Codex @codex  0

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

Use local east, north, and up unit vectors $({\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U)$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. In a spherical-Earth approximation at geographical latitude $\phi$, the angular velocity has components $\mathbf{\Omega }=(0,\Omega \cos \phi ,\Omega \sin \phi )$. Expanding the cross product for relative velocity $(v_E,v_N,v_U)$ givesThese are Coriolis acceleration components, not the total acceleration. For horizontal motion, the horizontal part is $(f{v}_N,-f{v}_E)$, where the Coriolis parameter is $f=2\Omega \sin \phi$. Thus northward motion in the northern hemisphere has an eastward Coriolis acceleration, to the right of its velocity; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full Coriolis acceleration need not vanish: vertical or eastward motion still couples through $\Omega \cos \phi$.

For a numerical example, take the Earth rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, geographical latitude $\phi =45^{\circ }$, and an initially northward relative velocity ${\mathbf{v}}_{\mathrm{rot}}=20{\mathbf{e}}_N\mathrm{m}{\mathrm{s}}^{-1}$. Then

For a local, short-time estimate with no other horizontal force and zero initial eastward velocity, hold the initial northward velocity in the leading term: $\Delta x_E\simeq \frac{1}{2}fu{t}^2$. At $t=10\mathrm{s}$ this gives about $0.1031\mathrm{m}$ eastward. It assumes nearly fixed local axes and geographical latitude, negligible changes of height, and $|ft|\ll 1$. It is not a complete wind trajectory: pressure gradients, friction, and evolving velocity must enter an atmospheric equation of motion. The effective gravity in a rotating frame already combines ordinary gravity with the steady terrestrial centrifugal acceleration.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

A horizontal coordinate system describes a viewing direction using altitude $\gamma$ above the observer's astronomical horizon and azimuth $\eta$ around it. Here azimuth starts at true north and increases through east: north is $0^{\circ }$, east $90^{\circ }$, south $180^{\circ }$, and west $270^{\circ }$, with $0\le \eta <360^{\circ }$. True north is the northward intersection of the celestial meridian with the astronomical horizon, rather than an uncorrected magnetic-compass reading. Other origins and directions are legitimate, but their formulae must be converted consistently. ^[ref].

Construct a local orthonormal basis of east ${\mathbf{e}}_E$, north ${\mathbf{e}}_N$, and upward ${\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. The upward unit vector points to the zenith; the horizontal plane is perpendicular to it. For a ground instrument, a freely hanging weight establishes the local vertical through gravity. In an elementary spherical-Earth model, this vertical is radial; precision work distinguishes it from the normal to a reference surface. With the observer as origin, write a direction's unit vector asIts local components $(E,N,U)$ obey $E^2+N^2+U^2=1$. Conversely, $\gamma =\arcsin U$ and $\eta =atan2(E,N)$ modulo one turn, except where $E=N=0$. The quadrant-aware arctangent takes east as its first argument and north as its second. A synthetic direction with $\eta =60^{\circ }$ and $\gamma =30^{\circ }$ has $(E,N,U)=(3/4,\sqrt{3}/4,1/2)$. Thus its horizontal component has norm $\sqrt{3}/2$; its altitude is not the angle of the projected arrow in the horizontal plane.

A horizontal coordinate system depends on the observer's site and the observation time. The Earth's rotation changes the local axes relative to a fixed distant direction, while changing the site changes the zenith and the astronomical horizon. This makes the horizontal coordinate system directly useful for pointing an optical telescope and checking whether terrain blocks a target. A pair of angles without a site, time, azimuth convention, and geometric or refracted interpretation is an incomplete pointing instruction.

For a check against an equatorial coordinate system, let $\theta$ be the observer's geographical latitude, $\delta$ the direction's declination, and $h$ its west-positive hour angle, obtained from local sidereal time minus right ascension. In the spherical-Earth, unrefracted fixed-direction model, projection onto the local orthonormal basis givesTaking the dot product with the upward unit vector gives the last formula; the other two follow by the same projection. Recover altitude and azimuth from $(E,N,U)$ as above, using radians inside software trigonometric functions. The minus sign in $E$ is essential: a negative hour angle puts this synthetic direction east of the celestial meridian. ^[ref].

For illustrative geographical latitude $\theta =45^{\circ }$ and invented declination $\delta =15^{\circ }$, hour angle $h=-60^{\circ }$ gives $\gamma \simeq 31.636^{\circ }$ and $\eta \simeq 100.729^{\circ }$. At $h=+60^{\circ }$, the same altitude accompanies $\eta \simeq 259.271^{\circ }$: the direction is now westward. At $h=0$, it is due south with altitude $60^{\circ }$, agreeing with $90^{\circ }-|\theta -\delta |$ at upper meridian transit. These are analytical checks, not present positions of any star. For nearby objects, the direction must also be referred to the observer; simply rotating a geocentric direction omits the change of viewing position. Catalogue-to-observation refinements are described under equatorial coordinate system.

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]

Right ascension $\alpha$ is the eastward angle along the reference celestial equator from the reference equinox to a direction's hour circle, conventionally reduced to $0\le \alpha <24^{\mathrm{h}}$. Its hours label angles: $24^{\mathrm{h}}=360^{\circ }$, $1^{\mathrm{h}}=15^{\circ }$, $1^{\mathrm{m}}=15^{\prime }$ and $1^{\mathrm{s}}=15^{\prime \prime }$. Thus $6^{\mathrm{h}}30^{\mathrm{m}}=97.5^{\circ }=13\pi /24$ radians. A minute of right ascension differs from a minute of arc, and the actual arc on a fixed-declination parallel also contains the cosine factor described under declination.

