Perspective Transform (Bird's-Eye View)

cv2.getPerspectiveTransform computes the 3x3 homography matrix that maps 4 source points to 4 destination points, letting you warp a skewed view (e.g. a road seen at an angle) into a top-down “bird’s-eye” view.

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Compute the Matrix

import numpy as np
import cv2
 
src = np.float32([
    (sx1, sy1),
    (sx2, sy2),
    (sx3, sy3),
    (sx4, sy4),
])
dst = np.float32([
    (dx1, dy1),
    (dx2, dy2),
    (dx3, dy3),
    (dx4, dy4),
])
 
map_matrix = cv2.getPerspectiveTransform(src, dst)

Apply it with cv2.warpPerspective(image, map_matrix, (width, height)).

The Math

Each source point (x_i, y_i) maps to a destination point (x_i', y_i') through the homogeneous matrix, with t_i a per-point scale factor that keeps the third coordinate normalized to 1:

$$\left[\begin{array}{c}{t}_i\cdot x_i^{\prime }\\ t_i\cdot y_i^{\prime }\\ t_i\end{array}\right]=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& j\end{array}\right]\cdot \left[\begin{array}{c}{x}_i\\ y_i\\ 1\end{array}\right]\forall i\in \{1,2,3,4\}$$

Expanding the matrix product gives 3 equations per point:

$$\{\begin{array}{l}{t}_i\cdot x_i^{\prime }=a{x}_i+b{y}_i+c\\ t_i\cdot y_i^{\prime }=d{x}_i+e{y}_i+f\\ t_i=g{x}_i+h{y}_i+j\end{array}$$

Substituting t_i out turns this into the standard projective transform, non-linear in the unknowns because of the shared denominator:

$$\{\begin{array}{l}{x}_i^{\prime }=\frac{a{x}_i+b{y}_i+c}{g{x}_i+h{y}_i+j}\\ y_i^{\prime }=\frac{d{x}_i+e{y}_i+f}{g{x}_i+h{y}_i+j}\end{array}$$

With 4 point pairs (8 equations) and 8 unknowns (j is fixed to 1 by convention, since the matrix is only defined up to scale), the system is fully determined — this is why getPerspectiveTransform requires exactly 4 points, no more, no less.

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Cheatsheet

Compute the Matrix

map_matrix = cv2.getPerspectiveTransform(src, dst)  # src, dst: np.float32 arrays of 4 points

Apply the Matrix

warped = cv2.warpPerspective(image, map_matrix, (width, height))