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A heliocentric orbit describes a body's trajectory relative to the centre of the Sun. For a low-mass body whose motion is dominated by solar gravity, a useful first approximation is a Kepler orbit withwhere $\mathbf{r}$ is the body's position relative to the Sun, $r=\Vert \mathbf{r}\Vert$, $G$ the gravitational constant and $M_{\odot }$ the solar mass. A heliocentric description specifies an origin; the two-body equation additionally assumes that other gravitational forces can be neglected.^[ref]

For a bound elliptic orbit with semi-major axis $a$ and orbital eccentricity $e$, the specific orbital energy and orbital period obeyFor example, a hypothetical solar orbit with $a=2\mathrm{AU}$ and $e=1/2$ ranges from one to three astronomical units from the Sun. Its orbital period is $2^{3/2}=2\sqrt{2}$ times that of a circular solar orbit of radius one astronomical unit, independent of its orbital eccentricity in this approximation.^[ref]

The Sun itself accelerates as other bodies attract it. Accurate Solar system calculations must therefore include the relevant orbital perturbations and specify the reference frame. A spacecraft orbiting Earth also travels around the Sun, but solar-centred coordinates alone do not make a solar two-body problem an adequate model of that spacecraft's local motion.

Kris Hauser is a roboticist whose research concerns robot motion planning and control. The University of Illinois identifies him as a professor working across computing and engineering, with research on decision-making for physical robots under limits such as uncertain dynamics and bounded computation.^[ref]

His identified university profile provides an external portrait of Kris Hauser.

His author-hosted Robotic Systems by Kris Hauser brings geometry, motion selection and contact mechanics into one technical account. Its contact-planning treatment explains why selecting contact and selecting motion are coupled: making or breaking contact changes the constraints on the robot.^[ref] A collision-free trajectory alone does not establish that the available contacts can produce the required forces.

An original static planar example makes that distinction quantitative. Suppose two nonadhesive point contacts support a body of gravitational weight $W=200\mathrm{N}$ at horizontal positions $x_1=0$ and $x_2=0.5\mathrm{m}$. With center of mass coordinate $x_c$ and no other applied force or moment, vertical force and torque balance giveAt $x_c=0.20\mathrm{m}$, the normal reactions are $F_1=120\mathrm{N}$ and $F_2=80\mathrm{N}$. Moving the center to $x_c=0.60\mathrm{m}$ while retaining those same contacts would require $F_1=-40\mathrm{N}$ and $F_2=240\mathrm{N}$. A negative normal reaction violates the nonadhesive contact model. A planner must change the motion, contacts or assumed external forces; a path that avoids obstacles need not satisfy support constraints.

This is a derived example of the modeling question, not an experiment attributed to Hauser. His text supplies the explicit unilateral-contact and equilibrium framework.^[ref] A proposed robot test compares predicted reactions with calibrated contact measurements while shifting a controlled load, recording any change of contact rather than forcing the original static model onto a changed apparatus.

For equal masses and relative position $\mathbf{r}$, the individual positions are ${\mathbf{r}}_A=\mathbf{R}-\mathbf{r}/2$ and ${\mathbf{r}}_B=\mathbf{R}+\mathbf{r}/2$, where the center of mass has uniform velocity in an isolated system. Bound inverse-square force motion makes $\mathbf{r}$ an ellipse; each mass follows a half-size ellipse in the center-of-mass frame. Adding a nonzero center-of-mass drift generally makes their laboratory trajectories nonclosed. The relative orbital period is not necessarily a period of either laboratory position.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

Motion is a change of position with time, described relative to a reference frame. A book resting on a train has constant position in the train's frame while its position changes in a frame attached to the track. Neither description is complete until the frame and the measured time have been specified.

A record of successive positions gives a trajectory. Comparing nearby measurements gives velocity, and changes of velocity give acceleration. A camera recording a rolling ball at known time intervals provides a simple measurement: unequal gaps between successive images reveal a changing speed. The description of this record is kinematics; explaining the change using forces belongs to particle dynamics.

