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For a circular arc of radius $R$ carrying clockwise electric current $I$ and subtending angle $\theta$, with two tangent semi-infinite leads extending on the crossing side of the arc, the axial magnetic field at height $h$ above the centre isThe crossing is insulated. The positive term is from the two leads and the negative term from the arc, with $+z$ out of the clockwise-current plane. The Biot-Savart law yields the lead integral ${\int }_0^{\infty }R(R^2+h^2+s^2{)}^{-3/2}ds=R/(R^2+h^2)$. Cancellation of the axial component gives the title formula, expanding to $\theta =2+h^2/R^2+O(h^4/R^4)$ in radians. This is cancellation of one component away from the plane, not necessarily the entire magnetic field.

Consider an original model at fixed vacuum wavelength $600\mathrm{nm}$ and temperature $20^{\circ }\mathrm{C}$. Stipulate a transparent sample with Verdet constant $V=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, length $L=0.050\mathrm{m}$, uniform adjustable $B_z$, and a field-independent analyzer-zero offset ${\theta }_0=0.020\mathrm{rad}$. These are chosen values, not measured glass properties. On a continuously tracked small-angle branch,Taking half the difference gives ${\theta }_{\mathrm{odd}}=0.100\mathrm{rad}$; half the sum recovers ${\theta }_0=0.020\mathrm{rad}$. Hence $V={\theta }_{\mathrm{odd}}/(LB)=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. A field-reversal difference of $0.200\mathrm{rad}$ is twice the single-field rotation, not that rotation itself. More generally this subtraction removes every contribution that is even under $B\mapsto -B$, provided the sample and apparatus remain otherwise matched.

An ideal polarimeter analyzer at laboratory angle $a$ reads normalized light intensity $I(a,B)/I_0={\cos }^2[a-{\theta }_{\mathrm{obs}}(B)]$. Here $I_0$ is the light intensity immediately before the analyzer, separately calibrated for each condition. A scan locates the maximum at $a={\theta }_{\mathrm{obs}}$ modulo $\pi$. At $a=\pi /4$, expansion about zero gives $I/I_0=1/2+{\theta }_{\mathrm{obs}}+O({\theta }_{\mathrm{obs}}^3)$, with angles in radians. At crossed extinction, the response instead begins as ${\theta }_{\mathrm{obs}}^2$ and loses the sign near zero. The favorable $45^{\circ }$ slope follows directly by differentiating Malus's law; its sign reverses if the angular convention is reversed.

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.

How much does a small directional uncertainty help? Consider an original idealized model: a source direction fixed in advance, uniform background directions over the full sphere, and circular acceptance radius $\psi$. The accepted solid angle and single-event background probability areThe derivation integrates the area element on a unit sphere. For small $\psi$ in radians, $q\simeq {\psi }^2/4$: halving the radius quarters the accidental acceptance, to leading order. At $psi=0$ acceptance vanishes; at $\psi =\pi$ it covers the full sphere. The solid angle has units of steradians; $q$ is dimensionless.

For $0<\psi \le \pi$, take a deliberately crude point spread function: stipulate that every signal direction lies uniformly inside that cap. The signal and background probability density functions per steradian are $S=1/\Omega$ inside the cap and zero outside, and $B=1/(4\pi )$ everywhere. Both integrate to one. For an event inside the cap their likelihood ratio isThis ratio compares two specified directional models; it is not the probability that the event came from the source. Such a conditional probability would additionally require prior source odds and the other selection information. For fixed source fraction $f$ among selected events, a normalized mixture distribution has density $(1-f)B+fS$. For independent events the likelihood is the product of those densities; its ratio to background alone isOutside-cap events contribute factors $1-f$ rather than being silently discarded. At $f=0$, $\Lambda =1$; for a full-sky cap, $S=B$ and every $f$ gives $\Lambda =1$. A directional model that cannot distinguish source from background carries no directional evidence.

