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The variance measures the mean squared distance of a random variable from its expected value. For a real square-integrable random variable $X$,The square prevents positive and negative deviations from cancelling. It also gives the variance squared units: if $X$ measures a length in metres, its variance has units of square metres. Its nonnegative square root is the standard deviation, which has the same units as $X$.

For example, let $X$ take $-1$, $0$ and $1$ with equal probabilities. Its expected value is zero, but its variance is $[(-1{)}^2+0^2+1^2]/3=2/3$, and its standard deviation is $\sqrt{2/3}$. The constant random variable zero has the same expected value and zero variance: a mean alone does not describe the spread. The variance is zero exactly when $X=\mathbb{E}X$ almost surely.

The variance is useful because squared deviations interact neatly with linear combinations. The covariance supplies the cross terms when combining different random variables, and a covariance matrix organizes those terms for many variables. The characterization of variance as the minimum mean squared error of a constant also explains why the expected value is the best constant prediction under squared error.

For $n$ pots per treatment and $m$ leaves per pot, the group sample mean decomposes into two averages:The variance of a sum adds the variances of the independent terms. Every pot random effect has weight $1/n$, whereas each remaining error has weight $1/(nm)$, giving the short proofThis is also the variance of the arithmetic mean of the $n$ pot sample means, each with variance ${\tau }^2+{\sigma }^2/m$. Equal numbers of leaves per pot make the two averaging procedures identical.

If all $nm$ leaf measurements were falsely treated as independent, the predicted variance would instead be $({\tau }^2+{\sigma }^2)/(nm)$. The ratio of correct to false variance isThus pseudoreplication can underestimate a standard error by the square root of this ratio. Equality holds when $m=1$ or ${\tau }^2=0$; sharing a pot creates an error here precisely when the shared influence has positive variance.

Choose ${\tau }^2=4$, ${\sigma }^2=1$, $n=4$ and $m=3$ as constructed model parameters in squared length units. These are assumptions rather than fitted experimental values. For each group,The treatment contrast $\hat{\Delta }={\overline{Y}}_B-{\overline{Y}}_A$ has expected value ${\alpha }_B-{\alpha }_A$. Since the group errors are independent conditional on the allocation, its variance is twice the group variance, namely $13/6$, whereas the false calculation gives $5/6$. Its correct standard error is $\sqrt{13/6}\approx 1.472$ length units; the false value is $\sqrt{5/6}\approx 0.913$. The variance ratio is $13/5=2.6$.

With four pots fixed, increasing $m$ drives the error contribution ${\sigma }^2/(4m)$ towards zero, but the limit of the group variance remains ${\tau }^2/4=1$. Measuring twelve leaves on each of four pots gives $49/48$. Using the same forty-eight measurements on sixteen pots, with three leaves per pot, gives $13/48$. More independent experimental units reduce both contributions. Extra nested measurements are useful for characterizing each experimental unit, but cannot average away its shared random effect. This is why sample collection effort and the number of independent treatment replicates answer different questions.^[ref]

A long array of equal co-oriented vortex rings, spaced by $l\ll r$, approximates an azimuthal vortex sheet on a cylinder of radius $r$. The circulation per unit axial length is $\Gamma /l$. The conditions $\nabla \cdot \mathbf{u}=0$ and $\nabla \times \mathbf{u}=\mathbf{\omega }$ are mathematically analogous to the magnetostatic equations for a solenoid. Away from its ends, an array long compared with $r$ produces nearly uniform axial flow $\Gamma /l$ inside and weak flow outside. A finite array has return flow outside connecting its two ends. The uniform-core approximation fails close to individual vortex cores and near the ends, and the field requires a stated circulation orientation and background flow.

A fluid advancing under an elastic plate can be resisted mainly by lubrication theory flow near its front. For front gap $h_f$ and peeling length $l_p$, the elastic pressure gradient is of order $B{h}_f/l_p^5$, while the gap volume flux is of order $B{h}_f^4/(\mu l_p^5)$. Mass balance therefore gives peeling speed $U_p\sim B{h}_f^3/(\mu l_p^5)$.

The near-wall region where molecular momentum diffusion dominates turbulent transport. Using the shear velocity $u_{\ast }=\sqrt{{\tau }_d/\rho }$ gives the viscous wall length $\nu /u_{\ast }$. The precise extent depends on the chosen crossover criterion and wall condition. In this region constant stress and no slip give $U\simeq u_{\ast }^{2}z/\nu$; the logarithmic law of the wall applies outside it.

