Spherical Tractrix

Question

A pantoon P initially at North pole NP heading to Greenwich meridian intersection with equator, $( 0^{\circ} E, 0^{\circ} N $ ) towards point G moving down and east. It is dragged by a ship S moving east along the equator of a sphere ( unit radius ) connected by a very long floating chain of length $ PS =a=\pi/2, $ i.e., the chain is part of a full geodesic circle always kept taut.

P has spherical coordinates $ (\theta, \phi)$ for longitude & latitude somewhat like in the rough sketch shown with the spherical triangle PQS.

Assume the globe has no land masses.

The same locus would result if P is a small magnet stuck on a steel sphere and S moves in an equatorial groove pulling thread PS.

Please help with the parametrization of 3d tractrix (red) on the sphere.

EDIT1: The parametrization can be also given for starting latitude $\beta<\pi/2$ if it helps for an easier formulation.

Answer 1

tl; dr: Perhaps surprisingly, a "spherical tractrix" (whose chain length is one-quarter the circumference of the sphere) is a latitude line traced at constant speed:

Particularly, the point $P$ starting at the north pole does not move! As $S$ sails around the equator, the distance from $S$ to the north pole stays the same, so nothing pulls $P$ away from its initial position.

If instead $P$ starts on the prime meridian but not at a pole, the "tugboat" $S$ necessarily starts at longitude $90^{\circ}$, and $P$ gets pulled around its latitude as shown.

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Let $\theta(t)$ and $\phi(t)$ denote the longitude and latitude of $P$, so the Cartesian space coordinates of $P$ at time $t$ are $$ p(t) = \bigl(\cos\theta(t) \cos\phi(t), \sin\theta(t) \cos\phi(t), \sin\phi(t)\bigr). $$ Let $s(t) = \bigl(\cos(t + t_{0}), \sin(t + t_{0}), 0\bigr)$ denote the position of the dragging ship $S$, initially at longitude $t_{0}$ and proceeding eastward around the equator at unit angular speed. The equations of motion are:

1. The velocity $p'$ lies in the plane spanned by $p(t)$ and $s(t)$. Physically, $P$ is being pulled, so its direction of motion lies in the plane containing the origin, $P$, and $S$.
2. The distance from $p(t)$ to $s(t)$ is $\pi/2$ for all time. Physically, the chain from $S$ to $P$ has constant length. Because we're on a unit sphere, and the chain has length $\pi/2$, the dot product $p(t) \cdot s(t)$ is identically $0$.

The first condition says there exist functions $u$ and $v$ such that $$ p'(t) = u(t)p(t) + v(t)s(t). $$ Dotting both sides with $p(t)$ and using $p(t) \cdot p(t) = 1$ (because $p(t)$ lies on a unit sphere), $p(t) \cdot p'(t) = 0$ (because $p$ moves on a sphere about the origin), and $p(t) \cdot s(t) = 0$ (the second equation of motion) tells us $u(t) = 0$, and therefore $p'(t) = v(t)s(t)$.

Now, writing the second condition in components gives \begin{align*} 0 &= p(t) \cdot s(t) \\ &= \bigl(\cos\theta(t) \cos\phi(t), \sin\theta(t) \cos\phi(t), \sin\phi(t)\bigr) \cdot \bigl(\cos(t + t_{0}), \sin(t + t_{0}), 0\bigr) \\ &= \bigl(\cos\theta(t) \cos(t + t_{0}) + \sin\theta(t)\sin(t + t_{0})\bigr) \cos\phi(t) \\ &= \cos\bigl[\theta(t) - (t + t_{0})\bigr] \cos\phi(t). \end{align*} If $\cos\phi(0) = 0$, i.e., $P$ starts at a pole, then as noted above $P$ stays at that pole and this equation is satisfied. If $P$ does not start at a pole, then $\cos\phi$ is non-vanishing and the first factor vanishes identically. That is, $$ \theta(t) - (t + t_{0}) $$ is $\frac{\pi}{2}$ plus an integer multiple of $\pi$. Since $\theta(0) = 0$, we may assume $t_{0} = \frac{\pi}{2}$, which gives $$ s(t) = (-\sin t, \cos t, 0). $$ Consequently, $p'(t) = v(t)s(t)$ has vanishing third component, $\phi(t)$ is constant, and $P$ moves along a latitude circle.

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I haven't carefully checked what happens if the chain is shorter than one-quarter of the sphere's circumference, i.e., if $$ p(t) \cdot s(t) = \cos\bigl[\theta(t) - (t + t_{0})\bigr] \cos\phi(t) $$ is a positive constant, but in that case the diagram in the post looks qualitatively accurate: $P$ follows a spiral path asymptotic to the equator, analogously to a plane tractrix.

Answer 2

tl; dr: This answer analyses a chain whose length is less than $\pi/2$ (less than one-quarter the circumference of the sphere). Although not literally asked, this seems more along the intuitive motivation of the question.

