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Stereoscopy supplies depth because different cameras see different projections of the same track. A functional reconstruction first matches the same interaction vertex and track segments across views. A calibrated lens and window model maps each measured image coordinate back to a ray inside the liquid. Refraction at the windows must be included. Intersecting the rays, or finding their closest agreement when measurements have errors, estimates three-dimensional coordinates. Several points then determine a trajectory, rather than only a projected arc.

An original ideal-camera model makes the geometry explicit. Take two parallel pinhole cameras with centers $(0,0,0)$ and $(b,0,0)$, focal length $f$, and optical axes along $z$. Use virtual image planes in front of the pinholes so that image coordinates have the same orientation as the scene. A point $(X,Y,Z)$ then givesSubtracting the horizontal image coordinates gives disparity $d=u_L-u_R=fb/Z$, henceFor stipulated $b=0.20\mathrm{m}$, $f=0.050\mathrm{m}$, $u_L=0.010\mathrm{m}$, $u_R=0.005\mathrm{m}$ and $v_L=0.002\mathrm{m}$, the reconstructed point is $(0.40,0.08,2.0)\mathrm{m}$. These are ideal optical inputs, not dimensions of Alvarez's apparatus. This model omits refraction and lens distortion; actual calibration replaces its simple ray geometry. Since $|dZ/dd|=fb/d^2$, the same disparity error gives worse depth precision for distant points.

The Lorentz force then supplies a momentum scale. With magnetic field $B$ uniform and negligible stopping power over a fitted segment, bubble-chamber track readout gives $p_{\perp }=|q|Br$. The local three-dimensional helix tangent supplies the missing direction: if $\alpha$ is its angle to the field, $p=p_{\perp }/\sin \alpha$. In practical units,Here $e$ is the elementary charge and $c$ the speed of light. A short, almost straight arc poorly constrains its radius of curvature; a visually sharp track can therefore have poor momentum precision.

Stereoscopy enables three-dimensional track reconstruction.^[ref] Calibrated camera positions and optical geometry are needed before a photographed arc becomes a physical radius of curvature. Neutral particles do not directly leave the same continuous ionization trail as charged particles; an interaction or decay can instead reveal them through charged secondary tracks. A photon's inferred path between interactions is not a photographed luminous line.

In a uniform magnetic field, with negligible stopping power and elastic scattering over the fitted segment, the transverse projection is a circle. The Lorentz force turns transverse momentum at angular rate $v_{\perp }/r$, givingHere $r$ is measured in the plane perpendicular to the magnetic field. The relation remains valid with relativistic physical momentum; it does not require $p=mv$. Thus curvature determines magnetic rigidity, not mass or kinetic energy alone. For chosen $|q|=e$, $B=1.5\mathrm{T}$ and $r=0.40\mathrm{m}$, $p_{\perp }\simeq 0.180\mathrm{GeV}/c$. This is an invented example, not a fit to a historical photograph. Reconstructing a three-dimensional helix also determines the component parallel to the magnetic field; the transverse radius of curvature alone does not. Determining the sign of electric charge additionally requires the field direction and the direction of travel, as in the charge-sign ambiguity of a curved track.

In a uniform magnetic field, a nonrelativistic particle of mass $m$ and nonzero charge $q$ completes one circular transverse orbit in time $T_c=2\pi /{\omega }_c$, where ${\omega }_c=|q|B/m$. Parallel velocity changes the helix pitch but not this period. In relativistic motion the constant Lorentz factor multiplies the period: $T_c=2\pi \gamma m/(|q|B)$. The direction of rotation depends on the charge sign; the period is positive.

Local resistive-force theory for a helix gives $AD-B^2={\ell }^2{a}^2{\xi }_{\parallel }{\xi }_{\perp }$. Positive local drag therefore guarantees a positive-definite axial resistance matrix and positive viscous power loss. The identity also keeps the denominator in counterrotating two-helix balances positive.

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.

The gyroradius is the radius of the transverse circular motion of a charged particle in a uniform magnetic field. In nonrelativistic classical mechanics, with zero electric field and no other forces, it is $m{v}_{\perp }/(|q|B)$, where $v_{\perp }$ is the speed perpendicular to the magnetic field. The parallel component of velocity changes the advance of the helix, rather than its transverse radius.

Mirroring the handedness of a helix reverses its axial translation-rotation coupling coefficients while preserving its pure translation and rotation resistances. With $\mathbf{t}=\cos \theta {\mathbf{e}}_z-\sin \theta {\mathbf{e}}_{\phi }$ for a left-handed helix and the force-on-body convention, $B=C=\ell a({\xi }_{\perp }-{\xi }_{\parallel })\sin \theta \cos \theta$.