As an original numerical example, radii $1^{\circ }$ and $10^{\circ }$ give $q\simeq 7.615\times 10^{-5}$ and $7.596\times 10^{-3}$, respectively: nearly a hundredfold difference. For a fixed number $N$ of independent background events, the binomial distribution givesWith chosen $N=100$, those radii give approximately $0.00759$ and $0.5335$. A small single-event acceptance can become a common coincidence after many opportunities. This probability concerns a fixed cap, rather than a scan for the most favorable cap anywhere. For $m$ independent searches with common local tail probability $p$, the corresponding global tail probability is $1-(1-p{)}^m$; overlapping sky searches generally violate that independence. Actual source tests must include directional response, nonuniform exposure, energy information and the look-elsewhere effect in their null calibration.

Use a fixed right-handed Cartesian basis in three dimensions and active rotations in three dimensions through $\pi /2$ radians. The right-hand rule gives the following rotation matrices about the positive $x$ and $z$ axes:For the vector $v=(1,1,0{)}^T$, the two orders giveBoth outputs have Euclidean norm $\sqrt{2}$, but they point in different directions. Each arrow below is a vector from the origin; the curved paths track its endpoint during each rotation in three dimensions.

The axes in this example remain fixed throughout. A command to turn about an object's own current axis describes a different operation because that axis can move after the first turn. With an object's orthonormal basis represented by columns of $Q$, a fixed-axis rotation matrix $R$ acts by $RQ$, whereas the corresponding rotation expressed in the object's own basis acts by $QR$.^[ref]

For this single-tone AM model, overmodulation means $m>1$. The signed factor $q(t)$ becomes negative whenever $\cos (\Omega t)<-1/m$. On such intervals,The sign reversal is a $\pi$ radian shift of the carrier's wave phase. At each zero of $q$, the wave envelope vanishes and that phase is undefined. The waveform itself is continuous: the formula has not clipped it or created extra spectral lines.

At $m=1.4=7/5$, sign changes occur at message phases $\Omega t=\arccos (-5/7)$ and $2\pi -\arccos (-5/7)$ within each cycle. At the trough $\Omega t=\pi$, the desired signed factor is $-0.4$, whereas its nonnegative wave envelope is $0.4A$. Ideal envelope recovery followed by subtraction of $A$ would give $-0.6A$ there, rather than the desired message voltage $-1.4A$. This is loss of polarity in the detector, not loss of the message from the full modulated signal. Actual transmitter clipping is a separate signal distortion that can generate additional spectral components.

The parsec is a distance unit defined by the exact conventionThe astronomical unit is exactly $149597870700\mathrm{m}$; one arcsecond is exactly $\pi /648000$ radians. The IAU resolution explicitly records the parsec convention. A megaparsec is one million parsecs. ^[ref]. ^[ref].

The familiar description "the distance giving one arcsecond of stellar parallax" uses the small-angle approximation. Exact right triangle geometry instead gives $\mathrm{au}/\tan (\pi /648000)$, slightly shorter than the conventional parsec. The relative difference is approximately $(\pi /648000{)}^2/3\simeq 7.83\times 10^{-12}$. This is negligible for the examples here, but distinguishing an exact unit convention from an approximate geometric interpretation prevents a misleading equality.

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.

An additional atomic input is needed: the total Case B recombination coefficient does not specify the number of H-alpha photons per net radiative recombination. Let $y_{\alpha }={\alpha }_{\mathrm{H}\alpha }^{\mathrm{eff}}/{\alpha }_r$, where ${\alpha }_{\mathrm{H}\alpha }^{\mathrm{eff}}$ is the effective recombination coefficient. At the approximately $10^4\mathrm{K}$ temperature implicit in the supplied ${\alpha }_r$, take $y_{\alpha }\simeq 0.45$. The coefficient 2.2 used later in the paper corresponds to $y_{\alpha }=1/2.2\simeq 0.455$, which we use consistently here.

For an unextinguished, fully filled Strömgren sphere with negligible central cavity, the H-alpha luminosity and observed radiative flux satisfySubstitution eliminates $r_S$ and givesConvert the angular diameter to radians: $\theta =56\times 10^{-3}/206265=2.7150\times 10^{-7}$. Also $hc/{\lambda }_{\mathrm{H}\alpha }=3.0268\times 10^{-19}\mathrm{J}$. ThusThe implied hydrogen-ionizing photon production rate is $Q\simeq 1.00\times 10^{39}{\mathrm{s}}^{-1}$. Using $y_{\alpha }=0.45$ instead gives $D\simeq 505\mathrm{pc}$. These values use the printed milliarcseconds and integrated line observed radiative flux. Assuming one H-alpha photon per recombination would give a different, unjustified distance. An appreciable cavity or a nonunit filling factor must also enter the emitting volume.

