Image Transformations: Affine, Rotation, Homography

Why Does This Exist?

Photos never line up by themselves. A phone tilts, a drone drifts, and a scanned page skews, so pixels from two views sit in different coordinate frames. Without a warp model there is no panorama, no stabilization, and no measurement from images.

All warps share one language. Points become homogeneous triples $[x, y, 1]$, warps become $3 \times 3$ matrices, and the output is the matrix times the point, divided by the last coordinate. That language feeds homogeneous coordinates and camera models, and resampling the warped grid needs interpolation.

The one idea you need first

Append a $1$ to every point. Translation is not linear in $[x, y]$, but it is linear in $[x, y, 1]$ because the extra column carries the shift. The last row decides the family: $[0, 0, 1]$ keeps parallels parallel, anything else lets them meet.

Think of It Like This

How It Actually Works

Write a point as $[x, y, 1]$ and a warp as $x' = H x$, then divide by the third entry. Translation uses ones on the diagonal and shifts in the last column. Rotation by angle $\theta$ uses $c = \cos\theta$ and $s = \sin\theta$ in the top left $2 \times 2$ block. Scale multiplies axes by $s_x$ and $s_y$. Affine adds shear with six free numbers and keeps its last row $[0, 0, 1]$. Homography frees the last row, so division by $w$ bends lines.

In plain words: $t_x$ shifts right, $t_y$ shifts down, $c$ and $s$ turn the axes, and $h_{31}$ and $h_{32}$ control perspective tilt. Every symbol is a knob with units in pixels except the rotation entries, which are unitless cosines and sines.

Worked example

Translate $[10, 20]$ by $t_x = 5$ and $t_y = -3$: $x' = 10 + 5 = 15$, $y' = 20 + (-3) = 17$. The point lands at $[15, 17]$.

Rotate $[3, 1]$ by $90$ degrees counterclockwise. Here $c = 0$ and $s = 1$, so $x' = c x - s y = 0 \times 3 - 1 \times 1 = -1$ and $y' = s x + c y = 1 \times 3 + 0 \times 1 = 3$. The point lands at $[-1, 3]$, exactly a quarter turn.

Apply perspective $H$ with rows $[1, 0, 0]$, $[0, 1, 0]$, $[0.1, 0, 1]$ to $[2, 1]$. In homogeneous form $[2, 1, 1]$ maps to $[2, 1, 0.1 \times 2 + 1] = [2, 1, 1.2]$. Divide: $x' = 2 / 1.2 \approx 1.67$, $y' = 1 / 1.2 \approx 0.83$. An affine matrix would leave the denominator at $1$ and keep parallels parallel. The $1.2$ is the whole difference, and the diagram draws this $1.67$ beside the trapezoid.

Code

The point is the divide. Forgetting the homogeneous normalization applies the tilt to the numerator and skips it in the denominator.

1import numpy as np2
3
4def warp_point(H: np.ndarray, pt: np.ndarray) -> np.ndarray:5    p = np.array([pt[0], pt[1], 1.0])6    q = H @ p7    return q[:2] / q[2]  # divide by w8
9
10H_affine = np.array([[0.0, -1.0, 5.0], [1.0, 0.0, -3.0], [0.0, 0.0, 1.0]])11print(warp_point(H_affine, np.array([3.0, 1.0])))  # -> [ 4. -0.]12H_persp = np.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.1, 0.0, 1.0]])13print(np.round(warp_point(H_persp, np.array([2.0, 1.0])), 2))  # -> [1.67 0.83]

The first line chains the worked rotation with the worked translation: $[3, 1]$ turns to $[-1, 3]$ then shifts to $[4, 0]$. The second line reproduces $1.67$ and $0.83$ from the hand computation.

Watch Out For

The Quick Version

• Points are $[x, y, 1]$, warps are $3 \times 3$, output is $Hx$ divided by $w$.
• Translation adds: $10, 20$ with $5, -3$ gives $15, 17$.
• Rotation by 90 degrees sends $3, 1$ to minus $1, 3$.
• Affine keeps last row $[0, 0, 1]$ and preserves parallels.
• Homography frees the last row: $2, 1$ with tilt $0.1$ gives $1.67, 0.83$ and bends parallels.
• Divide by $w$ every time, and interpolate rather than round fractional positions.

What to Read Next

• Homogeneous Coordinates and Projective Geometry
• Camera Models: Pinhole, Intrinsic/Extrinsic
• Interpolation Methods