A helical gear is a cylindrical gear whose teeth follow a helix around the reference cylinder. The gear helix angle $\beta$ measures their inclination to the axis. In a conventional external pair on parallel axes, mating gears have opposite hands with matching helix-angle magnitudes. Inclination makes engagement progress across the face rather than beginning everywhere along a straight tooth at once, but also creates a gear axial force that the supports must carry.^[ref]

A helical gear must specify its reference plane. The gear normal module $m_n$ is based on a section normal to the tooth direction, whereas gear transverse module $m_t$ is based on the plane perpendicular to the gear axis. Their geometric relation isSimilarly, reference pressure angles satisfy $\tan {\alpha }_t=\tan {\alpha }_n/\cos \beta$. Equal numbers labeled “module” in different planes therefore need not specify interchangeable gears.^[ref]

For an original example, take $m_n=2\mathrm{mm}$, $N=20$ and $\beta =30{}^{\circ }$. Reference diameter is $46.188\mathrm{mm}$, rather than $40\mathrm{mm}$. With ${\alpha }_n=20{}^{\circ }$, the transverse angle is about $22.796{}^{\circ }$.

At standard mounting, a transmitted torque $2\mathrm{N}\mathrm{m}$ gives gear tangential force $F_t\approx 86.60\mathrm{N}$. Ideal contact-force resolution gives axial magnitudeIts direction depends on helix hand and transmitted-torque direction. The corresponding radial component uses the transverse gear pressure angle. This illustrates the design trade: distributed engagement can improve smoothness, but the same transmission requires axial-load support. The example concerns parallel-axis cylindrical gearing; it should not be applied unchanged to crossed-axis arrangements.

In a uniform magnetic field, the Lorentz force leaves the component of velocity parallel to the field constant and rotates the perpendicular component at the nonrelativistic cyclotron frequency $|q|B/m$. The resulting path is a helix around a field line.

For a positive refractive index $\mu (r)$ in cylindrical coordinates, a light ray with constant radius $R>0$ and varying axial coordinate has the form $\theta (z)={\theta }_0+\omega z$ when it exists. Fermat principle makes $L=\mu (r)\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$ stationary. Conservation of the momentum conjugate to the cyclic angle makes ${\theta }^{\prime }$ constant along a constant-radius path. The radial Euler-Lagrange equation is thengiving the displayed formula. A nonzero real helix requires $0<-R{\mu }^{\prime }(R)/\mu (R)<1$. If ${\mu }^{\prime }(R)=0$, $\omega =0$ gives a straight axial light ray; a generic radial profile need not support a constant-radius helix. Its helical pitch is $2\pi /|\omega |$. This ray model is useful for a graded-index optical fiber when the refractive index varies smoothly and geometrical optics applies; it does not by itself impose wave-mode quantization or boundary confinement.

A helical symmetry combines an axial rotation and a proportional axial translation. A helix with $z=b\theta$ is unchanged as a set by this transformation. If a mechanical Lagrangian is invariant, the Noether conserved quantity for a mechanical point symmetry is $L_z+b{P}_z$, where the two momenta include all moving components. Pure axial rotation and pure axial translation need not separately be symmetries. The helical symmetry of a potential is a related special case in which the constraint is replaced by an invariant potential.^[ref]

A general helix is a regular space curve whose unit tangent vector makes a constant angle with a fixed axis direction $u$. In arc length parameter $s$, this means ${\gamma }^{\prime }(s)\cdot u$ is constant. Equivalently, the axial coordinate grows at a constant rate per unit arc length.

The familiar circular helix has constant distance from its axis:Its speed is $\sqrt{R^2+a^2}$, and its tangent's axial component is $a/\sqrt{R^2+a^2}$. Its signed pitch of a helix is $2\pi a$. A general helix need not have constant radius or constant axial displacement per turn: the logarithmic conical helix has a varying radius while retaining a constant tangent-axis angle.

As the length ratio of an opposite-handed reaction helix tends to infinity, its translation and rotation tend to zero as $1/n$, while the finite first helix rotates at almost the full motor rate. Small motions of the long rotor balance finite force and torque. The common propulsion vanishes in this limit of the additive model.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

A nucleic acid double helix consists of two nucleic acid strands winding around a common axis. In the familiar DNA arrangement, sugar–phosphate backbones lie outside and paired nucleobases lie inside, stacked along the axis. The strands are antiparallel nucleic acid strands. The flattened base pairing ladder above displays partners and direction; winding that ladder into a helix adds spatial organization. RNA can also form double-helical paired regions, so the geometry is not exclusive to DNA. See Alberts et al., Molecular Biology of the Cell: DNA structure.

