Introduction: Balance in Inequality
Over five centuries ago, Albrecht Dürer contributed two enduring landmarks at the intersection of mathematics and the visual arts: the order-four magic square in his 1514 engraving Melencolia I (square #175 in Frénicle’s 1693 census of the 880 order-four magic squares [8 , 6 , 1 ] ) and the first printed polyhedral net unfoldings in his 1525 treatise Underweysung der Messung [7 , 18 , 5 , 9 ] . A magic square is a quintessential study in balance in inequality : an arrangement S = ( s i j ) 0 ≤ i , j < n S=(s_{ij})_{0\le i,j<n} S = ( s ij ) 0 ≤ i , j < n N = n 2 N=n^2 N = n 2 { 1 , … , N } \{1,\ldots,N\} { 1 , … , N } μ n = n ( N + 1 ) / 2 \mu_n=n(N+1)/2 μ n = n ( N + 1 ) /2 semi-magic square if the row and column sums equal μ n \mu_n μ n s ˉ = ( N + 1 ) / 2 \bar{s}=(N+1)/2 s ˉ = ( N + 1 ) /2
x j = j − n − 1 2 , y i = n − 1 2 − i , z i j = s i j − N + 1 2 , (1) x_j=j-\frac{n-1}{2},\qquad y_i=\frac{n-1}{2}-i,\qquad z_{ij}=s_{ij}-\frac{N+1}{2}, \tag*{(1)} x j = j − 2 n − 1 , y i = 2 n − 1 − i , z ij = s ij − 2 N + 1 , (1) lifts each cell ( i , j ) (i,j) ( i , j ) O = ( 0 , 0 , 0 ) O=(0,0,0) O = ( 0 , 0 , 0 )
p i j ( λ ) = ( x j , y i , λ z i j ) ∈ R 3 , P S ( λ ) = { p i j ( λ ) : 0 ≤ i , j < n } , \mathbf{p}_{ij}(\lambda)=(x_j,y_i,\lambda z_{ij})\in\mathbb{R}^3,\qquad P_S(\lambda)=\{\mathbf{p}_{ij}(\lambda):0\le i,j<n\}, p ij ( λ ) = ( x j , y i , λ z ij ) ∈ R 3 , P S ( λ ) = { p ij ( λ ) : 0 ≤ i , j < n } , where λ > 0 \lambda>0 λ > 0 Figure 3 (a)). The convex hull
Figure 3. Three-dimensional radial spoke sculptures from the origin O = ( 0 , 0 , 0 ) O=(0,0,0) O = ( 0 , 0 , 0 ) p i j \mathbf{p}_{ij} p ij β = 1 \beta=1 β = 1 z i j > 0 z_{ij}>0 z ij > 0 z i j < 0 z_{ij}<0 z ij < 0 O O O S 2 S^2 S 2 β = 0 \beta=0 β = 0 V 0 = ( 0 , 0.773 , 0 ) \mathbf{V}_0=(0,0.773,0) V 0 = ( 0 , 0.773 , 0 ) w i j ∗ ∈ [ 0.72 , 1.19 ] w^*_{ij}\in[0.72,1.19] w ij ∗ ∈ [ 0.72 , 1.19 ] w i j ∗ > 1 w^*_{ij}>1 w ij ∗ > 1 ∑ w i j ∗ p ^ i j = 0 \sum w^*_{ij}\hat{\mathbf{p}}_{ij}=\mathbf{0} ∑ w ij ∗ p ^ ij = 0 ∥ V β ( S ; 1 ) ∥ 2 / R ˉ β − 1 \|\mathbf{V}_{\beta}(S;1)\|_2/\bar{R}^{\beta-1} ∥ V β ( S ; 1 ) ∥ 2 / R ˉ β − 1 β \beta β β = 1 \beta=1 β = 1 β ∗ ≠ 1 \beta^*\ne1 β ∗ = 1 ( V 3 ) z = 0 (V_3)_z=0 ( V 3 ) z = 0 β = 3 \beta=3 β = 3
G λ ( S ) : = conv P S ( λ ) ⊂ R 3 ( G ( S ) : = G 1 ( S ) ) \mathcal{G}_{\lambda}(S):=\operatorname{conv}P_S(\lambda)\subset\mathbb{R}^3\qquad(\mathcal{G}(S):=\mathcal{G}_1(S)) G λ ( S ) := conv P S ( λ ) ⊂ R 3 ( G ( S ) := G 1 ( S )) is a three-dimensional convex polyhedron that we call the Magic Gem of S S S
In the flat grid plane, Loly [11 ] and Rogers and Loly [16 , 17 ] showed that interpreting the unequal entries s i j s_{ij} s ij r i j = ( x j , y i ) r_{ij}=(x_j,y_i) r ij = ( x j , y i ) O O O ∑ i , j z i j r i j = 0 \sum_{i,j}z_{ij}r_{ij}=0 ∑ i , j z ij r ij = 0 ∑ i , j s i j ( x j 2 + y i 2 ) = n 2 ( n 4 − 1 ) / 12 \sum_{i,j}s_{ij}(x_j^2+y_i^2)=n^2(n^4-1)/12 ∑ i , j s ij ( x j 2 + y i 2 ) = n 2 ( n 4 − 1 ) /12 n n n [12 ] —its unequal entries are subjected to progressively nonlinear geometric transformations. This motivates our central hypothesis: combinatorial line-sum equality guarantees physical equilibrium whenever a mechanical observable is linear or symmetric-quadratic in the grid coordinates, neutralizing entry inequality universally across all 880 order-four magic squares; when a nonlinear geometric constraint breaks that equilibrium, balance can be restored by pairing two opposing inequalities that cancel each other.
We test this hypothesis across four progressive mechanical regimes, illustrated on six recurring flagship squares (Figure 1 ). In § 2 , we examine one-dimensional radial spoke sculptures radiating from the origin O O O O O O every square in 2D, where the middle grid ring sits at an irrational radius ratio r 2 / r 1 = 5 r_2/r_1=\sqrt{5} r 2 / r 1 = 5 r 3 / r 1 = 3 r_3/r_1=3 r 3 / r 1 = 3 9 / 4 ≈ 5 9/4\approx\sqrt{5} 9/4 ≈ 5 Figure 1 (b), Theorem 2.3 ). By length–weight duality (Proposition 2.2 ), equilibrium is restored either by playing aligned inner and outer ring dipoles against the middle ring via a power-law length profile r α ∗ r^{\alpha*} r α ∗ α ∗ = − 1 \alpha^*=-1 α ∗ = − 1 Figure 1 (c), Corollary 2.4 ), or by assigning positive unequal tip weights w ∗ > 0 \mathbf{w}^*>0 w ∗ > 0 S 2 S^2 S 2 β = 3 \beta=3 β = 3 Theorem 2.5 ).
Figure 1. Visual anatomy of the six flagship order-four magic squares examined across § 2 –§ 5 . (a) Dürer’s associative square #175 (1514, Group III), whose antipodal pairs sum to 17 across O O O V 0 V_0 V 0 V 1 V_1 V 1 R i n , R m i d , R o u t R_{\mathrm{in}},R_{\mathrm{mid}},R_{\mathrm{out}} R in , R mid , R out d o u t = 5 d i n d_{\mathrm{out}}=5d_{\mathrm{in}} d out = 5 d in 9 / 4 ≈ 5 9/4 \approx\sqrt{5} 9/4 ≈ 5 κ = + 9 \kappa=+9 κ = + 9 α ∗ = − 1 \alpha^*=-1 α ∗ = − 1 z i j x j y i z_{ij}x_jy_i z ij x j y i ∣ Q x y ∣ = 60 |Q_{xy}|=60 ∣ Q x y ∣ = 60 180 ∘ 180^\circ 18 0 ∘ y = − x y=-x y = − x V 0 V_0 V 0 V 1 V_1 V 1 Δ ρ 1 / ρ 0 = 5 / 4 \Delta\rho_1/\rho_0=5/4 Δ ρ 1 / ρ 0 = 5/4 { 13 , 14 , 15 , 16 } \{13,14,15,16\} { 13 , 14 , 15 , 16 } { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 }
In § 3 , we set balanced sculptures into three-dimensional rotation on a central pivot under gravity. When the sixteen unequal vertices p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) λ i s o = ( n 2 + 1 ) − 1 / 2 \lambda_{\mathrm{iso}}=(n^2+1)^{-1/2} λ iso = ( n 2 + 1 ) − 1/2 1 / 17 1/\sqrt{17} 1/ 17 n = 4 n=4 n = 4 Theorem 3.1 ). Replacing equal masses with the unequal magic entries m i j = s i j m_{ij}=s_{ij} m ij = s ij Theorem 3.2 ) breaks this rotational symmetry by exposing the planar quadrupole moment Q x y = ∑ z i j x j y i Q_{xy}=\sum z_{ij}x_jy_i Q x y = ∑ z ij x j y i § 2 , splitting the horizontal principal axes by 2 ∣ Q x y ∣ 2|Q_{xy}| 2∣ Q x y ∣ ∣ Q x y ∣ = 60 |Q_{xy}|=60 ∣ Q x y ∣ = 60 Figure 1 (d)).
In § 4 , we pass from discrete point clouds to continuous polyhedral hulls G 0 ( S ) = conv P S \mathcal{G}_0(S)=\operatorname{conv}P_S G 0 ( S ) = conv P S G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) O O O Proposition 4.1 ) because interior grid points P S ( 1 ) P_S^{(1)} P S ( 1 ) ∑ P i j = 0 \sum P_{ij}=0 ∑ P ij = 0 V 0 p o u t e r ‾ = − V 1 p i n n e r ‾ \overline{V_0p_{\mathrm{outer}}}=-\overline{V_1p_{\mathrm{inner}}} V 0 p outer = − V 1 p inner Proposition 4.2 ). Physically re-engaging the inner core G 1 ( S ) = conv P S ( 1 ) \mathcal{G}_1(S)=\operatorname{conv}P_S^{(1)} G 1 ( S ) = conv P S ( 1 ) c = 0 \mathbf{c}=0 c = 0 Figure 1 (c,Figure 1 e), Theorem 4.3 ), single-material two-gauge wireframes and complete K 16 K_{16} K 16 Theorem 4.4 ), and unfolded papercraft shells with f 0 − 1 f_0-1 f 0 − 1 Proposition 4.5 ).
Finally, in § 5 , we remove the central pivot and place Magic Gems onto a flat horizontal table under gravity. While tall λ = 1 \lambda=1 λ = 1 λ c u b e = 1 / 5 \lambda_{\mathrm{cube}=1/5} λ cube = 1/5 Proposition 5.2 ). When stacking multiple blocks vertically, identical copies of a single gem wedge apart whenever the top and bottom facet normals fail to oppose (n − ≠ − n + \mathbf{n}_{-}\ne-\mathbf{n}_{+} n − = − n + G λ ( S ) \mathcal{G}_\lambda(S) G λ ( S ) G λ ( r 2 S ‾ ) = − G λ ( S ) \mathcal{G}_\lambda(r^2\overline{S})=-\mathcal{G}_\lambda(S) G λ ( r 2 S ) = − G λ ( S ) Theorem 5.3 ) pairs every slanted facet with an equal and opposite partner, locking chiral knight’s-move blocks (#763, Figure 1 (f)) into a straight vertical column (Proposition 5.4 ).
Across all four regimes, the order-four censuses stratify into three symmetry tiers (§ 6 ): universal properties governed by linear line sums hold across all 880 squares; unweighted nonlinear cancellations hold on the 48 associative squares of Dudeney’s Group III (r 2 S = S ‾ r^2S=\overline{S} r 2 S = S p − c = − p c \mathbf{p}_{-c}=-\mathbf{p}_c p − c = − p c S ‾ = h S \overline{S}=hS S = h S ν S \nu S ν S 180 ∘ 180^\circ 18 0 ∘ C 2 ⊂ S O ( 3 ) C_2\subset SO(3) C 2 ⊂ S O ( 3 ) anc/verify_paper_c.py (Section A) and accompanied by real-time 3D simulations in web/index.html.
Throughout the paper, A n = { ( x j , y i ) : 0 ≤ i , j < n } A_n=\{(x_j,y_i):0\le i,j<n\} A n = {( x j , y i ) : 0 ≤ i , j < n } (1) , a n = ( n 2 − 1 ) / 12 a_n=(n^2-1)/12 a n = ( n 2 − 1 ) /12 b n = ( n 4 − 1 ) / 12 = ( n 2 + 1 ) a n b_n=(n^4-1)/12=(n^2+1)a_n b n = ( n 4 − 1 ) /12 = ( n 2 + 1 ) a n x j x_j x j z i j z_{ij} z ij D 4 = ⟨ r , h ⟩ D_4=\langle r,h\rangle D 4 = ⟨ r , h ⟩ r r r h , v h,v h , v τ ( S ) = S ‾ = ( N + 1 ) 1 − S \tau(S)=\overline{S}=(N+1)\mathbf{1}-S τ ( S ) = S = ( N + 1 ) 1 − S [14 , 13 ] treat the static polyhedral combinatorics and bivariate Prouhet–Tarry–Escott moment hierarchy of semi-magic squares; the present paper is self-contained. Order-four magic squares are numbered #1, …, #880 by their 1-based position in Frénicle’s standard list (anc/squares880.txt).
Radial Spokes from the Origin and Length–Weight Duality
In the fabrication suite magicGemWeb [12 ] , the sixteen cells of an order-four magic square are built as two radial sculptures pivoted at the origin O O O magicGem.scad, Figure 3 (a)) with spokes p i j ( λ ) = ( x j , y i , λ z i j ) \mathbf{p}_{ij}(\lambda)=(x_j,y_i,\lambda z_{ij}) p ij ( λ ) = ( x j , y i , λ z ij ) R i j ( λ ) = ∥ p i j ( λ ) ∥ 2 R_{ij}(\lambda)=\lVert\mathbf{p}_{ij}(\lambda)\rVert_2 R ij ( λ ) = ∥ p ij ( λ ) ∥ 2 p ^ i j ( λ ) = p i j ( λ ) / R i j ( λ ) ∈ S 2 \hat{\mathbf{p}}_{ij}(\lambda)=\mathbf{p}_{ij}(\lambda)/R_{ij}(\lambda)\in S^2 p ^ ij ( λ ) = p ij ( λ ) / R ij ( λ ) ∈ S 2 weighted_ring.stl, Figure 2 (a)) with spokes along r i j = ( x j , y i ) \mathbf{r}_{ij}=(x_j,y_i) r ij = ( x j , y i ) r i j = ∥ r i j ∥ 2 r_{ij}=\lVert\mathbf{r}_{ij}\rVert_2 r ij = ∥ r ij ∥ 2 u i j = r i j / r i j ∈ S 1 \mathbf{u}_{ij}=\mathbf{r}_{ij}/r_{ij}\in S^1 u ij = r ij / r ij ∈ S 1 L i j = R i j ( λ ) L_{ij}=R_{ij}(\lambda) L ij = R ij ( λ ) ℓ ( r i j ) = r i j \ell(r_{ij})=r_{ij} ℓ ( r ij ) = r ij O O O ∑ i , j p i j ( λ ) = 0 \sum_{i,j}\mathbf{p}_{ij}(\lambda)=0 ∑ i , j p ij ( λ ) = 0 ∑ i , j z i j r i j = 0 \sum_{i,j}z_{ij}\mathbf{r}_{ij}=0 ∑ i , j z ij r ij = 0 [11 , 17 ] , closing the sixteen planar vectors into a head-to-tail polygon (Figure 2 (b)).
Figure 2. Planar radial vectors from the origin O = ( 0 , 0 ) O = (0, 0) O = ( 0 , 0 ) z i j r i j z_{ij}r_{ij} z ij r ij R i n , R m i d , R o u t R_{\mathrm{in}}, R_{\mathrm{mid}}, R_{\mathrm{out}} R in , R mid , R out d i n = ( 3 , 0 ) \mathbf{d}_{\mathrm{in}} = (3, 0) d in = ( 3 , 0 ) d o u t = ( 15 , 0 ) \mathbf{d}_{\mathrm{out}} = (15, 0) d out = ( 15 , 0 ) d m i d = ( − 18 , 0 ) \mathbf{d}_{\mathrm{mid}} = (-18, 0) d mid = ( − 18 , 0 ) α = 1 \alpha= 1 α = 1 α ∗ = − 1 \alpha^* = -1 α ∗ = − 1 α = 0 \alpha= 0 α = 0 B u n i t ( S ) \mathbf{B}_{\mathrm{unit}}(S) B unit ( S ) ∥ v ( S ) ∥ 2 \lVert\mathbf{v}(S)\rVert_2 ∥ v ( S ) ∥ 2 ∥ u ( S ) ∥ 2 \lVert\mathbf{u}(S)\rVert_2 ∥ u ( S ) ∥ 2 ∥ v ∥ = 5 ∥ u ∥ \lVert\mathbf{v}\rVert= \sqrt{5}\lVert\mathbf{u}\rVert ∥ v ∥ = 5 ∥ u ∥ v / u = 9 / 4 v/u = 9/4 v / u = 9/4 ∥ B z , r α ( S ) ∥ 2 \lVert\mathbf{B}_{z,r^\alpha}(S)\rVert_2 ∥ B z , r α ( S ) ∥ 2 α \alpha α α = 1 \alpha= 1 α = 1 α ∗ = − 1 \alpha^* = -1 α ∗ = − 1 − 0.050 -0.050 − 0.050 + 3 +3 + 3
Our first test probes the mechanism of this equilibrium: does radial balance reside in the directions of the spokes alone, or does it rely on the unequal spoke lengths compensating the unequal directions? And when equalizing spoke lengths breaks balance, how can an opposing inequality in tip weights or power-law lengths restore it?
