Find the Bounding Rectangle of Rotated Rectangle [duplicate]

Question

I have rectangle with co-ordinates(x1,y1) and (x2,y2) and I have to rotate the rectangle an amount of θ about it centre using Rotation Matrix

 |  cosθ  sinθ |
 | -sinθ  cosθ |

I need to find the co-ordinates of bounding rectangle after rotation.

Before rotation

0,0
 |"""""""""""""""""""""""""""""""""""""""""""|
 |                                           |
 |  x1,y1                                    |
 |       |"""""""""""""|                     |
 |       |             |                     |
 |       |             |                     | 
 |       |             |                     |
 |       """""""""""""" x2,y2                |
 |                                           |
 |                                           |
  """"""""""""""""""""""""""""""""""""""""""" W,H

After rotation

 0,0
     |"""""""""""""""""""""""""""""""""""""""""""|
     |           ?,?                             |
     |            |""""/\"""""|                  |      
     |            |   /   \   |                  |
     |            |  /      \ |                  |
     |            | /        /|                  |
     |            |/        / |                  |
     |            |\       /  |                  |
     |            |  \    /   |                  |
     |            |    \ /    |                  |
     |             """""""""""  ?,?              |
     |                                           |
     |                                           |
      """"""""""""""""""""""""""""""""""""""""""" W,H

Is there any general equation for finding the co-ordinates of bounding rectangle?.

Thanks....

Haris.

Answer 1

Just mark all Fi angles on your drawing, and you can see that

Old_Width = X2_Old - X1_Old, Old_Height = Y2_Old - Y1_Old
New_Height = Old_Width * Abs(Sin(Fi)) + Old_Height * Abs(Cos(Fi))
New_Width = Old_Width * Abs(Cos(Fi)) + Old_Height * Abs(Sin(Fi))
X1_New = X1_Old - (New_Width - OldWidth) / 2 = 
         (X1_Old + X2_Old - New_Width) / 2

Delphi test:

procedure TForm1.DrawRotatedRectWithFrame(X0, Y0, X1, Y1: Integer; Fi: Double);
var
  P: array[0..3] of TPoint;
  CX, CY, WX, WY, NW, NH : Integer;
  CF, SF: Double;
begin
  CX := (X0 + X1) div 2;  //Center point
  CY := (Y0 + Y1) div 2;
  WX := (X1 - X0) div 2;  //Half-width
  WY := (Y1 - Y0) div 2;
  SinCos(Fi, SF, CF);
  //calculate vertices of rotated rectangle
  P[0] := Point(Round(CX  -WX*CF + WY * SF), Round(CY - WX * SF - WY * CF));
  P[1] := Point(Round(CX  +WX*CF + WY * SF), Round(CY + WX * SF - WY * CF));
  P[2] := Point(Round(CX  +WX*CF - WY * SF), Round(CY + WX * SF + WY * CF));
  P[3] := Point(Round(CX  -WX*CF - WY * SF), Round(CY - WX * SF + WY * CF));
  Canvas.Polygon(P); //draw rotated rectangle

  Canvas.Rectangle(CX - 2, CY - 2, CX + 3, CY + 3); //mark center point
  NH := Round(Abs(WX * SF) + Abs(WY * CF));  //boundrect half-height
  NW := Round(Abs(WX * CF) + Abs(WY * SF));  //boundrect half-width
  Canvas.Brush.Style := bsClear;
  Canvas.Rectangle(CX - NW, CY - NH, CX + NW, CY + NH); //draw bound rectangle
end;

Output example:

Answer 2

Point (x₁, y₁) rotates to (x₁ cos θ - y₁ sin θ, x₁ sin θ + y₁ cos θ), while point (x₂, y₂) rotates to (x₂ cos θ - y₂ sin θ, x₂ sin θ + y₂ cos θ). The other two points can be calculated accordingly.

The coordinates of the bounding reactangle are (x₃, y₃) and (x₄, y₄), where x₃ is the smallest of all new x coordinates, y₃ the smallest of all new y coordinates, x₄ the greatest of all new x coordinates and y₄ the greatest of all new y coordinates.

Which of the corners produces the smallest x (and so on) depends on your angle or rotation. For angles from 0° to 90°, x₃ will come from (x₁, y₁), so x₃ = x₁ cos θ - y₁ sin θ. For angles from 90° to 180°, it will come from (x₂, y₁), and so on. So either you decide which points to use based on your angle of rotation, or you just take the smallest and greatest of all x's and y's.

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But I think you should probably ask this on https://math.stackexchange.com/