Multi-sample consistency of distance-query intervals asks whether every bounded distance reading can belong to one nonnegative clearance function with the claimed Lipschitz constant. The same motion history must explain all samples. Checking each adjacent pair in isolation can select a different intermediate distance for each check and miss a contradiction.

Let strictly increasing times $t_0,\dots ,t_n$ have nonempty intervals $[{\ell }_i,u_i]$, with $0\le {\ell }_i\le u_i$, and let $L\ge 0$ bound clearance change per unit time. Necessary and sufficient conditions areNecessity follows by comparing any feasible distances $d(t_i),d(t_j)$ using Lipschitz continuity. For sufficiency, constructEach envelope has Lipschitz constant at most $L$. The pairwise conditions and the triangle inequality imply $f_{-}(t)\le f_{+}(t)$ everywhere. At each sample time, ${\ell }_i\le f_{-}(t_i)\le u_i$, so $f_{-}$ itself is an admissible nonnegative clearance history. This proves existence, not that this constructed history is the robot's actual history.

The intervals $[f_{-}(t_i),f_{+}(t_i)]$ are exact feasible projections at sample times. Both bounding functions are feasible and convex combinations of them remain feasible, so every intermediate value at one fixed time is attainable. The associated multi-sample clearance envelope also gives between-sample lower bounds for a clearance certificate with bounded distance-query error.

For an original counterexample to adjacent-only testing, let successive times differ by one second, take $L=4mm/s$, and useThe first adjacent pair can use middle distance $0$, and the second can use middle distance $10$; each pair is compatible. Yet the endpoints require a $10\mathrm{mm}$ change in two seconds, exceeding the allowed $8\mathrm{mm}$. No common history exists.

An equivalent linear-time feasibility test propagates a reachable interval. Start $J_0=I_0$ and recursively setStop if an interval becomes empty. In the example, $J_1=[0,4]$ and the next propagated upper bound is $8$, so $J_2$ is empty. Retaining the tightened interval is the key difference from independent adjacent-pair checks. If all intervals remain nonempty, select a final value and work backward through the reachable intervals; piecewise linear interpolation yields a feasible history.

Incompatibility diagnoses inconsistent readings or assumptions, not a demonstrated collision. Compatibility proves only the existence of a scalar clearance history; it does not prove that some rigid robot trajectory realizes that history or that the true clearance stays positive. Continuous collision avoidance remains a constraint along the physical path, as in the primary motion-planning formulation.^[ref] A proposed implementation tests synthetic histories with known bounds, deliberately inconsistent triples and timestamp errors, keeping model inconsistency separate from insufficient clearance.

Multistability is coexistence of several stable long-term behaviors at the same fixed parameter values. Those behaviors can be equilibrium points, periodic orbits or more complicated attractors. Different basins of attraction select which behavior a trajectory reaches from its initial state.

For example, $\dot{x}=x-x^3$ has attracting equilibria at $\pm 1$ and a repelling equilibrium at zero. A sufficiently large perturbation can move a state from one basin to the other, leaving a persistent change after the perturbation ends. This gives a simple mechanism for dynamical memory without changing the governing parameters. Bistability is the two-state case. Hysteresis additionally describes history-dependent switching under a changed control parameter; coexistence alone does not specify a complete hysteresis loop or its switching thresholds.

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

A node is a planar hyperbolic equilibrium whose two linearized eigenvalues are real and have the same sign. Negative eigenvalues give a stable node; positive eigenvalues give an unstable node. Nearby motion approaches or recedes without the repeated rotation characteristic of a focus.

For $\dot{x}=-ax$, $\dot{y}=-by$ with $a,b>0$, solutions are $x=x_0{e}^{-at}$ and $y=y_0{e}^{-bt}$. Each coordinate decays, while unequal decay rates usually make a trajectory align with the slower direction as it nears the origin. Reversing both signs gives outward motion. Repeated eigenvalues require attention to the number of independent eigenvectors and can add polynomial factors to the exponential response. The node classification describes local behavior near an equilibrium; it does not determine the geometry of remote orbits.

