Skewness and aspect ratio of cells

Equiangular skewness: $S=\max\left[\dfrac{\theta_{max}-\theta_e}{180-\theta_e},\ \dfrac{\theta_e-\theta_{min}}{\theta_e}\right]$, with $\theta_e=60^\circ$ (triangle) and $90^\circ$ (quad). Aspect ratio = longest edge / shortest edge.

Triangle

Edge lengths:

• A(0,0)-B(2,0): $2.000$ mm
• B(2,0)-C(0.5,1.2): $\sqrt{1.5^2+1.2^2}=1.921$ mm
• C-A: $\sqrt{0.5^2+1.2^2}=1.300$ mm

Angles from the dot product:

• At A: vectors AB=(2,0), AC=(0.5,1.2): $\cos\theta=\dfrac{1.0}{2\times1.3}=0.3846$, $\theta_A=67.4^\circ$
• At B: BA=(-2,0), BC=(-1.5,1.2): $\cos\theta=\dfrac{3.0}{2\times1.921}=0.7809$, $\theta_B=38.7^\circ$
• At C: $180-67.4-38.7=74.0^\circ$

$$S=\max\left[\frac{74.0-60}{120},\ \frac{60-38.7}{60}\right]=\max[0.116,\ 0.356]=0.356$$ $$AR=\frac{2.000}{1.300}=1.54$$

Skewness 0.36 falls in the "good" range (0.25 to 0.5), and AR is small, so the triangle is acceptable.

Quadrilateral (a leaning parallelogram)

Edges: bottom $4.0$ mm, right side $\sqrt{0.6^2+0.5^2}=0.781$ mm, top $4.0$ mm, left side $0.781$ mm.

Angles: at (0,0) between (4,0) and (0.6,0.5): $\cos\theta=\dfrac{2.4}{4\times0.781}=0.768$, $\theta=39.8^\circ$. The adjacent angle is $180-39.8=140.2^\circ$.

$$S=\max\left[\frac{140.2-90}{90},\ \frac{90-39.8}{90}\right]=0.558$$ $$AR=\frac{4.0}{0.781}=5.12$$

Comment

The quadrilateral is both strongly skewed and stretched. A stretched cell is fine in a boundary layer if aligned with the wall, but the sheared shape (not rectangular) reduces orthogonality and accuracy; it should be improved by reducing the shear (making the angles closer to 90°).

Answer: Triangle $S=0.36$, $AR=1.54$ (good); quad $S=0.56$, $AR=5.1$ (fair).