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Cylindrical Projection

MediumStitching 📚 CV: Image Alignment and Stitching

You are given a point in planar image coordinates (centered at optical axis) and need to project it onto a cylindrical surface.

Cylindrical projection wraps the image around a cylinder of radius equal to the focal length:

$x' = f \cdot \arctan\left(\frac{x}{f}\right)$ $y' = \frac{f \cdot y}{\sqrt{x^2 + f^2}}$

Where:

$(x, y)$ is the input point (relative to image center)
$f$ is the focal length in pixels
$(x', y')$ is the cylindrical coordinate

Cylindrical projection has two key benefits:

Rotation around the camera's vertical axis becomes pure horizontal translation
This simplifies panorama stitching to 1D alignment

Example:

Input:

point = (100, 50)
f = 500

Output:

(98.6939, 49.0098)

Reasoning:

Applying cylindrical projection formulas:

x' = f × arctan(x/f) = 500 × arctan(100/500) = 500 × arctan(0.2) = 500 × 0.1974 (radians) = 98.6939
y' = f × y / √(x² + f²) = 500 × 50 / √(100² + 500²) = 25000 / √(10000 + 250000) = 25000 / √260000 = 25000 / 509.902 = 49.0098

Result: (98.6939, 49.0098)

The x coordinate shrinks slightly (100 → 98.69) and y also adjusts.

Constraints:

point is (x, y) in planar coordinates, centered at image center
f is the focal length in pixels
Return (x', y') in cylindrical coordinates
Round to 4 decimal places

More from CV: Image Alignment and Stitching

Jacobian for Bundle AdjustmentHard Cylindrical Projection for PanoramasHard Essential Matrix EstimationHard Procrustes AnalysisMedium Gradient Descent UpdateMedium

You are mapping a 2D image point $(x,y)$ (with origin at the image center) to a cylindrical projection using the given formulas, with the cylinder’s radius equal to the focal length $f$ . This is a direct coordinate transform problem that you’ll apply to each pixel when warping an image for panorama stitching.

1. Background Knowledge (Concepts & Theory)

Camera and image coordinates A pinhole camera maps 3D scene points onto a 2D image plane. If you place the origin at the optical axis (image center) and measure in pixels, then $(x,y)$ is the offset from the center and $f$ is the focal length in pixels. Under a simple pinhole model, a 3D point on a ray is projected as:

X_\text{image} = f \cdot \frac{X_\text{cam}}{Z_\text{cam}},\quad Y_\text{image} = f \cdot \frac{Y_\text{cam}}{Z_\text{cam}}

Cylindrical projection Instead of projecting onto a plane, you can imagine projecting rays from the camera onto a cylinder around the camera with radius $f$ . A ray direction can be parameterized by its horizontal angle $\theta$ and vertical elevation relative to the optical axis. Cylindrical projection stores $\theta$ along the horizontal axis and a scaled vertical coordinate along the vertical axis. This leads to:

x' = f \cdot \arctan\left(\frac{x}{f}\right),\quad y' = \frac{f \cdot y}{\sqrt{x^2 + f^2}}

This “unwraps” rotations around the vertical axis into roughly linear horizontal shifts.

Why panoramas use cylinders When you rotate the camera around its vertical axis, the same scene point moves mostly horizontally on the cylindrical image—essentially as a pure translation in $x'$ . That makes image alignment easier: instead of a full 2D homography, alignment reduces mainly to a 1D horizontal shift, simplifying panorama stitching.

2. Algorithm / Approach Pattern

For this problem type (single-point cylindrical projection):

Normalize by focal length: Interpret $\frac{x}{f}$ as the tangent of a viewing angle.
Compute the horizontal angle: Use $\theta = \arctan(x / f)$ , then scale by $f$ to get $x'$ .
Compute vertical scaling: Account for how vertical rays intersect the cylinder using the length of the ray’s horizontal component $\sqrt{x^2 + f^2}$ , then scale $y$ accordingly to get $y'$ .
Return transformed coordinates: $(x',y')$ relative to the cylindrical image’s center.

In a full image warping context, you typically:

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