Around the World in a Day
I visited Iceland with my father in May of 2017. Iceland is close to the Arctic circle, so the sun barely sets on summer nights, slipping beneath the horizon for a few hours of twilight. Inspired by the (almost) midnight sun we started to wonder whether it would be possible to spend an entire year in only daylight.
North of the Arctic Circle, the sun never sets in the Arctic summer; south of the Antarctic Circle, the sun never sets in the Antarctic summer. On an idealized earth daylight would arrive at the North Pole on the vernal (spring) equinox, remain for six months, and set on the autumnal equinox just in time to arrive at the South Pole. So the real challenge is to make it from the North Pole to the South Pole on the equinox without ever leaving daylight.
The earth and its seasons. Yellow lines mark the polar circles. The North Pole is lit throughout the northern summer and is dark during the northern winter. The opposite is true for the South Pole. Both poles are lit simultaneously during the equinox. Click and drag the earth to change your view.
The Puzzle
That conversation motivated the following questions:
 What is the minimum speed needed to fly from the North Pole to the South Pole (on the equinox) to remain entirely in daylight?
 What flight path must be taken to complete this flight?
This puzzle turned out to be one of the most surprising math puzzles I’ve ever solved. We talked about it for hours in Iceland but never came up with a satisfying answer. The question was tantalizingly geometric, but remained beyond what I could do in my head. After three years of absentminded visualization I returned to the problem armed with paper and a pen, and started working it in earnest. Even then I didn’t see the whole picture until discussing it with my brother, Jack. So, if you enjoy math challenges, this is a gem—try to solve it before reading further.
Necessary Assumptions
To solve this problem, we need to make some assumptions:
 The earth is a perfect sphere and rotates about a fixed axis passing through the North and South Poles.
 The day–night boundary is discrete (no refraction of light).
 The rate at which the earth’s tilt changes relative to the sun is slow enough that we can assume that the earth’s tilt is constant and fixed for each day. We set the tilt of the earth to its equinoctial orientation, with the northernmost and southernmost boundaries of night passing through the North and South Poles.
Possible Solutions
Let’s begin with a little brainstorming. How fast do you think we need to fly, and do you think you’ve ever traveled that fast? Has anyone? What is a relevant reference speed?
The speed we need to travel depends on the path we take. So, another way to phrase the question is what paths should we take?
You can try different paths in the game below. The goal is to fly your plane from the North Pole to the South Pole without crossing into night. The slower your plane, the harder the task.
The obvious first thought is to fly due south. After all, flying due south minimizes the distance travelled. What then?
The distance from the North Pole to the South Pole, flying due south, is 12,437 miles. To make the trip entirely in daylight, we start at dawn and have 12 hours to reach the South Pole before dusk, which requires a plane that can travel at 1,036.4 mph (mach 1.35). The top speed of a Concorde jet was 1,354 mph, more than fast enough to make the journey—if only its range were more than 4,505 miles. Short of finding three functioning Concordes, or requisitioning a highspeed military jet that can be refueled midflight, the trip due south requires going too far too fast.
If you haven’t tried it yet, fly due south in the game. It is impossible to win to win this way. Nevertheless, the game is winnable along other routes. We could fly a little southwest at the beginning and end. That way we chase the daylight at the start and run from the night at the end.
Note that the speed at which the day–night boundary moves depends on your latitude. Near the poles the boundaries travel slowly and near the equator they travel quickly. This means that, for any plane, there is a range of latitude near the poles where you can fly faster than the daynight boundaries. Within these ranges you can outrun the sunset, so, as long as you don’t fly into the night you could stay in daylight indefinitely.
Denote the speed of your plane $v$ and the latitude at which your plane’s speed equals the westward motion of the day–night boundary $\phi_*(v)$. The real challenge is to get from $\phi_*(v)$ degrees north to $\phi_*(v)$ degrees south without seeing the sunset. If you can do that, you can make it from the North Pole to the South Pole entirely in daylight.
This leads to an alternative proposal: for a plane with a max speed $v$, fly south from the North Pole to the latitude $\phi_*(v)$ along some path that avoids flying into the night and ensures that you reach latitude $\phi_*(v)$ at the dawn boundary. Then fly due south until you reach $\phi_*(v)$ degrees south. As long as you arrive at $\phi_*(v)$ degrees south before dusk, you can take any path you like to the South Pole as long as you don’t fly into the night.
This is a reasonable proposal, but we can do better. Let’s think about a different strategy.
Suppose that instead of flying south we fly southwest. For every degree longitude we travel west, we recover four minutes of solar time. For every fifteen degrees, we recover an hour. If we try to spiral south around the globe, then we may be able to recover enough time to make it with a slower plane.
Close enough to the poles our plane is fast enough to outrun dusk and catch dawn. But as we head south, dawn and dusk travel faster and faster, and we eventually can’t outrun dusk. Now we need to go southwest. The farther south we point the shorter our trip but the faster dusk approaches. The more west we point the slower dusk approaches but the longer our trip takes. It would be reasonable to head southwest, starting out pointing mostly west when we are fast relative to dusk (near the poles) and turning more and more south as we become slower and slower relative to dusk (near the equator).
To implement this strategy we need to know how far west from south we should point at each latitude. To find the optimal declination (angle west from south), consider the following analogy:
An Aquatic Analogy
Imagine that you need to swim across a river that flows faster than you can swim. You will drift downstream even if you swim straight into the current. Your goal is to make it across without drifting too far downstream. In the flight path problem, the more you fly west the slower the sunset approaches but the longer it takes to travel south. The more southward you fly the shorter your journey but the faster the sunset approaches. In the river problem, the more you swim upstream the slower you drift downstream but the longer it takes to cross. The more directly you swim across the less time it takes but the faster you drift downstream.
An example set of swimmers on four different rivers are shown below. The swimmers all swim at the same speed but at different angles against the current. The rivers increase in speed from left to right.
Let $V$ denote the speed of the river and $v < V$ denote your speed. If you swim at a constant velocity and with a constant orientation, then you will follow a straight line across. To minimize the distance you drift downstream you should minimize the slope of the line you travel (distance drifted downstream divided by the width of the river). So, at each instant your goal is to minimize the amount you drift downstream for each bodylength you move across the river. This is accomplished by swimming upstream at an angle that keeps your declination (orientation of your body) perpendicular to the direction in which you are drifting. Notice that in Figure 2 the most successful swimmer always swims perpendicular to the line marking his or her path.
Let $x$ denote the direction across the river and $y$ denote downstream direction. Let $\theta$ denote the angle at which you swim across the river relative to the upstream direction. Your drift velocity is the sum of your downstream drift due to the river, which is represented by the velocity vector $[0,V]$, and the velocity at which you swim, $v [\sin(\theta), \cos(\theta)]$. Vectors add tiptotail, so we can represent your drift velocity as a triangle with a downstream leg of length $V$ plus a shorter leg of length $v$ oriented in the direction you swim. These are the blue and red vectors shown in Figure 3.
