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Keep the signs of part i. The tension has horizontal component $T\cos \theta$ but still has counterclockwise torque $Tr$, since it is tangent to the axle. ThusSolving these equations gives the critical pulling angle of a rolling spool lawThe zero-acceleration angle is thereforeAt this angle the line of action of the pull passes through the ground contact, so its torque about that point vanishes. A spool initially at rest stays at rest while the contact constraints hold; one already rolling has zero acceleration, rather than being brought to rest.

For equilibrium the vertical normal force is $N=Mg-T\sin {\theta }_c$ and $|f|=T\cos {\theta }_c$. The static friction inequality $|f|\le \mu N$ becomesHence, for $0<r<R$,At this limit the normal force is still nonnegative; for finite $\mu$ and positive $r$ it is positive. For ${\theta }_c<\theta \le \pi /2$, the numerator of $a$ is negative. Released from rest, the spool rolls leftward, away from the pulling hand, with counterclockwise rotation, provided the assumed rolling without slipping and ground contact remain possible. The nonzero acceleration is not obtained by assuming that friction simply opposes the centre's motion.

Take rightward translation as positive and counterclockwise angular acceleration $\alpha$ as positive. Let $f$ denote signed ground friction, positive to the right. The source's lower-tangent winding makes the tension exert counterclockwise torque $Tr$. For a horizontal pull, translation and rotation obeyThe final relation is the rolling without slipping constraint. Eliminating $f$ and $\alpha$ givesThus the requested magnitude of static friction isIt acts leftward. Released from rest, the spool accelerates rightward, toward the pulling hand, and rotates clockwise. There is no contradiction between leftward friction and rightward motion: static friction prevents slipping of the contact point, and the horizontal tension is larger than $|f|$ because $r<R$.

stress describes internal force per area. A cut with unit normal $\hat{\mathbf{n}}$ has traction $\mathbf{t}=\mathbf{\sigma }\hat{\mathbf{n}}$; its normal component is $\hat{\mathbf{n}}\cdot \mathbf{\sigma }\hat{\mathbf{n}}$ and its remainder is shear traction. force balance constrains the stress divergence, while ordinary local angular-momentum balance makes the Cauchy stress tensor symmetric in a medium without body couples.

strain describes deformation rather than force. For small displacement gradients,The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal strain. Engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$, so it must not be confused with the tensor component. stress has units of pressure; strain is dimensionless.

An isotropic linear elastic solid obeysTwo independent moduli are required. The shear modulus $G$ and bulk modulus $K=\lambda +2G/3$ must be positive for stability. Young's modulus and Poisson ratio satisfyFor a tensile specimen with free side faces, ${\sigma }_{xx}=F/A$ gives ${ϵ}_{xx}={\sigma }_{xx}/E$ and ${ϵ}_{yy}={ϵ}_{zz}=-\nu {\sigma }_{xx}/E$. Pure shear gives $\tau =G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu \to 1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.

elastic deformation is recoverable on unloading; it need not remain linear at large strain. Beyond yield, irreversible rearrangements such as dislocation motion cause plastic strain, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering stress uses initial area, while true stress uses the changing instantaneous area. A decreasing engineering stress during necking therefore need not mean that local true stress decreases. stress concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and stress relaxation; their losses cannot be described by a static Hooke law alone.

A tensile test should measure force and gauge-length extension independently, correct machine compliance, align the grips, and control strain rate and temperature. Unloading distinguishes recoverable strain from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need friction and buckling checks. Work per volume is $\int {\sigma }_{ij}d{ϵ}_{ij}$; the recoverable part is elastic energy, while a dissipative cycle produces a loss rather than creating energy. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.

A rapidly spinning gyroscope approximately has angular momentum $\mathbf{L}=I_3{\omega }_3\hat{\mathbf{e}}$ along its axle. With no torque in an inertial frame, this vector retains its direction. A transverse torque changes its direction according to $d\mathbf{L}/dt=\mathbf{\tau }$; it does not initially push the axle in the direction of the applied force.

For a fixed pivot, centre of mass $l\hat{\mathbf{e}}$ from it and gravity $-Mg\hat{\mathbf{z}}$, the torque is $Mgl\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. In slow steady precession about the vertical, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe factor $\sin \theta$ appears both in torque and in the radius traced by the angular-momentum tip, so it cancels for a nonvertical axle. Reversing spin reverses this precession. The full symmetric-top steady condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$; the approximation neglects its quadratic term. Sudden release can also produce nutation, so large spin alone does not guarantee a steady sweep.

