectangle corner?

enter image description here

example php function:

    function getPositionWithoutRotation(

        float $rotation,

        float $x,

        float $y,

        float $width,

        float $height

    ) {

        $angleRadian = ($rotation * pi()) / 180;



        $xRelativeToCenter = ($width / 2) * cos($angleRadian) - ($height / 2) * sin($angleRadian);

        $yRelativeToCenter = ($width / 2) * sin($angleRadian) + ($height / 2) * cos($angleRadian);



        $cx = $x - $xRelativeToCenter;

        $cy = $y - $yRelativeToCenter;



        $x0 = $cx + ($width / 2);

        $y0 = $cy + ($height / 2);



        return [

            'x0' => $x0,

            'y0' => $y0,

        ];

    }

result:

enter image description here

data:

x = 453

y = 244

w = 139

h = 139

rotation = 16



angleRadian = 0.27925268031909



xRelativeToCenter = 47.650891638432

yRelativeToCenter = 85.964484096995



cx = 405.34910836157

cy = 158.03551590301



x0 = 474.84910836157

y0 = 227.53551590301

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

add a comment |

up vote
-2
down vote

favorite

I have a rotated rectangle on a 2d surface

enter image description here

Known values are:

x,y (upper left corner)

width/height of rectangle

rotation

How can I rotate the rectangle back to zero with the point of origin being the center of the rectangle...

... and then get the new x,y values of the upper left rectangle corner?

enter image description here

example php function:

    function getPositionWithoutRotation(

        float $rotation,

        float $x,

        float $y,

        float $width,

        float $height

    ) {

        $angleRadian = ($rotation * pi()) / 180;



        $xRelativeToCenter = ($width / 2) * cos($angleRadian) - ($height / 2) * sin($angleRadian);

        $yRelativeToCenter = ($width / 2) * sin($angleRadian) + ($height / 2) * cos($angleRadian);



        $cx = $x - $xRelativeToCenter;

        $cy = $y - $yRelativeToCenter;



        $x0 = $cx + ($width / 2);

        $y0 = $cy + ($height / 2);



        return [

            'x0' => $x0,

            'y0' => $y0,

        ];

    }

result:

enter image description here

data:

x = 453

y = 244

w = 139

h = 139

rotation = 16



angleRadian = 0.27925268031909



xRelativeToCenter = 47.650891638432

yRelativeToCenter = 85.964484096995



cx = 405.34910836157

cy = 158.03551590301



x0 = 474.84910836157

y0 = 227.53551590301

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

I have a rotated rectangle on a 2d surface

enter image description here

Known values are:

x,y (upper left corner)

width/height of rectangle

rotation

How can I rotate the rectangle back to zero with the point of origin being the center of the rectangle...

... and then get the new x,y values of the upper left rectangle corner?

enter image description here

example php function:

    function getPositionWithoutRotation(

        float $rotation,

        float $x,

        float $y,

        float $width,

        float $height

    ) {

        $angleRadian = ($rotation * pi()) / 180;



        $xRelativeToCenter = ($width / 2) * cos($angleRadian) - ($height / 2) * sin($angleRadian);

        $yRelativeToCenter = ($width / 2) * sin($angleRadian) + ($height / 2) * cos($angleRadian);



        $cx = $x - $xRelativeToCenter;

        $cy = $y - $yRelativeToCenter;



        $x0 = $cx + ($width / 2);

        $y0 = $cy + ($height / 2);



        return [

            'x0' => $x0,

            'y0' => $y0,

        ];

    }

result:

enter image description here

data:

x = 453

y = 244

w = 139

h = 139

rotation = 16



angleRadian = 0.27925268031909



xRelativeToCenter = 47.650891638432

yRelativeToCenter = 85.964484096995



cx = 405.34910836157

cy = 158.03551590301



x0 = 474.84910836157

y0 = 227.53551590301

php math geometry

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

edited Nov 12 at 10:59

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

asked Nov 12 at 7:53

Adnan Mujkanovic

781112

add a comment |

1 Answer
1

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oldest

votes

up vote
0
down vote

Edit: rechecked signs for left top corner

Essentially you need to determine rectangle (and rotation) center.