The purple circle in Figure 3 represents all possible drift velocities given all possible declinations $\theta$. Our goal is to minimize the angle between the drift velocity and the horizontal direction $x$—to minimize the slope of the purple line. This is achieved when the purple line is tangent to the outside of the circle as shown by the solid purple line. The outside of a circle is always perpendicular to a radial ray. Therefore, when the purple line is tangent to the circle it must be perpendicular to the red radial vector representing the swimmer’s velocity. This means that the point on the purple circle that minimizes this slope is given by choosing a declination so that the swimmers swims perpendicular to their resulting drift.
Therefore, the triangle formed by the river velocity $V$, swimmer velocity $v$, and drift velocity is a right triangle. It follows that $\cos(\theta) = v/V$, so the optimal declination is given by:
$\theta(v,V) = \cos^{1} \left( \frac{v}{V} \right).$Let’s prove that this is really the optimal declination with some calculus. Your drift velocity is $[v \sin(\theta), V  v \cos(\theta)]$. Our goal is to minimize the slope:
$$ \frac{\text{drift velocity in } y}{\text{drift velocity in } x} = \frac{V  v \cos(\theta)}{v \sin(\theta)} = \frac{V/v  \cos(\theta)}{\sin(\theta)}. $$
The only variable we can change in this ratio is the declination angle $\theta$, since $\theta$ controls the orientation of the swimmer. To minimize the ratio, take its derivative with respect to $\theta$ and set the derivative to zero. Use the quotient rule to take the derivative:
$$ \begin{aligned} \frac{d}{d\theta} \frac{V/v  \cos(\theta)}{\sin(\theta)} & = \frac{\sin(\theta)^2  (V/v  \cos(\theta)) \cos(\theta)}{\sin(\theta)^2} \\ & = \frac{\sin(\theta)^2 + \cos(\theta)^2  (V/v) \cos(\theta)}{\sin(\theta)^2} \\ & = \frac{1  (V/v) \cos(\theta)}{\sin(\theta)^2} = 0. \end{aligned} $$
This equality requires that $(V/v) \cos(\theta) = 1$, or $\theta(v,V) = \cos^{1}\left( \frac{v}{V} \right)$. We know this value for $\theta$ a minimizer since it is the only extremum of the ratio, the ratio is defined for all $\theta$ between 0 and 180 degrees, and the ratio diverges to positive infinity as $\theta$ goes to 0 or 180 degrees (the denominator goes to zero).
Therefore the optimal angle at which you should swim across the river relative to the upstream direction is the arccosine of the ratio of your speed to the river speed. The faster you are relative to the river the larger the ratio $v/V$, so the smaller the angle. More simply, the faster we are, the more we should swim upstream.
The optimal declination of the swimmer relative to the upstream direction as a function of the ratio of the swimmer’s speed $v$ to the speed of the river $V$. The upstream direction is shown by the vertical arrow, and the swimmer’s orientation denotes the swimmer’s direction.
This result is the key to solving the daylight problem. It will tell us how to orient the plane to minimize the amount of daylight lost per distance traveled southward.
The Solution
In order to solve the daylight problem we need to solve essentially the same optimization problem. The difference when working with daylight is that the speed at which daylight progresses depends on the latitude of the plane. Daylight progresses westward the fastest at the equator and slowest at the poles. This would be analogous to a river that is faster in the middle and slower near the shore.
Let’s work out how fast the dawn boundary moves as a function of our latitude $\phi$. Let $R$ be the radius of the earth. Let $r(\phi)$ be the distance between us and the earth’s axis of rotation. Then:
$r(\phi) = R \cos(\phi).$The earth spins at $\omega = \frac{2 \pi}{24} \text{rad/hr}$. Therefore the speed of the day–night boundary at latitude $\phi$ is:
$V(\phi) = \omega R \cos(\phi).$Recall that there are regions near the poles where the plane is faster than the daynight boundary. The bounding latitude, $\phi_*$, is:
$\phi_*(v) = \cos^{1}\left( \frac{v}{\omega R}\right).$Let $\rho$ be the ratio $v/(\omega R)$. Then $\rho$ is the ratio of our velocity to the velocity we would need to fly to keep up with daylight at the equator and $\phi_* = \cos^{1}(\rho)$.
Because we can outrun night indefinitely anywhere within $\phi_*$ of either pole, it is enough to be able to fly from $\phi_*$ to $ \phi_*$ in daylight in order to fly from the North Pole to the South Pole in daylight.
Between $\phi_*$ and $\phi_*$, we are slower than the daynight boundary. No matter what direction we fly, night is gaining on us. This is analogous to the swimmer who, no matter what direction he swims, is carried downstream.
As we fly, we don't care about the time on our watch, only the position of the sun. That is, we only care about solar time. At dawn, solar time is zero; at dusk, solar time is twelve hours. Our goal is to keep the elapsed solar time over the course of the flight to less than twelve hours (assuming we start at the dawn boundary). So, for a given velocity $v$ we seek the trajectory that minimizes the elapsed solar time in traveling from latitude $\phi_*$ to $\phi_*$.
Let $\theta$ denote the angle south from due west that the plane flies—its declination. The more we point the plane west, the slower solar time elapses but the longer our journey takes. Notice that fixing $\theta$ as a function of latitude $\phi$ uniquely specifies a trajectory. The declination $\theta(\phi)$ defines this trajectory implicitly and can be thought of as a bearing given to the pilot at each latitude (fly $\theta$ degrees south of west when at latitude $\phi$).
The solar time elapsed constantly increases while we are between $\phi_*(v)$ and $\phi_*(v)$ since we are slower than the dawndusk boundaries in that range. Let $S(t)$ denote the solar time elapsed at the actual time $t$ since we passed $\phi_*(v)$.
Let $\phi(t)$ denote our latitude as a function of time. There is never any reason to fly north, since any distance traveled north would have to be traversed south again. Thus the declination should be chosen so that our latitude never increases. Since $\phi(t)$ and $S(t)$ are both monotonic (one always decreases, the other always increases), we can define a function $S(\phi)$, which is the solar time elapsed in going from latitude $\phi_*(v)$ to latitude $\phi$ over the chosen trajectory.
Then, using the Fundamental Theorem of Calculus, we can express the total solar time elapsed over the trip from $\phi_*(v)$ to $\phi_*(v)$ as an integral of the rate at which solar time elapses per change in latitude:
$S(\phi_*(v)) = \int_{\phi_*(v)}^{\phi_*(v)} \frac{d}{d \phi} S(\phi) d\phi.$To find the rate at which solar time elapses per change in latitude, apply the chain rule:
$\frac{d}{d \phi} S(\phi) = \frac{\frac{d}{dt} S(t)}{\frac{d}{dt} \phi(t)}.$Our goal now is to work out how quickly our latitude changes given a declination $\theta$ and how quickly solar time elapses.
