Aligned-strain laboratory

Open the live laboratory to change face velocities and inspect the strain, force and dissipation they produce. The controls evaluate a fixed finite operator. They keep the geometry, active face space and stationary wall traces fixed.

The specimen is a $3\times3\times3$ grid of one-metre cells, with the central cell occupied by the box $[1,2]^3$. Density is $1\,\mathrm{kg/m^3}$. Its source-bound packet contains 48 active velocity faces, 210 retained strain rows and 6 structurally zero rows. Independent exact rational assembly gives $B=17\,\mathrm{m/kg}$. The packet binds the native records, executable and source digests to a qualification receipt and independent oracle comparison. It is the same stationary no-slip obstacle and sealed free-slip outer domain specified in the reconstruction derivation.

Follow one row

A row gathers $s_r=\sum_f E_{rf}u_f$. Three normal families measure $\partial_a u_a$ at fluid cell centres, each with weight $2V$. Three engineering shear families measure $\gamma_{ab}=\partial_b u_a+\partial_a u_b$ at internal grid edges. Each actual fluid quadrant contributes its own volume weight $v_q$. Here the centred unit geometry gives $v_q=1/4\,\mathrm{m^3}$; the general construction uses represented centre-to-edge lengths and cell widths.

Stationary obstacle and sealed outer normal traces are prescribed zeros and eliminated from the coefficients. The six empty normal rows remain quadrature samples, even though their strain is always zero. Outer engineering shear vanishes analytically under sealed-normal and free-slip conditions before edge enumeration. Zero rows on included internal samples remain present.

The same coefficients scatter $Ku=E^TWEu$ back to the active faces. With scalar dynamic viscosity $\mu$, the force and positive loss are

$$F=-\mu Ku,\qquad D=\mu\sum_r w_rs_r^2,\qquad \sum_f u_fF_f=-D.$$

Inspect the row’s coefficients, gathered strain, weighted loss and support in the slice. The total ledger includes all rows, including those outside the selected slice. The exact transpose-work statement and nonnegative-loss statement explain this ledger.

Lean’s strainLoss excludes $\mu$ and has units $\mathrm{m^3/s^2}$; the browser multiplies it by $\mu$ to display $D$ in watts. The force-work ledger is power, not a change of energy over a timestep.

See why a corner couples components

Reflect coordinates and vector components so the solid quadrant is southwest. For the two surviving samples $U,V$, choose $u_x=U\max(y,0)/\ell_y$ and $u_y=V\max(x,0)/\ell_x$. These reconstructions reach zero on both solid half-walls. On the unit grid, $\ell_x=\ell_y=1/2\,\mathrm m$, so the northeast, northwest and southeast rows are $2U+2V$, $2U$ and $2V$. Their three quarter-volume contributions give

$$D_{\mathrm{corner}}/\mu =\tfrac14(2U+2V)^2+\tfrac14(2U)^2+\tfrac14(2V)^2 =2U^2+2UV+2V^2.$$

Thus the local block is $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ in these unit, reflected coordinates. The displayed formula uses numerical SI values; the block entries have units of metres. The shared quadrant supplies the cross term. The corner selector retains the actual global component directions: reflections change the coefficient signs shown by the inspector. This piecewise corner choice is newly derived; the unit-corner identity checks its quadratic algebra.

Try the controls

• Corner samples: vary the two speeds in $\mathrm{m/s}$ and compare their individual sector losses with the shared sector.
• Normal stretch and cross-component shear: vary rates in $\mathrm{s^{-1}}$. These are affine samples on active faces; prescribed wall traces stay zero.
• Local rotation: the selected full-fluid shear rows cancel when the two derivatives oppose. Surrounding fixed traces can still produce global loss. A rigid rotation across the whole domain generally violates the stationary obstacle and sealed outer walls, so it is not a global null-mode claim.
• Viscosity, slice and row: viscosity scales force and loss; slice and row selection reveal their support. None of these controls advances time or performs pressure projection.

Read the conditional proof correctly

The coefficient-force bound derives the finite quadratic bound from fixed nonnegative weights, positive masses and explicit per-face coefficient bounds. The exact Euler theorem requires exact arithmetic, no forcing, nonnegative time and viscosity, and $\Delta t\,\mu B\le2$. The projection-composition theorem also requires the matched pressure masses, positive density, areas and distances, positive time, and an exact full pressure-residual solve.

The browser’s nearest-rounded $B=17$ estimate is unenclosed and authorizes no timestep. A small displayed work-identity residual is a numerical diagnostic. These exact-real statements do not certify native assembly, JavaScript/IEEE arithmetic, approximate pressure solves or continuum convergence.

The original variational motivation is Batty and Bridson (2008), PDF pp. 5–6, equations (9)–(11) and §5.1, which places normal and shear stress at MAC cell centres and edges. The mass-weighted projection reference is Batty, Bertails and Bridson (2007), PDF pp. 3–4, equations (1), (3)–(6). The reconstruction derivation separates those source principles from the new restricted stencil and corner choice.