An orbit is the set of states reached from one initial state under a dynamical system's evolution. A forward orbit restricts attention to future times; a complete orbit includes backward times when the model admits them. Keeping the time ordering gives a trajectory, while forgetting the parameter emphasizes the geometry of the visited set.

For a harmonic oscillator, the phase plane orbit is an ellipse determined by its mechanical energy. The particle repeatedly visits the same positions, but its full state distinguishes motion in opposite directions. A damped oscillator instead approaches an equilibrium point along a spiralling or nonoscillatory orbit, depending on its parameters. The state-space description is therefore broader than a spatial orbital path: the variables can describe velocities, temperatures or populations as well as physical positions.

Assume the vector field and multiplier are continuously differentiable, as required for Green's theorem and uniqueness of trajectories. A nonconstant periodic trajectory of a planar autonomous system is a simple closed orbit: uniqueness prevents crossings, and one traversal can be taken over its least period. Let $\Omega$ be its bounded interior. Along the orbit $dx=fdt$, $dy=gdt$, soGreen's theorem identifies this with ${\iint }_{\Omega }[{\partial }_x(\varphi f)+{\partial }_y(\varphi g)]dxdy$, which is strictly positive by hypothesis, a contradiction. Thus there is no nonconstant periodic orbit. This standard periodic-orbit criterion does not by itself exclude stationary solutions.

For the first concrete system choose $\varphi =-1$. ThenThere are also no equilibria: $f=0$ requires $y=1$, but then $g=-e^{-x}-x^2<0$. Consequently it has no periodic solution, including constant ones.

For the second system the origin is outside the domain of the printed vector field. For $r>0$,Thus for an admissible initial point with $\rho =\sqrt{a^2+b^2}>0$,where $a=\rho \cos \alpha$, $b=\rho \sin \alpha$. These formulas are valid while $r>0$, in particular for every $t\ge 0$. Trajectories rotate counterclockwise, spiral outward when $0<r<1$, and inward when $r>1$, approaching the unit circle. On $r=1$ they give $x(t)=\cos (\alpha +t)$, $y(t)=\sin (\alpha +t)$, a periodic orbit with least period $2\pi$. It is an attracting limit cycle. There is no contradiction with the preceding criterion: this vector field is singular at the origin and has the exhibited closed orbit.

Let $\mathsf{R}(\alpha )$ denote the planar rotation matrix, and choose the initial cup center on the positive $x$ axis, as in the requested equations. Its trajectory is $\mathbf{C}(t)=R(\cos \Omega t,\sin \Omega t)$. The handle rotates relative to the radial line at signed rate $\varphi$, so its absolute angular velocity is $\nu =\Omega +\varphi$. A positive recurrence time is $2\pi /|\varphi |$; the sign records rotation direction.

Writing $\mathbf{q}=(x_1,y_1)$, the relation isor explicitlyOne obtains the same result by rotating cup coordinates through $\varphi t$ into the radially aligned axes, then rotating through $\Omega t$ into the inertial axes.

At $\varphi =-\Omega$, $\nu =0$, andThe axes maintain a fixed orientation in inertial space while their origin accelerates around a circle. This is translation without rotation of the cup frame; it is not an inertial frame because the origin accelerates. In particular its inertial correction is a spatially uniform fictitious force, rather than a Coriolis force from rotation of these axes.

Treat the marble as a frictionless point particle; no size, rolling constraint or friction law is supplied. Neglect its vertical kinetic energy as requested, but retain the gravitational potential energy $mgz=m{\omega }^2(x_1^2+y_1^2)/2$.

For completeness keep the general constant cup rotation rate $\nu =\Omega +\varphi$. In cup components the center's velocity is $\mathbf{w}=R\Omega (\sin \varphi t,\cos \varphi t)$. With $\mathsf{J}(x,y)=(-y,x)$, the horizontal velocity norm is $|\dot{\mathbf{q}}+\nu \mathsf{J}\mathbf{q}+\mathbf{w}|$. ThusIn the nonrotating translating case $\varphi =-\Omega$, this reduces toThe time-only constant $m{R}^2{\Omega }^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent LagrangianIt differs by a total derivative and a time-only term, so gives the same trajectories but different expressions for the canonical momenta. The small-vertical-velocity approximation is typically consistent at small surface slope ${\omega }^2|\mathbf{q}|/g$; it need not remain valid during unlimited resonant growth.