If we fly at a declination $\theta$ south from due west, our southward velocity is $v \sin(\theta)$. Then our rate of change in latitude is:
$\frac{d}{dt} \phi = \frac{v \sin(\theta)}{R}.$If we fly at a declination $\theta$ south from due west, our westward velocity is $v \cos(\theta)$. The dawndusk boundary moves west at speed $V(\phi) = \omega R \cos(\phi)$, so solar time thus elapses at rate:
$\frac{d}{dt} S(t) = \frac{V(\phi)  v \cos(\theta)}{V(\phi)}.$Therefore, the rate at which solar time elapses per unit change in latitude is:
$\frac{d}{d \phi} S(\phi) = \frac{R}{V(\phi)} \frac{V(\phi)  v \cos(\theta)}{v \sin(\theta)}$and the total solar time elapsed over the course of the journey is:
$S(\phi_*(v)) = \int_{\phi_*(v)}^{\phi_*(v)}\frac{R}{V(\phi)} \frac{V(\phi)/v  \cos(\theta(\phi))}{\sin(\theta(\phi))} d \phi$Here the declination $\theta$ is expressed as a function of our latitude, since our optimal declination may change depending on our latitude.
An alternative approach to this optimization problem is to assume that our declination $\theta$ is constant for a small interval $[\phi,\phi  \Delta \phi]$. It would then take us $R \Delta \phi/(v \sin(\theta))$ seconds to travel from $\phi$ to $\phi  \Delta \phi$. How much solar time passes during this leg?
Finding the elapsed solar time is a little trickier because the rate at which the dawn boundary moves west changes the farther south we go. However, if we pick a small enough interval of latitude, then we can reasonably approximate the rate at which the boundary moves west as a constant equal to $V(\phi)$ for the duration of the interval. (To do this rigorously express the solar time elapsed over the interval as an integral, use the Mean Value Theorem, and take the limit as $\Delta \phi$ goes to zero.)
Dawn moves away from you at speed $V(\phi)  v \cos(\theta)$, so solar time passes at rate $(V(\phi)  v \cos(\theta))/V(\phi)$. The total solar time elapsed over the interval is equal to the amount of time taken to fly from $\phi$ to $\phi  \Delta \phi$ times the rate at which solar time passed during the flight, which equals our friend from the chain rule:
$$ \frac{R}{V(\phi)} \frac{V(\phi)  v \cos(\theta)}{v \sin(\theta)}. $$
Our goal is to minimize this integral by choosing an optimal declination function $\theta(\phi)$. Finding the optimal declination is easy since the only part of the integral we change by changing the declination $\theta$ is the integrand (i.e., the rate at which solar time elapses per unit change in latitude). Thus, if we can choose $\theta(\phi)$ to minimize the integrand at each latitude $\phi$, we will have minimized the integral.
Here we should recognize the swimmer problem introduced before. Our goal is to minimize the drift in solar time per unit latitude traveled, just as our goal before was to minimize our drift downstream per bodylength traveled across the river. Moreover, minimizing the integrand is easy because the part of the integrand that depends on $\theta$ has exactly the same form as the slope of the drifting swimmer (see the "Proof using Calculus" section). Therefore, the optimal declination of the plane is given by the same formula we derived earlier:
$\theta(\phi) = \cos^{1} \left( \frac{v}{V(\phi)} \right) = \cos^{1} \left( \frac{v}{\omega R} \sec(\phi) \right) = \cos^{1} \left( \rho \sec(\phi) \right).$The optimal declination as a function of latitude is shown in Figure 6.
Now we know what direction to point the plane given our velocity and latitude. Near $\phi_*(v)$ or $\phi_*(v)$, where we are fast relative to dawn, we fly mostly west, but near the equator, where we are slow relative to dawn, we fly mostly south. What remains is to work out how much solar time elapses for a given velocity.
To find the total solar time elapsed, substitute the optimal declination into the integral for the total solar time elapsed. With some clever rearranging the integral becomes:
$\begin{aligned} S = \frac{1}{\omega} \int_{\cos^{1}(\rho)}^{\cos^{1}(\rho)} \sqrt{ \rho^{2}  \cos(\phi)^{2}} d \phi. \end{aligned}$To get the form shown above, substitute $\theta(\phi) = \cos^{1}(\rho \sec(\phi))$ inside the integral for the solar time elapsed:
$$ \begin{aligned} S & = \int_{\phi_*}^{\phi_*} \frac{R}{V(\phi)} \frac{\frac{V(\phi)}{v}  \cos(\theta(\phi))}{\sin(\theta(\phi))} d \phi \\ & = \int_{\phi_*}^{\phi_*} \frac{R}{V(\phi)} \frac{\frac{V(\phi)}{v}  \frac{v}{V(\phi)} }{\sqrt{1  \left( \frac{v}{V(\phi)} \right)^2}} d \phi \\ & = \int_{\phi_*}^{\phi_*} \frac{R}{V(\phi)} \frac{V(\phi)}{v}\frac{1  \left( \frac{v}{V(\phi)} \right)^2 }{\sqrt{1  \left( \frac{v}{V(\phi)} \right)^2}} d \phi \\ & = \frac{R}{v} \int_{\phi_*}^{\phi_*} \sqrt{1  \left( \frac{v}{V(\phi)} \right)^2} d \phi . \end{aligned} $$
Then substitute in for $V(\phi)$ and write the bounds of integration explicitly. These steps give:
$$ \begin{aligned} S & = \frac{R}{v} \int_{\phi_*}^{\phi_*} \sqrt{1  \left( \frac{v}{\omega R} \right)^2 \sec(\phi)^2} d \phi \\ & = \frac{R}{v} \int_{\phi_*}^{\phi_*} \frac{v}{\omega R}\sqrt{ \left(\frac{\omega R}{v} \right)^2  \sec(\phi)^2} d \phi \\ & = \frac{1}{\omega} \int_{\phi_*}^{\phi_*} \sqrt{ \left(\frac{\omega R}{v} \right)^2  \sec(\phi)^2} d \phi \\ & = \frac{1}{\omega} \int_{\cos^{1}(\rho)}^{\cos^{1}(\rho)} \sqrt{ \rho^{2}  \cos(\phi)^{2}} d \phi. \end{aligned} $$
The angular velocity $\omega = 2 \pi/T$ where $T$ is the period of the earth’s rotation (24 hours). Hence the total solar time elapsed is:
$S =\frac{T}{2 \pi} \int_{\cos^{1}(\rho)}^{\cos^{1}(\rho)} \sqrt{ \rho^{2}  \cos(\phi)^{2}} d \phi.$Our goal is to keep the solar time elapsed to less than 12 hours, which is $T/2$, which requires finding the smallest $\rho$ such that:
$\int_{\cos^{1}(\rho)}^{\cos^{1}(\rho)} \sqrt{ \rho^{2}  \cos(\phi)^{2}} d \phi \leq \pi.$We have now arrived at the nondimensionalized version of the problem. If we can solve for the ratio $\rho$ that sets this integral equal to $\pi$, we can solve for the minimum speed $v$ needed to go from the North to South Pole entirely in daylight for any spherical planet (any radius $R$ and period $T$). Specifically, we are looking for $\rho$ such that:
$\int_{\cos^{1}(\rho)}^{\cos^{1}(\rho)} \sqrt{ \rho^{2}  \cos(\phi)^{2}} d \phi = \pi.$This integral can be approximated numerically using quadrature (numerical integration). The ratio $\rho$ is at least zero and at most one, so we can perform a bisection search over ratios to find the ratio $\rho$ such that the numerical approximation to this integral is close to $\pi$. This search gives the estimate $\rho = 1/2$.