Let $C$ be the cosine of the pontoon's initial latitude. The latitude $\phi$ and longitude $\theta$ satisfy the ODEs \begin{align*} \phi'(t) &= -\frac{C\tan\phi(t) \sqrt{\cos^{2}\phi(t) - C^{2}}}{1 - C^{2}}, \\ \theta'(t) &= \frac{\sin^{2}\bigl(t - \theta(t)\bigr)}{1 - C^{2}}. \end{align*} Each can in principle be integrated explicitly (though the prospect of inverting the integrals to get an explicit parametrization looks unlikely); the animation loop instead shows several numerical solutions (heavy paths), including the tow chains (lighter arcs), with different chain lengths unlike in the "quarter-circumference" case):

(The remaining case, where the chain is longer than one-quarter the circumference but shorter than half, differs substantially from both this case and the prior "quarter-circumference" case. I have not carried out the details (and do not plan to, though they should be pursuable along the lines below), but qualitatively: If the pontoon starts on the prime meridian, the tow ship must start somewhere to the east in order for the chain to be initially taut; for a given chain length not all initial latitudes are possible; the pontoon's latitude initially increases, reaches a maximum, then decreases asymptotically toward $0$.)

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We'll assume the tow ship $S$ starts on the prime meridian and travels eastward at unit speed around the equator, so at time $t$ its position is $$ s(t) = (\cos t, \sin t, 0). $$ Further, the pontoon $P$ starts on the prime meridian at latitude $\varphi_{0}$, so the tow chain has length $\varphi_{0}$. Put $C = \cos\varphi_{0}$.

Let $\theta(t)$ and $\phi(t)$ denote the longitude and latitude of $P$ at time $t$. It will be useful to introduce the (rotating) orthonormal frame relative to $P$: $$ e_{1}(t) = \left[\begin{array}{@{}c@{}} \cos\theta(t) \\ \sin\theta(t) \\ 0 \\ \end{array}\right],\qquad e_{2}(t) = \left[\begin{array}{@{}c@{}} -\sin\theta(t) \\ \phantom{-}\cos\theta(t) \\ 0 \\ \end{array}\right],\qquad e_{3}(t) = \left[\begin{array}{@{}c@{}} 0 \\ 0 \\ 1 \\ \end{array}\right]. $$ For later use, note that these satisfy the trigonometric identities \begin{equation} \left\{ \begin{aligned} e_{1}(t) \cdot s(t) &= \cos\bigl(t - \theta(t)\bigr), \\ e_{2}(t) \cdot s(t) &= \sin\bigl(t - \theta(t)\bigr), \\ e_{3}(t) \cdot s(t) &= 0; \end{aligned} \right. \end{equation} and the first-order differential equations \begin{equation} \left\{ \begin{aligned} e_{1}'(t) &= \phantom{-} \theta'(t)\, e_{2}(t), \\ e_{2}'(t) &= -\theta'(t)\, e_{1}(t), \\ e_{3}'(t) &= 0. \end{aligned} \right. \end{equation} Particularly, the position of the tow ship in the rotating frame is \begin{align*} s(t) &= (s \cdot e_{1})\, e_{1}(t) + (s \cdot e_{2})\, e_{2}(t) \\ &= \cos\bigl(t - \theta(t)\bigr)\, e_{1}(t) + \sin\bigl(t - \theta(t)\bigr)\, e_{2}(t). \end{align*}

Because the tow chain has positive length shorter than one-fourth the circumference of the sphere and the initial latitude is positive, the trig arguments $\phi(t)$ and $t - \theta(t)$ lie in the first quadrant. Particularly, we may divide freely by sine and cosine, and write $\sin = \sqrt{1 - \cos^{2}}$.

The Cartesian space coordinates of $P$ at time $t$ are \begin{align*} p(t) &= \bigl(\cos\theta(t) \cos\phi(t), \sin\theta(t) \cos\phi(t), \sin\phi(t)\bigr) \\ &= \cos\phi(t) \left[\begin{array}{@{}c@{}} \cos\theta(t) \\ \sin\theta(t) \\ 0 \\ \end{array}\right] + \sin\phi(t) \left[\begin{array}{@{}c@{}} 0 \\ 0 \\ 1 \\ \end{array}\right] \\ &= \cos\phi(t)\, e_{1}(t) + \sin\phi(t)\, e_{3}(t). \end{align*} As before, the equations of motion are:

1. At each instant, the velocity $p'(t)$ lies in the plane spanned by $p(t)$ and $s(t)$. Geometrically, $P$ is being pulled, so at each instant its direction of motion lies in the plane containing the origin, $P$, and $S$. Analytically, there exist functions $u$ and $v$ such that $$ p'(t) = u(t)p(t) + v(t)s(t). $$
2. The distance from $p(t)$ to $s(t)$ is $\varphi_{0}$ for all time, so for all $t$ we have \begin{align*} C &= \cos\varphi_{0} \\ &= p(t) \cdot s(t) \\ &= \cos\phi(t)\, e_{1}(t) \cdot s(t) \\ &= \cos\phi(t)\cos\bigl(t - \theta(t)\bigr). \end{align*}