Assume positive, pairwise distinct masses and generic nonzero $\omega ,\lambda$. Let $p_{x_i}=m_i{\dot{x}}_i$, and similarly for $y,z$. The three uncoupled $x$ oscillators each conserveUniform translation of all $y_i$ conserves $P_y={\sum }_i{p}_{y_i}$. The $z_1,z_2$ interaction permits their common translation, while $z_3$ is entirely free, giving two independent momenta $P_{z,12}=p_{z_1}+p_{z_2}$ and $p_{z_3}$. The interacting $z_1,z_2$ subsystem is autonomous and has its own conserved energy. A useful basic collection is thereforeThe total energy is generally not conserved. With $\kappa =\lambda (m_1-m_2)$, its derivative is $\dot{H}=-\kappa \sin (\kappa t)U_{12}(y_1-y_2)$, which is generically nonzero.

Unequal masses do not remove every rotation charge in the oscillator sector. Put $X_i=\sqrt{m_i}{x}_i$, $P_i=p_{x_i}/\sqrt{m_i}$. This is a canonical transformation and makes that sector an isotropic harmonic oscillator. Its rotations conserveThese mass-weighted oscillator symmetries exist already at distinct masses. The energies and three rotation charges are not all functionally independent; the three-oscillator sector has at most five independent time-independent integrals on regular trajectories.

There are further simple explicitly time-dependent constants, such as $x_i\cos \omega t-({\dot{x}}_i/\omega )\sin \omega t$ and $x_i\sin \omega t+({\dot{x}}_i/\omega )\cos \omega t$, and center-of-mass boost constants. For example $MY-t{P}_y$ is constant for $M={\sum }_i{m}_i$ and $Y={\sum }_i{m}_i{y}_i/M$. These illustrate why an unrestricted raw number of conserved functions is not meaningful.

A constant of motion is a function $I(\mathbf{q},\mathbf{p},t)$ whose total derivative vanishes along every physical trajectory. It may depend explicitly on time; arbitrary functions of conserved quantities are also conserved, so a count normally concerns a specified independent set or symmetry generators.

For a fixed-time infinitesimal variation $\delta q_i=ϵ{\xi }_i(\mathbf{q},t)$, suppose $\delta L=ϵdB/dt$. On a solution of the Euler-Lagrange equations,Therefore Noether theorem givesFor a transformation also changing time by $\delta t=ϵ\tau$, the action variation includes $Ld(\delta t)/dt$, and the transformed velocity varies by $ϵ({\dot{\xi }}_i-{\dot{q}}_i\dot{\tau })$. The same integration by parts gives the general point-symmetry chargeThus a time-translation symmetry gives the energy, a spatial translation gives the corresponding total momentum, and a rotation gives its angular-momentum generator. The required symmetry is a symmetry of the action, allowing a boundary term; demanding that the numerical value of the Lagrangian remain strictly unchanged is unnecessarily restrictive. Conversely, explicit time dependence of a Lagrangian generally prevents conservation of its energy even while other charges survive.^[ref]

For $n$ degrees of freedom, phase space is the $2n$-dimensional space of canonical coordinates $(q_1,\dots ,q_n,p_1,\dots ,p_n)$. Initial data at one point determine a trajectory under the Hamilton equations.

Liouville's theorem in Hamiltonian mechanics states that this flow preserves the canonical volume element $d\Gamma ={\prod }_{i}d{q}_{i}d{p}_i$. Its divergence isIf $J(t)$ is the Jacobian determinant of the flow from initial data, the determinant evolution formula gives $\dot{J}=J\nabla \cdot (\dot{q},\dot{p})=0$. Since $J(0)=1$, it remains one, proving preservation of every advected phase volume. The proof also applies to smooth explicitly time-dependent Hamiltonians.

