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A proposed transport system using a cable extending from a planetary surface beyond a synchronous orbit. In a co-rotating model, the outward effective force on the outer portion supports the inner portion. The endpoint loads, distribution of mass, and material strength determine a feasible tension profile; the uniform unloaded space-elevator cable is an ideal limit.

Consider two equal balls of mass $m$, four light rods each of length $L$, a fixed upper pivot and a light sliding lower collar. Each upper and lower rod makes angle $\theta$ with the spindle. A spring of spring constant $k$ is uncompressed at $\theta =0$ and is compressed by collar motion. Its compression is $\delta =2L(1-\cos \theta )$. Collar balance gives $2T_l\cos \theta =k\delta$, while vertical ball balance gives $(T_u-T_l)\cos \theta =mg$. Radial Newton's second law gives $(T_u+T_l)\sin \theta =m{\omega }^{2}L\sin \theta$. Eliminating the rod tensions proves the displayed relation for nonzero $\theta$.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

For an elastic filament with reference arclength $s$, curvature $c$, axial strain $ϵ$ and excess twist rate $\Omega$, a harmonic energy density is $(K_b{c}^2+S{ϵ}^2+K_t{\Omega }^2)/2$. Bending and torsional rigidities have units of energy times length; the stretch modulus has units of force. Factoring out $k_{B}T$ therefore makes $A,C$ lengths but $B$ an inverse length. Axial strain under tension is approximately $F/S$. Twist is controlled by torque and rotational boundary conditions, not force alone. A three-dimensional thermally fluctuating worm-like chain has persistence length $K_b/(k_{B}T)$.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.

A uniform string with fixed ends $x=\pm \ell$, tension $T$, linear mass density $\rho$, and attached central point mass $m>0$ obeys the wave equation away from zero, continuous displacement at zero, and $m{y}_{tt}(0,t)=T[y_x(0^{+},t)-y_x(0^{-},t)]$. The normal modes split into antisymmetric modes that leave the mass at rest and symmetric modes that move it.

To prove the spectral equations, use $X=A\sin (k(x+\ell ))$ on the left and $X=B\sin (k(\ell -x))$ on the right, with ${\omega }^2=(T/\rho )k^2$. Continuity gives $(A-B)\sin (k\ell )=0$ and the force law gives $Tk(A+B)\cos (k\ell )=m{\omega }^{2}A\sin (k\ell )$. If $\sin (k\ell )=0$, then $k\ell =n\pi$, $n\ge 1$, and $B=-A$. Otherwise $A=B$ and cancellation gives $\cot (k\ell )=mk/(2\rho )$. In dimensionless variables $z=k\ell$, $\alpha =m/(2\rho \ell )$, strict monotonicity gives one symmetric root in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$. The apparent zero root of the sine equation has only the zero profile and is excluded.

For $m/(\rho \ell )\gg 1$, the lowest root tends to zero. The Taylor series $\cot z=1/z-z/3+O(z^3)$ gives $z\sim {\alpha }^{-1/2}$ and hence $\omega \sim \sqrt{2T/(m\ell )}$. The two nearly straight half-strings supply a restoring force $-2TY/\ell$, explaining this harmonic oscillator limit.

For a uniform string with tension $T>0$, linear mass density $\rho >0$, left end fixed, and a freely transverse-moving point mass $M\ge 0$ at $x=L$, the wave equation and endpoint law areThe last equation is Newton's second law: the string pulls the mass back toward the interior, giving the minus sign. With wave speed $c=\sqrt{T/\rho }$, a nonzero normal mode has spatial part $X(x)=\sin (\omega x/c)$ and satisfiesFor $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous Neumann boundary condition. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.

Multiplying the spatial equation by $X$ and using integration by parts givesFor two different modes, the same calculation gives the weighted inner product orthogonality $\rho {\int }_0^L{X}_m{X}_{n}dx+M{X}_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.

A taut spring of equilibrium length $L$ and initial tension $T_0$ whose ends acquire a small relative transverse displacement $y$ tilts through $y/L$. Its leading transverse force is therefore $-T_{0}y/L$. The length change is quadratic in $y$, so using only incremental axial extension misses this first-order force. A Hooke spring with natural length $l_0$ has $T_0=k(L-l_0)$ and $k_{\perp }=k(1-l_0/L)$. The unstretched limit has no linear transverse stiffness under this geometry.

A radial co-rotating cable of constant linear mass density $\lambda$ extends from planetary radius $R$ to $b={\eta }^{\prime }R$, with zero applied endpoint tension. If the geostationary orbit has radius $\eta R$, integration of the effective force gives ${\eta }^{\prime }({\eta }^{\prime }+1)=2{\eta }^3$ for a nonzero cable. At position $r$ its tension is $\lambda [GM(1/R-1/r)-{\omega }^2(r^2-R^2)/2]$. Differentiation shows the maximum at the geostationary orbit. An anchor load or counterweight alters the integrated condition.

An elastic beam above a smooth obstacle cannot penetrate it, and a nonadhesive obstacle can only push. The displayed constraints express this when $R$ is an upward distributed normal contact force; concentrated point forces must also be nonnegative. In the small-slope Euler-Bernoulli beam equation,Integrating across a contact point with no applied couple gives jump conditions $[w]=[w^{\prime }]=[B{w}^{\prime \prime }]=0$ and $[B{w}^{\prime \prime \prime }]=R_i$, provided the tension is continuous. For a flat obstacle, no distributed load and zero tension, an open contact interval has $R=0$; its entire normal contact force can therefore be concentrated at its endpoints.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.

