计算环绕原点的N个3D卡片状平面的最小环绕半径

Alright, let's walk through this problem step by step—no trial-and-error needed, just good old trigonometry and geometric reasoning.

First, let's define our key variables to keep things clear:

• N: Number of 3D planar cards
• W: Width of each card (this is the dimension of the plane that will sit adjacent to neighboring cards; it's perpendicular to the card's normal vector, which points directly toward the origin)
• R: Minimal circumradius we need to calculate (distance from the origin to the center of each card)

Core Geometric Insight #

When all cards are arranged to face the origin and just touch their neighbors without overlapping, we can simplify the problem to a 2D top-down view:

• Each card's center lies on a circle of radius R around the origin.
• The angle between the normal vectors (or the lines connecting the origin to adjacent card centers) is θ = 2π/N (since we're evenly spacing N items around a full 360° circle).
• For adjacent cards to just touch, the half-width of each card (W/2) forms the opposite side of a right triangle, where:
  1. The adjacent side is our target radius R
  2. The angle at the origin is θ/2 = π/N (half the angle between two adjacent card centers)

Deriving the Formula #

Using basic trigonometry for that right triangle:

tan(π/N) = (W/2) / R

Rearranging to solve for R:

R = (W/2) / tan(π/N)

Or using the cotangent function (since cot(x) = 1/tan(x) for cleaner notation):

R = (W/2) * cot(π/N)

Quick Sanity Check #

Let's test this with intuitive cases to confirm it works:

• For N=4 (four cards arranged in a square): cot(π/4) = 1, so R = W/2. This makes perfect sense—each card's center is half the card width away from the origin, so their edges meet exactly at the square's midpoints.
• For N=6 (hexagonal arrangement): cot(π/6) = √3 ≈ 1.732, so R ≈ 0.866 * W. This aligns with hexagonal packing logic, where the radius needs to be slightly larger than half the card width to fit six adjacent cards without overlap.

Notes for Implementation #

• If your cards are rectangular (not square), use the width dimension oriented perpendicular to the card's normal (the one that will be adjacent to neighboring cards).
• This derivation assumes cards are thin (we ignore thickness) and oriented perfectly with normals pointing directly at the origin.

内容的提问来源于stack exchange，提问作者RectangleEquals