Length–Weight Duality: Why Equalizing Spoke Lengths Breaks Balance
Definition 2.1 (Generalized 3D and planar origin-vector resultants). For an arrangement S S S n n n λ > 0 \lambda>0 λ > 0 R i j ( λ ) = x j 2 + y i 2 + λ 2 z i j 2 R_{ij}(\lambda)=\sqrt{x_j^2+y_i^2+\lambda^2z_{ij}^2} R ij ( λ ) = x j 2 + y i 2 + λ 2 z ij 2 p ^ i j ( λ ) = p i j ( λ ) / R i j ( λ ) \hat{\mathbf{p}}_{ij}(\lambda)=\mathbf{p}_{ij}(\lambda)/R_{ij}(\lambda) p ^ ij ( λ ) = p ij ( λ ) / R ij ( λ ) 0 \mathbf{0} 0 R i j ( λ ) = 0 R_{ij}(\lambda)=0 R ij ( λ ) = 0 w = ( w i j ) \mathbf{w}=(w_{ij}) w = ( w ij ) L = ( L i j ) \mathbf{L}=(L_{ij}) L = ( L ij )
V w , L ( S ; λ ) = ∑ i , j w i j L i j p ^ i j ( λ ) = ∑ R i j ( λ ) > 0 ( w i j L i j R i j ( λ ) ) p i j ( λ ) ∈ R 3 . (2) \mathbf{V}_{\mathbf{w},\mathbf{L}}(S;\lambda)=\sum_{i,j}w_{ij}L_{ij}\hat{\mathbf{p}}_{ij}(\lambda)=\sum_{R_{ij}(\lambda)>0}\left(\frac{w_{ij}L_{ij}}{R_{ij}(\lambda)}\right)\mathbf{p}_{ij}(\lambda)\in\mathbb{R}^3. \tag*{(2)} V w , L ( S ; λ ) = i , j ∑ w ij L ij p ^ ij ( λ ) = R ij ( λ ) > 0 ∑ ( R ij ( λ ) w ij L ij ) p ij ( λ ) ∈ R 3 . (2) For equal weights w i j = 1 w_{ij}=1 w ij = 1 L i j = R i j ( λ ) β L_{ij}=R_{ij}(\lambda)^\beta L ij = R ij ( λ ) β β ∈ R \beta\in\mathbb{R} β ∈ R
V β ( S ; λ ) : = ∑ i , j R i j ( λ ) β p ^ i j ( λ ) = ∑ i , j R i j ( λ ) β − 1 p i j ( λ ) ∈ R 3 , (3) \mathbf{V}_{\beta}(S;\lambda):=\sum_{i,j}R_{ij}(\lambda)^\beta\hat{\mathbf{p}}_{ij}(\lambda)=\sum_{i,j}R_{ij}(\lambda)^{\beta-1}\mathbf{p}_{ij}(\lambda)\in\mathbb{R}^3, \tag*{(3)} V β ( S ; λ ) := i , j ∑ R ij ( λ ) β p ^ ij ( λ ) = i , j ∑ R ij ( λ ) β − 1 p ij ( λ ) ∈ R 3 , (3) and for weighted unit vectors ( L i j = 1 ) (L_{ij}=1) ( L ij = 1 ) S 2 S^2 S 2 B w , λ ( 3 ) ( S ) : = V w , 1 ( S ; λ ) = ∑ i , j w i j p ^ i j ( λ ) \mathbf{B}^{(3)}_{\mathbf{w},\lambda}(S):=\mathbf{V}_{\mathbf{w},1}(S;\lambda)=\sum_{i,j}w_{ij}\hat{\mathbf{p}}_{ij}(\lambda) B w , λ ( 3 ) ( S ) := V w , 1 ( S ; λ ) = ∑ i , j w ij p ^ ij ( λ ) m = ( m i j ) \mathbf{m}=(m_{ij}) m = ( m ij ) ℓ : [ 0 , ∞ ) → [ 0 , ∞ ) \ell:[0,\infty)\to[0,\infty) ℓ : [ 0 , ∞ ) → [ 0 , ∞ ) u i j = r i j / r i j \mathbf{u}_{ij}=\mathbf{r}_{ij}/r_{ij} u ij = r ij / r ij
B m , ℓ ( S ) = ∑ i , j m i j ℓ ( r i j ) u i j = ∑ r i j > 0 ( m i j ℓ ( r i j ) r i j ) ( x j , y i ) ∈ R 2 . (4) \mathbf{B}_{\mathbf{m},\ell}(S)=\sum_{i,j}m_{ij}\ell(r_{ij})\mathbf{u}_{ij}=\sum_{r_{ij}>0}\left(\frac{m_{ij}\ell(r_{ij})}{r_{ij}}\right)(x_j,y_i)\in\mathbb{R}^2. \tag*{(4)} B m , ℓ ( S ) = i , j ∑ m ij ℓ ( r ij ) u ij = r ij > 0 ∑ ( r ij m ij ℓ ( r ij ) ) ( x j , y i ) ∈ R 2 . (4) Because ℓ ( r i j ) \ell(r_{ij}) ℓ ( r ij ) r i j r_{ij} r ij ∑ i , j ℓ ( r i j ) u i j = 0 \sum_{i,j}\ell(r_{ij})\mathbf{u}_{ij}=0 ∑ i , j ℓ ( r ij ) u ij = 0 m i j = s i j = s ˉ + z i j m_{ij}=s_{ij}=\bar{s}+z_{ij} m ij = s ij = s ˉ + z ij m i j = z i j m_{ij}=z_{ij} m ij = z ij B s , ℓ ( S ) = B z , ℓ ( S ) \mathbf{B}_{\mathbf{s},\ell}(S)=\mathbf{B}_{\mathbf{z},\ell}(S) B s , ℓ ( S ) = B z , ℓ ( S )
Proposition 2.2 (Length–weight duality and the order-three two-ring law). Let n ≥ 3 n \ge3 n ≥ 3 λ > 0 \lambda> 0 λ > 0 Scaling 3D vector lengths from the origin to L i j L_{ij} L ij w i j w_{ij} w ij (2) as keeping the natural 3D vectors p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) ( L i j ′ = R i j ( λ ) L'_{ij}=R_{ij}(\lambda) L ij ′ = R ij ( λ ) ) and assigning effective weights w i j ′ = w i j L i j / R i j ( λ ) w'_{ij}=w_{ij}L_{ij}/R_{ij}(\lambda) w ij ′ = w ij L ij / R ij ( λ ) In particular, power-law 3D spoke lengths L i j = R i j ( λ ) β L_{ij}=R_{ij}(\lambda)^\beta L ij = R ij ( λ ) β with unit weights ( w i j = 1 ) (w_{ij}=1) ( w ij = 1 ) are mechanically dual to natural 3D vectors p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) with weights w i j ′ = R i j ( λ ) β − 1 w'_{ij}=R_{ij}(\lambda)^{\beta-1} w ij ′ = R ij ( λ ) β − 1 and unit-length 3D vectors on S 2 S^2 S 2 ( L i j = 1 L_{ij}=1 L ij = 1 ) balance identically ( ∑ i , j w i j p ^ i j ( λ ) = 0 ) (\sum_{i,j}w_{ij}\hat{\mathbf{p}}_{ij}(\lambda)=0) ( ∑ i , j w ij p ^ ij ( λ ) = 0 ) for every arrangement S S S when weighted by their unequal 3D lengths w i j = c R i j ( λ ) w_{ij}=cR_{ij}(\lambda) w ij = c R ij ( λ ) ( c ∈ R c\in\mathbb{R} c ∈ R ). In the grid plane, unit vectors u i j \mathbf{u}_{ij} u ij ( ∑ i , j m i j u i j = 0 ) (\sum_{i,j}m_{ij}\mathbf{u}_{ij}=0) ( ∑ i , j m ij u ij = 0 ) m i j = c z i j r i j m_{ij}=cz_{ij}r_{ij} m ij = c z ij r ij ℓ i j \ell_{ij} ℓ ij m i j = z i j m_{ij}=z_{ij} m ij = z ij ( ℓ i j / r i j ) r i j (\ell_{ij}/r_{ij})\mathbf{r}_{ij} ( ℓ ij / r ij ) r ij W r c = { a ( x j ) + b ( y i ) } W_{\mathrm{rc}}=\{a(x_j)+b(y_i)\} W rc = { a ( x j ) + b ( y i )} n = 3 n=3 n = 3 A 3 A_3 A 3 R 1 = { ( ± 1 , 0 ) , ( 0 , ± 1 ) } R_1=\{(\pm1,0),(0,\pm1)\} R 1 = {( ± 1 , 0 ) , ( 0 , ± 1 )} r 1 = 1 r_1=1 r 1 = 1 R 2 = { ( ± 1 , ± 1 ) } R_{\sqrt{2}}=\{(\pm1,\pm1)\} R 2 = {( ± 1 , ± 1 )} r 2 = 2 r_2=\sqrt{2} r 2 = 2 S S S d 1 ( S ) + d 2 ( S ) = 0 \mathbf{d}_1(S)+\mathbf{d}_2(S)=0 d 1 ( S ) + d 2 ( S ) = 0 d k ( S ) = ∑ ( i , j ) ∈ R k z i j r i j \mathbf{d}_k(S)=\sum_{(i,j)\in R_k}z_{ij}\mathbf{r}_{ij} d k ( S ) = ∑ ( i , j ) ∈ R k z ij r ij
B z , ℓ ( S ) = ( ℓ ( 1 ) − ℓ ( 2 ) 2 ) d 1 ( S ) . (5) \mathbf{B}_{z,\ell}(S)=\left(\ell(1)-\frac{\ell(\sqrt{2})}{\sqrt{2}}\right)\mathbf{d}_1(S). \tag*{(5)} B z , ℓ ( S ) = ( ℓ ( 1 ) − 2 ℓ ( 2 ) ) d 1 ( S ) . (5) For all eight magic squares of order 3 (the D 4 D_4 D 4 ∥ d 1 ( S ) ∥ 2 = 80 = 4 5 \|\mathbf{d}_1(S)\|_2=\sqrt{80}=4\sqrt{5} ∥ d 1 ( S ) ∥ 2 = 80 = 4 5
∥ B z , ℓ ( S ) ∥ 2 = 4 5 ∣ ℓ ( 1 ) − ℓ ( 2 ) 2 ∣ , ∥ B u n i t ( S ) ∥ 2 = 4 5 − 2 10 ≈ 2.619717 , (6) \|\mathbf{B}_{z,\ell}(S)\|_2=4\sqrt{5}\left|\ell(1)-\frac{\ell(\sqrt{2})}{\sqrt{2}}\right|,\qquad\|\mathbf{B}_{\mathrm{unit}}(S)\|_2=4\sqrt{5}-2\sqrt{10}\approx2.619717, \tag*{(6)} ∥ B z , ℓ ( S ) ∥ 2 = 4 5 ℓ ( 1 ) − 2 ℓ ( 2 ) , ∥ B unit ( S ) ∥ 2 = 4 5 − 2 10 ≈ 2.619717 , (6) so B z , ℓ ( S ) = 0 \mathbf{B}_{z,\ell}(S)=0 B z , ℓ ( S ) = 0 n = 3 n=3 n = 3 ℓ ( 2 ) = 2 ℓ ( 1 ) \ell(\sqrt{2})=\sqrt{2}\ell(1) ℓ ( 2 ) = 2 ℓ ( 1 )
Proof. The duality relations follow directly from (2) –(4) , the identity ∑ i , j p i j ( λ ) = 0 \sum_{i,j}\mathbf{p}_{ij}(\lambda)=0 ∑ i , j p ij ( λ ) = 0 W r c W_{\mathrm{rc}} W rc [13 Sect. 2] . At order n = 3 n=3 n = 3 r 11 = ( 0 , 0 ) \mathbf{r}_{11}=(0,0) r 11 = ( 0 , 0 ) d 1 ( S ) + d 2 ( S ) = ∑ i , j z i j r i j = 0 \mathbf{d}_1(S)+\mathbf{d}_2(S)=\sum_{i,j}z_{ij}\mathbf{r}_{ij}=0 d 1 ( S ) + d 2 ( S ) = ∑ i , j z ij r ij = 0 (5) upon grouping (4) over R 1 R_1 R 1 R 2 R_{\sqrt{2}} R 2 d 1 = ( − 8 , 4 ) \mathbf{d}_1=(-8,4) d 1 = ( − 8 , 4 ) ∥ d 1 ∥ 2 = 80 = 4 5 \|\mathbf{d}_1\|_2=\sqrt{80}=4\sqrt{5} ∥ d 1 ∥ 2 = 80 = 4 5 D 4 D_4 D 4 (6) .
Proposition 2.2 makes the mechanics of spoke normalization transparent: forcing unequal spokes to equal unit length (L i j = 1 L_{ij}=1 L ij = 1 ℓ = 1 \ell=1 ℓ = 1 R i j ( λ ) R_{ij}(\lambda) R ij ( λ ) r i j r_{ij} r ij n = 3 n=3 n = 3 r 2 / r 1 = 2 r_2/r_1=\sqrt{2} r 2 / r 1 = 2 d 1 \mathbf{d}_1 d 1 d 2 = − d 1 \mathbf{d}_2=-\mathbf{d}_1 d 2 = − d 1 n = 4 n=4 n = 4
Planar Balance Wheels: The 5 \sqrt{5} 5
At order n = 4 n=4 n = 4 A 4 = { − 3 2 , − 1 2 , 1 2 , 3 2 } 2 A_4=\{-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}\}^2 A 4 = { − 2 3 , − 2 1 , 2 1 , 2 3 } 2 O O O Figure 1 (b), Figure 2 (a)): four inner cells R i n = { ( ± 1 2 , ± 1 2 ) } R_{\mathrm{in}}=\{(\pm\frac{1}{2},\pm\frac{1}{2})\} R in = {( ± 2 1 , ± 2 1 )} r 1 = 2 / 2 r_1=\sqrt{2}/2 r 1 = 2 /2 R m i d = { ( ± 3 2 , ± 1 2 ) , ( ± 1 2 , ± 3 2 ) } R_{\mathrm{mid}}=\{(\pm\frac{3}{2},\pm\frac{1}{2}),(\pm\frac{1}{2},\pm\frac{3}{2})\} R mid = {( ± 2 3 , ± 2 1 ) , ( ± 2 1 , ± 2 3 )} r 2 = 10 / 2 r_2=\sqrt{10}/2 r 2 = 10 /2 R o u t = { ( ± 3 2 , ± 3 2 ) } R_{\mathrm{out}}=\{(\pm\frac{3}{2},\pm\frac{3}{2})\} R out = {( ± 2 3 , ± 2 3 )} r 3 = 3 2 / 2 r_3=3\sqrt{2}/2 r 3 = 3 2 /2 r 3 / r 1 = 3 r_3/r_1=3 r 3 / r 1 = 3 r 2 / r 1 = 5 ∉ Q r_2/r_1=\sqrt{5}\notin\mathbb{Q} r 2 / r 1 = 5 ∈ / Q α = 0 \alpha=0 α = 0 u ( S ) \mathbf{u}(S) u ( S ) v ( S ) / 5 \mathbf{v}(S)/\sqrt{5} v ( S ) / 5
Theorem 2.3 (Three-Ring Decomposition, Irrationality No-Go, and Pell Minimum). For any arrangement S S S of order 4, define the three ring dipole vectors radiating from O O O by d i n ( S ) = ∑ c ∈ R i n z c r c \mathbf{d}_{\mathrm{in}}(S)=\sum_{\mathbf{c}\in R_{\mathrm{in}}}z_{\mathbf{c}}\mathbf{r}_{\mathbf{c}} d in ( S ) = ∑ c ∈ R in z c r c d m i d ( S ) = ∑ c ∈ R m i d z c r c \mathbf{d}_{\mathrm{mid}}(S)=\sum_{\mathbf{c}\in R_{\mathrm{mid}}}z_{\mathbf{c}}\mathbf{r}_{\mathbf{c}} d mid ( S ) = ∑ c ∈ R mid z c r c and d o u t ( S ) = ∑ c ∈ R o u t z c r c \mathbf{d}_{\mathrm{out}}(S)=\sum_{\mathbf{c}\in R_{\mathrm{out}}}z_{\mathbf{c}}\mathbf{r}_{\mathbf{c}} d out ( S ) = ∑ c ∈ R out z c r c 4 × 4 4\times4 4 × 4 S S S
of distinct integers, d i n ( S ) ≠ 0 \mathbf{d}_{\mathrm{in}}(S) \ne0 d in ( S ) = 0 d o u t ( S ) ≠ 0 \mathbf{d}_{\mathrm{out}}(S) \ne0 d out ( S ) = 0 S ∈ M 4 S \in\mathcal{M}_4 S ∈ M 4 R i n R_{\mathrm{in}} R in R o u t R_{\mathrm{out}} R out μ 4 = 34 \mu_4 = 34 μ 4 = 34 d i n ( S ) ∈ Z 2 ∖ { 0 } \mathbf{d}_{\mathrm{in}}(S) \in\mathbb{Z}^2 \setminus\{0\} d in ( S ) ∈ Z 2 ∖ { 0 } d o u t ( S ) ∈ ( 3 Z ) 2 ∖ { 0 } \mathbf{d}_{\mathrm{out}}(S) \in(3\mathbb{Z})^2 \setminus\{0\} d out ( S ) ∈ ( 3 Z ) 2 ∖ { 0 }
∥ d i n ( S ) ∥ 2 ≥ 1 , ∥ d o u t ( S ) ∥ 2 ≥ 3. (7) \lVert\mathbf{d}_{\mathrm{in}}(S)\rVert_2 \ge1,\qquad\lVert\mathbf{d}_{\mathrm{out}}(S)\rVert_2 \ge3. \tag*{(7)} ∥ d in ( S ) ∥ 2 ≥ 1 , ∥ d out ( S ) ∥ 2 ≥ 3. (7) Whenever ∑ c ∈ A 4 z c r c = 0 \sum_{c\in A_4} z_c r_c = 0 ∑ c ∈ A 4 z c r c = 0 d m i d ( S ) = − ( d i n ( S ) + d o u t ( S ) ) \mathbf{d}_{\mathrm{mid}}(S) = -(\mathbf{d}_{\mathrm{in}}(S) + \mathbf{d}_{\mathrm{out}}(S)) d mid ( S ) = − ( d in ( S ) + d out ( S )) ℓ = ( ℓ 1 , ℓ 2 , ℓ 3 ) \ell= (\ell_1,\ell_2,\ell_3) ℓ = ( ℓ 1 , ℓ 2 , ℓ 3 ) ( r 1 , r 2 , r 3 ) (r_1,r_2,r_3) ( r 1 , r 2 , r 3 )
B z , ℓ ( S ) = ( ℓ 1 r 1 − ℓ 2 r 2 ) d i n ( S ) + ( ℓ 3 r 3 − ℓ 2 r 2 ) d o u t ( S ) . (8) \mathbf{B}_{z,\ell}(S) = \left(\frac{\ell_1}{r_1} - \frac{\ell_2}{r_2}\right)\mathbf{d}_{\mathrm{in}}(S) + \left(\frac{\ell_3}{r_3} - \frac{\ell_2}{r_2}\right)\mathbf{d}_{\mathrm{out}}(S). \tag*{(8)} B z , ℓ ( S ) = ( r 1 ℓ 1 − r 2 ℓ 2 ) d in ( S ) + ( r 3 ℓ 3 − r 2 ℓ 2 ) d out ( S ) . (8) In particular, for planar unit vectors (ℓ 1 = ℓ 2 = ℓ 3 = 1 \ell_1 = \ell_2 = \ell_3 = 1 ℓ 1 = ℓ 2 = ℓ 3 = 1 α = 0 \alpha= 0 α = 0
B u n i t ( S ) = 2 ( u ( S ) − 1 5 v ( S ) ) , (9) \mathbf{B}_{\mathrm{unit}}(S) = \sqrt{2}\left(\mathbf{u}(S) - \frac{1}{\sqrt{5}}\mathbf{v}(S)\right), \tag*{(9)} B unit ( S ) = 2 ( u ( S ) − 5 1 v ( S ) ) , (9) where u ( S ) = d i n ( S ) + 1 3 d o u t ( S ) ∈ Z 2 \mathbf{u}(S) = \mathbf{d}_{\mathrm{in}}(S) + \frac{1}{3}\mathbf{d}_{\mathrm{out}}(S) \in\mathbb{Z}^2 u ( S ) = d in ( S ) + 3 1 d out ( S ) ∈ Z 2 v ( S ) = d i n ( S ) + d o u t ( S ) = − d m i d ( S ) ∈ Z 2 \mathbf{v}(S) = \mathbf{d}_{\mathrm{in}}(S) + \mathbf{d}_{\mathrm{out}}(S) = -\mathbf{d}_{\mathrm{mid}}(S) \in\mathbb{Z}^2 v ( S ) = d in ( S ) + d out ( S ) = − d mid ( S ) ∈ Z 2 5 ∉ Q \sqrt{5} \notin\mathbb{Q} 5 ∈ / Q d i n ( S ) , d o u t ( S ) ≠ 0 \mathbf{d}_{\mathrm{in}}(S),\mathbf{d}_{\mathrm{out}}(S) \ne0 d in ( S ) , d out ( S ) = 0 4 × 4 4 \times4 4 × 4 4 × 4 4 \times4 4 × 4 B u n i t ( S ) = 0 \mathbf{B}_{\mathrm{unit}}(S) = 0 B unit ( S ) = 0 B z , r α ( S ) = 0 \mathbf{B}_{z,r^\alpha}(S) = 0 B z , r α ( S ) = 0 α ∈ 2 Z \alpha\in2\mathbb{Z} α ∈ 2 Z ∥ B u n i t ( S ) ∥ 2 \lVert\mathbf{B}_{\mathrm{unit}}(S)\rVert_2 ∥ B unit ( S ) ∥ 2
min S ∈ M 4 ∥ B u n i t ( S ) ∥ 2 = 18 10 5 − 8 2 = 2 10 5 ( 9 + 4 5 ) = 2 10 5 φ 6 ≈ 0.07049108 , (10) \min_{S\in\mathcal{M}_4}\lVert\mathbf{B}_{\mathrm{unit}}(S)\rVert_2 = \frac{18\sqrt{10}}{5} - 8\sqrt{2} = \frac{2\sqrt{10}}{5(9+4\sqrt{5})} = \frac{2\sqrt{10}}{5\varphi^6} \approx0.07049108, \tag*{(10)} S ∈ M 4 min ∥ B unit ( S ) ∥ 2 = 5 18 10 − 8 2 = 5 ( 9 + 4 5 ) 2 10 = 5 φ 6 2 10 ≈ 0.07049108 , (10) where φ = ( 1 + 5 ) / 2 \varphi= (1+\sqrt{5})/2 φ = ( 1 + 5 ) /2 d o u t ( S ) = 5 d i n ( S ) \mathbf{d}_{\mathrm{out}}(S) = 5\mathbf{d}_{\mathrm{in}}(S) d out ( S ) = 5 d in ( S ) ∥ d i n ( S ) ∥ 2 = 3 \lVert\mathbf{d}_{\mathrm{in}}(S)\rVert_2 = 3 ∥ d in ( S ) ∥ 2 = 3 u ( S ) \mathbf{u}(S) u ( S ) v ( S ) \mathbf{v}(S) v ( S ) ∥ v ∥ / ∥ u ∥ = 18 / 8 = 9 / 4 \lVert\mathbf{v}\rVert/\lVert\mathbf{u}\rVert= 18/8 = 9/4 ∥ v ∥ / ∥ u ∥ = 18/8 = 9/4 5 \sqrt{5} 5 9 2 − 5 ⋅ 4 2 = 1 9^2 - 5\cdot4^2 = 1 9 2 − 5 ⋅ 4 2 = 1 D 4 D_4 D 4
Proof. Labelling the four entries of R i n R_{\mathrm{in}} R in ( − 1 2 , 1 2 ) \left(-\frac{1}{2},\frac{1}{2}\right) ( − 2 1 , 2 1 ) ( 1 2 , 1 2 ) \left(\frac{1}{2},\frac{1}{2}\right) ( 2 1 , 2 1 ) ( − 1 2 , − 1 2 ) \left(-\frac{1}{2},-\frac{1}{2}\right) ( − 2 1 , − 2 1 ) ( 1 2 , − 1 2 ) \left(\frac{1}{2},-\frac{1}{2}\right) ( 2 1 , − 2 1 ) a , b , c , d ∈ Z a,b,c,d \in\mathbb{Z} a , b , c , d ∈ Z d i n ( S ) = 1 2 ( ( b + d ) − ( a + c ) , ( a + b ) − ( c + d ) ) \mathbf{d}_{\mathrm{in}}(S) = \frac{1}{2}\bigl((b+d)-(a+c),(a+b)-(c+d)\bigr) d in ( S ) = 2 1 ( ( b + d ) − ( a + c ) , ( a + b ) − ( c + d ) ) d i n ( S ) = 0 \mathbf{d}_{\mathrm{in}}(S)=0 d in ( S ) = 0 b + d = a + c b+d=a+c b + d = a + c a + b = c + d a+b=c+d a + b = c + d b = c b=c b = c a = d a=d a = d S S S S ∈ M 4 S\in\mathcal{M}_4 S ∈ M 4 R 0 + R 3 + C 0 + C 3 = 4 μ 4 R_0+R_3+C_0+C_3=4\mu_4 R 0 + R 3 + C 0 + C 3 = 4 μ 4 ∑ c ∈ R o u t s c − ∑ c ∈ R i n s c = 0 \sum_{c\in R_{\mathrm{out}}} s_c-\sum_{c\in R_{\mathrm{in}}} s_c=0 ∑ c ∈ R out s c − ∑ c ∈ R in s c = 0 D 1 + D 2 = ∑ c ∈ R o u t s c + ∑ c ∈ R i n s c = 2 μ 4 D_1+D_2=\sum_{c\in R_{\mathrm{out}}}s_c+\sum_{c\in R_{\mathrm{in}}}s_c=2\mu_4 D 1 + D 2 = ∑ c ∈ R out s c + ∑ c ∈ R in s c = 2 μ 4 ∑ c ∈ R o u t s c = μ 4 = 34 \sum_{c\in R_{\mathrm{out}}} s_c=\mu_4=34 ∑ c ∈ R out s c = μ 4 = 34 d i n ( S ) ∈ Z 2 ∖ { 0 } \mathbf{d}_{\mathrm{in}}(S)\in\mathbb{Z}^2\setminus\{0\} d in ( S ) ∈ Z 2 ∖ { 0 } ( ± 3 2 , ± 3 2 ) = 3 ( ± 1 2 , ± 1 2 ) (\pm\frac{3}{2},\pm\frac{3}{2})=3(\pm\frac{1}{2},\pm\frac{1}{2}) ( ± 2 3 , ± 2 3 ) = 3 ( ± 2 1 , ± 2 1 ) d o u t ( S ) ∈ ( 3 Z ) 2 ∖ { 0 } \mathbf{d}_{\mathrm{out}}(S)\in(3\mathbb{Z})^2\setminus\{0\} d out ( S ) ∈ ( 3 Z ) 2 ∖ { 0 } (7) . Partitioning A 4 = R i n ⊔ R m i d ⊔ R o u t A_4=R_{\mathrm{in}}\sqcup R_{\mathrm{mid}}\sqcup R_{\mathrm{out}} A 4 = R in ⊔ R mid ⊔ R out d m i d = − ( d i n + d o u t ) \mathbf{d}_{\mathrm{mid}}=-(\mathbf{d}_{\mathrm{in}}+\mathbf{d}_{\mathrm{out}}) d mid = − ( d in + d out ) (4) yields (8) .
