Knowing the center point, length, width and rotation angle, find the coordinates of the four vertices of the rectangle
- theoretical basis
-
- Case 1:
θ
∈
[
0
,
π
/
2
]
\theta \in [0, \pi/2]
θ∈[0,π/2]
- Case two:
θ
∈
[
π
/
2
,
π
]
\theta \in [\pi/2,\pi]
θ∈[π/2,π]
- Case 1:
- Python code implementation
Theoretical basis
There are several prerequisites for this implementation:
- The known information is as follows: [x_center, y_center, w, h, angle], where the default
w
w
w is the longest side of the rectangle, i.e.
w
>
h
w > h
w>h.
- known rotation angle
θ
\theta
θ is the longest side of the rectangle
w
w
w relative to
x
x
The rotation angle of the x-coordinate axis
- Rotation angle
θ
\theta
The rotation interval of θ is in
[
0
,
π
]
[0, \pi]
[0,π]
The situation can be divided into two types, namely
θ
∈
[
0
,
π
/
2
]
\theta \in [0, \pi/2]
θ∈[0,π/2] and
θ
∈
[
π
/
2
,
π
]
\theta \in [\pi/2, \pi]
θ∈[π/2,π]
Situation 1:
θ
∈
[
0
,
π
/
2
]
\theta \in [0, \pi/2]
θ∈[0,π/2]
Look at the first case first
θ
∈
[
0
,
π
/
2
]
\theta \in [0, \pi/2]
θ∈[0,π/2]:
The center point of the known rectangle
(
x
,
the y
)
(x, y)
(x,y), rotation angle
θ
\theta
θ is marked in orange in the figure.
come first
(
x
1
,
the y
1
)
(x_1,y_1)
(x1?,y1?), the green auxiliary line in the figure is used, and the angles of the triangles used are marked
θ
\theta
θ:
-
x
1
=
x
+
c
o
the s
θ
?
w
/
2
?
the s
i
no
θ
?
h
/
2
x_1=x + cos\theta*w/2-sin\theta*h/2
x1?=x + cosθ?w/2?sinθ?h/2
-
the y
1
=
the y
+
the s
i
no
θ
?
w
/
2
+
c
o
the s
θ
?
h
/
2
y_1=y + sin\theta*w/2 + cos\theta*h/2
y1?=y + sinθ?w/2 + cosθ?h/2
ask again
(
x
2
,
the y
2
)
(x_2,y_2)
(x2?,y2?), using the purple auxiliary line in the figure:
-
x
2
=
x
+
c
o
the s
θ
?
w
/
2
+
the s
i
no
θ
?
h
/
2
x_2=x + cos\theta*w/2 + sin\theta*h/2
x2?=x + cosθ?w/2 + sinθ?h/2
-
the y
2
=
the y
+
the s
i
no
θ
?
w
/
2
?
c
o
the s
θ
?
h
/
2
y_2=y + sin\theta*w/2-cos\theta*h/2
y2?=y + sinθ?w/2?cosθ?h/2
Next
(
x
3
,
the y
3
)
(x_3,y_3)
(x3?,y3?) and
(
x
4
,
the y
4
)
(x_4,y_4)
(x4?,y4?) is
(
x
1
,
the y
1
)
(x_1,y_1)
(x1?,y1?) and
(
x
2
,
the y
2
)
(x_2,y_2)
(x2?,y2?) about
(
x
,
the y
)
(x, y)
The symmetry point of (x, y) only needs to change the positive term into a negative term, and the negative term into a positive term:
-
x
3
=
x
?
c
o
the s
θ
?
w
/
2
+
the s
i
no
θ
?
h
/
2
x_3=x-cos\theta*w/2 + sin\theta*h/2
x3?=x?cosθ?w/2 + sinθ?h/2
-
the y
3
=
the y
?
the s
i
no
θ
?
w
/
2
?
c
o
the s
θ
?
h
/
2
y_3=y-sin\theta*w/2-cos\theta*h/2
y3?=y?sinθ?w/2?cosθ?h/2
-
x
4
=
x
?
c
o
the s
θ
?
w
/
2
?
the s
i
no
θ
?
h
/
2
x_4=x-cos\theta*w/2-sin\theta*h/2
x4?=x?cosθ?w/2?sinθ?h/2
-
the y
4
=
the y
?
the s
i
no
θ
?
w
/
2
+
c
o
the s
θ
?
h
/
2
y_4=y-sin\theta*w/2 + cos\theta*h/2
y4?=y?sinθ?w/2 + cosθ?h/2
Situation 2:
θ
∈
[
π
/
2
,
π
]
\theta \in [\pi/2,\pi]
θ∈[π/2,π]
Similarly, the rotation angle
θ
\theta
θ is greater than
π
/
2
\pi/2
π/2, so the angles of the auxiliary triangles used are marked as
π
?