Our goal is a pair of differential equations for the unknown coordinates $\theta$ and $\phi$. Differentiate $$ p(t) = \cos\phi(t)\, e_{1}(t) + \sin\phi(t)\, e_{3}(t) $$ using the product rule and frame ODEs above: $$ p'(t) = -\sin\phi(t)\phi'(t)\, e_{1}(t) + \cos\phi(t) \theta'(t)\, e_{2}(t) + \cos\phi(t) \phi'(t)\, e_{3}(t). $$ On the other hand, the first equation of motion $p' = up + vs$ becomes \begin{align*} p'(t) &= u(t) \cos\phi(t)\, e_{1}(t) + u(t)\sin\phi(t)\, e_{3}(t) + v(t)s(t) \\ &= \bigl[u(t) \cos\phi(t) + v(t)\cos\bigl(t - \theta(t)\bigr)\bigr]\, e_{1}(t) + v(t)\sin\bigl(t - \theta(t)\bigr)\, e_{2}(t) + u(t)\sin\phi(t)\, e_{3}(t). \end{align*} Equating components in the two preceding equations gives \begin{alignat*}{2} -\sin\phi(t)\phi'(t) &= u(t) \cos\phi(t) &+ v(t)\cos\bigl(t - \theta(t)\bigr), \\ \cos\phi(t) \theta'(t) &= &\phantom{+} v(t)\sin\bigl(t - \theta(t)\bigr), \\ \cos\phi(t) \phi'(t) &= u(t)\sin\phi(t). \end{alignat*} Solving the second for $v$ and the third for $u$ and substituting in the first gives $$ \frac{1}{\sin\phi(t)}\phi'(t) + \frac{C}{\sin\bigl(t - \theta(t)\bigr)} \theta'(t) = 0. $$

Logarithmically differentiating the constraint $C = \cos\phi(t)\cos\bigl(t - \theta(t)\bigr)$ gives a second ODE: $$ \frac{\sin\phi(t)}{\cos\phi(t)} \phi'(t) + \frac{\sin\bigl(t - \theta(t)\bigr)}{\cos\bigl(t - \theta(t)\bigr)}\bigl(1 - \theta'(t)\bigr) = 0. $$

Algebra (a bit lengthy, sketched below) allows us to decouple these equations, yielding \begin{align*} \phi'(t) &= -\frac{C\tan\phi(t) \sqrt{\cos^{2}\phi(t) - C^{2}}}{1 - C^{2}}, \\ \theta'(t) &= \frac{\sin^{2}\bigl(t - \theta(t)\bigr)}{1 - C^{2}}. \end{align*} The diagram below shows several numerical solutions (heavy paths), including the tow chain (lighter arcs) at regularly-spaced time intervals.

Interestingly, and importantly for numerical plotting, the ODE for latitude $\phi$ is non-Lipschitz at its initial condition (because of the square root in the numerator), and has non-unique solutions. Contrary to my initial claim, the constant solution $\cos\phi(t) = C$ does satisfy the constraint $C = \cos\phi(t) \cos\bigl(t - \theta(t)\bigr)$ provided the pontoon and the tow ship stay at the same longitude forever! This, however, is "not physical" unless we construe the tow chain as a rigid, northward-pointing linkage. The "decaying" solutions shown correspond to the tow ship starting to the south and infinitesimally east of the pontoon.

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Our system has the form \begin{align*} \alpha\phi' + \beta\theta' &= 0, \\ \gamma\phi' + \delta\theta' &= \epsilon, \end{align*} for $$ \alpha = \frac{1}{\sin\phi},\quad \beta = \frac{C}{\sin(t - \theta)},\qquad \gamma = \frac{\sin\phi}{\cos\phi},\quad \delta = \epsilon = -\frac{\sin(t - \theta)}{\cos(t - \theta)}. $$

The linear system above is readily solved for $\phi'$ and $\theta'$: $$ \phi' = -\frac{\beta\epsilon}{\alpha\delta - \beta\gamma},\qquad \theta' = \frac{\alpha\epsilon}{\alpha\delta - \beta\gamma}. $$ It suffices to write $\phi'$ as a function of $\phi$ and $\theta'$ as a function of $t - \theta$.

The constraint $C = \cos\phi \cos(t - \theta)$ gives two pairs of formulas that convert trig functions of $\phi$ to trig functions of $t - \theta$: \begin{align*} \cos\phi &= \frac{C}{\cos(t - \theta)}, \\ \sin\phi &= \frac{1}{\cos(t - \theta)}\sqrt{\cos^{2}(t - \theta) - C^{2}}; \end{align*} and \begin{align*} \cos(t - \theta) &= \frac{C}{\cos\phi}, \\ \sin(t - \theta) &= \frac{1}{\cos\phi}\sqrt{\cos^{2}\phi - C^{2}}. \end{align*}

Finally, calculation gives \begin{align*} \alpha\delta - \beta\gamma &= -\frac{(1 - C^{2})\cos^{2}\phi}{C\sin\phi\sqrt{\cos^{2}\phi - C^{2}}} \\ &= -\frac{(1 - C^{2})}{C\sin(t - \theta)\sqrt{\cos^{2}(t - \theta) - C^{2}}}. \end{align*} The asserted ODEs (eventually) follow.