Solar time elapsed (in fractions of a day) over the optimal trajectory given the velocity ratio. The elapsed time crosses half a day when the velocity ratio (ratio of our velocity to the velocity of the day–night boundary at the equator) equals a half.
If we can prove that when $\rho = 1/2$ the integral equals $\pi$, then we have proved that the smallest value of the ratio $\rho$ for which it is possible to circumnavigate the globe in daylight is $1/2$. Therefore we evaluate the integral at $\rho = 1/2$, where, as predicted, it equals $\pi$. The details are provided in the expandable below.
To prove that $\rho = 1/2$ is the slowest ratio needed to make the journey in daylight we perform the nondimensionalized integral with $\rho$ set to $1/2$:
$$ \int_{\cos^{1}(0.5)}^{\cos^{1}(0.5)} \sqrt{ 4  \cos(\phi)^{2}} d \phi = \int_{\pi/3}^{\pi/3} \sqrt{ 4  \sec(\phi)^{2}} d \phi. $$
The indefinite integral—using a computer algebra system—is:
$$ \int \sqrt{ 4  \sec(\phi)^{2}} d \phi = 2 \sin^{1}\left( \frac{2 \sin(\phi)}{\sqrt{3}} \right)  \tan^{1} \left( \frac{\sin(\phi)}{\sqrt{2 \cos(2\phi)+1}}\right). $$
Substituting in $\pi/3$, one term at a time:
$$ \begin{aligned} & \sin( \pm \pi/3) = \pm \sqrt{3}/2 \\ & \sin^{1}\left(\frac{2 \sin(\pm \pi/3)}{\sqrt{3}} \right) = \sin^{1}(\pm 1) = \pm \pi/2 \\ & \cos(\pm 2 \pi/3) = 1/2 \\ & 2 \cos(\pm 2 \pi/3) + 1 = 1 + 1 = 0 \\ & \tan^{1} \left( \frac{\sin(\pm \pi/3)}{\sqrt{2 \cos(\pm 2 \pi/3)+1}}\right) = \tan^{1}(\pm \infty) = \pm \pi/2. \\ \end{aligned} $$
Therefore:
$$ \int_{\pi/3}^{\pi/3} \sqrt{ 4  \sec(\phi)^{2}} d \phi = (2 \pi/2  \pi/2)  (2 \pi/2 + \pi/2) = \pi/2 + \pi/2 = \pi. $$
Thus if $\rho = 1/2$, then the integral equals $\pi$—exactly half a day elapses in solar time. The minimum speed must then be:
$$ v = \frac{1}{2} \omega R. \quad \blacksquare $$
We can now answer our first question. The slowest we can go and still make it from the North to South Pole entirely in daylight is half the speed of the day–night boundary at the equator. That is:
$v = \frac{1}{2} \omega R.$What a surprisingly simple answer!
It's worth highlighting the numerical component of this proof to show how one might work out an inspired guess in the first place. There are a variety of problems in math that are solved by ansatz solutions, or educated guesses. Ansatz solutions often feel like cheating since it is often unclear where the ansatz came from. The key here is that we can work out an educated guess numerically. We didn’t need to pull a miraculous answer out of thin air. Instead we trusted our machinery to lead us to a good guess before checking it.
To summarize:
 The minimum speed required to travel from the North Pole to the South Pole entirely in daylight is half the speed at which the day–night boundary moves west at the equator: $v = \frac{1}{2} \omega R$.
 The optimal trajectory for a plane traveling at speed $v = \frac{1}{2} \omega R$ is uniquely specified by settings its declination south from west at latitude $\phi$ to: $\theta(\phi) = \cos^{1}((1/2) \sec(\phi))$.
 The latitude $\phi_*(v)$ at which the plane is faster than the day–night boundary is $\cos^{1}(1/2) = \pi/3 = 60$ degrees.
In Numbers
What is $v = \frac{1}{2} \omega R$ in miles per hour? How fast do we need to fly?
The earth’s radius is 3,958.8 miles and it completes a full rotation once every 24 hours, so the day–night boundary moves at 1,036.4 mph at the equator. The minimum speed needed to fly from the North Pole to the South Pole entirely in daylight is half that speed: 518.2 mph.
Recall that if we flew due south from the North Pole we needed to travel 1,036.4 mph to make it within daylight. By optimally picking our trajectory, we halved the minimum speed needed, a reduction of over 500 mph!
What is the minimum speed needed so that flying due south from $\phi_*(v)$ to $\phi_*(v)$ is a viable trajectory?
Well, given a plane with speed $v$, the distance from $\phi_*(v)$ to $\phi_*(v)$ flying due south is $2 R \phi_*(v) = 2 R \cos^{1}(\rho)$ where $\rho = v/(\omega R)$ is the ratio of the speed of your plane to the speed of dawn at the equator. You fly at speed $v$ so the time it takes is:
$$ \frac{2 R \cos^{1}(\rho)}{v} = \frac{2}{\omega} \frac{\cos^{1}(\rho)}{\rho} $$
In order to make it before night arrives this traversal time must be less than twelve hours, or $T/2$ where $T = 24$ hours for the period of a day. Recalling $\omega = 2 \pi/T$ gives the inequality:
$$ \frac{T}{\pi} \frac{\cos^{1}(\rho)}{\rho} \leq \frac{T}{2} $$
Or:
$$ \cos^{1}(\rho) \leq \frac{\pi}{2} \rho. $$
Therefore, the minimum speed required is given by the ratio $\rho$ that satisfies $\rho = \cos(\frac{\pi}{2} \rho)$. This equation is solved (approximately) by $\rho = 0.5946$. Then the minimum speed needed to fly from the North Pole to the South Pole by flying due south between latitudes $\phi_*(v)$ and $\phi_*(v)$ is 617.3 mph, almost exactly 100 mph faster than the minimum speed needed along the optimal trajectory.
Note that the minimum speed along the optimal trajectory is exactly half the minimum speed given by flying the entire way due south, while the minimum speed given by flying due south for the range of latitudes in which our plane is slower than the dawn boundary is about sixtenths the minimum speed needed to fly the whole way due south. About fourfifths of the reduction in required speed comes from catching up with the dawn boundary at $\phi_*(v)$ before traversing the middle latitudes, then avoiding nighttime by flying away from it when south of $\phi_*(v)$. The remaining onefifth reduction in speed when using the optimal trajectory is associated with using the optimal declination to minimize the loss in solar time during the transit of the middle latitudes.