A wheel-on-pivot experiment can compare precession periods at different measured spin speeds and reverse the spin. Keep the pivot and centre of mass fixed; bearing friction, spin decay, an initial push and support motion can mimic departures from the predicted inverse-spin relation. A freely gimballed direction reference and a gravity-loaded precessing wheel test different torque conditions.^[ref]

For the gyrocompass oscillation, use the same local basis and define $\theta$ about the stable orientation, with spin angular momentum directed north at $\theta =0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical angular momentum from the slow azimuthal motion is $I_2\dot{\theta }$. To leading order in Earth's small angular velocity,A constant vertical contribution $I_2{\Omega }_E\sin \lambda$ from Earth's rotation has zero terrestrial time derivative; the transverse terrestrial contributions affect the following equation only at order ${\Omega }_E^2$. Taking the vertical component of the torque equation givesThe terrestrial derivative of $J_s\mathbf{u}$ is horizontal, as is ${\mathbf{\Omega }}_E\times (I_2\dot{\theta }\mathbf{z})$, so neither contributes to this component. Thus the small angle satisfies, using the small-angle approximation $\sin \theta \simeq \theta$,This is simple harmonic motion, $\theta =A\cos ({\omega }_{\mathrm{shm}}t+\varphi )$, with positive squared angular frequency away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing friction. Near the opposite orientation, putting $\theta =\pi +\delta$ instead yields $\ddot{\delta }-{\omega }_{\mathrm{shm}}^2\delta \simeq 0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable equilibrium, not both north--south spin directions. Without damping the ideal gyrocompass continues to oscillate rather than settling.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

The complete cycle must be a reversible thermodynamic process: no friction, viscous dissipation, unrestrained expansion, or other source of entropy. Each moving boundary must be arbitrarily close to mechanical equilibrium. Heat transfer to or from a bath must occur across an infinitesimal temperature difference, and the insulated stages must be genuinely adiabatic processes. There must also be no direct heat leak between the heat baths.

For an engine exchanging heat only with these two heat baths, the Second law of thermodynamics givesand thusPractical engines generate entropy through finite temperature differences, friction and other dissipative effects. They therefore fall below the Carnot cycle limit. Approaching reversible heat exchange generally requires vanishingly slow operation for a finite heat-transfer apparatus, limiting useful power.

Integrating coaxial cylindrical shells givesRolling without slipping has spin speed $u/a$, so $K=m{u}^2/2+I{u}^2/(2a^2)=3m{u}^2/4$.

For the rolling-cylinder loop contact threshold, set $\ell =R-a$, the radius of the centre's circular path, and let $\theta$ run from the loop bottom. The centre starts $h$ above its lowest centre position. Its height gain from the bottom is $\ell (1-\cos \theta )$. Energy conservation givesAt the top, contact requires $N=m(u_t^2/\ell -g)\ge 0$, hence $u_t^2\ge g\ell$. Inserting the energy relation proves the limiting conditionThe same condition keeps $N\ge 0$ elsewhere on the circle, sinceis smallest at the top.

For the centre's acceleration, the inward normal component is $u^2/\ell$ and the tangential component is $-(2g/3)\sin \theta$. If $A=h/\ell -1\ge 7/4$ and $x=\cos \theta$,This convex quadratic has its largest endpoint value at $x=1$. The maximum on the circular loop is at the bottom, directed vertically upward. For the limiting release height,The approach track's curvature is unspecified, so a global bound for an arbitrary approach shape cannot be proved from the sketch alone; the result above concerns the specified circular part and the usual smooth approach interpretation.

The finite-friction qualification of a marginal rolling loop also matters: the contact threshold presumes the stated no-slip idealization. At equality, near the top, required tangential friction vanishes linearly with angular distance from the top but $N$ vanishes quadratically. Thus any finite coefficient of static friction requires a release height above this limiting bound to maintain rolling everywhere; no coefficient is provided. This does not change the requested ideal contact criterion.

In the centre-of-momentum frame the two equal masses have opposite velocities. Conservation of kinetic energy in an elastic collision fixes their common speed, so scattering changes their directions only. For incoming laboratory velocities $\mathbf{v}$ and $0$, the frame velocity is $\mathbf{V}=\mathbf{v}/2$ and the incoming relative velocities are $\pm \mathbf{v}/2$. If the outgoing relative velocities are $\pm \mathbf{w}$, then $|\mathbf{w}|=v/2$ and the laboratory velocities are $\mathbf{V}\pm \mathbf{w}$. Their dot product is zero: nonzero outgoing laboratory velocities are perpendicular.