The useful surprise is that a structure explains a copying principle. Complementary base pairs allow each DNA strand to specify the other's nucleic acid sequence; separate strands can each act as templates in DNA replication. This is a structural constraint, while the enzymes and stages of DNA replication supply the actual process. A nucleic acid sequence specifies letter order without, by itself, giving every detail of the three-dimensional helix.

On 25 April 1953, James Watson and Francis Crick published a model with two oppositely directed chains, outside phosphate groups, and specific base pairs. Their short paper explicitly connected this pairing to a possible copying mechanism, while describing the model as requiring further experimental tests. See the original ^[ref], with an accessible ^[ref].

Rosalind Franklin and Raymond Gosling's paper in the same issue had a different role: it presented X-ray diffraction evidence from DNA fibres and its structural interpretation, including the helical character of the wetter B form and evidence for phosphate groups on the outside. They distinguished forms produced at different water contents and assessed compatibility with the Watson–Crick model. Read their original ^[ref]. The model and the experimental analysis are distinct contributions; a diffraction pattern is evidence constraining a molecular geometry, rather than a direct picture of letter-by-letter base pairing.

Two coaxial opposite-handed helices linked by an internal motor share a translation speed but counterrotate. In an additive local-drag model, a length ratio $n$ multiplies the second helix's diagonal resistances by $n$ and reverses its coupling. With ${\Omega }_1-{\Omega }_2=\omega$, force and torque balance give $U=-2nBD\omega /[AD(1+n{)}^2-B^2(1-n{)}^2]$.

The osculating plane of a regular space curve at a point of nonzero curvature is the affine plane through the curve point spanned by its unit tangent vector $T$ and principal normal vector $N$. Its normal is the binormal vector $B=T\times N$, so the plane is described byEquivalently, any regular parameter gives the span of ${\gamma }^{\prime }(t_0)$ and ${\gamma }^{\prime \prime }(t_0)$ when these vectors are linearly independent. The second-order Taylor expansion then lies in this plane, which explains the name: it captures the curve's instantaneous bending.

A planar circle has its own containing plane as its osculating plane everywhere. For the helix $\gamma (t)=(\cos t,\sin t,t)$ at $t=0$, the point is $(1,0,0)$ and a normal is $(0,-1,1)$, giving $z=y$. At zero curvature the ordinary Frenet frame does not define a unique osculating plane.

The velocity vector of the helix is $r^{\prime }(t)=(-\sin t,\cos t,1)$, whose Euclidean norm is $\sqrt{2}$. Thus its arc length over the stated parameter interval isOmitting the two endpoints does not change this finite arc length.

The arc length of the helix follows from its parametrization:Thus ${\ell }_h=15\pi \sqrt{5}\mathrm{m}$. Its endpoint is $(0,20,15\pi )$, at height $h=15\pi \mathrm{m}$.

Under the usual intended interpretation that the descending quarter of a circle drops through one radius, its radius is $R=h$. Its arc length is ${\ell }_c=\pi R/2=15{\pi }^2/2\mathrm{m}$, givingThe printed smoothness condition is inconsistent. The terminal tangent vector of the helix is ${\mathbf{r}}^{\prime }(3\pi )=(-10,0,5)$. Every tangent vector to a curve confined to the $yz$ plane has zero $x$ component, so no such regular curve can join this helix with a common tangent line. Moreover, the helix is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.

Even after allowing a corner, a quarter of a circle reaching the ground does not alone determine its radius: the orientation of its endpoint radii must also be specified. For example, in the $yz$ plane, the circular arctraversed toward decreasing $\theta$, descends monotonically for $0\le {\theta }_0\le \pi /2$ and reaches $z=0$ when $R=h/(\sin {\theta }_0+\cos {\theta }_0)$. Choosing ${\theta }_0=\pi /4$ gives $R=h/\sqrt{2}$, whereas ${\theta }_0=\pi /2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.

For a regular differentiable curve $\mathbf{x}(t)$, its speed is $ds/dt=|{\mathbf{x}}^{\prime }(t)|>0$, where $s$ is arc length. The oriented unit tangent vector is $\hat{\mathbf{T}}={\mathbf{x}}^{\prime }/|{\mathbf{x}}^{\prime }|$. The curvature of a space curve is $\kappa =|d\hat{\mathbf{T}}/ds|$. Equivalently, for a twice differentiable regular curve,For the helix, differentiation gives ${\mathbf{x}}^{\prime }=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. ConsequentlyIndeed $d\hat{\mathbf{T}}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne (0,0)$; the excluded constant curve has no well-defined unit tangent vector. For $a=0,b\ne 0$ the curve is a straight line with zero curvature of a space curve.