θ
\pi -\theta
π?θ.
come first
(
x
1
,
the y
1
)
(x_1,y_1)
(x1?,y1?), the green auxiliary line in the figure is used, and the angles of the triangles used are marked
π
?
θ
\pi -\theta
π?θ:
-
x
1
=
x
?
c
o
the s
(
π
?
θ
)
?
w
/
2
+
the s
i
no
(
π
?
θ
)
?
h
/
2
=
x
+
c
o
the s
θ
?
w
/
2
+
the s
i
no
θ
?
h
/
2
x_1=x-cos(\pi-\theta)*w/2 + sin(\pi-\theta)*h/2=x + cos\theta*w/2 + sin\theta* h/2
x1?=x?cos(π?θ)?w/2 + sin(π?θ)?h/2=x + cosθ?w/2 + sinθ?h/2
-
the y
1
=
the y
?
the s
i
no
(
π
?
θ
)
?
w
/
2
?
c
o
the s
(
π
?
θ
)
?
h
/
2
=
the y
?
the s
i
no
θ
?
w
/
2
+
c
o
the s
θ
?
h
/
2
y_1=y-sin(\pi-\theta)*w/2-cos(\pi-\theta)*h/2=y-sin\theta*w/2 + cos\theta* h/2
y1?=y?sin(π?θ)?w/2?cos(π?θ)?h/2=y?sinθ?w/2 + cosθ?h/2
-
x
2
=
x
?
c
o
the s
(
π
?
θ
)
?
w
/
2
?
the s
i
no
(
π
?
θ
)
?
h
/
2
=
x
+
c
o
the s
θ
?
w
/
2
?
the s
i
no
θ
?
h
/
2
x_2=x-cos(\pi-\theta)*w/2-sin(\pi-\theta)*h/2=x + cos\theta*w/2-sin\theta* h/2
x2?=x?cos(π?θ)?w/2?sin(π?θ)?h/2=x + cosθ?w/2?sinθ?h/2
-
the y
2
=
the y
+
the s
i
no
(
π
?
θ
)
?
w
/
2
?
c
o
the s
(
π
?
θ
)
?
h
/
2
=
the y
+
the s
i
no
θ
?
w
/
2
+
c
o
the s
θ
?
h
/
2
y_2=y + sin(\pi-\theta)*w/2-cos(\pi-\theta)*h/2=y + sin\theta*w/2 + cos\theta* h/2
y2?=y + sin(π?θ)?w/2?cos(π?θ)?h/2=y + sinθ?w/2 + cosθ?h/2
(
x
3
,
the y
3
)
(x_3,y_3)
(x3?,y3?) and
(
x
4
,
the y
4
)
(x_4,y_4)
(x4?,y4?) is
(
x
1
,
the y
1
)
(x_1,y_1)
(x1?,y1?) and
(
x
2
,
the y
2
)
(x_2,y_2)
(x2?,y2?) about
(
x
,
the y
)
(x, y)
The symmetry point of (x,y).
It can be seen that in the two cases, the obtained values of the four points are the same, for example, in case 1
(
x
1
,
the y
1
)
(x_1,y_1)
(x1?,y1?) and case two
(
x
2
,
the y
2
)
(x_2,y_2)
(x2?, y2?) are the same, so it can be discussed regardless of the situation in the code implementation, without affecting the final result.
python code implementation
def get_corners(box): #Here my project yaw [-pi/4, 3*pi/4), needs to be mapped to [0, pi)
box = box.detach().cpu().numpy()
x = box[0]
y = box[1]
w = box[2]
l = box[3]
yaw = box[4]
if yaw <0: #used for mapping
yaw = yaw + np.pi
bev_corners = np.zeros((4, 2), dtype=np.float32)
cos_yaw = np.cos(yaw)
sin_yaw = np. sin(yaw)
bev_corners[0, 0] = (w / 2) * cos_yaw - (l / 2) * sin_yaw + x
bev_corners[0, 1] = (w / 2)* sin_yaw + (l / 2) * cos_yaw + y
bev_corners[1, 0] = (l / 2) * sin_yaw + (w / 2) * cos_yaw + x
bev_corners[1, 1] = (w / 2)* sin_yaw - (l / 2) * cos_yaw + y
bev_corners[2, 0] = (-w / 2) * cos_yaw - (-l / 2) * sin_yaw + x
bev_corners[2, 1] = (-w / 2)* sin_yaw + (-l / 2) * cos_yaw + y
bev_corners[3, 0] = (-l / 2) * sin_yaw + (-w / 2) * cos_yaw + x
bev_corners[3, 1] = (-w / 2)* sin_yaw - (-l / 2) * cos_yaw + y
return bev_corners