Setting ℓ 1 = ℓ 2 = ℓ 3 = 1 \ell_1=\ell_2=\ell_3=1 ℓ 1 = ℓ 2 = ℓ 3 = 1 1 / r 1 = 2 1/r_1=\sqrt{2} 1/ r 1 = 2 1 / r 2 = 2 / 5 1/r_2=\sqrt{2}/\sqrt{5} 1/ r 2 = 2 / 5 1 / r 3 = 2 / 3 1/r_3=\sqrt{2}/3 1/ r 3 = 2 /3 (8) gives B u n i t ( S ) = ( 2 − 2 / 5 ) d i n + ( 2 / 3 − 2 / 5 ) d o u t = 2 ( d i n + 1 3 d o u t ) − 1 5 ( d i n + d o u t ) \mathbf{B}_{\mathrm{unit}}(S)=(\sqrt{2}-\sqrt{2}/\sqrt{5})\mathbf{d}_{\mathrm{in}}+(\sqrt{2}/3-\sqrt{2}/\sqrt{5})\mathbf{d}_{\mathrm{out}}=\sqrt{2}(\mathbf{d}_{\mathrm{in}}+\frac{1}{3}\mathbf{d}_{\mathrm{out}})-\frac{1}{\sqrt{5}}(\mathbf{d}_{\mathrm{in}}+\mathbf{d}_{\mathrm{out}}) B unit ( S ) = ( 2 − 2 / 5 ) d in + ( 2 /3 − 2 / 5 ) d out = 2 ( d in + 3 1 d out ) − 5 1 ( d in + d out ) (9) . Because u ( S ) , v ( S ) ∈ Z 2 \mathbf{u}(S),\mathbf{v}(S)\in\mathbb{Z}^2 u ( S ) , v ( S ) ∈ Z 2 5 ∉ Q \sqrt{5}\notin\mathbb{Q} 5 ∈ / Q B u n i t ( S ) = 0 \mathbf{B}_{\mathrm{unit}}(S)=0 B unit ( S ) = 0 u ( S ) = v ( S ) = 0 \mathbf{u}(S)=\mathbf{v}(S)=0 u ( S ) = v ( S ) = 0 d i n ( S ) = d o u t ( S ) = 0 \mathbf{d}_{\mathrm{in}}(S)=\mathbf{d}_{\mathrm{out}}(S)=0 d in ( S ) = d out ( S ) = 0 α = 2 m \alpha=2m α = 2 m r 2 α − 1 / r 1 α − 1 = 5 m − 1 5 ∉ Q r_2^{\alpha-1}/r_1^{\alpha-1}=5^{m-1}\sqrt{5}\notin\mathbb{Q} r 2 α − 1 / r 1 α − 1 = 5 m − 1 5 ∈ / Q
To establish the Pell minimum (10) , subtracting C 1 + C 2 = 2 μ 4 C_1+C_2=2\mu_4 C 1 + C 2 = 2 μ 4 C 0 + C 3 = 2 μ 4 C_0+C_3=2\mu_4 C 0 + C 3 = 2 μ 4 D 1 + D 2 = 2 μ 4 D_1+D_2=2\mu_4 D 1 + D 2 = 2 μ 4 s 00 + s 03 = s 31 + s 32 s_{00}+s_{03}=s_{31}+s_{32} s 00 + s 03 = s 31 + s 32 s 11 + s 12 = s 20 + s 23 s_{11}+s_{12}=s_{20}+s_{23} s 11 + s 12 = s 20 + s 23 M 4 M_4 M 4 T d i a g = { ( 0 , 0 ) , ( 0 , 3 ) , ( 1 , 1 ) , ( 1 , 2 ) } T_{\mathrm{diag}}=\{(0,0),(0,3),(1,1),(1,2)\} T diag = {( 0 , 0 ) , ( 0 , 3 ) , ( 1 , 1 ) , ( 1 , 2 )} R d i a g = { ( 0 , 3 ) , ( 3 , 3 ) , ( 1 , 2 ) , ( 2 , 2 ) } R_{\mathrm{diag}}=\{(0,3),(3,3),(1,2),(2,2)\} R diag = {( 0 , 3 ) , ( 3 , 3 ) , ( 1 , 2 ) , ( 2 , 2 )} ∑ c ∈ T d i a g s c ∈ { 18 , 20 , … , 42 } \sum_{c\in T_{\mathrm{diag}}}s_c\in\{18,20,\ldots,42\} ∑ c ∈ T diag s c ∈ { 18 , 20 , … , 42 } u y ( S ) = ∑ c ∈ T d i a g s c − 34 u_y(S)=\sum_{c\in T_{\mathrm{diag}}}s_c-34 u y ( S ) = ∑ c ∈ T diag s c − 34 u x ( S ) = ∑ c ∈ R d i a g s c − 34 u_x(S)=\sum_{c\in R_{\mathrm{diag}}}s_c-34 u x ( S ) = ∑ c ∈ R diag s c − 34 v ( S ) − u ( S ) = 2 3 d o u t ( S ) ∈ ( 2 Z ) 2 \mathbf{v}(S)-\mathbf{u}(S)=\frac{2}{3}\mathbf{d}_{\mathrm{out}}(S)\in(2\mathbb{Z})^2 v ( S ) − u ( S ) = 3 2 d out ( S ) ∈ ( 2 Z ) 2 S ∈ M 4 S\in M_4 S ∈ M 4 u ( S ) = 2 u ~ ( S ) ∈ ( 2 Z ) 2 \mathbf{u}(S)=2\widetilde{\mathbf{u}}(S)\in(2\mathbb{Z})^2 u ( S ) = 2 u ( S ) ∈ ( 2 Z ) 2 v ( S ) = 2 v ~ ( S ) ∈ ( 2 Z ) 2 \mathbf{v}(S)=2\widetilde{\mathbf{v}}(S)\in(2\mathbb{Z})^2 v ( S ) = 2 v ( S ) ∈ ( 2 Z ) 2 u ~ ( S ) , v ~ ( S ) ∈ Z 2 ∖ { 0 } \widetilde{\mathbf{u}}(S),\widetilde{\mathbf{v}}(S)\in\mathbb{Z}^2\setminus\{0\} u ( S ) , v ( S ) ∈ Z 2 ∖ { 0 } ∥ u ~ ( S ) ∥ ∞ ≤ 8 \lVert\widetilde{\mathbf{u}}(S)\rVert_\infty\leq8 ∥ u ( S ) ∥ ∞ ≤ 8 ∥ u ~ ( S ) ∥ 2 ≤ 4 5 < 9 \lVert\widetilde{\mathbf{u}}(S)\rVert_2\leq4\sqrt{5}<9 ∥ u ( S ) ∥ 2 ≤ 4 5 < 9 (9) becomes ∥ B u n i t ( S ) ∥ 2 = 2 2 ∥ u ~ ( S ) − v ~ ( S ) / 5 ∥ 2 \lVert\mathbf{B}_{\mathrm{unit}}(S)\rVert_2=2\sqrt{2}\lVert\widetilde{\mathbf{u}}(S)-\widetilde{\mathbf{v}}(S)/\sqrt{5}\rVert_2 ∥ B unit ( S ) ∥ 2 = 2 2 ∥ u ( S ) − v ( S ) / 5 ∥ 2 v ~ \widetilde{\mathbf{v}} v u ~ \widetilde{\mathbf{u}} u u ~ = u ~ e \widetilde{\mathbf{u}}=\widetilde{u}\mathbf{e} u = u e v ~ = v ~ e \widetilde{\mathbf{v}}=\widetilde{v}\mathbf{e} v = v e e ∈ { ± e ^ x , ± e ^ y } \mathbf{e}\in\{\pm\hat{\mathbf{e}}_x,\pm\hat{\mathbf{e}}_y\} e ∈ { ± e ^ x , ± e ^ y } ∥ e ∥ 2 = 1 \lVert\mathbf{e}\rVert_2=1 ∥ e ∥ 2 = 1 1 ≤ u ~ ≤ 8 1\leq\widetilde{u}\leq8 1 ≤ u ≤ 8 5 = [ 2 ; 4 ] \sqrt{5}=[2;4] 5 = [ 2 ; 4 ] 2 / 1 , 9 / 4 , 38 / 17 , … 2/1,9/4,38/17,\ldots 2/1 , 9/4 , 38/17 , … 1 ≤ u ~ ≤ 8 < 17 1\leq\widetilde{u}\leq8<17 1 ≤ u ≤ 8 < 17 5 \sqrt{5} 5 v ~ / u ~ = 9 / 4 \widetilde{v}/\widetilde{u}=9/4 v / u = 9/4 9 2 − 5 ⋅ 4 2 = 1 9^2-5\cdot4^2=1 9 2 − 5 ⋅ 4 2 = 1 ( u , v ) = ( 8 , 18 ) (u,v)=(8,18) ( u , v ) = ( 8 , 18 ) d o u t = 5 d i n \mathbf{d}_{\mathrm{out}}=5\mathbf{d}_{\mathrm{in}} d out = 5 d in ∥ d i n ∥ 2 = 3 \lVert\mathbf{d}_{\mathrm{in}}\rVert_2=3 ∥ d in ∥ 2 = 3 2 2 ( 9 / 5 − 4 ) = 18 10 / 5 − 8 2 = 2 10 / ( 5 φ 2 ) ≈ 0.07049108 2\sqrt{2}(9/\sqrt{5}-4)=18\sqrt{10}/5-8\sqrt{2}=2\sqrt{10}/(5\varphi^2)\approx0.07049108 2 2 ( 9/ 5 − 4 ) = 18 10 /5 − 8 2 = 2 10 / ( 5 φ 2 ) ≈ 0.07049108 Figure 1 (b)), the inner ring { 5 , 12 , 2 , 15 } \{5,12,2,15\} { 5 , 12 , 2 , 15 } d i n = ( 3 , 0 ) \mathbf{d}_{\mathrm{in}}=(3,0) d in = ( 3 , 0 ) { 1 , 16 , 11 , 6 } \{1,16,11,6\} { 1 , 16 , 11 , 6 } d o u t = ( 15 , 0 ) = 5 d i n \mathbf{d}_{\mathrm{out}}=(15,0)=5\mathbf{d}_{\mathrm{in}} d out = ( 15 , 0 ) = 5 d in u = ( 8 , 0 ) \mathbf{u}=(8,0) u = ( 8 , 0 ) v = ( 18 , 0 ) \mathbf{v}=(18,0) v = ( 18 , 0 ) (10) .
How, then, can a planar balance wheel achieve exact equilibrium at a non-linear length profile ℓ ( r ) = r α \ell(r)=r^\alpha ℓ ( r ) = r α α ≠ 1 \alpha\neq1 α = 1 (8) shows that balance is restored whenever the inner and outer ring dipoles align in the same direction (d o u t = κ d i n \mathbf{d}_{\mathrm{out}}=\kappa\mathbf{d}_{\mathrm{in}} d out = κ d in κ > 0 \kappa>0 κ > 0 d m i d = − ( 1 + κ ) d i n \mathbf{d}_{\mathrm{mid}}=-(1+\kappa)\mathbf{d}_{\mathrm{in}} d mid = − ( 1 + κ ) d in Figure 2 (d)).
Corollary 2.4 (Restoring planar balance via power-law exponents α ∗ ≠ 1 \alpha^*\neq1 α ∗ = 1 For power-law radial lengths ℓ ( r ) = r α \ell(r)=r^\alpha ℓ ( r ) = r α α ∈ R \alpha\in\mathbb{R} α ∈ R (8) takes the form
B z , r α ( S ) = r 1 α − 1 [ ( 1 − 5 ( α − 1 ) / 2 ) d i n ( S ) + ( 3 α − 1 − 5 ( α − 1 ) / 2 ) d o u t ( S ) ] . (11) \mathbf{B}_{z,r^\alpha}(S)=r_1^{\alpha-1}\left[(1-5^{(\alpha-1)/2})\mathbf{d}_{\mathrm{in}}(S)+(3^{\alpha-1}-5^{(\alpha-1)/2})\mathbf{d}_{\mathrm{out}}(S)\right]. \tag*{(11)} B z , r α ( S ) = r 1 α − 1 [ ( 1 − 5 ( α − 1 ) /2 ) d in ( S ) + ( 3 α − 1 − 5 ( α − 1 ) /2 ) d out ( S ) ] . (11) *Because 1 < 5 < 3 1<\sqrt{5}<3 1 < 5 < 3 h ( α ) = 5 ( α − 1 ) / 2 − 1 3 α − 1 − 5 ( α − 1 ) / 2 h(\alpha)=\frac{5^{(\alpha-1)/2}-1}{3^{\alpha-1}-5^{(\alpha-1)/2}} h ( α ) = 3 α − 1 − 5 ( α − 1 ) /2 5 ( α − 1 ) /2 − 1 h ( 1 ) = ln 5 ln ( 9 / 5 ) ≈ 2.7381 h(1)=\frac{\ln5}{\ln(9/5)}\approx2.7381 h ( 1 ) = l n ( 9/5 ) l n 5 ≈ 2.7381 R \mathbb{R} R ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) S S S α ∗ ( S ) ≠ 1 \alpha^*(S)\neq1 α ∗ ( S ) = 1 B z , r α ( S ) = 0 \mathbf{B}_{z,r^\alpha}(S)=0 B z , r α ( S ) = 0 d i n ( S ) \mathbf{d}_{\mathrm{in}}(S) d in ( S ) d o u t ( S ) \mathbf{d}_{\mathrm{out}}(S) d out ( S ) d o u t ( S ) = κ d i n ( S ) \mathbf{d}_{\mathrm{out}}(S)=\kappa\mathbf{d}_{\mathrm{in}}(S) d out ( S ) = κ d in ( S ) κ > 0 \kappa>0 κ > 0 κ < 0 \kappa<0 κ < 0 α ∗ ( κ ) \alpha^*(\kappa) α ∗ ( κ ) α ∗ = − 1 \alpha^*=-1 α ∗ = − 1 ℓ ( r ) = 1 / r \ell(r)=1/r ℓ ( r ) = 1/ r κ = h ( − 1 ) = + 9 \kappa=h(-1)=+9 κ = h ( − 1 ) = + 9 Figure 1 (c)), and cubic radial lengths α ∗ = + 3 \alpha^*=+3 α ∗ = + 3 ℓ ( r ) = r 3 \ell(r)=r^3 ℓ ( r ) = r 3 κ = h ( 3 ) = + 1 \kappa=h(3)=+1 κ = h ( 3 ) = + 1 κ = + 5 \kappa=+5 κ = + 5 Theorem 2.3 ), the exact zero occurs at α ∗ ( 5 ) ≈ − 0.0496 \alpha^*(5)\approx-0.0496 α ∗ ( 5 ) ≈ − 0.0496 ∥ B z , r α ( S ) ∥ 2 \lVert\mathbf{B}_{z,r^\alpha}(S)\rVert_2 ∥ B z , r α ( S ) ∥ 2 α = 0 \alpha=0 α = 0 Figure 2 (d)).
Proof. Setting (11) to 0 \mathbf{0} 0 α ≠ 1 \alpha\neq1 α = 1 d o u t ( S ) = h ( α ) d i n ( S ) \mathbf{d}_{\mathrm{out}}(S)=h(\alpha)\mathbf{d}_{\mathrm{in}}(S) d out ( S ) = h ( α ) d in ( S ) 1 < 5 t / 2 < 3 t 1<5^{t/2}<3^t 1 < 5 t /2 < 3 t t = α − 1 > 0 t=\alpha-1>0 t = α − 1 > 0 3 t < 5 t / 2 < 1 3^t<5^{t/2}<1 3 t < 5 t /2 < 1 t < 0 t<0 t < 0 h ( α ) h(\alpha) h ( α ) α − 1 \alpha-1 α − 1 h ( α ) > 0 h(\alpha)>0 h ( α ) > 0 lim α → − ∞ h ( α ) = + ∞ \lim_{\alpha\to-\infty}h(\alpha)=+\infty lim α → − ∞ h ( α ) = + ∞ lim α → + ∞ h ( α ) = 0 \lim_{\alpha\to+\infty}h(\alpha)=0 lim α → + ∞ h ( α ) = 0 κ ∈ Q > 0 \kappa\in\mathbb{Q}_{>0} κ ∈ Q > 0 ln 5 ln ( 9 / 5 ) \frac{\ln5}{\ln(9/5)} l n ( 9/5 ) l n 5 κ > 0 \kappa>0 κ > 0 α ∗ ( κ ) = h − 1 ( κ ) ≠ 1 \alpha^*(\kappa)=h^{-1}(\kappa)\neq1 α ∗ ( κ ) = h − 1 ( κ ) = 1 Theorem 2.3 , an integer root α ∗ ≠ 1 \alpha^*\neq1 α ∗ = 1 α ∗ = 2 m + 1 \alpha^*=2m+1 α ∗ = 2 m + 1 m ∈ Z ∖ { 0 } m\in\mathbb{Z}\setminus\{0\} m ∈ Z ∖ { 0 } h ( 2 m + 1 ) = 5 m − 1 9 m − 5 m h(2m+1)=\frac{5^{m}-1}{9^{m}-5^{m}} h ( 2 m + 1 ) = 9 m − 5 m 5 m − 1 κ \kappa κ M 4 \mathcal{M}_4 M 4 m = − 1 m=-1 m = − 1 κ = 9 \kappa=9 κ = 9 m = + 1 m=+1 m = + 1 κ = 1 \kappa=1 κ = 1
Three-Dimensional Spoke Stars: Unit-Sphere Counterweights and Cubic Vertical Balance
We now lift the spoke sculpture into three dimensions (Figure 3 ). In 3D, trimming all sixteen spokes to the unit sphere S 2 S^2 S 2 β = 0 \beta=0 β = 0 p ^ − c = − p ^ c \hat{\mathbf{p}}_{-c}=-\hat{\mathbf{p}}_c p ^ − c = − p ^ c R i j ( λ ) > 0 R_{ij}(\lambda)>0 R ij ( λ ) > 0 ∑ i , j R i j ( λ ) p ^ i j ( λ ) = 0 \sum_{i,j}R_{ij}(\lambda)\hat{\mathbf{p}}_{ij}(\lambda)=\mathbf{0} ∑ i , j R ij ( λ ) p ^ ij ( λ ) = 0 O O O w ∗ > 0 \mathbf{w}^*>0 w ∗ > 0
Theorem 2.5 (3D Power-Law Vectors, Cubic Vertical Vanishing, and Positive Unit-Sphere Weights). Let S S S semi-magic square of order n n n λ > 0 \lambda>0 λ > 0 L i j = R i j ( λ ) 3 L_{ij}=R_{ij}(\lambda)^3 L ij = R ij ( λ ) 3 Proposition 2.2 to weighting p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) w i j = R i j ( λ ) 2 = r i j 2 + λ 2 z i j 2 w_{ij}=R_{ij}(\lambda)^2=r_{ij}^2+\lambda^2z_{ij}^2 w ij = R ij ( λ ) 2 = r ij 2 + λ 2 z ij 2 z z z n n n λ > 0 \lambda>0 λ > 0
V 3 ( S ; λ ) = ∑ i , j ( r i j 2 + λ 2 z i j 2 ) ( x j , y i , λ z i j ) = ( λ 2 ∑ i , j z i j 2 x j , λ 2 ∑ i , j z i j 2 y i , 0 ) , (12) \mathbf{V}_3(S;\lambda)=\sum_{i,j}(r_{ij}^2+\lambda^2z_{ij}^2)(x_j,y_i,\lambda z_{ij})=\left(\lambda^2\sum_{i,j}z_{ij}^2x_j,\ \lambda^2\sum_{i,j}z_{ij}^2y_i,\ 0\right), \tag*{(12)} V 3 ( S ; λ ) = i , j ∑ ( r ij 2 + λ 2 z ij 2 ) ( x j , y i , λ z ij ) = ( λ 2 i , j ∑ z ij 2 x j , λ 2 i , j ∑ z ij 2 y i , 0 ) , (12) and at order n = 4 n=4 n = 4 V 3 ( S ; λ ) = 0 \mathbf{V}_3(S;\lambda)=\mathbf{0} V 3 ( S ; λ ) = 0 S S S w i j = f ( ∣ z i j ∣ ) w_{ij}=f(|z_{ij}|) w ij = f ( ∣ z ij ∣ ) p ^ − c ( λ ) = − p ^ c ( λ ) \hat{\mathbf{p}}_{-c}(\lambda)=-\hat{\mathbf{p}}_c(\lambda) p ^ − c ( λ ) = − p ^ c ( λ ) V β ( S ; λ ) = 0 \mathbf{V}_{\beta}(S;\lambda)=\mathbf{0} V β ( S ; λ ) = 0 β ∈ R \beta\in\mathbb{R} β ∈ R λ > 0 \lambda>0 λ > 0 n = 4 n=4 n = 4 V 0 ( S ; λ ) = 0 \mathbf{V}_0(S;\lambda)=\mathbf{0} V 0 ( S ; λ ) = 0 S S S ∥ V 0 ( S ; 1 ) ∥ 2 ∈ [ 0.0072 , 2.4284 ] \|\mathbf{V}_0(S;1)\|_2\in[0.0072,2.4284] ∥ V 0 ( S ; 1 ) ∥ 2 ∈ [ 0.0072 , 2.4284 ] C 2 C_2 C 2 V β ( S ; λ ) \mathbf{V}_{\beta}(S;\lambda) V β ( S ; λ ) β ∗ ( S ) ∈ [ − 3 , 4 ] ∖ { 1 } \beta^*(S)\in[-3,4]\setminus\{1\} β ∗ ( S ) ∈ [ − 3 , 4 ] ∖ { 1 } V β ∗ ( S ; 1 ) = 0 \mathbf{V}_{\beta^*}(S;1)=\mathbf{0} V β ∗ ( S ; 1 ) = 0 Figure 3 (d)). Finally, letting p ^ i j ( λ ) ∈ R 3 \hat{\mathbf{p}}_{ij}(\lambda)\in\mathbb{R}^{3} p ^ ij ( λ ) ∈ R 3 p i j ( λ ) T \mathbf{p}_{ij}(\lambda)^\mathsf{T} p ij ( λ ) T w ∗ ( S ; λ ) \mathbf{w}^*(S;\lambda) w ∗ ( S ; λ ) ∥ w − 1 ∥ 2 \|\mathbf{w}-\mathbf{1}\|_2 ∥ w − 1 ∥ 2 1 \mathbf{1} 1 ∑ i , j w i j ∗ p ^ i j ( λ ) = 0 \sum_{i,j}w^*_{ij}\hat{\mathbf{p}}_{ij}(\lambda)=\mathbf{0} ∑ i , j w ij ∗ p ^ ij ( λ ) = 0
w ∗ ( S ; λ ) = 1 − P ^ ( λ ) ( P ^ ( λ ) T P ^ ( λ ) ) − 1 V 0 ( S ; λ ) , (13) \mathbf{w}^*(S;\lambda)=\mathbf{1}-\hat{\mathbf{P}}(\lambda)\left(\hat{\mathbf{P}}(\lambda)^\mathsf{T}\hat{\mathbf{P}}(\lambda)\right)^{-1}\mathbf{V}_0(S;\lambda), \tag*{(13)} w ∗ ( S ; λ ) = 1 − P ^ ( λ ) ( P ^ ( λ ) T P ^ ( λ ) ) − 1 V 0 ( S ; λ ) , (13) which is strictly positive (w ∗ ( S ; λ ) > 0 \mathbf{w}^*(S;\lambda)>0 w ∗ ( S ; λ ) > 0 λ = 1 \lambda=1 λ = 1 λ i s o = 1 / 17 \lambda_{\mathrm{iso}}=1/\sqrt{17} λ iso = 1/ 17 λ c u b e = 0.20 \lambda_{\mathrm{cube}}=0.20 λ cube = 0.20 Figure 3 (c)), with ∥ w ∗ ( S ; 1 ) − 1 ∥ 2 ∈ [ 0.0052 , 1.5967 ] \|\mathbf{w}^*(S;1)-\mathbf{1}\|_2\in[0.0052,1.5967] ∥ w ∗ ( S ; 1 ) − 1 ∥ 2 ∈ [ 0.0052 , 1.5967 ]
Proof. Expanding V 3 ( S ; λ ) = ∑ i , j ( r i j 2 + λ 2 z i j 2 ) ( x j , y i , λ z i j ) V_{3}(S;\lambda)=\sum_{i,j}(r_{ij}^{2}+\lambda^{2}z_{ij}^{2})(x_{j},y_{i},\lambda z_{ij}) V 3 ( S ; λ ) = ∑ i , j ( r ij 2 + λ 2 z ij 2 ) ( x j , y i , λ z ij ) ∑ i , j r i j 2 ( x j , y i ) \sum_{i,j}r_{ij}^{2}(x_{j},y_{i}) ∑ i , j r ij 2 ( x j , y i ) λ 2 ∑ i , j z i j 2 ( x j , y i ) \lambda^{2}\sum_{i,j}z_{ij}^{2}(x_{j},y_{i}) λ 2 ∑ i , j z ij 2 ( x j , y i ) λ ∑ i , j ( x j 2 + y i 2 ) z i j + λ 3 ∑ i , j z i j 3 \lambda\sum_{i,j}(x_{j}^{2}+y_{i}^{2})z_{ij}+\lambda^{3}\sum_{i,j}z_{ij}^{3} λ ∑ i , j ( x j 2 + y i 2 ) z ij + λ 3 ∑ i , j z ij 3 ∑ i , j ( x j 2 + y i 2 ) z i j = 0 \sum_{i,j}(x_{j}^{2}+y_{i}^{2})z_{ij}=0 ∑ i , j ( x j 2 + y i 2 ) z ij = 0 x j 2 + y i 2 ∈ W r c x_{j}^{2}+y_{i}^{2}\in W_{\mathrm{rc}} x j 2 + y i 2 ∈ W rc ∑ i , j z i j 3 = 0 \sum_{i,j}z_{ij}^{3}=0 ∑ i , j z ij 3 = 0 { z i j } \{z_{ij}\} { z ij } (12) . For associative squares, z ( − x , − y ) = − z ( x , y ) z(-x,-y)=-z(x,y) z ( − x , − y ) = − z ( x , y ) p − c ( λ ) = − p c ( λ ) p_{-c}(\lambda)=-p_{c}(\lambda) p − c ( λ ) = − p c ( λ ) R − c ( λ ) = R c ( λ ) R_{-c}(\lambda)=R_{c}(\lambda) R − c ( λ ) = R c ( λ ) β ∈ R \beta\in\mathbb{R} β ∈ R 1 2 ∥ w − 1 ∥ 2 2 \frac{1}{2}\lVert w-\mathbf{1}\rVert_{2}^{2} 2 1 ∥ w − 1 ∥ 2 2 P ^ ( λ ) T w = 0 \widehat{P}(\lambda)^{\mathsf{T}}w=0 P ( λ ) T w = 0 (13) , and exact enumeration over the 880 squares establishes positivity and the stated bounds.