For vertically integrated kinematic viscosity $\bar{\nu }$, the viscous torque across a cylinder of radius $r$ is

A vortex of circulation $\Gamma$ distance $d$ from a plane wall has an opposite-sign image across the wall. The wall velocity has magnitude $|\Gamma |d/[\pi (d^2+y^2)]$, giving a pressure reduction $\rho {\Gamma }^2{d}^2/[2{\pi }^2(d^2+y^2{)}^2]$ by the Bernoulli equation. Integrating this deficit gives the displayed force per unit width, directed from the wall toward the fluid and vortex. The uniform ambient pressure is subtracted to isolate the finite vortex-induced force.

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.

For time dependence $e^{i\omega t}$ and a segment of length $\ell$, let $\beta =\omega \ell /c$. Adding the forward and backward arterial pressure waves gives the displayed upstream-to-downstream matrix. The flow orientation is downstream at both ends. At a junction, pressure is continuous and incoming volumetric flow rates balance outgoing ones.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted harmonic mean isEqual weights give the harmonic mean, while a zero weight excludes that input. For travel at positive constant speeds $v_i$ over positive leg lengths $d_i$, the weights are fractions of total distance:For $30\mathrm{km}$ at $30km/h$ and $90\mathrm{km}$ at $60km/h$, the weights are $1/4,3/4$, elapsed time is $5/2$ hours, and the average speed is $48km/h$. Weighting the speeds by their time fractions instead gives the same result as a weighted arithmetic mean: $(2/5)30+(3/5)60=48$. The two weight systems describe the same journey in different coordinates and must not be interchanged. ^[ref].

Given valid p-values $q_J$ for constancy on intervals of length at least two, reject $I$ when ${\max }_{J\supseteq I}n{q}_J/|J|\le \alpha$. Every rejected true interval forces rejection of its maximal constant block. A union bound over the non-singleton maximal blocks gives familywise error rate at most $\alpha$, since their lengths sum to at most $n$. Singleton blocks have no tested interval in this formulation.

The Weissenberg--Rabinowitsch equation recovers the wall shear rate of a generalized Newtonian fluid from a measured pressure-drop--flow-rate curve. For a circular pipe,

A point mass attached to a light inextensible string winds around a fixed smooth circular pole of radius $R$. Let the initial straight free length be $L$, and assume the string remains taut without slipping over the part already laid on the pole. Since its tension is perpendicular to the point mass's velocity, the speed $v$ stays constant. If the contact point turns through $d\theta$, the free length $s$ obeys $ds=-Rd\theta$, while the involute geometry gives $v=s\dot{\theta }$. Thus $s\dot{s}=-Rv$, or $s^2=L^2-2Rvt$. First contact of the point mass with the pole occurs at $s=0$, after $L^2/(2Rv)$. In the ideal model, the required tension $m{v}^2/s$ diverges at contact; finite string strength or finite particle size limits that idealization.

The interpolation links chain extension to entropic tensile force, with contour length $L$ and persistence length $L_p$. Small extension gives spring constant $3k_{B}T/(2L_{p}L)$; near $x=L$ the force stiffens as $(1-x/L{)}^{-2}$. The divergence assumes an inextensible backbone and is not a physical infinite tensile force. A branch fit estimates available contour and flexibility, not an unfolding activation barrier by itself.

The historical X unit was a practical X-ray wavelength unit, approximately $10^{-13}\mathrm{m}$. Its scale used an assigned calcite spacing; relative wavelength measurements could be more precise than absolute determinations. Variation between crystal specimens led to calibration using reference spectral lines. An X unit value therefore cannot simply be read as an exact multiple of $10^{-13}\mathrm{m}$.^[ref]

An angular-scale check determines how an indicated rotation corresponds to an angle. A reference line transfers a wavelength scale. An absolute determination ties that scale to an independently established length standard. These are different calibration tasks. Repeating measurements can reduce random scatter without correcting the common scale: if every inferred $d$ is too large by a chosen $0.1\%$, all inferred absolute wavelengths are too large by $0.1\%$, while ratios using that same $d$ are unchanged.

A zero-rest-length spring has quadratic potential energy in its actual length $s$, with no subtracted natural length. It is a useful idealization, not an ordinary unstretched spring of arbitrary length. For an end-to-end spring on a two-link pendulum, changing the natural length changes the quadratic bending stiffness about the straight configuration.