Let half-width of rectangle is w, half-height is h. So corner coordinates relative to center after rotation by angle fi are

x' =  -w * cos(fi) - h * sin(fi)

y' =  -w * sin(fi) + h * cos(fi)

and center is

 cx = x - x'

 cy = y - y'

And corner coordinates without rotation:

x0 = cx - w

y0 = cy + h

Python code:

import math

def getPositionWithoutRotation(rotation, x, y, width, height):

    angleRadian = (rotation * math.pi) / 180



    xRelativeToCenter = - (width / 2) * math.cos(angleRadian) - (height / 2) * math.sin(angleRadian)

    yRelativeToCenter = -(width / 2) * math.sin(angleRadian) + (height / 2) * math.cos(angleRadian)



    cx = x - xRelativeToCenter

    cy = y - yRelativeToCenter



    x0 = cx - (width / 2)

    y0 = cy + (height / 2)



    return x0, y0



print(getPositionWithoutRotation(-30, -0.37, 1.37, 2, 2))

print(getPositionWithoutRotation(16, 453, 244, 139, 139))



(-1.0039745962155613, 1.0039745962155615)  //OK

(469.4644840969946, 265.84910836156826)    // ????

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

|
show 4 more comments

Your Answer

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1 Answer
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1 Answer
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oldest

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up vote
0
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Edit: rechecked signs for left top corner

Essentially you need to determine rectangle (and rotation) center.

Let half-width of rectangle is w, half-height is h. So corner coordinates relative to center after rotation by angle fi are

x' =  -w * cos(fi) - h * sin(fi)

y' =  -w * sin(fi) + h * cos(fi)

and center is

 cx = x - x'

 cy = y - y'

And corner coordinates without rotation:

x0 = cx - w

y0 = cy + h

Python code:

import math

def getPositionWithoutRotation(rotation, x, y, width, height):

    angleRadian = (rotation * math.pi) / 180



    xRelativeToCenter = - (width / 2) * math.cos(angleRadian) - (height / 2) * math.sin(angleRadian)

    yRelativeToCenter = -(width / 2) * math.sin(angleRadian) + (height / 2) * math.cos(angleRadian)



    cx = x - xRelativeToCenter

    cy = y - yRelativeToCenter



    x0 = cx - (width / 2)

    y0 = cy + (height / 2)



    return x0, y0



print(getPositionWithoutRotation(-30, -0.37, 1.37, 2, 2))

print(getPositionWithoutRotation(16, 453, 244, 139, 139))



(-1.0039745962155613, 1.0039745962155615)  //OK

(469.4644840969946, 265.84910836156826)    // ????

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

|
show 4 more comments

up vote
0
down vote

Edit: rechecked signs for left top corner

Essentially you need to determine rectangle (and rotation) center.

Let half-width of rectangle is w, half-height is h. So corner coordinates relative to center after rotation by angle fi are

x' =  -w * cos(fi) - h * sin(fi)

y' =  -w * sin(fi) + h * cos(fi)

and center is

 cx = x - x'

 cy = y - y'

And corner coordinates without rotation:

x0 = cx - w

y0 = cy + h

Python code:

import math

def getPositionWithoutRotation(rotation, x, y, width, height):

    angleRadian = (rotation * math.pi) / 180



    xRelativeToCenter = - (width / 2) * math.cos(angleRadian) - (height / 2) * math.sin(angleRadian)

    yRelativeToCenter = -(width / 2) * math.sin(angleRadian) + (height / 2) * math.cos(angleRadian)



    cx = x - xRelativeToCenter

    cy = y - yRelativeToCenter



    x0 = cx - (width / 2)

    y0 = cy + (height / 2)



    return x0, y0



print(getPositionWithoutRotation(-30, -0.37, 1.37, 2, 2))

print(getPositionWithoutRotation(16, 453, 244, 139, 139))



(-1.0039745962155613, 1.0039745962155615)  //OK

(469.4644840969946, 265.84910836156826)    // ????