If you are south of the Arctic Circle, then the sun is up at noon no matter the time of year. We will stay just barely south of the Arctic Circle and try to fly fast enough due west so that we stay at noon in solar time. The circumference of the earth at the Arctic Circle (66 degrees latitude) is 10,117 miles. To keep up with solar time we need to make it around this circle every 24 hours, which requires going only 421.5 mph, which is about 100 mph slower than what was needed to make the trip to the South Pole.
This is a less ambitious solution to the daylight problem. It is also less satisfying since it requires staying in a plane nonstop for the entire year, which is obviously impossible. There is no plane that can fly continuously at 421.5 miles an hour for 365 days.
An advantage of the simplicity of our answer is that this same calculation could be easily repeated for any other planet. For example, the radius of Mars is 2,106.1 miles, and a Sol (one Martian day) lasts 24.6 hours, so on Mars you would only need to fly 268.8 mph. In contrast, on Jupiter you would need a plane that could fly at an astounding 13,739 mph.
How does 518.2 mph (minimum speed required on earth) compare to existing airplanes?
The solar time elapsed along the optimal trajectory associated with velocities $v$ on the earth is shown in Figure 8, along with the speed of some reference airplanes. The cruising speed of Boeing 777 is 560 mph, a 747 is 580 mph, and a 737 is 583 mph. All of these planes fly fast enough to fly from the North Pole to the South Pole entirely in daylight. If you've flown in a standard passenger jet then you've almost certainly gone fast enough to make the journey.
Can we find a plane that can fly far enough to make the trip from the North Pole to the South Pole? In other words, how long is the optimal trajectory?
We have already solved for the optimal declination $\theta(\phi) = \cos^{1}((1/2) \sec(\phi))$ as a function of latitude. The optimal declination implicitly defines the optimal trajectory by fixing the orientation of the plane at each latitude. We can convert this implicitlydefined trajectory into an explicit trajectory by writing the position of the plane as the solution to a differential equation defined by the optimal declination.
The Optimal Trajectory
To solve for the optimal trajectory we will use the optimal declination to express the motion of the plane as the solution to a pair of ordinary differential equations. This requires solving for the rate of change in the latitude and longitude of the plane given its latitude.
The rate of change in the latitude of the plane is its southward velocity divided by the radius of the earth:
$\frac{d}{dt} \phi(t) = \frac{v \sin(\theta(\phi))}{R} = \frac{\omega}{4} \sqrt{4  \sec(\phi)^2}.$The rate of change in the longitude of the plane is its westward velocity divided by the circumference of the earth at the current latitude:
$\frac{d}{dt} \text{longitude}(t) = \frac{v \cos(\theta(\phi))}{R \cos(\phi)} = \frac{\omega}{4} \sec(\phi)^2.$These equations describe the motion of the plane between $\phi_*(v)$ and $\phi_*(v)$. We can use these equations to simulate the optimal flight path numerically. Here we need to pay attention to one finicky detail. When we start at $\phi_*(v) = 60$ degrees north, our optimal declination is due west, and the solution to the differential equation is to fly endlessly around the 60th parallel, with the tip of the plane in night and the tail of the plane in day. This is obviously not the solution we want, but it is a valid solution to our equations. A simple numerical fix is to start the plane ever so slightly south of the 60th parallel, but how do we know that solutions starting slightly south of the parallel connect to solutions starting at the parallel?
You might recognize that this is an initial value problem and that the 60th parallel is an unstable equilibrium. If we start at the 60th parallel, we stay at the 60th parallel, but if we start south of the 60th parallel, we fly away from it. Usually this would mean that if we start slightly south of the 60th parallel and walk backward in time, our trajectory would spiral closer and closer to the 60th parallel without ever reaching it, so it would take us infinitely long to leave the parallel. The same problem occurs in the south. The 60th parallel south is also an equilibrium, and if we start at the 60th parallel south, the optimal solution is to continue flying west. Then symmetry would imply that our optimal trajectory spirals closer and closer to the 60th parallel south without ever reaching it. So our optimal trajectory is infinitely long?
Thankfully, we can take a finite trajectory from the 60th parallel north to the 60th parallel south that is consistent with the ODE defined by the optimal declination. The trick is that, unlike most ODEs, this system of ODEs does not admit unique solutions. Instead, there are multiple solutions consistent with the initial condition of starting at the 60th parallel north that all branch away from each other. One solution is to fly west and never leave the parallel. Another is to start due west but at some point turn the plane an infinitesimally small amount south. As soon as the plane moves south of the parallel it is locked on a unique trajectory that will carry it all the way to the 60th parallel south in a finite time.
Hold on. What’s going on here?
An ordinary differential equation is defined by relating the rate of change of a quantity to its value. In our case, $\frac{d}{dt} \phi(t) = \frac{\omega}{4} \sqrt{4  \sec(\phi)^2}.$ This equation controls the dynamics of the quantity but not the actual solution, since the solution usually depends on where the quantity starts. The actual solution (trajectory) is fixed by introducing an initial value—in our case, the initial latitude $\phi = 60$ degrees.
Solutions to ODEs never cross each other, and when given an initial value, the solution trajectories are usually, but not always, unique. Sometimes it is possible to have multiple solutions that branch out of a single initial condition.
An initial value problem (IVP) has a unique solution if, among other things, it satisfies the following criterion: Let $\phi(t)$ be a quantity that changes in time; let $f(\phi)$ control the rate of change of the quantity so that $\frac{d}{dt} \phi(t) = f(\phi)$. In order for there to be a unique solution to the IVP, $\frac{d}{dt} \phi(t) = f(\phi)$, $\phi(0) = \phi_0$, the function $f(\phi)$ must be Lipschitz continuous at $\phi_0$.
Okay, but what does that mean? It means that $f(\phi)$ can’t change too quickly around $\phi_0$. If $f(\phi)$ changes extremely quickly about $\phi_0$, tiny perturbations off of the initial condition will lead to disproportionately large changes in the rate of change of $\phi$ and multiple solutions that all branch out of the same initial condition. For example, the differential equation $\frac{d}{dt} \phi(t) = \sqrt{\phi}(t)$ has solutions $\phi(t) = 0$ and $\phi(t) = (\frac{1}{2})^{2/3} t^2$ and both solutions have initial condition $\phi(0) = 0$. This is what is happening in our case. The function $f(\phi) = \frac{\omega}{4} \sqrt{4  \sec(\phi)^2}$ is not Lipschitz continuous at $\phi = \pi/3 = 60$ degrees. Let’s prove it.
