This deformed Kagome construction uses a rigid equilateral triangle with side length $1$ and a rigid scalene triangle with side lengths $(a,1,b)$. They meet at a hinge. The $a,b$ sliders change the triangle shown in Figures A and B, while the $\theta$ slider deforms Figure B only; Figure A remains at a fixed reference orientation. The admissible range of $b$ updates with $a$ to enforce the triangle inequality $\lvert a-1\rvert<b<a+1$.

Geometry of the unit cell

Pin the common hinge at $O=(0,0)$. For the equilateral triangle, take

\[P=(1,0),\qquad Q=\left(\frac12,-\frac{\sqrt3}{2}\right).\]

For the scalene triangle, the sides meeting at $O$ have lengths $a$ and $1$, with included angle $\alpha$. Its moving vertices are

\[R=\begin{pmatrix}a\cos\theta\\a\sin\theta\end{pmatrix}, \qquad S=\begin{pmatrix}\cos(\theta+\alpha)\\\sin(\theta+\alpha)\end{pmatrix}.\]

The third side has length $b$, so

\[b^2=a^2+1-2a\cos\alpha, \qquad \alpha=\cos^{-1}\!\left(\frac{a^2+1-b^2}{2a}\right).\]

Place the primitive lattice vectors into the columns of the deformation matrix. Written as a single matrix equation with the full $\theta$ dependence,

\[\boxed{ A(\theta) =\begin{bmatrix}\mathbf v_1&\mathbf v_2\end{bmatrix} =\begin{pmatrix} a\cos\theta-\cos(\theta+\alpha)+\frac12 & a\cos\theta-\frac12 \\ a\sin\theta-\sin(\theta+\alpha)+\frac{\sqrt3}{2} & a\sin\theta+\frac{\sqrt3}{2} \end{pmatrix} }.\]

It is worth noting that this one-parameter family of deformations with macroscopic deformation gradient $A(\theta)$ has not modulated the rigid body motions yet (since $\mathbf v_1$ is changing the orientation in this notation). One can modulate out the rigid body motions by fixing $\mathbf v_1$ in the horizontal direction as shown in the right figure of the above interactive figure.

Area of the unit cell

The unit cell is the parallelogram spanned by the two columns of $A(\theta)$, and its area is

\[\begin{aligned} \left|\det A(\theta)\right|&=\Bigg| \left(a\cos\theta-\cos(\theta+\alpha)+\frac12\right) \left(a\sin\theta+\frac{\sqrt3}{2}\right)\\ &\qquad- \left(a\sin\theta-\sin(\theta+\alpha)+\frac{\sqrt3}{2}\right) \left(a\cos\theta-\frac12\right) \Bigg|. \end{aligned}\]

After expansion, the terms $a^2\cos\theta\sin\theta$ and $\frac{\sqrt3}{2}a\cos\theta$ cancel, and we simplify the area as

\[\boxed{ \left|\det A(\theta)\right| =\left| \frac{\sqrt3}{2}+a\sin\alpha+a\sin\theta -\cos\left(\theta+\alpha-\frac\pi6\right) \right| }.\]

Angle of maximum area

For fixed $a$ and $\alpha$, write the signed determinant as

\[D(\theta)=\det A(\theta) =\frac{\sqrt3}{2}+a\sin\alpha +\left(a+\sin\left(\alpha-\frac\pi6\right)\right)\sin\theta -\cos\left(\alpha-\frac\pi6\right)\cos\theta.\]

Its maximum occurs at

\[\boxed{ \theta_{\max} =\frac\pi2+\operatorname{atan2}\!\left( \cos\left(\alpha-\frac\pi6\right), a+\sin\left(\alpha-\frac\pi6\right) \right) \pmod{2\pi} }.\]

At this angle,

\[\max_{\theta}\left|\det A(\theta)\right| =\frac{\sqrt3}{2}+a\sin\alpha +\sqrt{a^2+1+2a\sin\left(\alpha-\frac\pi6\right)}.\]

For $a=0.72$ and $b=0.57$, $\alpha\approx0.5938$ rad and $\theta_{\max}\approx2.4717$ rad $(141.62^\circ)$.

Ellipticity and hyperbolicity

Differentiate the lattice matrix with respect to the mechanism angle:

\[\boxed{ A'(\theta)=\frac{\mathrm dA}{\mathrm d\theta} = \begin{pmatrix} -a\sin\theta+\sin(\theta+\alpha) & -a\sin\theta \\ a\cos\theta-\cos(\theta+\alpha) & a\cos\theta \end{pmatrix} }.\]

The incremental deformation measured in the current lattice is

\[L(\theta)=A'(\theta)A(\theta)^{-1}.\]

Its antisymmetric part is an infinitesimal rigid rotation, so the strain associated with the soft mechanism is the symmetric part

\[\boxed{ \varepsilon(\theta) =\operatorname{sym}L(\theta) =\frac12\left[L(\theta)+L(\theta)^{\mathsf T}\right] }.\]

The sign of its determinant gives the long-wavelength compatibility type:

\[\begin{array}{ccl} \det\varepsilon(\theta)>0 &\Longrightarrow& \text{elliptic (the two principal strains have the same sign)},\\ \det\varepsilon(\theta)<0 &\Longrightarrow& \text{hyperbolic (the principal strains have opposite signs)},\\ \det\varepsilon(\theta)=0 &\Longrightarrow& \text{critical transition}. \end{array}\]

When the material is uniformly elliptic

Uniform ellipticity means that $\det\varepsilon(\theta)\geq0$ throughout the one-parameter family, so the structure has no interval of hyperbolic behavior. Isolated critical configurations, where $\det\varepsilon(\theta)=0$, may still occur. The figure below shows a small symmetric perturbation of the regular Kagome lattice with $a=0.95$ and $b=1.1$: both triangles are congruent, with side lengths $(0.95,1,1.1)$. For this perturbed family, $\det\varepsilon(\theta)$ remains nonnegative over the displayed mechanism range, so no hyperbolic region appears. Its isolated zero occurs at $\theta_c=\pi-\alpha\approx1.9439$ rad, where the incremental deformation is a pure infinitesimal rotation and $\varepsilon(\theta_c)=0$.

When the material changes from elliptic to hyperbolic

In general, when a deformed Kagome metamaterial is not uniform elliptic, it has two critical angles, $\theta_c^1$ and $\theta_c^2$ where the material transits between being elliptic and hyperbolic. The regions in which the structure is elliptic or hyperbolic are shown in the figure below.

Reference