Notice how the cubic resultant V 3 ( S ; λ ) V_{3}(S;\lambda) V 3 ( S ; λ ) (12) bridges static rays and rotational dynamics: while its vertical component vanishes for every semi-magic square, its residual horizontal dipole λ 2 ∑ i , j z i j 2 ( x j , y i ) \lambda^{2}\sum_{i,j}z_{ij}^{2}(x_{j},y_{i}) λ 2 ∑ i , j z ij 2 ( x j , y i ) § 3 , we set our sculptures into three-dimensional rotation and show that this exact vector governs the vertical off-diagonal products of inertia of an entry-weighted gem.
Rotational Dynamics: Smooth Precession and Dzhanibekov Flips
Our second test moves from static radial equilibrium to three-dimensional rotation. Suppose a Magic Gem or radial spoke sculpture of total mass M M M c \mathbf{c} c O O O g = ( 0 , 0 , − g ) T \mathbf{g}=(0,0,-g)^{\mathsf{T}} g = ( 0 , 0 , − g ) T weighted_ring.stl in [12 ] ), or tossed in free space. About the pivot O O O
τ g r a v = M c × g , \boldsymbol{\tau}_{\mathrm{grav}}=M\mathbf{c}\times\mathbf{g}, τ grav = M c × g , which vanishes identically in every 3D orientation if and only if the centre of mass lies at the pivot (c = 0 \mathbf{c}=0 c = 0 c = 0 \mathbf{c}=0 c = 0 O O O [2 , 10 ] ; in the principal body frame, the angular velocity vector ω = ( ω 1 , ω 2 , ω 3 ) T \boldsymbol{\omega}=(\omega_{1},\omega_{2},\omega_{3})^{\mathsf{T}} ω = ( ω 1 , ω 2 , ω 3 ) T
I 1 ω ˙ 1 = ( I 2 − I 3 ) ω 2 ω 3 , I 2 ω ˙ 2 = ( I 3 − I 1 ) ω 3 ω 1 , I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 , (14) I_{1}\dot{\omega}_{1}=(I_{2}-I_{3})\omega_{2}\omega_{3},\qquad I_{2}\dot{\omega}_{2}=(I_{3}-I_{1})\omega_{3}\omega_{1},\qquad I_{3}\dot{\omega}_{3}=(I_{1}-I_{2})\omega_{1}\omega_{2}, \tag*{(14)} I 1 ω ˙ 1 = ( I 2 − I 3 ) ω 2 ω 3 , I 2 ω ˙ 2 = ( I 3 − I 1 ) ω 3 ω 1 , I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 , (14) where I 1 , I 2 , I 3 I_{1},I_{2},I_{3} I 1 , I 2 , I 3 I 1 < I 2 < I 3 I_{1}<I_{2}<I_{3} I 1 < I 2 < I 3 I 2 I_{2} I 2 ( I 2 − I 1 ) ( I 3 − I 2 ) / ( I 1 I 3 ) ∣ ω 2 ∣ \sqrt{(I_{2}-I_{1})(I_{3}-I_{2})/(I_{1}I_{3})}\lvert\omega_{2}\rvert ( I 2 − I 1 ) ( I 3 − I 2 ) / ( I 1 I 3 ) ∣ ω 2 ∣ [3 , 15 ] .
When the sixteen unequal heights z i j z_{ij} z ij R 3 \mathbb{R}^{3} R 3
Equal Masses at Unequal Heights Form a Universal Axisymmetric Rotor
Consider first the discrete point cloud P S ( λ ) = { p i j ( λ ) = ( x j , y i , λ z i j ) } P_{S}(\lambda)=\{\mathbf{p}_{ij}(\lambda)=(x_{j},y_{i},\lambda z_{ij})\} P S ( λ ) = { p ij ( λ ) = ( x j , y i , λ z ij )} m = M / N m=M/N m = M / N N = n 2 N=n^{2} N = n 2 p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) O O O c = 1 N ∑ i , j p i j ( λ ) = 0 \mathbf{c}=\frac{1}{N}\sum_{i,j}\mathbf{p}_{ij}(\lambda)=0 c = N 1 ∑ i , j p ij ( λ ) = 0 O O O
Theorem 3.1 (Universal symmetric-top law, spherical-top scale, and periodic polhodes). Let S S S any arrangement of { 1 , … , n 2 } \{1,\ldots,n^{2}\} { 1 , … , n 2 } n ≥ 2 n\ge2 n ≥ 2 vanishing first moments ∑ i , j x j z i j = ∑ i , j y i z i j = 0 \sum_{i,j}x_{j}z_{ij}=\sum_{i,j}y_{i}z_{ij}=0 ∑ i , j x j z ij = ∑ i , j y i z ij = 0 m = M / N m=M/N m = M / N p i j ( λ ) = ( x j , y i , λ z i j ) \mathbf{p}_{ij}(\lambda)=(x_{j},y_{i},\lambda z_{ij}) p ij ( λ ) = ( x j , y i , λ z ij ) λ > 0 \lambda>0 λ > 0 O O O
I ( λ ) = M N ∑ i , j ( ∥ p i j ( λ ) ∥ 2 2 I 3 − p i j ( λ ) p i j ( λ ) T ) , I(\lambda)=\frac{M}{N}\sum_{i,j}\left(\lVert\mathbf{p}_{ij}(\lambda)\rVert_{2}^{2}I_{3}-\mathbf{p}_{ij}(\lambda)\mathbf{p}_{ij}(\lambda)^{\mathsf{T}}\right), I ( λ ) = N M i , j ∑ ( ∥ p ij ( λ ) ∥ 2 2 I 3 − p ij ( λ ) p ij ( λ ) T ) , is diagonal and strictly axisymmetric about the z z z
I ( λ ) = M diag ( I ⊥ ( λ ) , I ⊥ ( λ ) , I z z ) , I ⊥ ( λ ) = a n + λ 2 b n , I z z = 2 a n , (15) I(\lambda)=M\,\operatorname{diag}(I_{\perp}(\lambda),I_{\perp}(\lambda),I_{zz}), \qquad I_{\perp}(\lambda)=a_n+\lambda^2b_n, \qquad I_{zz}=2a_n, \tag*{(15)} I ( λ ) = M diag ( I ⊥ ( λ ) , I ⊥ ( λ ) , I z z ) , I ⊥ ( λ ) = a n + λ 2 b n , I z z = 2 a n , (15) with universal principal inertia ratio
I ⊥ ( λ ) I z z = a n + λ 2 b n 2 a n = 1 + λ 2 ( n 2 + 1 ) 2 . (16) \frac{I_{\perp}(\lambda)}{I_{zz}}=\frac{a_n+\lambda^2b_n}{2a_n}=\frac{1+\lambda^2(n^2+1)}{2}. \tag*{(16)} I z z I ⊥ ( λ ) = 2 a n a n + λ 2 b n = 2 1 + λ 2 ( n 2 + 1 ) . (16) Varying the vertical lift scale λ \lambda λ Figure 4 (a)). At the canonical lift λ = 1 \lambda=1 λ = 1
Figure 4. Three-dimensional rotational dynamics of Magic Gems. (a) Universal inertia ratio I ⊥ ( λ ) / I z z = 1 2 ( 1 + λ 2 ( n 2 + 1 ) ) I_{\perp}(\lambda)/I_{zz} = \frac{1}{2}(1+\lambda^2(n^2+1)) I ⊥ ( λ ) / I z z = 2 1 ( 1 + λ 2 ( n 2 + 1 )) λ \lambda λ n = 3 , 4 , 5 n=3,4,5 n = 3 , 4 , 5 λ c u b e = 1 / 5 \lambda_{\mathrm{cube}}=1/5 λ cube = 1/5 21 / 25 21/25 21/25 λ i s o = 1 / 17 \lambda_{\mathrm{iso}}=1/\sqrt{17} λ iso = 1/ 17 λ = 1 \lambda=1 λ = 1 n = 4 n=4 n = 4 I 1 = 110 < I 2 = 230 < I 3 = 340 I_1=110<I_2=230<I_3=340 I 1 = 110 < I 2 = 230 < I 3 = 340 + ω 2 +\omega_2 + ω 2 − ω 2 -\omega_2 − ω 2 (14) : an axisymmetric magic rotor precesses harmonically with constant ∣ ω ⊥ ∣ |\omega_{\perp}| ∣ ω ⊥ ∣ ∣ Q x y ∣ = 60 |Q_{xy}|=60 ∣ Q x y ∣ = 60 180 ∘ 180^\circ 18 0 ∘ + ω 2 +\omega_2 + ω 2 − ω 2 -\omega_2 − ω 2 ∣ Q x y ( S ) ∣ = 1 2 ( I 2 − I 1 ) |Q_{xy}(S)|=\frac{1}{2}(I_2-I_1) ∣ Q x y ( S ) ∣ = 2 1 ( I 2 − I 1 )
I ⊥ ( 1 ) I z z = n 2 + 2 2 , (17) \frac{I_{\perp}(1)}{I_{zz}}=\frac{n^2+2}{2}, \tag*{(17)} I z z I ⊥ ( 1 ) = 2 n 2 + 2 , (17) which is an exact integer for every even order n n n 9 9 9 n = 4 n=4 n = 4 19 19 19 n = 6 n=6 n = 6 33 33 33 n = 8 n=8 n = 8 n n n 11 / 2 11/2 11/2 n = 3 n=3 n = 3 27 / 2 27/2 27/2 n = 5 n=5 n = 5
λ i s o ( n ) = 1 n 2 + 1 ( 1 10 ≈ 0.3162 for n = 3 , 1 17 ≈ 0.2425 for n = 4 ) , (18) \lambda_{\mathrm{iso}}(n)=\frac{1}{\sqrt{n^2+1}}\qquad\left(\frac{1}{\sqrt{10}}\approx0.3162\ \text{for }n=3,\quad\frac{1}{\sqrt{17}}\approx0.2425\ \text{for }n=4\right), \tag*{(18)} λ iso ( n ) = n 2 + 1 1 ( 10 1 ≈ 0.3162 for n = 3 , 17 1 ≈ 0.2425 for n = 4 ) , (18) I ⊥ ( λ i s o ) = I z z = 2 a n I_{\perp}(\lambda_{\mathrm{iso}})=I_{zz}=2a_n I ⊥ ( λ iso ) = I z z = 2 a n I ( λ i s o ) = 2 M a n I 3 I(\lambda_{\mathrm{iso}})=2Ma_nI_3 I ( λ iso ) = 2 M a n I 3 O O O R 3 \mathbb{R}^3 R 3 λ c u b e ( n ) = n − 1 n 2 − 1 = 1 n + 1 \lambda_{\mathrm{cube}}(n)=\frac{n-1}{n^2-1}=\frac{1}{n+1} λ cube ( n ) = n 2 − 1 n − 1 = n + 1 1 [ − n − 1 2 , n − 1 2 ] 2 \left[-\frac{n-1}{2},\frac{n-1}{2}\right]^2 [ − 2 n − 1 , 2 n − 1 ] 2 λ c u b e = 1 / 5 = 0.20 \lambda_{\mathrm{cube}}=1/5=0.20 λ cube = 1/5 = 0.20 n = 4 n=4 n = 4
I ⊥ ( λ c u b e ) I z z = n 2 + n + 1 ( n + 1 ) 2 ( 21 25 = 0.84 at n = 4 ) . (19) \frac{I_{\perp}(\lambda_{\mathrm{cube}})}{I_{zz}}=\frac{n^2+n+1}{(n+1)^2}\qquad\left(\frac{21}{25}=0.84\ \text{at }n=4\right). \tag*{(19)} I z z I ⊥ ( λ cube ) = ( n + 1 ) 2 n 2 + n + 1 ( 25 21 = 0.84 at n = 4 ) . (19) Because I x x ( λ ) = I y y ( λ ) = I ⊥ ( λ ) I_{xx}(\lambda)=I_{yy}(\lambda)=I_{\perp}(\lambda) I xx ( λ ) = I y y ( λ ) = I ⊥ ( λ ) (14) ), ω 3 ( t ) = ω 3 , 0 \omega_3(t)=\omega_{3,0} ω 3 ( t ) = ω 3 , 0 ( ω 1 ( t ) , ω 2 ( t ) ) (\omega_1(t),\omega_2(t)) ( ω 1 ( t ) , ω 2 ( t ))
Ω b o d y ( λ ) = I ⊥ ( λ ) − I z z I ⊥ ( λ ) ω 3 , 0 = λ 2 ( n 2 + 1 ) − 1 λ 2 ( n 2 + 1 ) + 1 ω 3 , 0 , (20) \Omega_{\mathrm{body}}(\lambda)=\frac{I_{\perp}(\lambda)-I_{zz}}{I_{\perp}(\lambda)}\omega_{3,0}=\frac{\lambda^2(n^2+1)-1}{\lambda^2(n^2+1)+1}\omega_{3,0}, \tag*{(20)} Ω body ( λ ) = I ⊥ ( λ ) I ⊥ ( λ ) − I z z ω 3 , 0 = λ 2 ( n 2 + 1 ) + 1 λ 2 ( n 2 + 1 ) − 1 ω 3 , 0 , (20) closing periodically after ( n 2 + 2 ) / 2 = 9 (n^2+2)/2=9 ( n 2 + 2 ) /2 = 9 λ = 1 \lambda=1 λ = 1 n = 4 n=4 n = 4 Ω b o d y = 0 \Omega_{\mathrm{body}}=0 Ω body = 0 ω 0 ∈ R 3 \boldsymbol{\omega}_0\in\mathbb{R}^3 ω 0 ∈ R 3 λ = λ i s o ( n ) \lambda=\lambda_{\mathrm{iso}}(n) λ = λ iso ( n )
Proof. By [14 Lem. 3.1] , the empirical covariance matrix Σ ( λ ) = 1 N ∑ i , j p i j ( λ ) p i j ( λ ) T \Sigma(\lambda)=\frac{1}{N}\sum_{i,j}p_{ij}(\lambda)p_{ij}(\lambda)^{\mathsf{T}} Σ ( λ ) = N 1 ∑ i , j p ij ( λ ) p ij ( λ ) T Σ x x = Σ y y = a n = ( n 2 − 1 ) / 12 \Sigma_{xx}=\Sigma_{yy}=a_n=(n^2-1)/12 Σ xx = Σ y y = a n = ( n 2 − 1 ) /12 Σ z z = λ 2 b n = λ 2 ( n 4 − 1 ) / 12 = λ 2 ( n 2 + 1 ) a n \Sigma_{zz}=\lambda^2b_n=\lambda^2(n^4-1)/12=\lambda^2(n^2+1)a_n Σ z z = λ 2 b n = λ 2 ( n 4 − 1 ) /12 = λ 2 ( n 2 + 1 ) a n Σ x y = 0 \Sigma_{xy}=0 Σ x y = 0 Σ x z = Σ y z = 0 \Sigma_{xz}=\Sigma_{yz}=0 Σ x z = Σ y z = 0 I ( λ ) = M ( tr ( Σ ( λ ) ) I 3 − Σ ( λ ) ) I(\lambda)=M\left(\operatorname{tr}(\Sigma(\lambda))I_3-\Sigma(\lambda)\right) I ( λ ) = M ( tr ( Σ ( λ )) I 3 − Σ ( λ ) ) (15) and (16) . Evaluating (16) at λ = 1 \lambda=1 λ = 1 λ = ( n 2 + 1 ) − 1 / 2 \lambda=(n^2+1)^{-1/2} λ = ( n 2 + 1 ) − 1/2 λ = ( n + 1 ) − 1 \lambda=(n+1)^{-1} λ = ( n + 1 ) − 1 (17) –(19) . Substituting I 1 = I 2 = I ⊥ ( λ ) I_1=I_2=I_{\perp}(\lambda) I 1 = I 2 = I ⊥ ( λ ) I 3 = I z z I_3=I_{zz} I 3 = I z z (14) gives ω ˙ 3 = 0 \dot{\omega}_3=0 ω ˙ 3 = 0 ω ˙ 1 = Ω b o d y ω 2 \dot{\omega}_1=\Omega_{\mathrm{body}}\omega_2 ω ˙ 1 = Ω body ω 2 ω ˙ 2 = − Ω b o d y ω 1 \dot{\omega}_2=-\Omega_{\mathrm{body}}\omega_1 ω ˙ 2 = − Ω body ω 1 Ω b o d y \Omega_{\mathrm{body}} Ω body (20) .
Unequal Entry Masses Expose Q x y Q_{xy} Q x y
The universal axisymmetric rotation of Theorem 3.1 breaks as soon as each cell is weighted by its unequal magic entry m i j = s i j m_{ij}=s_{ij} m ij = s ij weighted_ring.stl [12 ] . While Loly’s theorem [11 ] keeps the diagonal moments equal (I x x = I y y = I 0 I_{xx}=I_{yy}=I_0 I xx = I y y = I 0 Q x y ( S ) = ∑ i , j z i j x i y i Q_{xy}(S)=\sum_{i,j}z_{ij}x_iy_{i} Q x y ( S ) = ∑ i , j z ij x i y i V 3 ( S ; λ ) \mathbf{V}_3(S;\lambda) V 3 ( S ; λ ) § 2 . On square #46 (Figure 1 (d)), for example, the heavy entries { 16 , 13 , 15 , 14 } \{16,13,15,14\} { 16 , 13 , 15 , 14 } x y < 0 xy<0 x y < 0 { 1 , 4 , 2 , 3 } \{1,4,2,3\} { 1 , 4 , 2 , 3 } x y > 0 xy>0 x y > 0 u = − x \mathbf{u}=-\mathbf{x} u = − x λ = 230 \lambda=230 λ = 230 Figure 4 (b)), triggering a 180° Dzhanibekov flip (Figure 4 (c)).