An ideal uniform, flexible string under constant tension $T>0$ has transverse displacement $y(x,t)$ satisfyingwhere $\mu >0$ is linear mass density, meaning mass per unit length of the straight reference string. The approximation assumes small slope $|y_x|\ll 1$, transverse motion, negligible bending stiffness, negligible damping, and no distributed external force. Constant tension and uniform linear mass density make $c$ constant. Some texts use $\rho$ for this same linear mass density, giving $c=\sqrt{T/\rho }$; here $\rho$ must have units $\mathrm{kg}{\mathrm{m}}^{-1}$, rather than units of volume mass density. ^[ref]

The units provide a useful check: $T/\mu$ has units $(\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2})/(\mathrm{kg}{\mathrm{m}}^{-1})={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For $T=100\mathrm{N}$ and $\mu =0.010\mathrm{kg}{\mathrm{m}}^{-1}$, the wave speed is $100\mathrm{m}{\mathrm{s}}^{-1}$, so a pulse crosses a one-metre string in $0.010\mathrm{s}$. Quadrupling tension doubles wave speed; quadrupling linear mass density halves it. These square-root laws explain how material and loading set the time scale of the vibration.

For a length-$L$ string with fixed endpoints, the homogeneous Dirichlet boundary conditions are $y(0,t)=y(L,t)=0$. Its nonzero normal modes haveThus $cn\pi /L$ is an angular frequency, in radians per second, whereas $cn/(2L)$ is the frequency in hertz. The wavelength is $2L/n$, and each node of a standing wave lies at $x=jL/n$ for some $j=0,\dots ,n$. For the numerical string above with $L=1\mathrm{m}$, the first three frequencies are $50$, $100$ and $150\mathrm{Hz}$, with angular frequencies $100\pi$, $200\pi$ and $300\pi {\mathrm{s}}^{-1}$. ^[ref]

Each normal mode is a standing wave, but it also has a travelling wave description:At a fixed endpoint an incident displacement pulse reflects with reversed sign, allowing incident and reflected displacements to cancel at the endpoint. Periodic reflection for a fixed-end string implements both endpoints through odd, $2L$-periodic extensions of the initial data. This is the bridge between moving pulses and the stationary Fourier sine basis. ^[ref]

A short derivation comes from Newton's second law. The transverse force on a small segment is approximately $T[y_x(x+\delta x,t)-y_x(x,t)]$, and its mass is $\mu \delta x$. Dividing the force balance by $\delta x$ and taking the limit gives $\mu y_{tt}=T{y}_{xx}$. Separation of variables then gives ${\varphi }^{\prime \prime }+k^2\varphi =0$; the fixed-end boundary conditions select $k=n\pi /L$, while the temporal equation is a harmonic oscillator equation with $\omega =ck$.

For the homogeneous one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$ with constant $c>0$, wave speed means the speed at which a travelling wave profile translates. In $F(x-ct)$, a feature initially at $x=s$ follows $x=s+ct$, giving signed velocity $+c$; in $G(x+ct)$ it follows $x=s-ct$, giving signed velocity $-c$. The characteristic lines for the one-dimensional wave equation therefore have slopes $dx/dt=\pm c$. The speed of either profile is $c$, even though the displacement at a fixed point can change sign or pass through zero.^[ref]

The same $c$ bounds the causal cone for the constant-speed wave equation. For a wave equation on a string, $c=\sqrt{T/\mu }$ depends on tension and linear mass density; for small acoustic waves in stationary uniform air it is the speed of sound. This interpretation assumes the stated ideal wave equation. In media with wave dispersion, phase velocity, group velocity and the earliest signal arrival have different meanings: a measured wave crest speed need not be the speed of a wave envelope, and neither alone establishes a causal cone for the constant-speed wave equation.

At distance $x$ below the support of a flexible uniform cable of length $L$, the equilibrium tension is $\mu g(L-x)$, with linear mass density $\mu$. The local transverse speed is $\sqrt{g(L-x)}$. Integrating its reciprocal from zero to $L$ gives the displayed travel time. The speed vanishes at the free end, but the travel-time singularity is integrable. With variable tension, the small-amplitude wave equation is $\mu y_{tt}={\partial }_x(T{y}_x)$; its characteristic speed is still $\sqrt{T/\mu }$.

A point mass attached to a light inextensible string winds around a fixed smooth circular pole of radius $R$. Let the initial straight free length be $L$, and assume the string remains taut without slipping over the part already laid on the pole. Since its tension is perpendicular to the point mass's velocity, the speed $v$ stays constant. If the contact point turns through $d\theta$, the free length $s$ obeys $ds=-Rd\theta$, while the involute geometry gives $v=s\dot{\theta }$. Thus $s\dot{s}=-Rv$, or $s^2=L^2-2Rvt$. First contact of the point mass with the pole occurs at $s=0$, after $L^2/(2Rv)$. In the ideal model, the required tension $m{v}^2/s$ diverges at contact; finite string strength or finite particle size limits that idealization.