The Brackett series consists of transitions ending at $n=4$, at infrared wavelengths. They suffer much less interstellar extinction than H-alpha, so their corrected observed radiative fluxes can give a more reliable version of the same inference. At fixed measured angular size and number density, the formula has $D\propto F$: uncorrected interstellar extinction therefore makes this particular distance estimate too small. Infrared lines are weaker and need their own effective recombination coefficients; their advantage is reduced sensitivity to dust.

Represent a point by the column of homogeneous coordinates $(X,Y,Z,1{)}^T$. The actions of the four matrices are:

The last operation produces $(X,Y,Z,Z/f{)}^T$. It requires a perspective divide: divide the first three components by the last. For $Z\ne 0$,This is why the last component of homogeneous coordinates cannot simply be discarded. Points with $Z=0$ have no finite projected position, and a visible scene normally uses $Z>0$ under this forward-axis convention. The first three transformations leave the last component equal to one and need no nontrivial divide. The displayed $K$ retains no depth variation after division; it describes projection onto the image plane, rather than a complete depth-buffer mapping.

Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the circle. The diameter bisects the angle at $A$, so the two right triangles obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp \sin \theta ,-\cos \theta )$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower circular segment of central angle $2\theta$. The triangle $ABC$ has area $\sin \theta (1+\cos \theta )$, while the segment has area $\theta -\sin \theta \cos \theta$, by subtracting the central triangle from its circular sector. Adding giveswith $\theta$ measured in radians. This is the unit-radius case of the area enclosed by two chords and a circular arc.

Powers of the imaginary unit repeat in blocks $i,-1,-i,1$, each with sum zero. The first $44$ terms comprise eleven complete blocks; the last is $i^{45}=i$. Applying Euler's formula to the resulting complex exponential gives $e^i=\cos 1+i\sin 1$. HenceThe angles here are in radians.

The magnitude of the gravitational force remains approximately $mg$ during a laboratory swing. Its tangential component changes with position, however. For simple pendulum length $\ell$ and angle $\theta$ measured from the downward vertical,This restoring component is zero at the lowest point and largest in magnitude at the turning points. It always accelerates the bob towards the lowest point. The radial gravitational component is $mg\cos \theta$; the string tension $S$ must additionally supply the centripetal acceleration, so $S-mg\cos \theta =m{v}^2/\ell$. Thus it is the radial resultant, not gravity alone, that bends the trajectory. The bob has maximum speed at the bottom even though its tangential acceleration is then zero.

For a small angular amplitude, the small-angle approximation gives simple harmonic motion with oscillation period $T=2\pi \sqrt{\ell /g}$. Measure $\ell$ from pivot to the bob's centre of mass, time many complete swings, divide the elapsed time by their number, and calculateRepeat with several lengths and plot $T^2$ against $\ell$; the ideal slope is $4{\pi }^2/g$. The first-order relative uncertainty obeys $\delta g/g\simeq \delta \ell /\ell -2\delta T/T$, with absolute uncertainty contributions combined according to their dependence. Keep the amplitude small: the finite-amplitude period is $T=T_0(1+{\theta }_0^2/16+\cdots )$, with ${\theta }_0$ in radians. Using the longer finite-amplitude period in the small-amplitude formula would underestimate $g$. Pivot friction, a finite bob and support motion are further experimental corrections.^[ref]

Let the relevant projected reciprocal lattice vectors have magnitudes $g_1=1/d_1$ and $g_2=1/d_2$, with angle $\theta$ between them. Subtracting their phases gives the slow spatial modulation $2\pi ({\mathbf{g}}_1-{\mathbf{g}}_2)\cdot \mathbf{r}$. The moiré fringe spacing formula is thereforeFor parallel planes, $\theta =0$ and $D=d_1{d}_2/|d_1-d_2|$. For equal spacings, $D=d/[2|\sin (\theta /2)|]\simeq d/|\theta |$ at a small angle measured in radians. The apparent divergence as both mismatch and rotation vanish means that there is no finite beat period; it does not predict an infinite number of visible fringes.