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

|
show 4 more comments

up vote
0
down vote

Edit: rechecked signs for left top corner

Essentially you need to determine rectangle (and rotation) center.

Let half-width of rectangle is w, half-height is h. So corner coordinates relative to center after rotation by angle fi are

x' =  -w * cos(fi) - h * sin(fi)

y' =  -w * sin(fi) + h * cos(fi)

and center is

 cx = x - x'

 cy = y - y'

And corner coordinates without rotation:

x0 = cx - w

y0 = cy + h

Python code:

import math

def getPositionWithoutRotation(rotation, x, y, width, height):

    angleRadian = (rotation * math.pi) / 180



    xRelativeToCenter = - (width / 2) * math.cos(angleRadian) - (height / 2) * math.sin(angleRadian)

    yRelativeToCenter = -(width / 2) * math.sin(angleRadian) + (height / 2) * math.cos(angleRadian)



    cx = x - xRelativeToCenter

    cy = y - yRelativeToCenter



    x0 = cx - (width / 2)

    y0 = cy + (height / 2)



    return x0, y0



print(getPositionWithoutRotation(-30, -0.37, 1.37, 2, 2))

print(getPositionWithoutRotation(16, 453, 244, 139, 139))



(-1.0039745962155613, 1.0039745962155615)  //OK

(469.4644840969946, 265.84910836156826)    // ????

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

Edit: rechecked signs for left top corner

Essentially you need to determine rectangle (and rotation) center.

Let half-width of rectangle is w, half-height is h. So corner coordinates relative to center after rotation by angle fi are

x' =  -w * cos(fi) - h * sin(fi)

y' =  -w * sin(fi) + h * cos(fi)

and center is

 cx = x - x'

 cy = y - y'

And corner coordinates without rotation:

x0 = cx - w

y0 = cy + h

Python code:

import math

def getPositionWithoutRotation(rotation, x, y, width, height):

    angleRadian = (rotation * math.pi) / 180



    xRelativeToCenter = - (width / 2) * math.cos(angleRadian) - (height / 2) * math.sin(angleRadian)

    yRelativeToCenter = -(width / 2) * math.sin(angleRadian) + (height / 2) * math.cos(angleRadian)



    cx = x - xRelativeToCenter

    cy = y - yRelativeToCenter



    x0 = cx - (width / 2)

    y0 = cy + (height / 2)



    return x0, y0



print(getPositionWithoutRotation(-30, -0.37, 1.37, 2, 2))

print(getPositionWithoutRotation(16, 453, 244, 139, 139))



(-1.0039745962155613, 1.0039745962155615)  //OK

(469.4644840969946, 265.84910836156826)    // ????

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

edited Nov 12 at 11:42

answered Nov 12 at 8:05

MBo

46.5k22848

answered Nov 12 at 8:05

MBo

46.5k22848

answered Nov 12 at 8:05

MBo

46.5k22848

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

|
show 4 more comments

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

actually the center part with the double equal signs is a bit confusing. can the center equation be simplified to single equal signs?
– Adnan Mujkanovic
Nov 12 at 10:01

I just subsituted x' from the first equation, so cx = x - x and cx = x - (w * cos(fi) - h * sin(fi)) are the same
– MBo
Nov 12 at 10:26

i am not sure what im doing wrong but im not getting the result i wanted. link
– Adnan Mujkanovic
Nov 12 at 10:44

Could you show example input data and code used. I might make mistake in signs (depends on coordinate system)
– MBo
Nov 12 at 10:49

I added a code example and result example in the question
– Adnan Mujkanovic
Nov 12 at 10:53

|
show 4 more comments

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