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The aerodynamic load factor $n=L/W$ compares lift with gravitational weight. The stated convention takes $L$ along the chosen positive lift direction, so $n$ is a signed force ratio, not simply the magnitude of total acceleration divided by $g$. A steady level condition can have $n=1$ with zero inertial acceleration.^[ref]

A focused example is a wings-level pull-up at the bottom of a vertical curved path. Neglect other vertical forces and take upward normal acceleration as positive. The force balance givesFor stipulated true airspeed $V=50m/s$, curvature radius $R=500\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the upward centripetal acceleration is $5m/{\mathrm{s}}^2$ and $n\simeq 1.510$. With $m=600\mathrm{kg}$, lift is $8886\mathrm{N}$, while the net upward force is $3000\mathrm{N}$. Neither is the other.

In the steady horizontal-turn model, the condition instead is $n=1/\cos \varphi$; bank angle derives that balance. It does not apply unchanged to the vertical pull-up. At fixed air mass density, wing area and usable maximum lift coefficient, the relation $L=\rho V^{2}S{C}_L/2$ gives limiting speed proportional to $\sqrt{n}$ for positive $n$. For this example, the factor is about $1.229$ relative to the corresponding one-load condition.

A force balance measures aerodynamic load; a trajectory measurement estimates inertial acceleration. Comparing them requires the gravitational weight term and correct directions. A reported load factor also needs its force-axis convention: thrust, side forces and coupled motion can matter to what an instrument actually records. The definition alone establishes neither structural capability nor available lift before aerodynamic stall.

An aircraft is a vehicle capable of flight through the atmosphere. Its support can involve aerodynamic lift or buoyancy. A fixed-wing aircraft obtains aerodynamic lift from its aircraft wings as it moves relative to the surrounding air. An engine can supply thrust, but propulsion and the aerodynamic support of gravitational weight are distinct functions.^[ref]

For straight, level flight at constant velocity, neglect vertical thrust and align the remaining thrust with the flight direction. The force balance isHere $L$ is lift, $D$ is drag, $T$ is thrust and $W$ is gravitational weight. For a stipulated $1000\mathrm{kg}$ vehicle with $g=9.81m/{\mathrm{s}}^2$ and lift-to-drag ratio $L/D=12$, this requires $L=9810\mathrm{N}$ and $T=D=817.5\mathrm{N}$. At an air-relative speed of $50m/s$, $TV=40.875\mathrm{kW}$ is the useful propulsive mechanical power; an engine's input power can be larger because of losses.

An unpowered glider can instead maintain a steady descending velocity. If its glide angle below the horizontal is $\gamma$, resolving gravitational weight perpendicular and parallel to the flight path givesConsequently an assumed constant $L/D=12$ corresponds to $\gamma \simeq 4.76{}^{\circ }$ and 12 units of horizontal travel per unit of height lost in still air.^[ref] The loss of gravitational potential energy supplies the work against drag; the kinetic energy need not be falling during that steady descent.

These balances concern translation of the vehicle. Maintaining its orientation also requires a balance of torques about its centre of mass. An aircraft can satisfy an instantaneous vertical force balance while still having an unwanted pitching moment.

In antiplane perfect plasticity, velocity $w=v_3$ and shear stresses $a={\sigma }_{13},b={\sigma }_{23}$ depend only on $x_1,x_2$. Force balance is $a_{,1}+b_{,2}=0$, the fixed yield surface is $f(a,b)=k$, and the associated flow rule is $\nabla w=2\dot{\lambda }(f_a,f_b)$ with $\dot{\lambda }\ge 0$. Differentiating the yield equation in each coordinate and combining it with equilibrium shows that $a,b,w$ are constant along characteristic curves tangent to $(-f_b,f_a)$. For example $-f_b{a}_{,1}+f_a{a}_{,2}=f_b{b}_{,2}+f_a{a}_{,2}=0$. Nonzero regular yield gradient is needed for a characteristic direction.