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

The central estimate is $z=3.2743$. propagation of uncertainty must include the occurrence of $x$ in both numerator and denominator. With angular derivatives measured per radian,Interpreting the two quoted errors as independent standard deviations, convert them to radians before calculatingIf the quoted errors are instead absolute bounds, the first-order worst-case estimate is $|\delta z|\lesssim |z_x|\delta x+|z_y|\delta y=0.60$. Correlated errors add $2z_x{z}_y\mathrm{Cov}(x,y)$ to the variance. Treating $x+y$ and $x$ as independent would lose their shared measurement uncertainty.

By dimensional analysis, power has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-3}$, mass density has dimensions $\mathsf{M}{\mathsf{L}}^{-3}$, and angular velocity has dimensions ${\mathsf{T}}^{-1}$; radians are dimensionless. ThusMatching the three independent dimensions gives $\alpha =1$, $\beta =3$ and $\gamma -3\alpha =2$, henceWith unchanged mass density, angular velocity and the same performance coefficient (a dimensionless quantity), the pump power affinity law givesThe required power therefore increases by $61.1\%$. This uses the stipulated dependence on only these three quantities, with comparable pump geometry and performance; dimensional analysis alone does not guarantee that unrelated pumps have the same performance coefficient.

For fixed $\ell$ and $g$, the exact conical pendulum oscillation period satisfies $P/P_0=\sqrt{\cos \theta }$. In the small-angle approximation, with $\theta$ in radians,Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order, $x(t)=r\cos (\omega t)$ and $y(t)=r\sin (\omega t)$ are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with $\omega \to \sqrt{g/\ell }$. At finite angle, however, the conical pendulum's oscillation period decreases with $\theta$, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint $\theta =0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero conical pendulum motion also requires $\omega >\sqrt{g/\ell }$, because $\cos \theta =g/(\ell {\omega }^2)<1$. At the other endpoint, $\theta \to \pi /2$ gives $h\to 0$, $P\to 0$, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

A point-mass gravitational lens is the ideal gravitational lens obtained by concentrating a positive mass $M$ at one location. Use geometrical optics, the weak-field approximation, small angles in radians, and a stationary lens whose deflecting region is short compared with the observer-lens and lens-source distances. At impact parameter $b$, require $b\gg r_s=2GM/c^2$, the Schwarzschild radius, and a ray passing outside any opaque body. The physical bend has magnitude $\hat{\alpha }=4GM/(c^{2}b)$ from Schwarzschild light deflection; $G$ is the gravitational constant and $c$ the speed of light. Multiple images can occur within this weak-field approximation: strong gravitational lensing does not require a ray to approach a black hole closely.^[ref]

Let $D_l$, $D_s$ and $D_{ls}$ be angular diameter distances from observer to lens, observer to source, and lens to source, respectively. They refer to the background geometry used in the thin gravitational lens equation. In nearby Euclidean geometry, $D_{ls}=D_s-D_l$; in cosmology, do not make that subtraction. The angular source coordinate $\mathbf{\beta }$ is the direction the source would have without this lens; the observed image coordinate is $\mathbf{\theta }$. Both are measured relative to the observer-lens optical axis.^[ref]

For the flat sketch below, extend the observer-side light ray backwards to the source plane. Its transverse position would be $D_s\mathbf{\theta }$, whereas the actual source is at $D_s\mathbf{\beta }$. Their separation is $D_{ls}\hat{\mathbf{\alpha }}$. Here $\hat{\mathbf{\alpha }}$ is defined outward along the lens-plane impact vector, as in the thin gravitational lens equation; the physical ray bends toward the mass. Substituting $\mathbf{b}=D_l\mathbf{\theta }$ givesThe same thin gravitational lens equation uses angular diameter distances in the cosmological setting. The factor $D_{ls}/D_s$ converts the physical bend into the reduced angular deflection; these are different angles.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.