A function $f(\phi)$ is Lipschitz continuous on an interval $[a,b]$ if there exists a constant $k$ such that, for any pair of values $\phi_1, \phi_2$ both inside of $[a,b]$:
$$ \frac{f(\phi_2)  f(\phi_1)}{\phi_2  \phi_1} \leq k. $$
A function is Lipschitz continuous if the secant line connecting $f(\phi_1)$ to $f(\phi_2)$ must be bounded on the interval $[a,b]$ for any $\phi_1,\phi_2$. It follows that if $f(\phi)$ is infinitely steep at some point in the interval, $f(\phi)$ is not Lipschitz continuous on the interval.
How steep is $f(\phi) = \frac{\omega}{4} \sqrt{4  \sec(\phi)^2}$ at $\phi = \pi/3$? Go back and look at Figure 6. The optimal declination changes infinitely fast at $\phi = \phi_*(v)$ (the curves are horizontal where they meet the vertical axis), so it would not be surprising if $f(\phi) = \frac{\omega}{4} \sqrt{4  \sec(\phi)^2}$ changes infinitely fast at $\phi = 60$ degrees.
To check, take the derivative of $f(\phi)$ with respect to $\phi$:
$$ \frac{d}{d \phi} f(\phi) =  \frac{\omega}{4} \frac{\tan(\phi) \sec^2(\phi)}{\sqrt{4  \sec^2(\phi)}}. $$
Now plug in $\phi = \pi/3$. At $\pi/3$, $\cos(\phi) = 1/2$, so $\sec(\phi) = 2$. But then $4  \sec^2(\phi) = 0$. The numerator is finite, so the derivative diverges. It follows that $\frac{d}{d \phi} f(\phi)$ is infinite, so $f(\phi)$ is not Lipschitz continuous at the initial condition $\phi = \pi/3 = 60$ degrees. Therefore the system of ODEs do not necessarily admit unique solutions when starting from the 60th parallel north.
A simulated trajectory is illustrated below.
When solved numerically, it takes our plane 24 hours to complete the optimal trajectory (within numerical error). Since we are traveling at half the speed of dawn at the equator, which is fixed by the speed of rotation of the planet, the length of the optimal trajectory is half the circumference of the earth, or 12,450.5 miles. (These are nice numbers, remember them.) In fact, when solved numerically, the plane crosses the 60 degree south finish line 180 degrees west of where it started. This means that the plane arrives at 60 degree south at the exact opposite point on the planet from where it started. (This is also remarkably nice. Add it to our growing catalogue of nice facts.)
For reference, the world record longest flight (as of 2019) for a commercial airline was an experimental flight staged by Qantas from London to Sydney covering about 11,000 miles. The longest available commercial flight, operated by Qantas from Newark to Singapore, covers 9,534 miles. The Boeing 777200LR boasts the longest range of any commercial airliner at 10,808 miles. No commercial airline flies a plane far enough to complete our optimal trajectory. Damn.
Worse, the optimal trajectory only gets us from 60 degrees north to 60 degrees south^{1}. The trip from the North to South Pole would require traveling the additional 30 degrees latitude south from the North Pole to the 60 degree latitude start line and the additional 30 degrees south from the 60 degree latitude finish line. This adds at least 4,146 miles to the flight—ignoring the need to avoid night. So, while it is plausible that a commercial airliner could come somewhat close to completing the essential middle leg of the flight, no existing airliner could complete the full flight if it followed the optimal (solar time minimizing) trajectory.
To complete our story, we need to choose a trajectory above and below $\phi_*(v)$. In both cases we are free to pick any flight path that does not involve flying into the night, but it would be nice to choose a flight path that is, in some sense, optimal. A natural objective is to fly so that we minimize the rate of change of solar time at each instant. That is, we try to complete the flight while keeping the solar time as close to constant as possible. Above $\phi_*(v)$ and below $\phi_*(v)$ this can be accomplished by flying southwest, keeping the plane oriented just enough toward the west so that no solar time elapses (the plane stays at the dawn boundary in the north and at the dusk boundary in the south). This keeps a nice symmetry with the solution between $\phi_*(v)$ and $\phi_*(v)$ where we can’t help but lose solar time and is a natural solution under other objectives.
For example, once beneath $\phi_*(v)$ we might like to get to the South Pole as quickly as possible (in actual time, thus actual distance covered), however we cannot fly due south since we necessarily arrive at $\phi_*(v)$ exactly at the same time as dusk. If we flew due south we would fly into dusk. We are thus forced to orient our plane at least partially westward. To minimize the time it takes to get to the South Pole we want to maximize the rate at which we travel south (rate of change in latitude). Therefore we want to maximize the southern component of our velocity, $v \sin(\theta)$, while keeping the plane pointed just enough west to keep up with dusk. This requires setting $v \cos(\theta)$ equal to the westward velocity of dusk $V(\phi)$. This means that, south of $\phi_*(v)$, we need to fly with declination:
$$ \theta(v,\phi) = \cos^{1} \left( \frac{V(\phi)}{v} \right) = \cos^{1} \left( \frac{1}{\rho} \cos(\phi) \right). $$
The same declination would keep us traveling along the dawn boundary when north of $\phi_*(v)$. Therefore, under the minimalchangeinsolartime objective our full trajectory is fixed by:
$$ \theta(v,\phi) = \left\{ \begin{aligned} & \cos^{1} \left( \frac{1}{\rho} \cos(\phi) \right) & \text{ if } & \phi \geq \phi_*(v) \\ & \cos^{1} \left( \frac{\rho}{ \cos(\phi)} \right) & \text{ if } & \phi \in [\phi_*(v),\phi_*(v)] \\ & \cos^{1} \left( \frac{1}{\rho} \cos(\phi) \right) & \text{ if } & \phi \leq \phi_*(v) \\ \end{aligned} \right\} $$
where $\phi_*(v) = \cos^{1}(\rho)$ and $\rho = \frac{v}{\omega R} \geq 1/2$. Notice the symmetry of the solution. Above $\phi_*(v)$ and below $\phi_*(v)$ we use the same heading, and between the two we use the same type of heading rule, only the fraction inside the inverse cosine is flipped.
The resulting trajectory is animated below.
We can now describe the trajectory in full. At the North Pole we fly south along the dawn boundary. We follow the boundary, turning farther west to keep up with the sun. By the time we reach $\phi_*(v) = 60$ degrees north we are flying due west. At this point we are forced to turn south, since we must reach the South Pole and there is no longer any hope of keeping up with the boundary. The farther south we go, the more we turn the nose of the plane southward, since, as we approach the equator, we become slower and slower relative to the westward motion of daytime. At the equator we reach a maximum southerly declination of $\cos^{1}(1/2) = \pi/3$ or 60 degrees. As we pass the equator we orient the plane west, until, at $\phi_*(v) = 60$ degrees south, we fly due west again, now with the tip of the plane in daylight and the tail in night. Now for the home stretch. To reach the South Pole we tip the nose south and fly along the dusk boundary until we reach the South Pole.
By now you might be wondering what this trajectory looks like on a globe. The flight is animated below.