For an ensemble density $\rho (q,p,t)$, probability conservation and the same vanishing divergence giveThus the fine-grained density is constant along the Hamiltonian trajectories, although the ensemble may shear, stretch and develop fine structure. This is not a claim that coordinate-space volume or every lower-dimensional projection is conserved.

Fix an initial event $(q_i,t_i)$. For each reachable final event $(q_f,t_f)$, evaluate the action on a classical path satisfying those endpoints. On a chosen local branch this produces Hamilton's principal function $S(q_f,t_f)$, rather than a functional of every possible path. Multiple classical paths may make this function multivalued, so the derivation is local to a branch.

Allow the final event to vary. The integrated interior variation vanishes by the Euler-Lagrange equations. The endpoint variation must distinguish the total coordinate change from the change at fixed time: $\delta q(t_f)=\delta q_f-{\dot{q}}_f\delta t_f$. ConsequentlyHence $p_i=\partial S/\partial q_i$ and $\partial S/\partial t=-H$. Eliminating the momenta yields the Hamilton-Jacobi equationA complete integral of the Hamilton-Jacobi equation depends on $n$ suitable independent constants for $n$ coordinates; differentiating it with respect to those constants supplies the conjugate integration constants needed to reconstruct trajectories.

The required Hamilton-Jacobi equation isBecause $x,y$ and time are cyclic in the Hamiltonian, use separation $S=ax+by-Et+W(z)$. The remaining equation isFor $g\ne 0$, each sign branch has the separated complete integralIt is real in the classically allowed region $Q\ge 0$, and $p_z=s\sqrt{Q}$. At a turning point one matches the ascending and descending branches. If $g=0$, the integral is simply $W=p_{z}z$ with constant $p_z$.

A convenient equivalent complete integral, expressed using initial-momentum constants $\mathbf{\alpha }$, is globally polynomial in time:Substitution verifies that its time derivative cancels the kinetic term and $mgz$. It avoids a square-root branch change when extracting the complete trajectory.

The nullcline is $\boxed{y=h(x)=x^3/3-x}$. Write $F=x-x^3/3+y$. Away from it at an order-one distance, $F$ is order one on the indicated finite phase-space scales. Thus $\dot{x}=O({ϵ}^{-1})$ while $\dot{y}=-x=O(1)$: the trajectory changes $x$ on the fast time in a singular perturbation $t/ϵ$, with only an $O(ϵ)$ change in $y$. Equivalently $dy/dx=-ϵx/F=O(ϵ)$, which explains the nearly horizontal motion.

For positive $ϵ$, $F_x=1-x^2$ makes the branches $|x|>1$ fast attracting and the middle branch $|x|<1$ fast repelling. The arrows point right where $y>h(x)$ and left where $y<h(x)$, carrying a generic off-nullcline point towards an attracting branch. The reduced description changes near a fold, where $F_x=0$; the rapid/slow distinction there needs a separate transition layer.

The left panel shows positive $ϵ$; the right panel shows negative $ϵ$, including the three specified starting points. Horizontal arrows indicate the leading fast motion.

For negative $ϵ$ the fast arrows reverse: the outer branches become repelling and the middle branch attracting. On the middle branch the slow equation is $\dot{x}=x/(1-x^2)$, so the slow motion goes away from the saddle equilibrium at zero towards a fold.

At $(3,1)$, $F=3-9+1=-5$, so $\dot{x}>0$: the trajectory immediately escapes to the right. At $(1,-2)$, $F=1-1/3-2=-4/3$, so it also moves rapidly right and escapes. In each case the large-$x$ balance is $\dot{x}\sim x^3/(3|ϵ|)$, with $y$ decreasing; it is not a bounded relaxation cycle.

At $(-1,-1/3)$, $F=-1$, so the initial fast motion is rightward. At nearly fixed $y=-1/3$ it approaches the attracting middle root of $x-x^3/3=1/3$, approximately $x=0.347$. It then moves slowly down/right along the middle branch towards $(1,-2/3)$, leaves that fold to the right, and eventually escapes. These are the small-negative-$ϵ$ limiting descriptions; the fold locations and fast landing point receive small corrections. The right panel of the figure distinguishes this slow excursion from the two immediate escapes.