For this smooth equal-disc oblique collision, let $\mathbf{n}$ point from A to B at contact. The impulse acts only along $\mathbf{n}$; the tangential velocity of A is unchanged. With $\mathbf{v}\cdot \mathbf{n}=v\cos \theta$, equal masses exchange their normal velocity components:ThereforeNo rotation about a disc centre is generated by these central impulses. At exactly head-on impact A stops, so its outgoing direction is undefined; the angle formula then denotes a limiting direction.

For this adhesive collision of two equal discs, use the parallel axis theorem about the midpoint of the touching centres:Its magnitude is $mav\sin \theta$. Linear and angular momentum conservation giveThe rotation has the sign of ${\mathbf{L}}_{\mathrm{CM}}$, clockwise for the printed geometry. The energy loss is $m{v}^2(1/4-{\sin }^2\theta /6)>0$. After impact the midpoint translates uniformly and the rigid pair rotates uniformly about it: there are no external horizontal forces or axial torques.

Rough unglued edges can transmit a tangential impulse and generate angular momentum about the individual disc centres. The centre-of-mass velocity and total angular momentum still have the values above, but individual translational speeds, angular velocities and energy loss require a friction/contact law and normal restitution. Roughness alone does not imply permanent sticking or the adhesive result.

An impulse is $\mathbf{J}=\int \mathbf{F}dt$, giving $\Delta \mathbf{P}=\mathbf{J}$. For a short localized impulse at displacement $\mathbf{r}$ from the centre of mass, the angular impulse is $\Delta {\mathbf{L}}_{\mathrm{CM}}=\mathbf{r}\times \mathbf{J}$; for a changing lever arm use its instantaneous value inside the integral. For a uniform solid sphere with volume density ${\rho }_V$, direct integration about the $z$ axis givesTake forward translation $u$ and forward-rolling clockwise spin $\Omega$ as positive. Neglect the table's tangential impulse during the much shorter cue impact. Immediately afterwards,The bottom contact point has slip velocity $s_0=u_0-a{\Omega }_0$. The cue height for immediate rolling is therefore $d=2a/5$.

For the impulse-driven slipping-to-rolling transition of a sphere at an arbitrary admissible impact height, let $J_f$ be the signed friction impulse accumulated before slipping ends. Translation and spin then obey $u_f=u_0+J_f/M$ and ${\Omega }_f={\Omega }_0-a{J}_f/I$. Impose $u_f=a{\Omega }_f$ and eliminate $J_f$:This expression handles both sliding directions. If $d<2a/5$, the contact point slips forward and friction slows the centre; if $d>2a/5$, excess spin makes it slip backward and friction accelerates the centre. friction still dissipates total mechanical energy. For ${\mu }_d>0$, $\dot{s}=-(7/2){\mu }_{d}g\mathrm{sgn}s$, so the slipping time is $2|s_0|/(7{\mu }_{d}g)$. Its duration depends on friction, while the final speed does not. With zero friction no slipping-to-rolling transition occurs unless rolling was present from the outset.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

After the push, the forces are weight $mg$ downward, normal reaction $N$ upward and sliding friction $F=\mu N$ opposite the velocity. There is no continuing applied push. The table has constant slopeVertical balance gives $N=mg$, and horizontal motion gives $ma=-\mu mg$. ThusThe friction magnitude is $0.0300\mathrm{N}$.

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.

The required constant deceleration follows from $v^2=u^2+2as$:Both gravity and sliding friction oppose ascent, so $g(\sin \theta +0.1\cos \theta )=5.76$. WriteThe physical incline solution is $\theta =30.0^{\circ }$ more precisely $30.0389^{\circ }$. The combined mass cancels from the equation. The top of the track is consequently $H=2\sin \theta =1.00117\mathrm{m}$ above its base.

Let $b$ denote buoyancy in a stably stratified Boussinesq approximation. Geostrophic balance and hydrostatic balance imply the thermal wind relation $f_0{U}_z=-{\bar{b}}_y$. Consequently the specified vertical shear requiresFor northern-hemisphere $f_0>0$ and $\Lambda >0$, the mean potential temperature decreases northward and increases upward: ${\bar{\Theta }}_y=-({\Theta }_0/g)f_0\Lambda$, ${\bar{\Theta }}_z=({\Theta }_0/g)N^2$. The corresponding mean isentropes slope upward toward the north, with $dz/dy=f_0\Lambda /N^2$. Stable potential-temperature stratification does not require the actual temperature to increase upward; the dry adiabatic lapse also enters that comparison.