A (Suspiciously) Beautiful Solution
This was where I originally stopped. After all, the questions we asked at the start are answered:
 The minimum speed is half the speed of the equatorial day–night boundary.
 The special latitudes $\phi_*(v)$ are at 60 degrees north and south.
 The maximum declination of the plane occurs at the equator and equals 60 degrees south from west.
 It takes exactly one day (24 hours in real time) to fly the optimal trajectory from 60 degrees north to 60 degrees south.
 The length of the middle segment of the optimal trajectory from 60 degrees north to 60 degrees south is exactly equal to half the circumference of the earth.
 The plane crosses the 60 degree south line 180 degrees west of where it crosses the 60 degrees north line (on the exact opposite side of the world).
These answers are almost too good to be true. Worked out one at a time they seem like a string of miraculous coincidences.
Enter Jack. If I’d found the pearls, Jack wanted the necklace. These results are each too nice to be unrelated. There has to be underlying logic that strings them together.
Toward the Big Picture
What’s missing here is a big picture understanding of the solution. Our answers are all based on optimizing our declination at a given instant in time, so our solution is always based on a local understanding of the best heading given our latitude. We can use this information to reconstruct an optimal trajectory, but we still don’t have an explicit formula for that trajectory. Instead we have an implicitlydefined trajectory in the form of a set of instructions for how to orient the plane. The full picture requires the optimal trajectory explicitly.
We could try to solve the pair of differential equations defining the optimal trajectory directly, but these are remarkably nasty differential equations. If we try to approach the problem head on we reach a dead end.
We need a different way to think about this problem. The calculusbased approach is powerful and straightforward, but it hasn’t provided a full understanding of the problem. We chose to start by presenting the solution using calculus since it didn’t require any singularly brilliant insights. All it took was grit and patience. But grit and patience can only get us so far. Now we need inspiration.
This is a common challenge in math puzzles. Good puzzles often have clever, fundamental solutions that require thinking outside the box. This sort of thinking can be difficult. New ideas don’t appear out of the ether like fictitious leopards; new ideas come from old ideas. The answers we’ve reached are clues, we just need to put them together. . . .
A Bit of Circular Logic
Let’s review our clues.
One: we know that the minimum speed needed is half the speed at which daylight moves around the equator. Why half? Unclear. We’ll come back to this.
Two: the upper and lower latitudes $\phi_*$ at which our plane can outrun daylight are 60 degrees. This is nice, but not informative by itself.
Three: our maximum declination is 60 degrees south of west. A nice fact but not too informative.
The next three facts are more useful. The middle segment of our optimal trajectory takes 24 hours to fly, has length equal to half the circumference of the earth, and travels exactly halfway around the world (180 degrees west in order to go from 60 north to 60 south). Those last two points are interesting. Imagine putting a pin where our optimal trajectory crosses the 60 degrees north line, and a pin where it reaches the 60 degrees south line. Then those two pins would be on exact opposite sides of the globe. The path we take between them has length equal to half the circumference of the globe, so it must be the shortest path between these points. The shortest path between any two points on a globe is a great circle, so our path must be a great circle!
A great circle is the circle formed by passing a plane through the center of a sphere. If you pick any two points on the globe, the shortest path between them will be a segment of the great circle defined by the plane passing through those two points and the center of the globe.
The great circle segment connecting two points. A globe is shown in grey. Two points on the globe are shown in blue, and the center of the globe is shown in red. A plane connecting the center to the two points on the globe is shown in purple. The intersection of the purple plane with the globe defines a great circle, shown in blue. The segment connecting the two blue points is the shortest path between them.
We have to be careful not to get ahead of ourselves here. The last three “facts” were not facts at all—they were observations based on numerical results. We don’t know for sure that the optimal path crosses 60 degrees south 180 degrees west of where it crossed 60 degrees north, nor do we know that it took exactly 24 hours to fly (and, as a consequence, had length equal to half the earth’s circumference). Thus we haven’t proven that the optimal path is a great circle. We just have a (very) good reason to think it is. In other words, we have another ansatz to check (proof below).
In order to prove that our trajectory is a great circle we need a formula for the latitude and longitude of a great circle starting at 60 degrees north and arriving at 60 degrees south 180 degrees west of where we started. Any great circle can be expressed by the formula:
$$ \text{latitude}(\text{longitude}) = \tan^{1}(\gamma \sin(\text{longitude}  l_0)) $$
where $\gamma$ controls the farthest north the great circle reaches and $l_0$ is the longitude at which it crosses the equator. Open the next expandable to see why this equation works. Since we don’t care about where our trajectory crosses the equator we can let $l_0 = 0$. We also know that at 60 degrees north our trajectory is tangent to a latitude line (points due west), and that it reaches 60 degrees south at the exact opposite side of the world. This means that, if the optimal trajectory is half of a great circle, the farthest north the great circle can reach is 60 degrees. This fixes $\gamma = \sqrt{3}$. If our trajectory is a great circle, the latitude and longitude must satisfy:
$$ \text{latitude}(\text{longitude}) = \tan^{1}(\sqrt{3} \sin(\text{longitude})). $$
Or, inverting this equation:
$$ \text{longitude}(\text{latitude}) = \sin^{1} \left(\frac{1}{\sqrt{3}} \tan(\text{latitude}) \right). $$
This is an explicit formula for the optimal trajectory, but it is only an educated guess. To prove that this is a solution, we need to show that it satisfies the system of ODEs defining the optimal trajectory.
The system of ODEs defining the optimal trajectory were:
$$ \begin{aligned} & \frac{d}{dt} \text{latitude}(t) = \frac{\omega}{4} \sqrt{4  \sec(\text{latitude})^2} \\ & \frac{d}{dt} \text{longitude}(t) = \frac{\omega}{4} \sec(\text{latitude})^2. \end{aligned} $$
Using the chain rule:
$$ \frac{d \text{longitude}}{ d \text{latitude}} = \frac{\sec(\text{latitude})^2}{\sqrt{4  \sec(\text{latitude})^2}}. $$
The question is, does the great circle satisfy this equation? Differentiating a piece at a time:
$$ \begin{aligned} & \frac{d}{dx} \sin^{1} (x) = \frac{1}{\sqrt{1  x^2}} \\ & \frac{d}{dx} \tan(x) = \sec^2(x). \end{aligned} $$
so:
$$ \frac{d}{d \text{latitude}} \sin^{1} \left(\frac{1}{\sqrt{3}} \tan(\text{latitude}) \right) = \frac{1}{\sqrt{1  \frac{1}{3} \tan(\text{latitude})^2}} \frac{1}{\sqrt{3}} \sec(\text{latitude})^2. $$
Simplifying:
$$ \begin{aligned} \frac{d}{d \text{latitude}} \sin^{1} \left(\frac{1}{\sqrt{3}} \tan(\text{latitude}) \right) & = \frac{\sec(\text{latitude})^2}{\sqrt{3  \tan(\text{latitude})^2}} \\ & = \frac{\sec(\text{latitude})^2}{\sqrt{3  (\sec(\text{latitude})^2  1)}} \\ & = \frac{\sec(\text{latitude})^2}{\sqrt{4  \sec(\text{latitude})^2}}. \end{aligned} $$
The great circle passing from 60 degrees north to 60 degrees south and traveling 180 degrees west in the process satisfies the system of ODEs defining the optimal trajectory and has the same initial conditions. This great circle is therefore an optimal trajectory. If we fix the longitude at which we leave 60 degrees north, then the great circle is the unique optimal trajectory since the system of ODEs admits unique solutions for all latitudes except 60 degrees north and south. $\blacksquare$
It is tempting to think that we might have been able to prove that the optimal trajectory was a great circle without relying on all this calculus. This is, at best, unlikely, since all of our work was based on locally optimizing our declination. The only way we could show that a great circle was the solution was to prove that it matched the optimal declination at each latitude. Had we assumed a priori that the optimal solution had to be a great circle we could have jumped straight to finding the great circle that worked at the slowest possible speed, but the only way to know that the best trajectory is a great circle is to work out what conditions the optimal trajectory must satisfy. These conditions are given by solving for the optimal declination.