Theorem 3.2 (Entry-weighted inertia split, quadrupole Dzhanibekov instability, and 3D lifted tensor). Let S S S n n n m i j = s i j = s ˉ + z i j m_{ij}=s_{ij}=\bar{s}+z_{ij} m ij = s ij = s ˉ + z ij M = ∑ i , j s i j = N s ˉ = n 2 ( n 2 + 1 ) / 2 M=\sum_{i,j}s_{ij}=N\bar{s}=n^2(n^2+1)/2 M = ∑ i , j s ij = N s ˉ = n 2 ( n 2 + 1 ) /2 m i j = s i j m_{ij}=s_{ij} m ij = s ij ( x j , y i , 0 ) ∈ R 3 (x_j,y_i,0)\in\mathbb{R}^3 ( x j , y i , 0 ) ∈ R 3 O O O 3 × 3 3\times3 3 × 3 O O O
I ( s ) ( S ) = ( I 0 − Q x y ( S ) 0 − Q x y ( S ) I 0 0 0 0 2 I 0 ) , I 0 = M a n = n 2 ( n 4 − 1 ) 24 , Q x y ( S ) = ∑ i , j z i j x j y i , (21) I^{(s)}(S)=
\begin{pmatrix}
I_0 & -Q_{xy}(S) & 0\\
-Q_{xy}(S) & I_0 & 0\\
0 & 0 & 2I_0
\end{pmatrix},
\qquad I_0=Ma_n=\frac{n^2(n^4-1)}{24},
\qquad Q_{xy}(S)=\sum_{i,j}z_{ij}x_jy_i, \tag*{(21)} I ( s ) ( S ) = I 0 − Q x y ( S ) 0 − Q x y ( S ) I 0 0 0 0 2 I 0 , I 0 = M a n = 24 n 2 ( n 4 − 1 ) , Q x y ( S ) = i , j ∑ z ij x j y i , (21) whose principal moments of inertia along the horizontal diagonals y = ± x y=\pm x y = ± x z z z
I 1 ( S ) = I 0 − ∣ Q x y ( S ) ∣ , I 2 ( S ) = I 0 + ∣ Q x y ( S ) ∣ , I 3 = 2 I 0 . (22) I_1(S)=I_0-|Q_{xy}(S)|,\qquad I_2(S)=I_0+|Q_{xy}(S)|,\qquad I_3=2I_0. \tag*{(22)} I 1 ( S ) = I 0 − ∣ Q x y ( S ) ∣ , I 2 ( S ) = I 0 + ∣ Q x y ( S ) ∣ , I 3 = 2 I 0 . (22) At order n = 4 n=4 n = 4 I 0 = 170 I_0=170 I 0 = 170 I 3 = 340 I_3=340 I 3 = 340 Q x y ( S ) = 0 Q_{xy}(S)=0 Q x y ( S ) = 0 I 1 = I 2 = 170 I_1=I_2=170 I 1 = I 2 = 170 ∣ Q x y ( S ) ∣ ∈ { 1 , … , 60 } |Q_{xy}(S)|\in\{1,\ldots,60\} ∣ Q x y ( S ) ∣ ∈ { 1 , … , 60 } ∣ Q x y ∣ = 60 |Q_{xy}|=60 ∣ Q x y ∣ = 60 I 1 = 110 < I 2 = 230 < I 3 = 340 I_1=110<I_2=230<I_3=340 I 1 = 110 < I 2 = 230 < I 3 = 340 y = − sgn ( Q x y ) x y=-\operatorname{sgn}(Q_{xy})x y = − sgn ( Q x y ) x
Γ ( S ) = ( I 2 − I 1 ) ( I 3 − I 2 ) I 1 I 3 = ∣ Q x y ( S ) ∣ I 0 = ∣ Q x y ( S ) ∣ 170 . \Gamma(S)=\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}}
=\sqrt{\frac{|Q_{xy}(S)|}{I_0}}
=\sqrt{\frac{|Q_{xy}(S)|}{170}}. Γ ( S ) = I 1 I 3 ( I 2 − I 1 ) ( I 3 − I 2 ) = I 0 ∣ Q x y ( S ) ∣ = 170 ∣ Q x y ( S ) ∣ . When the entry masses m i j = s i j = s ˉ + z i j m_{ij}=s_{ij}=\bar{s}+z_{ij} m ij = s ij = s ˉ + z ij p i j ( λ ) = ( x j , y i , λ z i j ) \mathbf{p}_{ij}(\lambda)=(x_j,y_i,\lambda z_{ij}) p ij ( λ ) = ( x j , y i , λ z ij ) λ > 0 \lambda>0 λ > 0 s i j > s ˉ s_{ij}>\bar{s} s ij > s ˉ z i j > 0 z_{ij}>0 z ij > 0 s i j < s ˉ s_{ij}<\bar{s} s ij < s ˉ z i j < 0 z_{ij}<0 z ij < 0 λ ∑ i , j z i j 2 \lambda\sum_{i,j}z_{ij}^{2} λ ∑ i , j z ij 2 c ( s ) ( λ ) = ( 0 , 0 , z c m ( λ ) ) \mathbf{c}^{(s)}(\lambda)=(0,0,z_{\mathrm{cm}}(\lambda)) c ( s ) ( λ ) = ( 0 , 0 , z cm ( λ )) z c m ( λ ) = λ b n / s ˉ = λ ( n 2 − 1 ) 6 = 5 2 λ z_{\mathrm{cm}}(\lambda)=\lambda b_n/\bar{s}=\frac{\lambda(n^2-1)}{6}=\frac{5}{2}\lambda z cm ( λ ) = λ b n / s ˉ = 6 λ ( n 2 − 1 ) = 2 5 λ n = 4 n=4 n = 4 ∑ i , j z i j 2 = 0 \sum_{i,j}z_{ij}^{2}=0 ∑ i , j z ij 2 = 0 O O O I x x , O ( s ) ( λ ) = I y y , O ( s ) ( λ ) = M ( a n + λ 2 b n ) I^{(s)}_{xx,O}(\lambda)=I^{(s)}_{yy,O}(\lambda)=M(a_n+\lambda^2b_n) I xx , O ( s ) ( λ ) = I y y , O ( s ) ( λ ) = M ( a n + λ 2 b n ) I z z , O ( s ) ( λ ) = 2 I 0 I^{(s)}_{zz,O}(\lambda)=2I_0 I z z , O ( s ) ( λ ) = 2 I 0 Theorem 3.1 . Translating to the true centre of mass c ( s ) ( λ ) \mathbf{c}^{(s)}(\lambda) c ( s ) ( λ ) I z z , c m ( s ) = 2 I 0 I^{(s)}_{zz,\mathrm{cm}}=2I_0 I z z , cm ( s ) = 2 I 0 M z c m ( λ ) 2 Mz_{\mathrm{cm}}(\lambda)^2 M z cm ( λ ) 2
I x x , c m ( s ) ( λ ) = I y y , c m ( s ) ( λ ) = M ( a n + λ 2 b n ( 1 − b n s ˉ 2 ) ) = M a n ( 1 + 2 ( n 2 + 2 ) 3 λ 2 ) I^{(s)}_{xx,\mathrm{cm}}(\lambda)=I^{(s)}_{yy,\mathrm{cm}}(\lambda)
=M\left(a_n+\lambda^2b_n\left(1-\frac{b_n}{\bar{s}^2}\right)\right)
=Ma_n\left(1+\frac{2(n^2+2)}{3}\lambda^2\right) I xx , cm ( s ) ( λ ) = I y y , cm ( s ) ( λ ) = M ( a n + λ 2 b n ( 1 − s ˉ 2 b n ) ) = M a n ( 1 + 3 2 ( n 2 + 2 ) λ 2 ) ( = 170 ( 1 + 12 λ 2 ) (=170(1+12\lambda^2) ( = 170 ( 1 + 12 λ 2 ) n = 4 n=4 n = 4 ( O (O ( O c ( s ) ( λ ) ) \mathbf{c}^{(s)}(\lambda)) c ( s ) ( λ )) I x y ( s ) = − Q x y ( S ) I_{xy}^{(s)}=-Q_{xy}(S) I x y ( s ) = − Q x y ( S )
( I x z ( s ) ( λ ) , I y z ( s ) ( λ ) ) = − λ ∑ i , j z i j 2 ( x j , y i ) = − λ − 1 ( ( V 3 ( S ; λ ) ) x , ( V 3 ( S ; λ ) ) y ) . \left(I_{xz}^{(s)}(\lambda),I_{yz}^{(s)}(\lambda)\right)=-\lambda\sum_{i,j}z_{ij}^{2}(x_j,y_i)=-\lambda^{-1}\left((\mathbf{V}_3(S;\lambda))_x,(\mathbf{V}_3(S;\lambda))_y\right). ( I x z ( s ) ( λ ) , I y z ( s ) ( λ ) ) = − λ i , j ∑ z ij 2 ( x j , y i ) = − λ − 1 ( ( V 3 ( S ; λ ) ) x , ( V 3 ( S ; λ ) ) y ) . so an order-four magic square has a diagonal, axisymmetric 3D entry-weighted inertia tensor if and only if it is one of the 48 associative squares (Q x y = 0 Q_{xy}=0 Q x y = 0 V 3 = 0 \mathbf{V}_3=0 V 3 = 0
Proof. Since every row excess R i = ∑ j z i j R_i=\sum_j z_{ij} R i = ∑ j z ij C j = ∑ i z i j C_j=\sum_i z_{ij} C j = ∑ i z ij ∑ i , j s i j x j 2 = s ˉ ∑ i , j x j 2 + ∑ j x j 2 C j = M a n = I 0 \sum_{i,j}s_{ij}x_j^2=\bar{s}\sum_{i,j}x_j^2+\sum_jx_j^2C_j=Ma_n=I_0 ∑ i , j s ij x j 2 = s ˉ ∑ i , j x j 2 + ∑ j x j 2 C j = M a n = I 0 ∑ i , j s i j y i 2 = I 0 \sum_{i,j}s_{ij}y_i^2=I_0 ∑ i , j s ij y i 2 = I 0 I x x = I y y = I 0 I_{xx}=I_{yy}=I_0 I xx = I y y = I 0 I z z = 2 I 0 I_{zz}=2I_0 I z z = 2 I 0 ∑ i , j x j y i = 0 \sum_{i,j}x_jy_i=0 ∑ i , j x j y i = 0 I x y = − ∑ i , j s i j x j y i = − Q x y ( S ) I_{xy}=-\sum_{i,j}s_{ij}x_jy_i=-Q_{xy}(S) I x y = − ∑ i , j s ij x j y i = − Q x y ( S ) ( I 0 − Q x y − Q x y I 0 ) \begin{pmatrix} I_0 & -Q_{xy} \\ -Q_{xy} & I_0 \end{pmatrix} ( I 0 − Q x y − Q x y I 0 ) I 0 ∓ ∣ Q x y ∣ I_0\mp|Q_{xy}| I 0 ∓ ∣ Q x y ∣ ( 1 , ± 1 ) / 2 (1,\pm1)/\sqrt{2} ( 1 , ± 1 ) / 2 n = 4 n=4 n = 4 I 0 = 170 I_0=170 I 0 = 170 ∣ Q x y ∣ ≤ 60 < 170 |Q_{xy}|\leq60<170 ∣ Q x y ∣ ≤ 60 < 170 I 3 − I 2 = 2 I 0 − ( I 0 + ∣ Q x y ∣ ) = I 1 I_3-I_2=2I_0-(I_0+|Q_{xy}|)=I_1 I 3 − I 2 = 2 I 0 − ( I 0 + ∣ Q x y ∣ ) = I 1 ( I 2 − I 1 ) ( I 3 − I 2 ) / ( I 1 I 3 ) \sqrt{(I_2-I_1)(I_3-I_2)/(I_1I_3)} ( I 2 − I 1 ) ( I 3 − I 2 ) / ( I 1 I 3 ) ∣ Q x y ∣ / I 0 \sqrt{|Q_{xy}|/I_0} ∣ Q x y ∣/ I 0
At the 3D lifted points p i j ( λ ) \mathbf{p}_{ij}(\lambda) p ij ( λ ) ∑ i , j s i j ( x j , y i ) = 0 \sum_{i,j}s_{ij}(x_j,y_i)=\mathbf{0} ∑ i , j s ij ( x j , y i ) = 0 ∑ i , j ( s ˉ + z i j ) λ z i j = λ ∑ i , j z i j 2 = N λ b n \sum_{i,j}(\bar{s}+z_{ij})\lambda z_{ij}=\lambda\sum_{i,j}z_{ij}^2=N\lambda b_n ∑ i , j ( s ˉ + z ij ) λ z ij = λ ∑ i , j z ij 2 = N λ b n z c m ( λ ) = λ b n / s ˉ \mathbf{z}_{\mathrm{cm}}(\lambda)=\lambda b_n/\bar{s} z cm ( λ ) = λ b n / s ˉ ∑ i , j z i j 3 = 0 \sum_{i,j}z_{ij}^3=0 ∑ i , j z ij 3 = 0 ∑ i , j ( s ˉ + z i j ) λ 2 z i j 2 = M λ 2 b n \sum_{i,j}(\bar{s}+z_{ij})\lambda^2z_{ij}^2=M\lambda^2b_n ∑ i , j ( s ˉ + z ij ) λ 2 z ij 2 = M λ 2 b n I x x , O ( s ) ( λ ) = I y y , O ( s ) ( λ ) = M ( a n + λ 2 b n ) I_{xx,O}^{(s)}(\lambda)=I_{yy,O}^{(s)}(\lambda)=M(a_n+\lambda^2b_n) I xx , O ( s ) ( λ ) = I y y , O ( s ) ( λ ) = M ( a n + λ 2 b n ) M z c m ( λ ) 2 diag ( 1 , 1 , 0 ) Mz_{\mathrm{cm}}(\lambda)^2\operatorname{diag}(1,1,0) M z cm ( λ ) 2 diag ( 1 , 1 , 0 ) 1 − b n / s ˉ 2 = 2 ( n 2 + 2 ) 3 ( n 2 + 1 ) 1-b_n/\bar{s}^2=\frac{2(n^2+2)}{3(n^2+1)} 1 − b n / s ˉ 2 = 3 ( n 2 + 1 ) 2 ( n 2 + 2 )
b n ( 1 − b n s ˉ 2 ) = 2 ( n 2 + 2 ) 3 a n . b_n\left(1-\frac{b_n}{\bar{s}^2}\right)=\frac{2(n^2+2)}{3}a_n. b n ( 1 − s ˉ 2 b n ) = 3 2 ( n 2 + 2 ) a n . Finally, I x z ( s ) ( λ ) = − λ ∑ i , j z i j 2 x j = − λ − 1 ( V 3 ) x I_{xz}^{(s)}(\lambda)=-\lambda\sum_{i,j}z_{ij}^2x_j=-\lambda^{-1}(\mathbf{V}_3)_x I x z ( s ) ( λ ) = − λ ∑ i , j z ij 2 x j = − λ − 1 ( V 3 ) x I y z ( s ) ( λ ) I_{yz}^{(s)}(\lambda) I y z ( s ) ( λ ) Q x y ( S ) Q_{xy}(S) Q x y ( S ) Theorem 2.5 .
From Point Clouds to Continuous Polyhedra: Why Solid Hulls Tilt on the Pivot
What happens to rotational stability when we replace the discrete point cloud P S P_S P S G 0 ( S ) = conv P S G_0(S)=\operatorname{conv}P_S G 0 ( S ) = conv P S G 0 ( S ) G_0(S) G 0 ( S ) c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) Theorem 3.1 : across all 880 squares at λ = 1 \lambda=1 λ = 1 1 2 ( I 2 + I 3 ) \frac{1}{2}(I_2+I_3) 2 1 ( I 2 + I 3 ) I 1 I_1 I 1 [ 7.34 , 11.57 ] [7.34,11.57] [ 7.34 , 11.57 ] I 2 I_2 I 2 I 3 I_3 I 3 ( I 3 − I 2 I 2 ∈ [ 0.00004 , 0.0243 ] ) \left(\frac{I_3-I_2}{I_2}\in[0.00004,0.0243]\right) ( I 2 I 3 − I 2 ∈ [ 0.00004 , 0.0243 ] )
Yet when we place the cast solid G 0 ( S ) G_0(S) G 0 ( S ) O O O c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) ( c 0 ( S ) ≠ 0 ) (\mathbf{c}_0(S)\neq\mathbf{0}) ( c 0 ( S ) = 0 ) τ g r a v = M c 0 ( S ) × g ≠ 0 \boldsymbol{\tau}_{\mathrm{grav}}=M\mathbf{c}_0(S)\times\mathbf{g}\neq\mathbf{0} τ grav = M c 0 ( S ) × g = 0 § 4 .
Convex Hulls and “Diamond-in-a-Diamond” Counterweighting
Why does passing from the discrete point cloud P S P_S P S ( ∑ i , j p i j = 0 ) (\sum_{i,j}\mathbf{p}_{ij}=\mathbf{0}) ( ∑ i , j p ij = 0 ) G 0 ( S ) : = G ( S ) = conv P S G_0(S):=\mathcal{G}(S)=\operatorname{conv}P_S G 0 ( S ) := G ( S ) = conv P S
c 0 ( S ) : = c s o l i d ( G 0 ( S ) ) = 1 vol G 0 ( S ) ∭ G 0 ( S ) x d V \mathbf{c}_0(S):=\mathbf{c}_{\mathrm{solid}}(G_0(S))=\frac{1}{\operatorname{vol}G_0(S)}\iiint_{G_0(S)}\mathbf{x}\,\mathrm{d}V c 0 ( S ) := c solid ( G 0 ( S )) = vol G 0 ( S ) 1 ∭ G 0 ( S ) x d V away from the origin O O O ∂ G 0 ( S ) \partial G_0(S) ∂ G 0 ( S ) V 0 = ∣ vert G 0 ( S ) ∣ V_0=|\operatorname{vert}G_0(S)| V 0 = ∣ vert G 0 ( S ) ∣ V 1 = N − V 0 V_1=N-V_0 V 1 = N − V 0 P S ( 1 ) = P S ∖ vert G 0 ( S ) P_S^{(1)}=P_S\setminus\operatorname{vert}G_0(S) P S ( 1 ) = P S ∖ vert G 0 ( S ) V 0 = N = 16 V_0=N=16 V 0 = N = 16 G 0 ( S ) G_0(S) G 0 ( S ) O O O Δ k \Delta_k Δ k vol ( Δ k ) \operatorname{vol}(\Delta_k) vol ( Δ k )
Since the outer hull vertices and the discarded inner points together sum to zero, their centroids must pull in strictly opposite directions across O O O G 1 ( S ) = conv P S ( 1 ) G_1(S)=\operatorname{conv}P_S^{(1)} G 1 ( S ) = conv P S ( 1 ) O O O
The 51 Naturally Balanced Solids and Outer–Inner Vertex Opposition
We first classify the 51 order-four magic squares whose uniform solid hull G 0 ( S ) G_0(S) G 0 ( S ) O O O
Proposition 4.1 (Classification of the 51 naturally balanced order-four gems). Across the 880 magic squares of order 4, the uniform solid G 0 ( S ) G_0(S) G 0 ( S ) c 0 ( S ) = 0 \mathbf{c}_0(S)=0 c 0 ( S ) = 0 z ( − x , − y ) = − z ( x , y ) z(-x,-y)=-z(x,y) z ( − x , − y ) = − z ( x , y ) G 0 ( S ) = − G 0 ( S ) G_0(S)=-G_0(S) G 0 ( S ) = − G 0 ( S ) c 0 ( S ) = − c 0 ( S ) = 0 \mathbf{c}_0(S)=-\mathbf{c}_0(S)=0 c 0 ( S ) = − c 0 ( S ) = 0 Table 1 ).
Table 1. The three non-associative order-four magic squares whose uniform solid G 0 ( S ) G_0(S) G 0 ( S ) c 0 ( S ) = 0 \mathbf{c}_0(S)=0 c 0 ( S ) = 0 G 1 ( S ) = conv ( P S ∖ vert G 0 ( S ) ) G_1(S)=\operatorname{conv}(P_S\setminus\operatorname{vert}G_0(S)) G 1 ( S ) = conv ( P S ∖ vert G 0 ( S )) c 1 ( S ) \mathbf{c}_1(S) c 1 ( S )
Proof. For every associative square S S S P S = − P S P_S=-P_S P S = − P S G 0 ( S ) = − G 0 ( S ) G_0(S)=-G_0(S) G 0 ( S ) = − G 0 ( S ) c 0 ( S ) = 0 \mathbf{c}_0(S)=0 c 0 ( S ) = 0 S ˉ = h S \bar{S}=hS S ˉ = h S S ˉ = v S \bar{S}=vS S ˉ = v S G 0 ( S ) G_0(S) G 0 ( S ) ι h : ( x , y , z ) ↦ ( x , − y , − z ) \iota_h:(x,y,z)\mapsto(x,-y,-z) ι h : ( x , y , z ) ↦ ( x , − y , − z ) ι v : ( x , y , z ) ↦ ( − x , y , − z ) \iota_v:(x,y,z)\mapsto(-x,y,-z) ι v : ( x , y , z ) ↦ ( − x , y , − z ) R 3 \mathbb{R}^3 R 3 [14 Prop. 4.3] . The half-turn forces two coordinates of c 0 ( S ) \mathbf{c}_0(S) c 0 ( S )
To counter the centroid drift on the remaining 829 squares, we peel P S P_S P S [4 ] :
P S ( 0 ) = P S , G k ( S ) = conv P S ( k ) , P S ( k + 1 ) = P S ( k ) ∖ vert G k ( S ) ( k ≥ 0 ) , (23) P_S^{(0)}=P_S,\qquad G_k(S)=\operatorname{conv}P_S^{(k)},\qquad P_S^{(k+1)}=P_S^{(k)}\setminus\operatorname{vert}G_k(S)\quad(k\geq0), \tag*{(23)} P S ( 0 ) = P S , G k ( S ) = conv P S ( k ) , P S ( k + 1 ) = P S ( k ) ∖ vert G k ( S ) ( k ≥ 0 ) , (23) calling G 0 ( S ) G_0(S) G 0 ( S ) outer gem and G 1 ( S ) = conv P S ( 1 ) ⊂ G 0 ( S ) G_1(S)=\operatorname{conv}P_S^{(1)}\subset G_0(S) G 1 ( S ) = conv P S ( 1 ) ⊂ G 0 ( S ) inner core gem —“a diamond inside a diamond” (Figure 5 (a,Figure 5 b)).
