Pinhole Camera Model

What is a pinhole camera?

The pinhole camera is the simplest camera model: light enters through a small hole (the pinhole) and forms an image on a screen on the opposite side. Because no lens is used, it serves as an idealized geometric model widely used in CV and CG (here the hole is treated as a single point; blur from the finite hole size and diffraction are ignored).

Figure 1. 3D-to-2D projection by the pinhole camera model. (Parallel lines converge to a vanishing point when they are not parallel to the image plane. The X axis points out of the page, giving a right-handed frame.)

Mathematics of perspective projection

The perspective projection of a 3D point $P = (X, Y, Z)$ to an image-plane point $p = (x, y)$ follows from the geometry of similar triangles:

Note on convention: Figure 1 shows the physical pinhole geometry (image plane behind the camera, at $-Z$). For the derivations that follow, we instead use the virtual image plane model (image plane in front of the camera, at $+Z$, i.e., $Z = f$) shown in Figure 2. The two models differ only in whether the image is flipped; the similar-triangle proportions are identical (with signs included, the physical image plane gives $x = -fX/Z,\ y = -fY/Z$). The virtual image plane is the standard convention in CV/CG textbooks because it removes the sign flips and is easier to manipulate.

Figure 2. Similar-triangle relation for perspective projection (virtual image plane model, Y-Z plane).

Perspective projection formulas

Projection of a 3D point $(X, Y, Z)$ given in camera coordinates (origin at the pinhole, $+Z$ along the optical axis toward the scene; $Z > 0$) onto the image-plane point $(x, y)$:

$$x = f \dfrac{X}{Z}, \quad y = f \dfrac{Y}{Z}$$

where $f$ is the distance from the pinhole to the image plane (called the focal length by convention; a pinhole has no light-gathering focus) and $Z$ is the depth.

Important: Because the formulas divide by $Z$, perspective projection is not a linear transformation in the usual sense (it cannot be written as a single matrix product). However, if we use homogeneous coordinates ($(X, Y, Z) \to (X, Y, Z, 1)$, covered in the next chapter), the transformation can be split into (1) a part expressible as a matrix product and (2) a final division by $Z$ (normalization of the homogeneous coordinate). The non-linearity is confined to step (2), and step (1) lets us treat rotation, translation, and projection in a unified matrix framework.

Field of View (FOV)

The field of view (FOV) is the angular extent of the scene that the camera captures. It is determined by the focal length and the sensor size.

Definition: Field of View (FOV)

The maximum angle of the scene captured by the camera. Centered on the optical axis, $\theta$ is the angle between the two rays from the pinhole $O$ to the two ends of the image plane (sensor).

Depending on whether the width $w$, height $h$, or diagonal $d$ of the sensor is used, one obtains the horizontal, vertical, or diagonal field of view (different standards and use cases prefer different ones).

Figure 3 illustrates how the FOV changes when the focal length $f$ varies.

Figure 3. Sensor height fixed; only the focal length $f$ varies. The shorter $f$ is (pinhole closer to the sensor), the dramatically wider the FOV $\theta$ becomes (half angles 45°→24°→12°, a 4× range).

Deriving the FOV formula

The FOV follows from a simple right triangle. Because the optical axis is perpendicular to the sensor, the points $O$ (pinhole), the sensor center, and the top edge of the sensor form a right triangle.

Figure 4. FOV derivation geometry — from the right triangle $O$-(sensor center)-(top sensor edge) we read off the half-angle $\theta/2$.

In Figure 4 the adjacent leg of the right triangle is the focal length $f$ and the opposite leg is the sensor half-width $w/2$ (the figure is drawn vertically, but the same formula holds for the width or the height; for the vertical direction read $w$ as the height $h$). For the half angle $\theta/2$ we therefore have

$$\tan\!\left(\dfrac{\theta}{2}\right) = \dfrac{w/2}{f} = \dfrac{w}{2f}.$$

Taking $\arctan$ on both sides and doubling gives the full FOV $\theta$.

FOV formula

$$\boxed{\;\theta = 2\,\arctan\!\left(\dfrac{w}{2f}\right)\;}$$

Substitute the sensor width for $w$ to get the horizontal FOV, the height for the vertical FOV, or the diagonal $d$ for the diagonal FOV.

Larger $f$ ⇒ smaller $\theta$ (telephoto); smaller $f$ ⇒ larger $\theta$ (wide angle).

This formula assumes the principal point is at the center of the sensor. If it is offset, compute the angles on each side separately and add them.

Principal point and skew

In real cameras the optical axis does not always pass through the center of the image.

Figure 5. Principal point and the optical axis. (The pixel origin is at the top-left of the image; $u$ points right and $v$ points down.)