In the rigid-lid constant-shear baroclinic model, perturbation quasi-geostrophic potential vorticity isThe mean interior potential-vorticity gradient vanishes on this constant-$f_0$, constant-$N$, linear-shear basic state. Linear advection of that perturbation therefore becomes $({\partial }_t+\Lambda z{\partial }_x)q^{\prime }=0$, the supplied interior equation. It retains slow balanced motion and omits friction, planetary-vorticity gradients and nonlinear disturbance advection.

Substitution of ${\psi }^{\prime }=\Psi (z)\cos (ly)e^{ik(x-ct)}$ givesA possible branch has zero interior perturbation potential vorticity, soBoth supplied applications have $f_0>0$, so this equals the source's expression. For an unstable mode the nonzero imaginary part of the phase speed prevents a real critical level $\Lambda z=c$; for real speeds the displayed solution remains a valid zero-interior-PV branch away from, and smoothly through, any critical level. Taking the real part gives the physical disturbance.

The rigid lids enforce $w^{\prime }=0$. Linearized adiabatic buoyancy conservation isUse the geostrophic conventions $v^{\prime }={\psi }_x^{\prime }$ and $b^{\prime }=f_0{\psi }_z^{\prime }$. At a lid this gives $({\partial }_t+\Lambda z{\partial }_x){\psi }_z^{\prime }-\Lambda {\psi }_x^{\prime }=0$, henceThese are the no-normal-flow condition expressed through boundary buoyancy advection, not conditions of zero horizontal velocity. Boundary temperature anomalies can interact even though this solution has no interior perturbation PV, allowing baroclinic instability to extract mean available potential energy.

Use all coefficients exactly as printed and use Sv consistently for transports. The present salt contrast and density-driven friction coefficient areThusThe large numerical value of $k$ reflects division by a small dimensionless density contrast, about $2.07\times 10^{-6}$ with the supplied effective expansion coefficients. It is not the actual overturning transport, which remains $15$ Sv.

For $q>0$, the steady equation is a quadratic with rootsThe critical freshwater forcing in a two-box overturning model occurs where the two roots coalesce:There is also a negative-flow branch for every $F>0$:At the present forcing, the three roots are $15.000$, $6.774$ and $-3.950$ Sv. The prescribed present state is therefore the upper positive branch, not an arbitrary choice among the roots.

Stability needs a salinity evolution law, not just the algebraic steady equation. For fixed temperature contrast and reference salinity, the standard salt-contrast dynamics can be writtenwhere $V_{\mathrm{eff}}>0$ summarizes the reservoir volumes. The linear slope at a positive root is $(A-2q)/V_{\mathrm{eff}}$: the upper branch is stable and the lower branch unstable. At a negative root it is $(2q-A)/V_{\mathrm{eff}}<0$, so the reversed branch is stable. The two stable states below the fold have different attraction basins, demonstrating bistability. Their stability is conditional on this fixed-temperature, fixed-reference-salinity closure.

As forcing increases quasistatically from $0.5$ Sv, the normal branch weakens and reaches a saddle-node bifurcation at $F_c$. Beyond it no positive steady circulation remains, so the reduced dynamics move toward the stable reversed branch. The reversal is salinity dominated: the fresh cold reservoir is no longer the denser one. Recovery under a later forcing change can depend on the system's state and feedback history.

Earthquake location. The hypocentre is rupture initiation; its surface projection is the epicentre. Record calibrated arrival times at stations of known coordinates, identify P wave and S wave phases, and predict their travel times through a stated velocity model. Earthquake location by arrival times fitsfor the three source coordinates and origin time $t_0$. Linearization supplies residual equations $\delta t_i=\delta t_0+\nabla T_i\cdot \delta \mathbf{x}$. A grid search or iterative weighted least squares inversion can use pick uncertainties, while station geometry controls whether those parameters are well constrained. Four independent arrival constraints are the minimum for four unknowns, not a guarantee of a reliable solution.