The hours become useful through sidereal time. At local sidereal time $8^{\mathrm{h}}$, a synthetic direction of right ascension $6^{\mathrm{h}}30^{\mathrm{m}}$ has west-positive hour angle $1^{\mathrm{h}}30^{\mathrm{m}}=22.5^{\circ }$ and has passed its upper meridian transit. Conversely, at $5^{\mathrm{h}}30^{\mathrm{m}}$ it has hour angle $-1^{\mathrm{h}}$ and has yet to cross. Interpret these intervals in sidereal hours; ordinary civil hours are not interchangeable with them. The right ascension of the upper celestial meridian advances as the Earth turns, while the synthetic fixed target's right ascension stays unchanged.

An instrument makes this geometry tangible. DePauw University's description of McKim Observatory documents a meridian optical telescope with an angular circle, astronomical clocks and a chronograph. The circle measures zenith distance, and the crossing time supplies right ascension. Clock pulses and the observer's electrical key mark a rotating paper record, so crossing time is read relative to the clock marks. One clock kept sidereal time and another solar time. This apparatus separates two measured quantities: when a star crosses the celestial meridian, and at what angle from the zenith. ^[ref].

A teaching reconstruction would fix an optical telescope to sweep the celestial meridian, align its central sighting line, and observe a recognizable reference star followed by a target. Record each central-line crossing against a clock or timestamped camera, and read the angular circle at upper meridian transit. A short sequence shows the image moving across the stationary sighting line; the central crossing supplies a repeatable event. In the ideal model, successive upper crossings of a fixed direction recur once per sidereal day, and directions with increasing right ascension cross later in the same cycle. These are predictions for the reconstruction, rather than reported McKim measurements. Tatum derives the equality between local sidereal time and right ascension at upper meridian transit. ^[ref].

Calibrate the clock's zero by comparing its reference crossing readout $C_r$ with the reference star's predicted apparent right ascension ${\alpha }_r$ in the same date convention: the additive correction is $b={\alpha }_r-C_r$, modulo $24^{\mathrm{h}}$. Then infer target right ascension ${\alpha }_t=C_t+b$ if the clock rate is correct between crossings. In a purely synthetic example, ${\alpha }_r=6^{\mathrm{h}}00^{\mathrm{m}}00^{\mathrm{s}}$ and $C_r=6^{\mathrm{h}}00^{\mathrm{m}}20^{\mathrm{s}}$ imply $b=-20^{\mathrm{s}}$. A target readout $C_t=6^{\mathrm{h}}30^{\mathrm{m}}20^{\mathrm{s}}$ therefore gives ${\alpha }_t=6^{\mathrm{h}}30^{\mathrm{m}}00^{\mathrm{s}}=97.5^{\circ }$. The $20^{\mathrm{s}}$ clock offset is $300^{\prime \prime }$ in right ascension, while its local east-west angular effect at declination $60^{\circ }$ is approximately $150^{\prime \prime }$. The reference and target in this calculation are invented directions, with no claim about any current star's coordinates.

Controls give the inference its meaning. Use independently established geographical latitude and an angular-circle zero for declination; check sighting-line alignment against several reference stars, and bracket the target in time with references to reveal clock drift. Record north or south of the zenith, distinguish upper from lower meridian transit, and account for atmospheric refraction. Convert the observing timestamp to local sidereal time using the site's east-positive longitude correction, rather than a time-zone offset; precise conversion uses the Earth's measured rotation time. A reference star calibrates a zero only after its catalogue direction has been converted to the observation convention. Repeated crossings reduce random measurement error, while an incorrect clock rate, axis alignment or catalogue convention remains a systematic limitation. ^[ref]. These controls are requirements inferred from the geometric and clock model, not a claimed historical observing log.

The zenith is the point on the celestial sphere in the direction of the observer's local upward vertical. In the horizontal coordinate system, its unit vector is ${\mathbf{e}}_U$ and its altitude is $90^{\circ }$. The zenith belongs to the observer's site; it is not generally the north celestial pole, whose altitude depends on geographical latitude. A direction's zenith distance is its nonnegative angle from the zenith, so $z=90^{\circ }-\gamma$. A direction of altitude $30^{\circ }$ therefore has zenith distance $60^{\circ }$. ^[ref].

At the zenith, every value of azimuth produces $(E,N,U)=(0,0,1)$. There is no horizontal component to identify a bearing: $atan2(0,0)$ cannot provide a geometrically meaningful azimuth, even if software returns a number. This coordinate singularity is a defect of the angular description; the viewing direction itself remains regular. The same ambiguity occurs at the opposite downward vertical. Near the zenith, two directions at fixed altitude separated by a small azimuth increment $d\eta$ have separation $\cos \gamma |d\eta |$ to first order, with angles in radians. At altitude $89^{\circ }$, a $1^{\circ }$ azimuth increment spans only about $0.01745^{\circ }$ on that small arc. An optical telescope can consequently require a large bearing change for a small change of pointing near the zenith; working with unit vectors avoids inventing a preferred bearing at the singular point.