Conversion to pixel coordinates

From normalized image coordinates $(x_n, y_n) = (X/Z,\ Y/Z)$ (the image-plane point $(x, y)$ divided by the focal length $f$) to pixel coordinates $(u, v)$:

$$u = f_x\, x_n + c_x, \quad v = f_y\, y_n + c_y$$

$f_x, f_y$: focal length in pixel units ($f_x = f / p_x$, $f_y = f / p_y$, where $p_x, p_y$ are the horizontal and vertical pixel sizes)
$(c_x, c_y)$: pixel coordinates of the principal point

Here the pixel axes are assumed orthogonal, so the skew (the (1,2) entry $s$ of $K$) is 0. If $s \neq 0$, then $u = f_x\, x_n + s\, y_n + c_x$. For modern cameras it can be treated as essentially 0.

Combined camera model

Combining perspective projection with the pixel coordinate transform:

$$\begin{pmatrix} u \\ v \\ 1 \end{pmatrix} = \dfrac{1}{Z} \begin{pmatrix} f_x & 0 & c_x \\ 0 & f_y & c_y \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \end{pmatrix}$$

The matrix product on the right evaluates to $(f_x X + c_x Z,\; f_y Y + c_y Z,\; Z)^\top$; multiplying by $1/Z$ (normalizing the homogeneous coordinate) sets the third component to 1. Reading off the first two rows gives the scalar form:

$$u = f_x \dfrac{X}{Z} + c_x, \quad v = f_y \dfrac{Y}{Z} + c_y$$

This matrix is called the camera's intrinsic matrix $K$; we treat it in detail in the next chapter.

Code example (Python)

import numpy as np

# Camera parameters
fx, fy = 800, 800  # focal length (pixel units)
cx, cy = 320, 240  # principal point
f = 50             # focal length (mm)

# 3D point (camera coordinates)
P = np.array([1.0, 0.5, 5.0])  # (X, Y, Z) in meters

# Perspective projection
x = P[0] / P[2]  # X/Z
y = P[1] / P[2]  # Y/Z

# Convert to pixel coordinates
u = fx * x + cx
v = fy * y + cy
print(f"Pixel coordinates: ({u:.1f}, {v:.1f})")

# Field of view
sensor_width = 36  # mm (full frame; an example value chosen independently of fx and f above)
fov = 2 * np.arctan(sensor_width / (2 * f))
print(f"Field of View: {np.degrees(fov):.1f}°")

# Matrix-form projection
K = np.array([
    [fx, 0, cx],
    [0, fy, cy],
    [0,  0,  1]
])

P_camera = P  # 3D point in camera coordinates (the homogeneous vector is the 2D image point on the next line)
p_homogeneous = K @ P_camera
p = p_homogeneous[:2] / p_homogeneous[2]
print(f"Pixel (matrix): ({p[0]:.1f}, {p[1]:.1f})")

Summary

The pinhole camera is the idealized model of perspective projection.
Perspective projection: $x = fX/Z$, $y = fY/Z$ (a non-linear transform).
Smaller focal length ⇒ wider field of view.
The principal point is the intersection of the optical axis and the image plane (often offset from the image center).
The intrinsic matrix $K$ packages the projection in matrix form.

Frequently Asked Questions

What is the pinhole camera model?

The pinhole camera is the simplest camera model: light enters through a small hole (the pinhole) and forms an image on a screen on the opposite side. Treating the hole as a single point, it serves as an idealized geometric model widely used in computer vision and computer graphics. A 3D point $P(X, Y, Z)$ in camera coordinates ($Z > 0$) is mapped to the image point $p(x, y)$ by $x = fX/Z,\ y = fY/Z$, where $f$ is the distance from the pinhole to the image plane (the focal length).

What is the difference between the virtual image plane model and physical pinhole geometry?

In the physical pinhole geometry the image plane lies behind the camera (at $-Z$) and the image is flipped both vertically and horizontally. In the virtual image plane model the image plane is placed in front of the camera (at $+Z$, i.e., $Z = f$), removing the inversion. The similar-triangle relations are identical, but with signs included the physical image plane gives $x = -fX/Z$. The virtual image plane is more convenient (no sign flips) and is therefore the standard convention for deriving formulas in CV/CG.

How is the perspective projection formula x = fX/Z derived?

Consider the ray from the camera center $O$ through a 3D point $P(X, Y, Z)$ and let $p(x, y)$ be its intersection with the image plane at $Z = f$. The two right triangles formed by $O$, the image point and the 3D point give the proportions $x : f = X : Z$ and $y : f = Y : Z$. Rearranging yields $x = fX/Z,\ y = fY/Z$.

Why does perspective projection lose depth information?

Perspective projection divides by $Z$ in $x = fX/Z,\ y = fY/Z$, so the points $(X, Y, Z)$ and $(kX, kY, kZ)$ for any $k > 0$ map to the same 2D image point: every point on a single ray collapses to the same image point. Therefore the original $Z$ (depth) cannot be recovered uniquely from a single image point alone. Recovering it needs extra information, such as images from other viewpoints (stereo, structure from motion), an object of known size, or a depth sensor.