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Schwarz lanterns can have mesh tending to zero while their polyhedral areas diverge

Statement refuted

Inscribed triangulated surfaces with mesh tending to zero need not have areas tending to the surface-integral area of the cylinder; their areas can diverge to infinity.

Facts & Assumptions

Given: A cylinder of radius $r > 0$ and height $H > 0$ ; integers $n \geq 2$ and $m \geq 1$ ; and the Schwarz lantern with $m$ horizontal bands, $n$ vertices per ring, and successive rings staggered by angle $π / n$ .

[L1]

An affine parametrization of a nondegenerate triangle over the standard parameter triangle is a regular patch with constant cross-product density; its area is the density times the standard triangle's content $1 / 2$ (Regular parametrized surface patches on compact Jordan parameter regions, Surface area and scalar surface integrals on a regular patch, The surface area density is the norm of the cross product of the parameter tangents, A triangle has content $\frac{1}{2} ∣ \det [B - A C - A] ∣$ , equal to half base times height when the chosen side is nonzero). The cross product has its coordinate formula (The cross product in $R^{3}$ ), and $1 - \cos x = 2 \sin^{2} (x / 2)$ follows from the trigonometric addition and Pythagorean identities (Parity and the Pythagorean identity for sine and cosine, The addition formulas for sine and cosine).

[L2]

One has $\sin x / x \to 1$ as $x \to 0$ , and the algebra and sequential criterion for limits transfer this to the sequences used below (The limit of sin x divided by x at zero is one, Algebra of limits: sums, scalar multiples, products and quotients, Heine criterion: $\lim_{x \to c} f (x) = L$ iff $f (x_{k}) \to L$ for every sequence in $A ∖ {c}$ converging to $c$ , For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L3]

The reciprocal of a positive null sequence diverges to $+ \infty$ in the stated sense (For positive terms, null and divergence to $+ \infty$ are reciprocal, Divergence to $+ \infty$ and to $- \infty$ ).

Counterexample

technique · direct

1.1

Every vertex lies on the cylinder. Parametrize each triangular face affinely over the standard parameter triangle. Its constant parameter tangents are two edge vectors, so [L1] makes its area one half of their cross-product norm. Each of the $m$ bands contains $2 n$ congruent triangles; expanding those edge-vector cross products gives total area $A_{m, n} = 2 r n \sin (π / n) \sqrt{H^{2} + m^{2} r^{2} (1 - \cos (π / n))^{2}} .$

givenL1algebra

1.2

Take $m = n^{3}$ . The maximum edge length is bounded by the sum of the vertical step $H / n^{3}$ and a circular chord of angle at most $2 π / n$ , so the mesh tends to zero by [L2].

givenL2algebra

2.1

By [L1], $n^{2} (1 - \cos (π / n)) = 2 (n \sin (π / (2 n)))^{2} \to π^{2} / 2$ using [L2]. Hence $n^{3} (1 - \cos (π / n))$ grows like $(π^{2} / 2) n$ and diverges by [L3].

step 1.2L1L2L3algebra

3.1

Also $n \sin (π / n) \to π$ by [L2]. Substitution in step 1.1 and step 2.1 shows $A_{n^{3}, n} \to + \infty$ .

step 1.1step 2.1L2L3algebra

4.1

Thus these inscribed lanterns have mesh tending to zero while their areas diverge, proving the stated counterexample.

step 1.2step 3.1∎

Depends on

Used by

FALSE: surface area is the supremum of inscribed polyhedral areas False statement

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Sources

L. Brewin, Curvature corrected estimates for geodesic arc-length, Section 3.2 (standard reference, not scraped)