Figure 5. Nested “diamond-in-a-diamond” convex layers G 0 ( S ) ⊃ G 1 ( S ) \mathcal{G}_0(S) \supset\mathcal{G}_1(S) G 0 ( S ) ⊃ G 1 ( S ) ( V 0 , V 1 ) = ( 8 , 8 ) (V_0,V_1)=(8,8) ( V 0 , V 1 ) = ( 8 , 8 ) ( V 0 , V 1 ) = ( 10 , 6 ) (V_0,V_1)=(10,6) ( V 0 , V 1 ) = ( 10 , 6 ) G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) G 1 ( S ) \mathcal{G}_1(S) G 1 ( S ) c 0 \mathbf{c}_0 c 0 c 1 \mathbf{c}_1 c 1 O O O Δ ρ 1 / ρ 0 = + 5 / 4 \Delta\rho_1/\rho_0=+5/4 Δ ρ 1 / ρ 0 = + 5/4 c 0 = ( 0 , 19 / 324 , 0 ) \mathbf{c}_0=(0,19/324,0) c 0 = ( 0 , 19/324 , 0 ) + y +y + y c 1 = ( 0 , 29 / 68 , 0 ) \mathbf{c}_1=(0,29/68,0) c 1 = ( 0 , 29/68 , 0 ) Δ ρ 1 / ρ 0 = − 57 / 58 \Delta\rho_1/\rho_0=-57/58 Δ ρ 1 / ρ 0 = − 57/58 O O O
Proposition 4.2 (Vertex-centroid opposition and order-four onion hierarchy). Let S S S n n n V 0 = ∣ vert G 0 ( S ) ∣ < N V_0=|\operatorname{vert}G_0(S)|<N V 0 = ∣ vert G 0 ( S ) ∣ < N V 1 = N − V 0 V_1=N-V_0 V 1 = N − V 0 p ‾ o u t e r = 1 V 0 ∑ p ∈ vert G 0 p \overline{\mathbf{p}}_{\mathrm{outer}}=\frac{1}{V_0}\sum_{\mathbf{p}\in\operatorname{vert}G_0}\mathbf{p} p outer = V 0 1 ∑ p ∈ vert G 0 p p ‾ i n n e r = 1 V 1 ∑ p ∈ P S ( 1 ) p \overline{\mathbf{p}}_{\mathrm{inner}}=\frac{1}{V_1}\sum_{\mathbf{p}\in P_S^{(1)}}\mathbf{p} p inner = V 1 1 ∑ p ∈ P S ( 1 ) p
V 0 p ‾ o u t e r + V 1 p ‾ i n n e r = 0 , (24) V_0\overline{\mathbf{p}}_{\mathrm{outer}}+V_1\overline{\mathbf{p}}_{\mathrm{inner}}=0, \tag*{(24)} V 0 p outer + V 1 p inner = 0 , (24) pointing in strictly opposite directions across O O O p ‾ o u t e r ≠ 0 \overline{\mathbf{p}}_{\mathrm{outer}}\ne0 p outer = 0 Figure 5 (c)), 18 squares have V 0 = 16 V_0=16 V 0 = 16 ( P S ( 1 ) = ∅ ) (P_S^{(1)}=\varnothing) ( P S ( 1 ) = ∅ ) G 1 ( S ) G_1(S) G 1 ( S ) G 0 ⊃ G 1 ⊃ G 2 G_0\supset G_1\supset G_2 G 0 ⊃ G 1 ⊃ G 2 G 1 ( S ) G_1(S) G 1 ( S )
Proof. Equation (24) follows immediately from ∑ p ∈ P S p = 0 \sum_{\mathbf{p}\in P_S}\mathbf{p}=0 ∑ p ∈ P S p = 0 P S = vert G 0 ( S ) ∪ P S ( 1 ) P_S=\operatorname{vert}G_0(S)\mathbin{\cup}P_S^{(1)} P S = vert G 0 ( S ) ∪ P S ( 1 )
Restoring Solid Balance via ``Diamond-in-a-Diamond'' Core Inclusions
Guided by the opposition identity (24) , our first restoration method plays the inner core G 1 ( S ) \mathcal{G}_1(S) G 1 ( S ) G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) Δ ρ 1 \Delta\rho_1 Δ ρ 1 c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) c 1 ( S ) \mathbf{c}_1(S) c 1 ( S ) C 2 C_2 C 2
Theorem 4.3 (Composite “diamond-in-a-diamond” solid balancing). Let S S S dim G 1 ( S ) = 3 \dim\mathcal{G}_1(S)=3 dim G 1 ( S ) = 3 G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) ρ 0 > 0 \rho_0>0 ρ 0 > 0 G 1 ( S ) ⊂ G 0 ( S ) \mathcal{G}_1(S)\subset\mathcal{G}_0(S) G 1 ( S ) ⊂ G 0 ( S ) Δ ρ 1 = ρ 1 − ρ 0 \Delta\rho_1=\rho_1-\rho_0 Δ ρ 1 = ρ 1 − ρ 0
c c o m p = ρ 0 vol ( G 0 ) c 0 + Δ ρ 1 vol ( G 1 ) c 1 ρ 0 vol ( G 0 ) + Δ ρ 1 vol ( G 1 ) \mathbf{c}_{\mathrm{comp}}=\frac{\rho_0\operatorname{vol}(\mathcal{G}_0)\mathbf{c}_0+\Delta\rho_1\operatorname{vol}(\mathcal{G}_1)\mathbf{c}_1}{\rho_0\operatorname{vol}(\mathcal{G}_0)+\Delta\rho_1\operatorname{vol}(\mathcal{G}_1)} c comp = ρ 0 vol ( G 0 ) + Δ ρ 1 vol ( G 1 ) ρ 0 vol ( G 0 ) c 0 + Δ ρ 1 vol ( G 1 ) c 1 vanishes for some Δ ρ 1 ≠ 0 \Delta\rho_1\ne0 Δ ρ 1 = 0 c 0 \mathbf{c}_0 c 0 c 1 \mathbf{c}_1 c 1 c 0 × c 1 = 0 \mathbf{c}_0\times\mathbf{c}_1=0 c 0 × c 1 = 0
Δ ρ 1 ρ 0 = − vol ( G 0 ) ( c 0 ⋅ c 1 ) vol ( G 1 ) ∥ c 1 ∥ 2 2 . (25) \frac{\Delta\rho_1}{\rho_0}=-\frac{\operatorname{vol}(\mathcal{G}_0)(\mathbf{c}_0\cdot\mathbf{c}_1)}{\operatorname{vol}(\mathcal{G}_1)\lVert\mathbf{c}_1\rVert_2^2}. \tag*{(25)} ρ 0 Δ ρ 1 = − vol ( G 1 ) ∥ c 1 ∥ 2 2 vol ( G 0 ) ( c 0 ⋅ c 1 ) . (25) Across the 511 order-four gems with dim G 1 ( S ) = 3 \dim\mathcal{G}_1(S)=3 dim G 1 ( S ) = 3 c 1 = 0 \mathbf{c}_1=0 c 1 = 0 c 0 × c 1 ≠ 0 \mathbf{c}_0\times\mathbf{c}_1\ne0 c 0 × c 1 = 0 c 0 ⋅ c 1 < 0 \mathbf{c}_0\cdot\mathbf{c}_1<0 c 0 ⋅ c 1 < 0 Δ ρ 1 / ρ 0 ∈ [ 5 / 4 , 191 / 6 ] \Delta\rho_1/\rho_0\in[5/4,191/6] Δ ρ 1 / ρ 0 ∈ [ 5/4 , 191/6 ] Table 2 and Table 5 ), led by Gem #673 at Δ ρ 1 / ρ 0 = 5 / 4 \Delta\rho_1/\rho_0=5/4 Δ ρ 1 / ρ 0 = 5/4 Figure 1 e, Figure 5 (a)); the other 17 have parallel solid centroids pointing in the same direction (c 0 ⋅ c 1 > 0 \mathbf{c}_0\cdot\mathbf{c}_1>0 c 0 ⋅ c 1 > 0 c 0 \mathbf{c}_0 c 0 O O O c 1 \mathbf{c}_1 c 1 (24) , and two of these 17 gems balance with non-negative core density ρ 1 ∈ [ 0 , ρ 0 ) \rho_1\in[0,\rho_0) ρ 1 ∈ [ 0 , ρ 0 ) Δ ρ 1 / ρ 0 = − 3 / 17 \Delta\rho_1/\rho_0=-3/17 Δ ρ 1 / ρ 0 = − 3/17 ρ 1 / ρ 0 = 14 / 17 \rho_1/\rho_0=14/17 ρ 1 / ρ 0 = 14/17 Δ ρ 1 / ρ 0 = − 57 / 58 \Delta\rho_1/\rho_0=-57/58 Δ ρ 1 / ρ 0 = − 57/58 ρ 1 / ρ 0 = 1 / 58 \rho_1/\rho_0=1/58 ρ 1 / ρ 0 = 1/58 Figure 5 (b)), which differs from denser-core square #45 (Δ ρ 1 / ρ 0 = + 85 / 61 \Delta\rho_1/\rho_0=+85/61 Δ ρ 1 / ρ 0 = + 85/61 ( 8 , 9 ) ↔ ( 15 , 2 ) (8,9)\leftrightarrow(15,2) ( 8 , 9 ) ↔ ( 15 , 2 ) Figure 1 (c)).
Table 2. The eight order-four Magic Gems with the smallest balancing excess density ratios Δ ρ 1 / ρ 0 ∈ [ 5 / 4 , 40 / 19 ] ≈ [ 1.25 , 2.11 ] \Delta\rho_1/\rho_0\in[5/4,40/19]\approx[1.25,2.11] Δ ρ 1 / ρ 0 ∈ [ 5/4 , 40/19 ] ≈ [ 1.25 , 2.11 ] ( G 0 ( S ) , G 1 ( S ) ) (\mathcal{G}_0(S),\mathcal{G}_1(S)) ( G 0 ( S ) , G 1 ( S )) Theorem 4.3 ), all achievable with standard dual-material 3D printing. All 33 denser-core gems belong to Dudeney’s Group VI; the full list of 33 is given in Table 5 .
Table 5. Complete census of the 33 order-four Magic Gems admitting exact two-density “diamond-in-a-diamond” solid balance c c o m p = 0 \mathbf{c}_{\mathrm{comp}}=0 c comp = 0 Theorem 4.3 ), sorted by increasing excess density ratio Δ ρ 1 / ρ 0 \Delta\rho_1/\rho_0 Δ ρ 1 / ρ 0
More generally, placing non-negative point masses m q ≥ 0 m_q\ge0 m q ≥ 0 q ∈ P S ( 1 ) q\in P_S^{(1)} q ∈ P S ( 1 ) G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) O O O − c 0 ( S ) ∈ cone ( P S ( 1 ) ) -\mathbf{c}_0(S)\in\operatorname{cone}(P_S^{(1)}) − c 0 ( S ) ∈ cone ( P S ( 1 ) ) G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) N N N M / N M/N M / N p i j \mathbf{p}_{ij} p ij N N N p i j \mathbf{p}_{ij} p ij c c o m p = 0 \mathbf{c}_{\mathrm{comp}} = \mathbf{0} c comp = 0 Theorem 3.1 for every semi-magic square of any order n n n
Proof. Setting ρ 0 vol ( G 0 ) c 0 + Δ ρ 1 vol ( G 1 ) c 1 = 0 \rho_0\operatorname{vol}(\mathcal{G}_0)\mathbf{c}_0+\Delta\rho_1\operatorname{vol}(\mathcal{G}_1)\mathbf{c}_1=\mathbf{0} ρ 0 vol ( G 0 ) c 0 + Δ ρ 1 vol ( G 1 ) c 1 = 0 dim G 1 ( S ) = 3 \dim\mathcal{G}_1(S)=3 dim G 1 ( S ) = 3 (25) and the four-class partition. Existence of non-negative inner point masses m q ≥ 0 m_q\geq0 m q ≥ 0 − c 0 ∈ cone ( P S ( 1 ) ) -\mathbf{c}_0\in\operatorname{cone}(P_S^{(1)}) − c 0 ∈ cone ( P S ( 1 ) ) ∑ i , j p i j = 0 \sum_{i,j}\mathbf{p}_{ij}=\mathbf{0} ∑ i , j p ij = 0 Theorem 3.1 .
Overcoming the Density Barrier: Complete K N K_N K N
While the eight gems in Table 2 have modest excess density ratios Δ ρ 1 / ρ 0 ∈ [ 5 / 4 , 40 / 19 ] \Delta\rho_1/\rho_0\in[5/4,40/19] Δ ρ 1 / ρ 0 ∈ [ 5/4 , 40/19 ] Table 5 have small core volumes vol ( G 1 ) \operatorname{vol}(\mathcal{G}_1) vol ( G 1 ) Δ ρ 1 / ρ 0 ∈ [ 47 / 12 , 191 / 6 ] \Delta\rho_1/\rho_0\in[47/12,191/6] Δ ρ 1 / ρ 0 ∈ [ 47/12 , 191/6 ] c s o l i d ( G 1 ) = 1 4 ∑ p q = 0 \mathbf{c}_{\mathrm{solid}}(\mathcal{G}_1)=\frac{1}{4}\sum\mathbf{p}_q=\mathbf{0} c solid ( G 1 ) = 4 1 ∑ p q = 0
Replacing solid volume with the skeletal and thin-shell architectures of magicGemWeb [12 ] (Figure 6 ) overcomes both barriers. Connecting all ( N 2 ) \binom{N}{2} ( 2 N ) K N ( P S ( λ ) ) K_N(P_S(\lambda)) K N ( P S ( λ )) Figure 6 (b)) restores equal participation of all N N N G ( G 0 ) ∪ G ( G 1 ) G(\mathcal{G}_0)\cup G(\mathcal{G}_1) G ( G 0 ) ∪ G ( G 1 ) Figure 6 (a)) replaces volumetric scaling with quadratic rod-diameter scaling (ρ i n ∝ d 2 \rho_{\mathrm{in}}\propto d^2 ρ in ∝ d 2 c w i r e ( G 1 ) ≠ 0 \mathbf{c}_{\mathrm{wire}}(\mathcal{G}_1)\ne\mathbf{0} c wire ( G 1 ) = 0 ∂ G λ ( S ) \partial\mathcal{G}_\lambda(S) ∂ G λ ( S ) Figure 6 (c)).
Figure 6. Wireframes, complete K 16 K_{16} K 16 [12 ] . (a) Nested ball-and-rod wireframe G ( G 0 ) ∪ G ( G 1 ) G(\mathcal{G}_0) \cup G(\mathcal{G}_1) G ( G 0 ) ∪ G ( G 1 ) d 1 / d 0 = 0.793 d_1/d_0 = 0.793 d 1 / d 0 = 0.793 λ = 1 \lambda= 1 λ = 1 K 16 K_{16} K 16 ( 16 2 ) = 120 \binom{16}{2} = 120 ( 2 16 ) = 120 c ( K 16 ) = 0 \mathbf{c}(K_{16}) = \mathbf{0} c ( K 16 ) = 0 ∑ r o d s = 29 5 Σ ( λ ) \sum_{\mathrm{rods}} = \frac{29}{5}\Sigma(\lambda) ∑ rods = 5 29 Σ ( λ ) λ = 0.20 \lambda= 0.20 λ = 0.20 f = ( 12 , 24 , 14 ) \mathbf{f} = (12, 24, 14) f = ( 12 , 24 , 14 ) ⋆ F 0 \star F_0 ⋆ F 0 f 2 − 1 = 13 f_2 - 1 = 13 f 2 − 1 = 13 N t a b s = f 0 − 1 = 11 N_{\mathrm{tabs}} = f_0 - 1 = 11 N tabs = f 0 − 1 = 11 ∥ c ( S ) ∥ 2 \|\mathbf{c}(S)\|_2 ∥ c ( S ) ∥ 2 λ = 1.0 \lambda= 1.0 λ = 1.0 λ = 0.20 \lambda= 0.20 λ = 0.20
Theorem 4.4 (Complete K N K_N K N S S S { 1 , … , N } \{1,\ldots,N\} { 1 , … , N } N = n 2 N=n^2 N = n 2 n ≥ 2 n\geq2 n ≥ 2 ∑ u = 1 N p u ( λ ) = 0 \sum_{u=1}^{N}\mathbf{p}_u(\lambda)=\mathbf{0} ∑ u = 1 N p u ( λ ) = 0 Σ ( λ ) = 1 N ∑ u = 1 N p u ( λ ) p u ( λ ) T \Sigma(\lambda)=\frac{1}{N}\sum_{u=1}^{N}\mathbf{p}_u(\lambda)\mathbf{p}_u(\lambda)^{\mathsf{T}} Σ ( λ ) = N 1 ∑ u = 1 N p u ( λ ) p u ( λ ) T d / L u w → 0 + d/L_{\mathrm{uw}}\to0^+ d / L uw → 0 + ( N 2 ) \binom{N}{2} ( 2 N ) p u ( λ ) \mathbf{p}_u(\lambda) p u ( λ ) p v ( λ ) \mathbf{p}_v(\lambda) p v ( λ ) 1 ≤ u < v ≤ N 1\leq u<v\leq N 1 ≤ u < v ≤ N m e d g e = M r o d s / ( N 2 ) m_{\mathrm{edge}}=M_{\mathrm{rods}}/\binom{N}{2} m edge = M rods / ( 2 N ) m n o d e = M n o d e s / N m_{\mathrm{node}}=M_{\mathrm{nodes}}/N m node = M nodes / N ( N − 1 ) (N-1) ( N − 1 ) K N K_N K N O O O S S S
c ( K N ( λ ) ) = 1 ( N 2 ) ∑ 1 ≤ u < v ≤ N p u ( λ ) + p v ( λ ) 2 = 1 N ∑ u = 1 N p u ( λ ) = 0 , (26) \mathbf{c}(K_N(\lambda))=\frac{1}{\binom{N}{2}}\sum_{1\leq u<v\leq N}\frac{\mathbf{p}_u(\lambda)+\mathbf{p}_v(\lambda)}{2}=\frac{1}{N}\sum_{u=1}^{N}\mathbf{p}_u(\lambda)=\mathbf{0}, \tag*{(26)} c ( K N ( λ )) = ( 2 N ) 1 1 ≤ u < v ≤ N ∑ 2 p u ( λ ) + p v ( λ ) = N 1 u = 1 ∑ N p u ( λ ) = 0 , (26) and the continuous second-moment covariance matrix of the ( N 2 ) \binom{N}{2} ( 2 N ) Σ ( λ ) \Sigma(\lambda) Σ ( λ )
Σ r o d s ( = m a s s ) ( K N ( λ ) ) = 2 N − 3 3 ( N − 1 ) Σ ( λ ) ( = 5 8 Σ ( λ ) at n = 3 , 29 45 Σ ( λ ) at n = 4 ) , (27) \Sigma_{\mathrm{rods}}^{(=\mathrm{mass})}(K_N(\lambda))=\frac{2N-3}{3(N-1)}\Sigma(\lambda)\qquad\left(=\frac{5}{8}\Sigma(\lambda)\text{ at }n=3,\quad\frac{29}{45}\Sigma(\lambda)\text{ at }n=4\right), \tag*{(27)} Σ rods ( = mass ) ( K N ( λ )) = 3 ( N − 1 ) 2 N − 3 Σ ( λ ) ( = 8 5 Σ ( λ ) at n = 3 , 45 29 Σ ( λ ) at n = 4 ) , (27) so that for every semi-magic square S S S K N ( P S ( λ ) ) K_N(P_S(\lambda)) K N ( P S ( λ )) I ⊥ ( λ ) / I z z = 1 2 ( 1 + λ 2 ( n 2 + 1 ) ) I_{\perp}(\lambda)/I_{zz}=\frac{1}{2}(1+\lambda^2(n^2+1)) I ⊥ ( λ ) / I z z = 2 1 ( 1 + λ 2 ( n 2 + 1 )) λ i s o = ( n 2 + 1 ) − 1 / 2 \lambda_{\mathrm{iso}}=(n^2+1)^{-1/2} λ iso = ( n 2 + 1 ) − 1/2 Theorem 3.1 .
When a subgraph G = ( V G , E G ) G=(V_G,E_G) G = ( V G , E G ) d d d L u v ( λ ) = ∥ p u ( λ ) − p v ( λ ) ∥ 2 L_{uv}(\lambda)=\lVert\mathbf{p}_u(\lambda)-\mathbf{p}_v(\lambda)\rVert_2 L uv ( λ ) = ∥ p u ( λ ) − p v ( λ ) ∥ 2 K 16 K_{16} K 16 G ( G 0 ) G(\mathcal{G}_0) G ( G 0 ) c l e n = 0 \mathbf{c}_{\mathrm{len}}=\mathbf{0} c len = 0 S S S G ( G 0 ) G(\mathcal{G}_0) G ( G 0 ) K 16 K_{16} K 16 ∥ c l e n ∥ 2 \lVert\mathbf{c}_{\mathrm{len}}\rVert_2 ∥ c len ∥ 2 M 4 \mathcal{M}_4 M 4 λ = 1 \lambda=1 λ = 1 λ = 0.20 \lambda=0.20 λ = 0.20 Figure 6 (d)). Furthermore, for the 848 order-four gems with dim G 1 ( S ) ≥ 1 \dim G_1(S)\ge1 dim G 1 ( S ) ≥ 1 E ( G 0 ) E(\mathcal{G}_0) E ( G 0 ) d 0 d_0 d 0 E ( G 1 ) E(\mathcal{G}_1) E ( G 1 ) d 1 d_1 d 1 c w i r e ( G 0 ∪ G 1 ) = 0 \mathbf{c}_{\mathrm{wire}}(\mathcal{G}_0\cup\mathcal{G}_1)=\mathbf{0} c wire ( G 0 ∪ G 1 ) = 0 c w i r e ( G 0 ) \mathbf{c}_{\mathrm{wire}}(\mathcal{G}_0) c wire ( G 0 ) c w i r e ( G 1 ) \mathbf{c}_{\mathrm{wire}}(\mathcal{G}_1) c wire ( G 1 )
d 1 d 0 = ρ l i n , 1 ρ l i n , 0 = − L ( G 0 ) ( c w i r e ( G 0 ) ⋅ c w i r e ( G 1 ) ) L ( G 1 ) ∥ c w i r e ( G 1 ) ∥ 2 , (28) \frac{d_1}{d_0}=\sqrt{\frac{\rho_{\mathrm{lin},1}}{\rho_{\mathrm{lin},0}}}=\sqrt{-\frac{L(\mathcal{G}_0)\left(\mathbf{c}_{\mathrm{wire}}(\mathcal{G}_0)\cdot\mathbf{c}_{\mathrm{wire}}(\mathcal{G}_1)\right)}{L(\mathcal{G}_1)\lVert\mathbf{c}_{\mathrm{wire}}(\mathcal{G}_1)\rVert^2}}, \tag*{(28)} d 0 d 1 = ρ lin , 0 ρ lin , 1 = − L ( G 1 ) ∥ c wire ( G 1 ) ∥ 2 L ( G 0 ) ( c wire ( G 0 ) ⋅ c wire ( G 1 ) ) , (28) balancing 158 Group VI gems at λ = 1 \lambda=1 λ = 1 d 1 / d 0 = 0.793 d_1/d_0=0.793 d 1 / d 0 = 0.793 d 1 / d 0 = 0.795 d_1/d_0=0.795 d 1 / d 0 = 0.795 λ = 0.20 \lambda=0.20 λ = 0.20
Proof. Integrating x ( t ) = ( 1 − t ) p u + t p v \mathbf{x}(t)=(1-t)\mathbf{p}_u+t\mathbf{p}_v x ( t ) = ( 1 − t ) p u + t p v t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] 1 2 ( p u + p v ) \frac{1}{2}(\mathbf{p}_u+\mathbf{p}_v) 2 1 ( p u + p v ) 1 3 ( p u p u T + p v p v T ) + 1 6 ( p u p v T + p v p u T ) \frac{1}{3}(\mathbf{p}_u\mathbf{p}_u^{\mathsf{T}}+\mathbf{p}_v\mathbf{p}_v^{\mathsf{T}})+\frac{1}{6}(\mathbf{p}_u\mathbf{p}_v^{\mathsf{T}}+\mathbf{p}_v\mathbf{p}_u^{\mathsf{T}}) 3 1 ( p u p u T + p v p v T ) + 6 1 ( p u p v T + p v p u T ) N − 1 N-1 N − 1 K N K_N K N ∑ u < v ( p u p v T + p v p u T ) = − N Σ ( λ ) \sum_{u<v}(\mathbf{p}_u\mathbf{p}_v^{\mathsf{T}}+\mathbf{p}_v\mathbf{p}_u^{\mathsf{T}})=-N\Sigma(\lambda) ∑ u < v ( p u p v T + p v p u T ) = − N Σ ( λ ) ( N 2 ) \binom{N}{2} ( 2 N ) (26) –(27) . Exact evaluation of the edge lengths and (28) across all 880 squares establishes the length-weighted and nested wireframe censuses (Section A).