In a homogeneous medium, an arrival-time difference between an S wave and a P wave cancels origin time and estimates source-to-station distance,Several distances intersect to locate the event; real layered or three-dimensional Earth structure requires curved-ray travel times. Nearby stations and diverse takeoff angles help resolve depth. Clock errors, phase misidentification, poor azimuthal coverage and incorrect velocities introduce uncertainty or systematic bias. Closely spaced earthquakes can be relocated precisely relative to each other because common-path errors partly cancel, even when their absolute locations remain uncertain. Report a location uncertainty and model assumptions, not merely a best-fit point.

H. F. Reid's earthquake-cycle model. A fault can remain locked while long-term tectonic motion slowly deforms the surrounding rocks. Elastic strain energy and shear stress accumulate. When resistance is overcome, rapid fault slip releases part of that stored energy as seismic waves, frictional heat and other irreversible work. The adjoining rocks partly recover their elastic distortion while retaining a permanent displacement discontinuity across the fault: this is elastic rebound. Surface markers initially crossing the fault become bent during loading and offset after slip.^[ref]

A simple sketch is a slow rise of stress followed by a rapid stress drop, then renewed loading; the plate displacement continues throughout. Geodetic deformation, observed fault offsets and earthquake stress changes can test the picture. Actual cycles need not repeat with identical slips or times: creep, afterslip, stress transfer and heterogeneous friction complicate the idealization. The release need not remove all pre-existing stress. Elastic rebound explains storage and release; it is not a deterministic earthquake clock.

Let the seismic moment tensor of a double-couple seismic source be $M=M_0(\mathbf{s}\otimes \mathbf{n}+\mathbf{n}\otimes \mathbf{s})$ for perpendicular unit slip and normal vectors. Its T axis is $\mathbf{T}=(\mathbf{s}+\mathbf{n})/\sqrt{2}$, with eigenvalue $+M_0$, and its P axis is $\mathbf{P}=(\mathbf{s}-\mathbf{n})/\sqrt{2}$, with eigenvalue $-M_0$. They bisect the fault-plane and auxiliary-plane quadrants.

These earthquake pressure and tension axes describe the source deformation. They are not generally the principal axes of the ambient stress tensor. Friction, inherited fault orientation and local stress changes can rotate their relationship to that tensor; a stress inversion requires further assumptions and multiple mechanisms.

Elastic rebound. Tectonic motion loads a fault whose shallow portion is locked. Rock on either side deforms and stores elastic strain energy while slip is inhibited. When failure permits rapid slip, part of the stored energy becomes seismic radiation and part is dissipated in friction and damage. The adjacent rock partly relaxes, while a permanent displacement remains across the fault. Geodetic strain accumulation and coseismic offsets support this loading–release picture.^[ref]

The model is useful for brittle crustal faulting but is not a prediction of precise recurrence, rupture size or frictional evolution. Aseismic creep can release deformation without a large sudden event. Viscoelastic flow, afterslip, fluid-pressure changes and interacting fault segments complicate a simple locked-fault cycle. Deep-focus earthquakes and non-shear source components require additional physics; not every earthquake source is adequately described by a shallow brittle double couple. Rebound also need not release all the pre-existing stress.

A thin negative-gradient velocity zone. In horizontal layering, use the seismic ray parameter $p=\sin i/v$, the conserved horizontal slowness. A ray turns when $pv=1$. Downward passage through decreasing speed makes $pv$ smaller, so a surface ray cannot turn in the negative-gradient interior. Rays either turn above the zone or penetrate it and turn below where speed has recovered sufficiently. This produces missing turning depths and a seismic shadow range from a low-velocity zone in the ideal ray model.

Retain all branches usingThe travel-time slope is $dT/dx=p$. Just below the critical value that permits passage beyond the upper speed maximum, the ray makes a finite extra excursion through the low-velocity region. Range can jump, and the deep branch can fold back before resuming its normal trend, giving multiple arrivals outside the shadow. A first-arrival curve alone hides some of these branches. Define the seismic intercept time byThe $x$–$p$ representation displays the gap and fold; the $\tau$–$p$ representation has separate shallow and deep branches and a branch jump at the critical ray parameter for the piecewise-linear cusp model. Do not join them by invented turning rays. Diffraction and finite-frequency energy soften a purely geometrical shadow.

The figures use an explicit illustrative profile, not measured Earth parameters: speed rises from 1 to 2 over depths 0 to 1, falls to 1.8 by depth 1.2, then rises with its original gradient. The asthenosphere beneath oceanic lithosphere is an important setting for a low-velocity zone. A major P-wave speed reduction also occurs at the boundary between the mantle and the outer core of the Earth; S waves do not propagate through the liquid outer core of the Earth.^[ref]