For an axisymmetric vector potential $\mathbf{A}=\nabla \alpha \times \nabla \varphi +\beta \nabla \varphi$, defineMagnetostatic azimuthal force balance gives $L\alpha =F(\beta )$. In a barotropic equilibrium the remaining balance integrates tofor flux functions $F$ and $G$.

For thickness $h(r,t)$, radial velocity $u(r,t)$ and external pressure $p_{\mathrm{ext}}$, the leading Newtonian fluid stress tensor and incompressibility giveRadial force balance on an annular sector includes the inward projection of hoop traction and external pressure on the sloping broad faces. Together with conservation of mass it yields

Consider a stipulated ideal cart modeled as a point mass of mass $2\mathrm{kg}$ on a horizontal track in a laboratory inertial frame. Its initial position is $x(0)=0$ and its initial velocity is $v_x(0)=1.5m/s$ to the right. Independent devices exert horizontal forces $+6\mathrm{N}$ and $-6\mathrm{N}$ on this same cart. Its upward support force balances its downward gravitational weight; no other forces contribute. Rotation and deformation are excluded from this model.After $4\mathrm{s}$, the cart is at $6\mathrm{m}$, still moving right at $1.5m/s$. Starting the same model at rest instead would leave it at rest. Force balance determines the zero acceleration, while the initial velocity determines whether the cart moves.

For the plotted model, both horizontal devices are switched off simultaneously at $t=2\mathrm{s}$, with no impulse. The support and gravitational weight remain balanced. The individual horizontal forces change, but their sum stays zero on both sides of the switch, so the velocity stays continuous and unchanged. Switching off just one device would be a different model with a nonzero net external force.

The bank angle measures an aircraft's lateral tilt relative to a local horizontal reference. In the ideal fixed-wing aircraft turning model, the lift vector tilts with the vehicle. For more general motion, vehicle attitude, flight-path direction and aerodynamic force direction must be distinguished; a banked vehicle need not follow a level circle.^[ref]

For steady horizontal circular motion with negligible vertical thrust and lift tilted by $\varphi$, the force balance developed in fixed-wing aircraft gives the aerodynamic load factorAt $\varphi =60{}^{\circ }$, the required lift is twice the gravitational weight. If the maximum usable lift coefficient is held fixed, $L=\rho V^{2}S{C}_L/2$ implies the model's limiting airspeed scales asA stipulated wings-level limit of $20m/s$ therefore becomes $28.28m/s$ at this loading. This is a comparison at fixed configuration and air mass density, not a claim that a particular aircraft can sustain the required condition. Aerodynamic stall and structural limits constrain the accessible turn.

A banked glider also needs a different loading balance from the shallow straight-glide model used to define minimum sink speed. Increasing bank angle cannot be analyzed merely by rotating a drawing of its still-air trajectory. Measured performance must specify the turn condition as well as airspeed; the FAA's performance discussion illustrates how bank and loading affect glider operation.^[ref]

Biological fluid dynamics studies fluid flows generated by and acting on organisms, biological cells, tissues and biological transport networks. Examples include blood flow through microcirculation, cytoplasmic streaming within a biological cell, flagellum propulsion and bioconvection in swimming suspensions.

The Reynolds number $\mathrm{Re}=\rho UL/\mu$ compares inertia with viscous forces, using density $\rho$, dynamic viscosity $\mu$, speed $U$ and length $L$. A microscopic swimmer can operate in Stokes flow, where viscous forces dominate and motion responds to the instantaneous force balance. Larger swimming animals can instead transfer appreciable momentum to the surrounding fluid, as modeled by Lighthill elongated-body theory. ^[ref]

Transport introduces another comparison: the Péclet number $\mathrm{Pe}=UL/D$ measures advection against diffusion with diffusion coefficient $D$. Thus the same flow can move slowly enough to have negligible inertia, yet carry a slowly diffusing substance effectively. Choosing the appropriate dimensionless quantities connects biological scale to the governing mechanical and transport models.