Proposition 4.5 (Papercraft cutout nets, glue-tab invariant, and nested thin-shell balance). Let G λ ( S ) \mathcal{G}_{\lambda}(S) G λ ( S ) f f f ( f 0 , f 1 , f 2 ) (f_0,f_1,f_2) ( f 0 , f 1 , f 2 ) ∂ G λ ( S ) \partial\mathcal{G}_{\lambda}(S) ∂ G λ ( S ) T ∗ T^* T ∗ G ∗ ( G ) G^*(\mathcal{G}) G ∗ ( G ) [7 , 18 , 5 , 9 ] partitions the f 1 f_1 f 1 f 2 − 1 f_2-1 f 2 − 1 N t a b s = f 1 − ( f 2 − 1 ) N_{\mathrm{tabs}}=f_1-(f_2-1) N tabs = f 1 − ( f 2 − 1 ) Figure 6 (c)). Because ∂ G λ ( S ) ≅ S 2 \partial\mathcal{G}_{\lambda}(S)\cong S^2 ∂ G λ ( S ) ≅ S 2 E ∖ E ( T ∗ ) E\setminus E(T^*) E ∖ E ( T ∗ ) T T T G ( G λ ) G(\mathcal{G}_{\lambda}) G ( G λ ) f 0 f_0 f 0 f 0 f_0 f 0
N t a b s = f 1 − ( f 2 − 1 ) = ∣ E ( T ) ∣ = f 0 − 1 , (29) N_{\mathrm{tabs}}=f_1-(f_2-1)=\lvert E(T)\rvert=f_0-1, \tag*{(29)} N tabs = f 1 − ( f 2 − 1 ) = ∣ E ( T )∣ = f 0 − 1 , (29) taking every integer value in { 7 , … , 15 } \{7,\ldots,15\} { 7 , … , 15 } M 4 \mathcal{M}_4 M 4 T ∗ T^* T ∗ F 0 F_0 F 0 ⋆ F 0 \star F_0 ⋆ F 0 Figure 6 (c)), all 880 order-four Magic Gems (100%) unfold into strictly non-overlapping 2D planar nets at both λ = 1 \lambda=1 λ = 1 λ = 0.20 \lambda=0.20 λ = 0.20 ∂ G λ ( S ) \partial\mathcal{G}_{\lambda}(S) ∂ G λ ( S ) σ 0 > 0 \sigma_0>0 σ 0 > 0 c s h e l l ( G 0 ) = 0 \mathbf{c}_{\mathrm{shell}}(\mathcal{G}_0)=\mathbf{0} c shell ( G 0 ) = 0 ∂ G 1 ( S ) \partial\mathcal{G}_1(S) ∂ G 1 ( S ) dim G 1 = 3 \dim G_1=3 dim G 1 = 3 σ 1 \sigma_1 σ 1 c s h e l l ( ∂ G 0 ∪ ∂ G 1 ) = 0 \mathbf{c}_{\mathrm{shell}}(\partial\mathcal{G}_0\cup\partial\mathcal{G}_1)=\mathbf{0} c shell ( ∂ G 0 ∪ ∂ G 1 ) = 0 λ = 1 \lambda=1 λ = 1 λ = 0.20 \lambda=0.20 λ = 0.20
Proof. By planar graph duality on ∂ G λ ( S ) ≅ S 2 \partial\mathcal{G}_{\lambda}(S)\cong S^2 ∂ G λ ( S ) ≅ S 2 T ∗ ⊂ G ∗ ( G λ ) T^*\subset G^*(\mathcal{G}_{\lambda}) T ∗ ⊂ G ∗ ( G λ ) ∣ E ( T ∗ ) ∣ = f 2 − 1 \lvert E(T^*)\rvert=f_2-1 ∣ E ( T ∗ )∣ = f 2 − 1 T ⊂ G ( G λ ) T\subset G(\mathcal{G}_{\lambda}) T ⊂ G ( G λ ) f 0 f_0 f 0 N t a b s = ∣ E ( T ) ∣ = f 0 − 1 N_{\mathrm{tabs}}=\lvert E(T)\rvert=f_0-1 N tabs = ∣ E ( T )∣ = f 0 − 1
Remark 4.6 (Physical realization of finite-diameter struts and core suspension). In a physical 3D print of finite rod diameter d > 0 d>0 d > 0 L u v L_{uv} L uv e ^ u v \hat{\mathbf{e}}_{uv} e ^ uv 1 2 ( p u + p v ) \frac{1}{2}(\mathbf{p}_u+\mathbf{p}_v) 2 1 ( p u + p v ) m u v d 2 16 ( I 3 + e ^ u v e ^ u v T ) m_{uv}\frac{d^2}{16}(I_3+\hat{\mathbf{e}}_{uv}\hat{\mathbf{e}}_{uv}^{\mathsf{T}}) m uv 16 d 2 ( I 3 + e ^ uv e ^ uv T ) O ( d 3 ) O(d^3) O ( d 3 ) G ( G 1 ) G(\mathcal{G}_1) G ( G 1 ) G ( G 0 ) G(\mathcal{G}_0) G ( G 0 ) C 2 C_2 C 2 O O O d 1 / d 0 d_1/d_0 d 1 / d 0 (28) invariant.
Having restored internal balance to individual polyhedra, we now remove the central pivot O O O § 5 and place Magic Gems directly on a flat horizontal table and into multi-block vertical stacks under gravity.
Freestanding Gravity and Mirror-Twin Towers
Our fourth and final test removes the central pivot and places Magic Gems directly onto a flat horizontal table under gravity—first as individual resting blocks, and then stacked face-to-face into multi-block vertical columns. Because the hull vertices sit at unequal heights, the boundary facets of G λ ( S ) \mathcal{G}_\lambda(S) G λ ( S ) F ⊂ ∂ G λ ( S ) F \subset\partial\mathcal{G}_\lambda(S) F ⊂ ∂ G λ ( S )
Single-Block Table Equilibrium: Cubic Aspect Scaling Stabilizes 74.5% of Facets
Let F ⊂ ∂ G λ ( S ) F \subset\partial\mathcal{G}_\lambda(S) F ⊂ ∂ G λ ( S ) n ^ F ( λ ) \hat{\mathbf{n}}_F(\lambda) n ^ F ( λ ) G λ ( S ) \mathcal{G}_\lambda(S) G λ ( S ) F F F n ^ F ( λ ) \hat{\mathbf{n}}_F(\lambda) n ^ F ( λ ) c ( λ ) \mathbf{c}(\lambda) c ( λ ) π F ( c ( λ ) ) \pi_F(\mathbf{c}(\lambda)) π F ( c ( λ )) c ( λ ) \mathbf{c}(\lambda) c ( λ ) F F F h F ( λ ) = ( p F − c ( λ ) ) ⋅ ( − n ^ F ( λ ) ) > 0 h_F(\lambda) = (\mathbf{p}_F-\mathbf{c}(\lambda))\cdot(-\hat{\mathbf{n}}_F(\lambda)) > 0 h F ( λ ) = ( p F − c ( λ )) ⋅ ( − n ^ F ( λ )) > 0
Definition 5.1 (Static stability margin and critical tipping angle). For a facet F ⊂ ∂ G λ ( S ) F \subset\partial\mathcal{G}_\lambda(S) F ⊂ ∂ G λ ( S ) stability margin d s t a b ( S , F ; λ ) d_{\mathrm{stab}}(S,F;\lambda) d stab ( S , F ; λ ) F F F π F ( c ( λ ) ) \pi_F(\mathbf{c}(\lambda)) π F ( c ( λ )) ∂ F \partial F ∂ F π F ( c ( λ ) ) \pi_F(\mathbf{c}(\lambda)) π F ( c ( λ )) F F F F F F statically stable if and only if d s t a b ( S , F ; λ ) > 0 d_{\mathrm{stab}}(S,F;\lambda)>0 d stab ( S , F ; λ ) > 0 critical tipping angle
θ t i p ( S , F ; λ ) = arctan ( d s t a b ( S , F ; λ ) h F ( λ ) ) . (30) \theta_{\mathrm{tip}}(S,F;\lambda)=\arctan\left(\frac{d_{\mathrm{stab}}(S,F;\lambda)}{h_F(\lambda)}\right). \tag*{(30)} θ tip ( S , F ; λ ) = arctan ( h F ( λ ) d stab ( S , F ; λ ) ) . (30) Proposition 5.2 (Order-four resting stability census). Across all 880 magic squares of order 4, the gems G λ ( S ) \mathcal{G}_{\lambda}(S) G λ ( S ) λ = 1 \lambda= 1 λ = 1 3 × 3 3 \times3 3 × 3 c 0 ( S ) \mathbf{c}_{0}(S) c 0 ( S ) O O O λ c u b e = 1 / 5 = 0.20 \lambda_{\mathrm{cube}} = 1/5 = 0.20 λ cube = 1/5 = 0.20 17.8 ∘ 17.8^\circ 17. 8 ∘ 47.3 ∘ 47.3^\circ 47. 3 ∘ Figure 7 (d)). *
Proof. For each facet F ⊂ ∂ G λ ( S ) F \subset\partial\mathcal{G}_{\lambda}(S) F ⊂ ∂ G λ ( S ) a X + b Y + c Z + d = 0 aX + bY + cZ + d = 0 a X + bY + c Z + d = 0 z ↦ λ z z \mapsto\lambda z z ↦ λ z n F ( λ ) ∝ ( a λ , b λ , c ) \mathbf{n}_{F}(\lambda) \propto(a\lambda, b\lambda, c) n F ( λ ) ∝ ( aλ , bλ , c ) c 0 ( λ ) = ( c 0 , x , c 0 , y , λ c 0 , z ) \mathbf{c}_{0}(\lambda) = (c_{0,x}, c_{0,y}, \lambda c_{0,z}) c 0 ( λ ) = ( c 0 , x , c 0 , y , λ c 0 , z ) c 0 ( λ ) \mathbf{c}_{0}(\lambda) c 0 ( λ ) F F F d s t a b ( S , F ; λ ) d_{\mathrm{stab}}(S,F;\lambda) d stab ( S , F ; λ ) (30) across all 14,824 facets yields the stated counts and angles.
Figure 7. Mirror-twin facet locking and periodic vertical towers. (a) Four-block alternating tower of square #763 at λ = 0.20 \lambda= 0.20 λ = 0.20 G λ ( S ) \mathcal{G}_{\lambda}(S) G λ ( S ) G λ ( r 2 S ‾ ) = − G λ ( S ) \mathcal{G}_{\lambda}(r^{2}\overline{S}) = -\mathcal{G}_{\lambda}(S) G λ ( r 2 S ) = − G λ ( S ) { 13 , 14 , 15 , 16 } \{13,14,15,16\} { 13 , 14 , 15 , 16 } { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 } z z z ( x , y ) (x,y) ( x , y ) A x y = 5 A_{xy} = 5 A x y = 5 r = 5 / 2 ≈ 1.12 r = \sqrt{5}/2 \approx1.12 r = 5 /2 ≈ 1.12 d s t a b = 0.969 d_{\mathrm{stab}} = 0.969 d stab = 0.969 λ = 0.20 \lambda= 0.20 λ = 0.20 2 θ j o i n t = 22.6 ∘ 2\theta_{\mathrm{joint}} = 22.6^\circ 2 θ joint = 22. 6 ∘ n + ∗ = − n + \mathbf{n}_{+}^{*} = -\mathbf{n}_{+} n + ∗ = − n + 11.3 ∘ 11.3^\circ 11. 3 ∘ θ t i p \theta_{\mathrm{tip}} θ tip λ = 0.20 \lambda= 0.20 λ = 0.20 λ = 1.0 \lambda= 1.0 λ = 1.0
Why Same-Gem Stacks Wedge Apart, and How Mirror-Twin Inversion Restores Verticality
Suppose we now stack multiple blocks vertically along an upper facet F + ⊂ G λ ( S ) F_{+} \subset\mathcal{G}_{\lambda}(S) F + ⊂ G λ ( S ) + n + +\mathbf{n}_{+} + n + n + , z > 0 n_{+,z} > 0 n + , z > 0 F − ⊂ G λ ( S ) F_{-} \subset\mathcal{G}_{\lambda}(S) F − ⊂ G λ ( S ) + n − +\mathbf{n}_{-} + n − n − , z < 0 n_{-,z} < 0 n − , z < 0 same gem G λ ( S ) \mathcal{G}_{\lambda}(S) G λ ( S ) n − = − n + \mathbf{n}_{-} = -\mathbf{n}_{+} n − = − n + n − ≠ − n + \mathbf{n}_{-} \ne-\mathbf{n}_{+} n − = − n + Figure 7 (c), left).
Equilibrium is restored by pairing a gem with its rotational complement (mirror twin)
S ∗ : = r 2 S ‾ = ( N + 1 ) 1 − r 2 S , (31) S^{*} := r^{2}\overline{S} = (N+1)\mathbf{1} - r^{2}S, \tag*{(31)} S ∗ := r 2 S = ( N + 1 ) 1 − r 2 S , (31) which inverts the polyhedron through the origin in R 3 \mathbb{R}^{3} R 3 Figure 7 (a,Figure 7 c)).
Theorem 5.3 (Mirror-Twin Inversion and Zero-Rotation Alternating Towers). Let S S S n n n S ∗ = r 2 S ‾ S^{*} = r^{2}\overline{S} S ∗ = r 2 S λ > 0 \lambda> 0 λ > 0 R 3 \mathbb{R}^{3} R 3
P S ∗ ( λ ) = − P S ( λ ) , G λ ( S ∗ ) = − G λ ( S ) , (32) P_{S^{*}}(\lambda) = -P_{S}(\lambda), \qquad\mathcal{G}_{\lambda}(S^{*}) = -\mathcal{G}_{\lambda}(S), \tag*{(32)} P S ∗ ( λ ) = − P S ( λ ) , G λ ( S ∗ ) = − G λ ( S ) , (32) so the map F ↦ F ∗ : = − F F \mapsto F^{*}:=-F F ↦ F ∗ := − F F + ⊂ G λ ( S ) F_{+}\subset\mathcal{G}_{\lambda}(S) F + ⊂ G λ ( S ) + n + +\mathbf{n}_{+} + n + F + ∗ ⊂ G λ ( S ∗ ) F_{+}^{*}\subset\mathcal{G}_{\lambda}(S^{*}) F + ∗ ⊂ G λ ( S ∗ ) − n + -\mathbf{n}_{+} − n + F − ⊂ G λ ( S ) F_{-}\subset\mathcal{G}_{\lambda}(S) F − ⊂ G λ ( S ) + n − +\mathbf{n}_{-} + n − F − ∗ = − F − ⊂ G λ ( S ∗ ) F_{-}^{*}=-F_{-}\subset\mathcal{G}_{\lambda}(S^{*}) F − ∗ = − F − ⊂ G λ ( S ∗ ) − n − -\mathbf{n}_{-} − n −
Consequently, given any upper facet F + ⊂ G λ ( S ) F_{+}\subset\mathcal{G}_{\lambda}(S) F + ⊂ G λ ( S ) F − ⊂ G λ ( S ) F_{-}\subset\mathcal{G}_{\lambda}(S) F − ⊂ G λ ( S ) R 3 \mathbb{R}^{3} R 3 c ( F + ) \mathbf{c}(F_{+}) c ( F + ) c ( F − ) \mathbf{c}(F_{-}) c ( F − ) B 0 , B 1 , B 2 , … B_{0},B_{1},B_{2},\ldots B 0 , B 1 , B 2 , …
B 2 m = G λ ( S ) + m T , B 2 m + 1 = − G λ ( S ) + 2 c ( F + ) + m T , T = 2 ( c ( F + ) − c ( F − ) ) (33) B_{2m}=\mathcal{G}_{\lambda}(S)+m\mathbf{T},\qquad B_{2m+1}=-\mathcal{G}_{\lambda}(S)+2\mathbf{c}(F_{+})+m\mathbf{T},\qquad\mathbf{T}=2\bigl(\mathbf{c}(F_{+})-\mathbf{c}(F_{-})\bigr) \tag*{(33)} B 2 m = G λ ( S ) + m T , B 2 m + 1 = − G λ ( S ) + 2 c ( F + ) + m T , T = 2 ( c ( F + ) − c ( F − ) ) (33) meets with 100% polygon and vertex overlap at every odd joint B 2 m ∩ B 2 m + 1 B_{2m}\cap B_{2m+1} B 2 m ∩ B 2 m + 1 F + + m T F_{+}+m\mathbf{T} F + + m T B 2 m B_{2m} B 2 m − F + + 2 c ( F + ) + m T -F_{+}+2\mathbf{c}(F_{+})+m\mathbf{T} − F + + 2 c ( F + ) + m T B 2 m + 1 B_{2m+1} B 2 m + 1 ± n + \pm\mathbf{n}_{+} ± n + B 2 m − 1 ∩ B 2 m B_{2m-1}\cap B_{2m} B 2 m − 1 ∩ B 2 m − F − + 2 c ( F + ) + ( m − 1 ) T -F_{-}+2\mathbf{c}(F_{+})+(m-1)\mathbf{T} − F − + 2 c ( F + ) + ( m − 1 ) T B 2 m − 1 B_{2m-1} B 2 m − 1 F − + m T F_{-}+m\mathbf{T} F − + m T B 2 m B_{2m} B 2 m ∓ n − \mp\mathbf{n}_{-} ∓ n − B 2 m ∪ B 2 m + 1 B_{2m}\cup B_{2m+1} B 2 m ∪ B 2 m + 1 c ( F + ) + m T \mathbf{c}(F_{+})+m\mathbf{T} c ( F + ) + m T c ( F + ) = ( 0 , 0 , z + ) \mathbf{c}(F_{+})=(0,0,z_{+}) c ( F + ) = ( 0 , 0 , z + ) c ( F − ) = ( 0 , 0 , z − ) \mathbf{c}(F_{-})=(0,0,z_{-}) c ( F − ) = ( 0 , 0 , z − ) z z z z + > 0 > z − z_{+}>0>z_{-} z + > 0 > z − T = ( 0 , 0 , 2 ( z + − z − ) ) \mathbf{T}=(0,0,2(z_{+}-z_{-})) T = ( 0 , 0 , 2 ( z + − z − )) 2 M 2M 2 M z z z c 0 ( S ) = 0 \mathbf{c}_{0}(S)=\mathbf{0} c 0 ( S ) = 0 Figure 7 (a,Figure 7 c)).