Place equal positive point charges at distances $d$ on opposite sides of a grounded conducting sphere of radius $a$, and set $x=d/a>1$. Each physical charge has an image $-qa/d$ at distance $a^2/d$ from the centre on its own side. The repulsion from the other physical charge balances attraction to both images precisely when the displayed equation holds. At large $x$, the two sides behave as $1/(4x^2)$ and $2/x^3$, explaining the approximate root $x\simeq 8$. A constant nonzero sphere potential adds a central image and changes the force balance.

Fibre-matrix load transfer describes how interfacial shear stress transfers axial force between a reinforcing fibre and its surrounding composite matrix. For a cylindrical fibre of diameter $d$, axial force balance gives $d{\sigma }_f/dx=4\tau /d$ when $\tau$ is the appropriately signed interfacial shear stress.

The fibre stress is therefore small near a free end and builds over a transfer length. Short fibres may never reach their tensile strength before the interface slips or the matrix yields. The critical fibre length quantifies this competition in a simplified constant-shear model.

A nearly confined cylinder in Couette flow is driven by the unequal lubrication resistances of its two gaps. If their minimum widths are proportional to $1+\lambda$ and $1-\lambda$, force balance weights the relative wall velocities by the inverse square roots of those widths.

A freely translating inner axle with prescribed positive angular velocity $\Omega$ and downward load $F>0$ has full-film journal bearing force balance equationswhere $R_r,R_t$ are the squeeze resistance of an eccentric journal bearing. Rotation supplies tangential fluid force $R_t\Omega r/2$. The equilibrium lies horizontally to the right and satisfies $F=R_t\Omega r/2$. Linearization there has imaginary eigenvalues: the prescribed rotating axle continuously supplies energy, so viscous dissipation does not imply settling. For an initially concentric axle, putting $e=r/\Delta$ and $K=6\pi \mu \Omega a^3/{\Delta }^2$ gives the closed orbitThis follows by eliminating time and integrating $x_e+3ex/[2(1-e^2)]=\Omega {\Delta }^2{R}_{0}e/[2F(1-e^2{)}^{3/2}]$. It applies to the leading inertia-free full-film model; adding cavitation, constraints or finite-length leakage changes the prediction.

The Hagen-Poiseuille equation describes steady, fully developed, laminar flow of an incompressible Newtonian fluid through a straight rigid circular pipe. With pipe radius $R$, axial coordinate $z$ and constant viscosity $\mu$, force balance reduces to $\mu r^{-1}\frac{d}{dr}(r\frac{dw}{dr})=dp/dz$.

Steady pressure-driven flow through a circular pipe has the parabolic profile

Regularity at the centre and the no-slip boundary condition at $r=R$ fix the displayed profile. Integrating over the cross-section gives $Q=-\pi R^4(dp/dz)/(8\mu )=\pi R^4\Delta p/(8\mu L)$ for length $L$ and positive pressure drop $\Delta p$. The fourth power makes volume flux strongly sensitive to radius; the assumptions exclude arbitrary turbulent or developing pipe flow. ^[ref]

For a small-slope inextensible filament with prescribed transverse coordinate $Y(s,t)$, tangential-to-normal resistance ratio $\gamma$, and head drag parameter $\delta$, transverse force balance gives $V={\int }_0^L{Y}_{t}ds/[L(1+\delta )]$. The mean axial swimming speed is $-(1-\gamma )⟨{\int }_0^L{Y}_s(Y_t-V)ds⟩/[L(\delta +\gamma )]$. The mean material longitudinal deformation velocity vanishes for a periodic inextensible stroke anchored longitudinally at its base. A moving base offset must be distinguished from a coordinate measured relative to the actual attachment point.

For identical no-slip spheres of radius $a$ linked by a zero-rest-length spring, with $R\gg a$ and $\zeta =6\pi \mu a$, the leading Oseen tensor interaction reduces their relative spring relaxation rate by $3a/(2R)$. The midpoint follows an imposed incompressible linear velocity field. Add the two bead force balance equations to prove midpoint advection; subtract them and use $\mathsf{G}(\mathbf{R})\mathbf{R}=\mathbf{R}/(4\pi \mu R)$ to obtain the displayed equation. This deterministic far-field model omits bead inertia, thermal forcing, connector drag and near-contact lubrication.

For locally bed-tangent drag force and an upslope angle $a$, a spherical-grain force balance gives ${\tau }_{\mathrm{th}}(a)={\tau }_{\mathrm{th},0}(\cos a+\sin a/{\mu }_s)$. The small-slope term is $\delta {\tau }_{\mathrm{th}}=({\tau }_{\mathrm{th},0}/{\mu }_s){\eta }_x$. A horizontal drag force instead changes the normal contact force and gives a different slope correction.

Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

For circular Couette flow with $v=Ar+B/r$, the rotating-frame reduction of circular viscoelastic shear gives shear rate $g=-2B/r^2$. The total shear stress is $\mu g$, and the azimuthal minus radial normal-stress difference is $2(\mu -{\mu }_r)\tau g^2$ in the stress convention where ${\sigma }_{rr}=-p$. Radial force balance without inertia is $p^{\prime }=-2(\mu -{\mu }_r)\tau g^2/r$, which integrates to the displayed pressure. The pressure constant is undetermined by these equations.

The lift-to-drag ratio $L/D$ compares the lift supporting an aircraft with the drag opposing its motion. For $L>0$ and $D>0$, a larger ratio means less drag for a given lift. If both coefficients use the same dynamic pressure $q$ and reference wing area $S$, then $L/D=C_L/C_D$. Ratios using mismatched reference areas do not have this interpretation.

A parabolic drag polar provides a useful worked model:Here $C_{D0}$ is the zero-lift drag coefficient. In a simple finite-wing model, $k=1/(\pi eAR)$ relates lift-induced drag to wing aspect ratio and a span efficiency factor.^[ref] A fitted whole-aircraft $k$ can also incorporate additional effects; it need not be determined by geometry alone.

The maximum lift-to-drag ratio follows by differentiation:At this point the two modeled contributions to drag coefficient are equal. With the stipulated $C_{D0}=0.020$, $e=0.8$ and $AR=8$, $k\simeq 0.049736$, $C_L^{\ast }\simeq 0.6341$ and $(L/D{)}_{\max }\simeq 15.85$. The optimum is usable only if the actual wing can attain that lift coefficient in the regime where the model is valid; aerodynamic stall can intervene.

For steady unpowered descent in still air, the force balance gives $\tan \gamma =D/L$, where $\gamma$ is the glide angle. Thus the stipulated optimum corresponds to $\gamma \simeq 3.61{}^{\circ }$ and about $15.85$ units of horizontal travel per unit of height lost.^[ref] Wind and changing flight conditions alter travel over the ground. An airfoil section's ratio also should not be substituted for that of an entire aircraft, which includes other sources of drag.

For a static equilibrium and a fluid Lagrangian displacement $\xi$, conservation of mass, adiabatic entropy and magnetic flux freezing give $\delta p=-\xi \cdot \nabla p-\gamma p\nabla \cdot \xi$ together with the displayed relations. Linearizing the pressure/tension form of the momentum equation yields $\rho \ddot{\xi }=-\nabla \delta \Pi -g\delta \rho e_z+{\mu }_0^{-1}(\delta B\cdot \nabla B+B\cdot \nabla \delta B)$, where $\delta \Pi =\delta p+B\cdot \delta B/{\mu }_0$. Static force balance removes the zeroth-order terms; the density multiplying acceleration is the equilibrium density.