Proof. In centred coordinates (1) , r 2 r^{2} r 2 ( x , y ) ↦ ( − x , − y ) (x,y)\mapsto(-x,-y) ( x , y ) ↦ ( − x , − y ) z ↦ − z z\mapsto-z z ↦ − z r 2 S ‾ r^{2}\overline{S} r 2 S − p i j ( λ ) -\mathbf{p}_{ij}(\lambda) − p ij ( λ ) (32) . Central symmetry of F ± F_{\pm} F ± c ( F ± ) \mathbf{c}(F_{\pm}) c ( F ± ) − F ± + 2 c ( F ± ) = F ± -F_{\pm}+2\mathbf{c}(F_{\pm})=F_{\pm} − F ± + 2 c ( F ± ) = F ± m T m\mathbf{T} m T c ( F + ) + m T \mathbf{c}(F_{+})+m\mathbf{T} c ( F + ) + m T B 2 m B_{2m} B 2 m B 2 m + 1 B_{2m+1} B 2 m + 1 z z z
Chiral Knight’s-Move Towers: Exemplar Square #763
Proposition 5.4 (Vertically centered tower census and the knight’s-move square #763). Across the 880 magic squares of order 4, exactly 48 squares possess both an upper facet F + F_{+} F + F − F_{-} F − ( 0 , 0 , z ± ) (0,0,z_{\pm}) ( 0 , 0 , z ± ) A x y = 5 A_{xy}=5 A x y = 5 5 / 9 ≈ 55.6 % 5/9\approx55.6\% 5/9 ≈ 55.6% 3 × 3 3\times3 3 × 3 Table 3 ; e.g. squares #603, #763, and #774) in which π x y ( F + ) \pi_{xy}(F_{+}) π x y ( F + ) 2 × 1 2\times1 2 × 1 ( ± 3 2 , ± 1 2 ) (\pm\frac{3}{2},\pm\frac{1}{2}) ( ± 2 3 , ± 2 1 ) ( ± 1 2 , ± 3 2 ) (\pm\frac{1}{2},\pm\frac{3}{2}) ( ± 2 1 , ± 2 3 ) Figure 7 (b)). Only two of the 17 (#774 and #792) have parallel top and bottom facets (n − = − n + \mathbf{n}_{-}=-\mathbf{n}_{+} n − = − n +
In particular, square #763 (Dudeney Group VI, S ‾ = h S \overline{S}=hS S = h S S ∗ = r 2 S ‾ = v S S^{*}=r^{2}\overline{S}=vS S ∗ = r 2 S = v S c 0 = 0 \mathbf{c}_{0}=\mathbf{0} c 0 = 0 Table 1 ) by swapping the single complementary pair ( 4 , 13 ) ↔ ( 16 , 1 ) (4,13)\leftrightarrow(16,1) ( 4 , 13 ) ↔ ( 16 , 1 ) Figure 1 (f)):
S # 763 = ( 5 2 15 12 16 11 6 1 4 7 10 13 9 14 3 8 ) , S # 763 ∗ = v S # 763 = ( 9 14 3 8 4 7 10 13 16 11 6 1 5 2 15 12 ) . S_{\#763}=
\begin{pmatrix}
5&2&15&12\\
16&11&6&1\\
4&7&10&13\\
9&14&3&8
\end{pmatrix},
\qquad
S^{*}_{\#763}=vS_{\#763}=
\begin{pmatrix}
9&14&3&8\\
4&7&10&13\\
16&11&6&1\\
5&2&15&12
\end{pmatrix}. S #763 = 5 16 4 9 2 11 7 14 15 6 10 3 12 1 13 8 , S #763 ∗ = v S #763 = 9 4 16 5 14 7 11 2 3 10 6 15 8 13 1 12 . This swap lifts { 13 , 14 , 15 , 16 } \{13,14,15,16\} { 13 , 14 , 15 , 16 } 4 x − 3 y + 5 z = 30 4x-3y+5z=30 4 x − 3 y + 5 z = 30 ( 0 , 0 , 6 λ ) (0,0,6\lambda) ( 0 , 0 , 6 λ ) { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 } − 4 x − 3 y − 5 z = 30 -4x-3y-5z=30 − 4 x − 3 y − 5 z = 30 ( 0 , 0 , − 6 λ ) (0,0,-6\lambda) ( 0 , 0 , − 6 λ ) c 0 = ( 0 , − 5 / 126 , 0 ) \mathbf{c}_{0}=(0,-5/126,0) c 0 = ( 0 , − 5/126 , 0 ) B 2 m ∪ B 2 m + 1 B_{2m}\cup B_{2m+1} B 2 m ∪ B 2 m + 1 5 1 + λ 2 5\sqrt{1+\lambda^{2}} 5 1 + λ 2 d s t a b = 0.969 d_{\mathrm{stab}}=0.969 d stab = 0.969 θ t i p ( 1 ) = 39.35 ∘ \theta_{\mathrm{tip}}^{(1)}=39.35^{\circ} θ tip ( 1 ) = 39.3 5 ∘ λ = 0.20 \lambda=0.20 λ = 0.20 n ± ( λ ) ∝ ( ± 4 λ , − 3 λ , ± 5 ) \mathbf{n}_{\pm}(\lambda)\propto(\pm4\lambda,-3\lambda,\pm5) n ± ( λ ) ∝ ( ± 4 λ , − 3 λ , ± 5 ) θ j o i n t ( λ ) = arctan ( λ ) \theta_{\mathrm{joint}}(\lambda)=\arctan(\lambda) θ joint ( λ ) = arctan ( λ ) e ^ z ≈ ( 11.31 ∘ \hat{\mathbf{e}}_{z}\approx(11.31^{\circ} e ^ z ≈ ( 11.3 1 ∘ λ = 0.20 \lambda=0.20 λ = 0.20 μ s ≥ λ = 0.20 \mu_{s}\ge\lambda=0.20 μ s ≥ λ = 0.20 ( θ t a b l e = 0 ) (\theta_{\mathrm{table}}=0) ( θ table = 0 ) M M M H M H_{M} H M min ( arctan ( μ s ) − θ j o i n t ( λ ) , arctan ( d s t a b / H M ) ) \min(\arctan(\mu_{s})-\theta_{\mathrm{joint}}(\lambda),\arctan(d_{\mathrm{stab}}/H_{M})) min ( arctan ( μ s ) − θ joint ( λ ) , arctan ( d stab / H M )) Figure 7 (c)).
Table 3. Representative order-four Magic Gems from the 17 classes (#603, #763, #768, #774, #779, #785, #788, #791, #792, #818, #824, #828, #836, #837, #839, #842, #844) achieving the maximum projected contact area A x y = 5 A_{xy} = 5 A x y = 5 F + F_+ F + F − F_- F − Proposition 5.4 ). Plane equations ( a , b , c , d ) (a,b,c,d) ( a , b , c , d ) a X + b Y + c Z + d = 0 aX + bY + cZ + d = 0 a X + bY + c Z + d = 0 ∥ \parallel ∥ n − = − n + \mathbf{n}_- = -\mathbf{n}_+ n − = − n + d s t a b d_{\mathrm{stab}} d stab θ t i p ( 1 ) \theta_{\mathrm{tip}}^{(1)} θ tip ( 1 ) Definition 5.1 ) are reported for the upper facet at λ = 0.20 \lambda= 0.20 λ = 0.20
Proof. Testing each facet F ⊂ ∂ G 1 ( S ) F \subset\partial G_1(S) F ⊂ ∂ G 1 ( S ) F = 2 c ( F ) − F F = 2c(F) - F F = 2 c ( F ) − F c x ( F ) = c y ( F ) = 0 c_x(F) = c_y(F) = 0 c x ( F ) = c y ( F ) = 0 A x y = 5 A_{xy} = 5 A x y = 5 { 13 , 14 , 15 , 16 } \{13,14,15,16\} { 13 , 14 , 15 , 16 } { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 } 4 x − 3 y + 5 ( z / λ ) = 30 4x - 3y + 5(z/\lambda) = 30 4 x − 3 y + 5 ( z / λ ) = 30 − 4 x − 3 y − 5 ( z / λ ) = 30 -4x - 3y - 5(z/\lambda) = 30 − 4 x − 3 y − 5 ( z / λ ) = 30 n ± ( λ ) ∝ ( ± 4 λ , − 3 λ , ± 5 ) \mathbf{n}_{\pm}(\lambda) \propto(\pm4\lambda,-3\lambda,\pm5) n ± ( λ ) ∝ ( ± 4 λ , − 3 λ , ± 5 ) 5 1 + λ 2 5\sqrt{1+\lambda^2} 5 1 + λ 2 tan θ j o i n t ( λ ) = λ \tan\theta_{\mathrm{joint}}(\lambda) = \lambda tan θ joint ( λ ) = λ □ \square □
Remark 5.5 (Three-dimensional chirality and why a single mold cannot build the #763 tower). Because square #763 satisfies S ‾ = h S \overline{S} = hS S = h S z ( − x , y ) = − z ( x , y ) z(-x,y) = -z(x,y) z ( − x , y ) = − z ( x , y ) G λ ( S # 763 ) G_{\lambda}(S_{\#763}) G λ ( S #763 ) y y y R y , π ( x , y , z ) = ( − x , y , − z ) ∈ S O ( 3 ) R_{y,\pi}(x,y,z) = (-x,y,-z) \in SO(3) R y , π ( x , y , z ) = ( − x , y , − z ) ∈ S O ( 3 ) F + : 4 x − 3 y + 5 z = 30 F_+ : 4x - 3y + 5z = 30 F + : 4 x − 3 y + 5 z = 30 F − : − 4 x − 3 y − 5 z = 30 F_- : -4x - 3y - 5z = 30 F − : − 4 x − 3 y − 5 z = 30 G λ ( S # 763 ) G_{\lambda}(S_{\#763}) G λ ( S #763 ) x x x R x , π ( x , y , z ) = ( x , − y , − z ) R_{x,\pi}(x,y,z) = (x,-y,-z) R x , π ( x , y , z ) = ( x , − y , − z ) F + F_+ F + ( + 4 λ , + 3 λ , − 5 ) ≠ − n + = ( − 4 λ , + 3 λ , − 5 ) (+4\lambda,+3\lambda,-5) \ne-\mathbf{n}_+ = (-4\lambda,+3\lambda,-5) ( + 4 λ , + 3 λ , − 5 ) = − n + = ( − 4 λ , + 3 λ , − 5 ) F + F_+ F + Aut O ( 3 ) ( G λ ( S # 763 ) ) = { I 3 , R y , π } ≅ C 2 ⊂ S O ( 3 ) \operatorname{Aut}_{O(3)}(G_{\lambda}(S_{\#763})) = \{I_3,R_{y,\pi}\} \cong C_2 \subset SO(3) Aut O ( 3 ) ( G λ ( S #763 )) = { I 3 , R y , π } ≅ C 2 ⊂ S O ( 3 ) G λ ( S # 763 ) G_{\lambda}(S_{\#763}) G λ ( S #763 ) G λ ( S # 763 ∗ ) = − G λ ( S # 763 ) G_{\lambda}(S^{*}_{\#763}) = -G_{\lambda}(S_{\#763}) G λ ( S #763 ∗ ) = − G λ ( S #763 ) Figure 7 (a) requires two opposite-handed molds.
Synthesis and Open Questions
Stepping back from the individual theorems, the mechanical tests of § 2 –§ 5 tell a unified story about how combinatorial balance in inequality behaves when lifted from a flat grid of numbers into three-dimensional physical space (Table 4 ). Across all four physical regimes—radial spokes, spinning rotors, convex polyhedra, and stacked columns—the 880 order-four magic squares stratify into three distinct mechanical tiers determined by how their symmetries interact with nonlinearity.
Table 4.
At the broadest level, whenever a physical observable is linear or symmetric-quadratic in the coordinates ( x j , y i , z i j ) (x_j,y_i,z_{ij}) ( x j , y i , z ij ) ( β = 1 , α = 1 ) (\beta= 1,\alpha= 1) ( β = 1 , α = 1 ) ( ( V 3 ) z = 0 ) ((V_3)_z = 0) (( V 3 ) z = 0 ) ( w ∗ > 0 ) (\mathbf{w}^* > 0) ( w ∗ > 0 ) λ i s o = 1 / 17 \lambda_{\mathrm{iso}} = 1/\sqrt{17} λ iso = 1/ 17 K 16 K_{16} K 16 ( V ‾ o u t e r = − V ‾ i n n e r ) (\overline{V}_{\mathrm{outer}} = -\overline{V}_{\mathrm{inner}}) ( V outer = − V inner ) N t a b s = f 0 − 1 N_{\mathrm{tabs}} = f_0 - 1 N tabs = f 0 − 1 ( G ( r 2 S ‾ ) = − G ( S ) ) (\mathcal{G}(r^2\overline{S}) = -\mathcal{G}(S)) ( G ( r 2 S ) = − G ( S ))
When a nonlinear geometric operation breaks this universal balance, it typically creates a three-component residual imbalance vector c = ( c x , c y , c z ) ∈ R 3 \mathbf{c} = (c_x,c_y,c_z) \in\mathbb{R}^3 c = ( c x , c y , c z ) ∈ R 3 S ‾ = h S \overline{S} = hS S = h S v S vS v S ( C 2 ⊂ S O ( 3 ) ) (C_2 \subset SO(3)) ( C 2 ⊂ S O ( 3 )) v / u = 9 / 4 ≈ 5 v/u = 9/4 \approx\sqrt{5} v / u = 9/4 ≈ 5 α ∗ = − 1 \alpha^\ast= -1 α ∗ = − 1 α ∗ = + 3 \alpha^\ast= +3 α ∗ = + 3 β ∗ ≠ 1 \beta^\ast\ne1 β ∗ = 1 d 1 / d 0 d_1/d_0 d 1 / d 0 σ 1 / σ 0 \sigma_1/\sigma_0 σ 1 / σ 0
Finally, when a nonlinear observable requires unweighted equilibrium without an opposing parameter—such as equal-weight unit-sphere spokes (V 0 = 0 V_0 = 0 V 0 = 0 V 3 = 0 V_3 = 0 V 3 = 0 Q x y = 0 Q_{xy} = 0 Q x y = 0 V 3 = 0 V_3 = 0 V 3 = 0 p − c ( λ ) = − p c ( λ ) \mathbf{p}_{-c}(\lambda) = -\mathbf{p}_c(\lambda) p − c ( λ ) = − p c ( λ ) r 2 S = S ‾ r^2 S = \overline{S} r 2 S = S
Table 4 : Synthesis of the four progressive mechanical tests across the 880 order-four magic squares, organized by Baseline Equilibrium, Nonlinear Symmetry Breaking, and Dual Restoration (certified in exact rational arithmetic; see Section A).
These findings suggest several natural directions for further investigation. In the planar spoke regime, the twenty-four non-central cells of A 5 A_5 A 5 n = 5 n = 5 n = 5 r ∈ { 1 , 2 , 2 , 5 , 2 2 } r \in\{1, \sqrt{2}, 2, \sqrt{5}, 2\sqrt{2}\} r ∈ { 1 , 2 , 2 , 5 , 2 2 } 2 \sqrt{2} 2 5 \sqrt{5} 5 B u n i t ( S ) = ∑ z i j u i j \mathbf{B}_{\mathrm{unit}}(S) = \sum z_{ij}\mathbf{u}_{ij} B unit ( S ) = ∑ z ij u ij n ≥ 5 n \ge5 n ≥ 5 min S ∈ M n ∥ B u n i t ( S ) ∥ 2 \min_{S\in\mathcal{M}_n}\lVert\mathbf{B}_{\mathrm{unit}}(S)\rVert_2 min S ∈ M n ∥ B unit ( S ) ∥ 2 n → ∞ n \to\infty n → ∞ I x x = I y y = I z z I_{xx} = I_{yy} = I_{zz} I xx = I y y = I z z λ i s o = ( n 2 + 1 ) − 1 / 2 \lambda_{\mathrm{iso}} = (n^2 + 1)^{-1/2} λ iso = ( n 2 + 1 ) − 1/2 G λ ( S ) \mathcal{G}_\lambda(S) G λ ( S ) ( I 3 − I 2 ) / I 2 ∈ [ 0.00004 , 0.0243 ] (I_3 - I_2)/I_2 \in[0.00004, 0.0243] ( I 3 − I 2 ) / I 2 ∈ [ 0.00004 , 0.0243 ] n = 4 n = 4 n = 4 n ≥ 5 n \ge5 n ≥ 5 ( G 0 , G 1 ) (\mathcal{G}_0,\mathcal{G}_1) ( G 0 , G 1 ) n = 4 n = 4 n = 4 I 3 I_3 I 3 G 0 ( S n ) \mathcal{G}_0(S_n) G 0 ( S n ) n ≥ 5 n \ge5 n ≥ 5 [14 ] (Thm. 6.4), leaving n 2 − 12 n^2 - 12 n 2 − 12 d o n i o n ( S n ) d_{\mathrm{onion}}(S_n) d onion ( S n ) f f f Theorem 5.3 ), it is natural to ask whether a Magic Gem G λ ( S ) \mathcal{G}_\lambda(S) G λ ( S ) D 4 ⋉ ⟨ τ ⟩ D_4 \ltimes\langle\tau\rangle D 4 ⋉ ⟨ τ ⟩
Acknowledgments
This work grew out of the author’s physical and computational explorations of 3D-printed Magic Gems, ball-and-rod wireframes, and folded papercraft models [12 ] , and forms the third part of a series alongside [14 , 13 ] . Large language models (OpenAI’s ChatGPT, Anthropic’s Claude, and Google’s Gemini) were used by the author as interactive assistants for symbolic checks, verification scripting, interactive simulator prototyping, and manuscript copy-editing; all mathematical formulations, proofs, and exact computational certificates were verified by the author, who takes full responsibility for the work.
Exact computational methods and certificates
All computations in this paper use Python 3.9 with NumPy 2.0, SciPy 1.13, and Python’s exact rational module fractions.Fraction. The complete verification script anc/verify-paper_c.py and figure generator anc/make_figures_c.py run in under 10 s on a single core and write the certified census archive anc/results_paper_c.json.
Exact simplex moments for continuous polyhedra. As in [14 App. A] , doubling the centred coordinates (1) produces integer vectors P = 2 p ∈ Z 3 P = 2p \in\mathbb{Z}^3 P = 2 p ∈ Z 3 a X + b Y + c Z + d = 0 aX+bY+cZ+d=0 a X + bY + c Z + d = 0 G 0 ( S ) \mathcal{G}_0(S) G 0 ( S ) G 1 ( S ) = conv P S ( 1 ) \mathcal{G}_1(S)=\operatorname{conv} P_S^{(1)} G 1 ( S ) = conv P S ( 1 ) vol ( G ) \operatorname{vol}(G) vol ( G ) c s o l i d ( G ) \mathbf{c}_{\mathrm{solid}}(G) c solid ( G ) Σ s o l i d ( G ) \boldsymbol{\Sigma}_{\mathrm{solid}}(G) Σ solid ( G ) G G G G G G Δ k = conv { v 0 , v k , 1 , v k , 2 , v k , 3 } \Delta_k=\operatorname{conv}\{\mathbf{v}_0,\mathbf{v}_{k,1},\mathbf{v}_{k,2},\mathbf{v}_{k,3}\} Δ k = conv { v 0 , v k , 1 , v k , 2 , v k , 3 } v 0 = 1 2 P 0 ∈ ( 1 2 Z ) 3 \mathbf{v}_0=\frac{1}{2}P_0\in\left(\frac{1}{2}\mathbb{Z}\right)^3 v 0 = 2 1 P 0 ∈ ( 2 1 Z ) 3
V k = 1 6 ∣ det ( v k , 1 − v 0 , v k , 2 − v 0 , v k , 3 − v 0 ) ∣ ∈ Q ≥ 0 , V_k=\frac{1}{6}\left|\det(\mathbf{v}_{k,1}-\mathbf{v}_0,\mathbf{v}_{k,2}-\mathbf{v}_0,\mathbf{v}_{k,3}-\mathbf{v}_0)\right|\in\mathbb{Q}_{\geq0}, V k = 6 1 ∣ det ( v k , 1 − v 0 , v k , 2 − v 0 , v k , 3 − v 0 ) ∣ ∈ Q ≥ 0 , exact first moment
∭ Δ k x d V = V k 4 ( v 0 + v k , 1 + v k , 2 + v k , 3 ) ∈ Q 3 , \iiint_{\Delta_k}\mathbf{x}\,\mathrm{d}V=\frac{V_k}{4}\left(\mathbf{v}_0+\mathbf{v}_{k,1}+\mathbf{v}_{k,2}+\mathbf{v}_{k,3}\right)\in\mathbb{Q}^3, ∭ Δ k x d V = 4 V k ( v 0 + v k , 1 + v k , 2 + v k , 3 ) ∈ Q 3 , and exact canonical second moment
∭ Δ k x i x j d V = V k 20 ( 2 ∑ a = 0 3 v a , i v a , j + ∑ 0 ≤ a ≠ b ≤ 3 v a , i v b , j ) ∈ Q , \iiint_{\Delta_k}x_i x_j\,\mathrm{d}V=\frac{V_k}{20}\left(2\sum_{a=0}^{3}v_{a,i}v_{a,j}+\sum_{0\leq a\ne b\leq3}v_{a,i}v_{b,j}\right)\in\mathbb{Q}, ∭ Δ k x i x j d V = 20 V k 2 a = 0 ∑ 3 v a , i v a , j + 0 ≤ a = b ≤ 3 ∑ v a , i v b , j ∈ Q , where ( v k , 0 , v k , 1 , v k , 2 , v k , 3 ) = ( v 0 , v k , 1 , v k , 2 , v k , 3 ) (\mathbf{v}_{k,0},\mathbf{v}_{k,1},\mathbf{v}_{k,2},\mathbf{v}_{k,3})=(\mathbf{v}_0,\mathbf{v}_{k,1},\mathbf{v}_{k,2},\mathbf{v}_{k,3}) ( v k , 0 , v k , 1 , v k , 2 , v k , 3 ) = ( v 0 , v k , 1 , v k , 2 , v k , 3 ) Δ k \Delta_k Δ k vol ( G ) = ∑ k V k ∈ Q \operatorname{vol}(G)=\sum_k V_k\in\mathbb{Q} vol ( G ) = ∑ k V k ∈ Q c s o l i d ( G ) = 1 vol ( G ) ∑ k ∭ Δ k x d V ∈ Q 3 \mathbf{c}_{\mathrm{solid}}(G)=\frac{1}{\operatorname{vol}(G)}\sum_k\iiint_{\Delta_k}\mathbf{x}\,\mathrm{d}V\in\mathbb{Q}^3 c solid ( G ) = vol ( G ) 1 ∑ k ∭ Δ k x d V ∈ Q 3 Σ s o l i d , i j = 1 vol ( G ) ∑ k ∭ Δ k x i x j d V − c s o l i d , i c s o l i d , j ∈ Q \Sigma_{\mathrm{solid},ij}=\frac{1}{\operatorname{vol}(G)}\sum_k\iiint_{\Delta_k}x_i x_j\,\mathrm{d}V-c_{\mathrm{solid},i}c_{\mathrm{solid},j}\in\mathbb{Q} Σ solid , ij = vol ( G ) 1 ∑ k ∭ Δ k x i x j d V − c solid , i c solid , j ∈ Q G G G
Complete list of the 33 two-density balanced gems. For completeness, Table 5 lists all 33 order-four magic squares for which dim G 1 ( S ) = 3 \dim G_1(S)=3 dim G 1 ( S ) = 3 c 0 ( S ) \mathbf{c}_0(S) c 0 ( S ) c 1 ( S ) \mathbf{c}_1(S) c 1 ( S ) Δ ρ 1 / ρ 0 = − vol ( G 0 ) c 0 / ( vol ( G 1 ) c 1 ) > 0 \Delta\rho_1/\rho_0=-\operatorname{vol}(G_0)c_0/(\operatorname{vol}(G_1)c_1)>0 Δ ρ 1 / ρ 0 = − vol ( G 0 ) c 0 / ( vol ( G 1 ) c 1 ) > 0 Theorem 4.3 ). Every one of the 33 squares belongs to Dudeney’s Group VI.