In the proof provided above we claimed that any great circle can be expressed using the formula:
$$ \text{latitude}(\text{longitude}) = \tan^{1}(\gamma \sin(\text{longitude}  l_0)) $$
Let’s work out why.
A great circle is defined as the intersection of a sphere with a plane passing through its center. So far we’ve been working entirely in spherical coordinates. Since we now need to add a plane we should introduce Cartesian coordinates as well. Put the origin of the coordinate system at the center of the sphere and let the $z$ axis point north. Any plane can then be written in the general form $a x + b y + c z = d$. Since the plane passes through the origin (center of the sphere), the coordinates $\langle 0,0,0 \rangle$ have to solve the plane equation. This requires $d = 0$. Since $d = 0$ we can rewrite the plane in the form:
$$ z = \frac{a}{c} x  \frac{b}{c} y = A x + By $$
for the appropriate choice of constants $A$ and $B$. Now convert from Cartesian coordinates to spherical coordinates with the caveat that, since we are working with latitude and longitude, the azimuthal angle is measured off of the $x$$y$ plane, not down from the $z$ axis:
$$ r \sin(\text{latitude}) = A r \cos(\text{latitude}) \cos(\text{longitude}) + B r \cos(\text{latitude}) \sin(\text{longitude}) $$
Cancelling the common factors of $r$ and dividing across by $\cos(\text{latitude})$ leaves:
$$ \tan(\text{latitude}) = A \cos(\text{longitude}) + B \sin(\text{longitude}). $$
Now, for any constants $A$ and $B$, the linear combination $A \cos(t) + B \sin(t)$ is still a trigonometric function and can be rewritten as a scaled and shifted sine curve: $\gamma \sin(t  t_0)$. It follows immediately that, for the right choice of $\gamma$ and $l_0$:
$$ \text{latitude} = \tan^{1} \left( \gamma \sin(\text{longitude}  l_0) \right). $$
The Big Picture
Now we can see the big picture explicitly. The flight consists of three parts: First you fly south from the North Pole until you reach 60 degrees north. Then you fly along the shortest possible (southwestardly) route to the opposite point on the planet. If you arrange to leave 60 degrees north at dawn you arrive at 60 degrees south at dusk. From there you fly along the dusk boundary to the South Pole.
We now have a much simpler understanding of the optimal flight path. For example, the whole business with uniqueness of solutions is clearly explained by the great circle solution. The great circle is tangent to the 60 degrees north and 60 degrees south lines where it meets them, so if we fly along the great circle starting from 60 degrees north we really do start by flying due west, then slowly peel away from the 60th parallel. We can use this understanding to string together each of our surprisingly nice results.
First, why does half the speed of dawn at the equator work? And why all the 60 degrees ($\phi_*$ and optimal declination at the equator)?
Suppose that instead of trying to fly from the North Pole to the South Pole in daylight we simply tried to fly between two points on the equator but on opposite sides of the world. Then, if we fly west at dawn we need to go a halfcircumference of the earth, but we have 24 hours to do it since we pick up 12 hours of daylight along the way. Therefore, if we fly at half the speed of dawn at the equator we can reach a point exactly opposite to where we started before being caught by nightfall.
Imagine that flight path. It’s half of a circle wrapping around the equator. Imagine tilting that circle $\phi$ degrees off the equator so that it starts at latitude $\phi$, and arrives at $\phi$ on the other side of the world. The half circle hasn’t changed length since all we’ve done is rotated it, and the total longitude traversed hasn’t changed either since the starting and ending points are still on opposite sides of the globe. If we fly at halfequatorial speed and leave from the northernmost point at dawn, we still make it to the southernmost point at dusk, no matter how far we tilt the circle off the equator.
Tilting a great circle. The equator and equatorial plane are shown in red. Two points on opposite sides of the earth are shown in blue. The two points start on the equator, and are contained in a blue plane. Half of a great circle corresponding to the shortest (south westward) path between the two points is also shown in blue. This represents a proposed flight path. Notice that the length of the flight path does not change when the plane is tilted up from the equator.
Thus, if we fly west along a great circle between any two points on opposite sides of the world at half equatorial speed and leave the northernmost point at dawn, we can arrive at the southernmost point at dusk. Making it to the other side by dusk is half the battle. We still need to connect the northernmost and southernmost points of the circle to the Poles. At half earth speed we know that we can make it from the North Pole to 60 degrees north at dawn, and from 60 degrees south we can make it to the South Pole provided we arrive at 60 degrees south before night. If we tilt our great circle 60 degrees up from the equator, we can follow the great circle from 60 degrees north to the opposite side of the world at 60 degrees south—leaving at dawn and arriving at dusk. Since we are always flying west slower than the dawn boundary we will never fly into the night, and since dusk doesn’t catch us until we reach 60 degrees south, we spend the entire great circle portion of the trip in the daylight.
This proves that it is possible to make the journey at half the speed of equatorial dawn. Since we were looking for the slowest speed possible, there is no need to look for faster speeds. It doesn’t answer why we can’t go slower. It also doesn’t answer why we can’t keep tipping the great circle north past 60 degrees.
Why can’t we tip the great circle past 60 degrees? Notice that, at the northernmost part of the great circle we are always flying due west. Thus if we start at the dawn boundary flying due west, we need to make sure that we don’t start by flying straight into the night. Below 60 degrees, dawn moves west faster than we fly. Above 60 degrees, we fly faster than dawn moves west, so if we fly due west at dawn we will fly straight into the night. In the extreme case where we tilt the great circle all the way up to 90 degrees (so it passes through the North Pole) we would still fly west, straight into midnight. Giving ourselves a head start by flying into night is cheating, so we can’t tip the great circle past 60 degrees if we fly at half earth speed. The same is true for the angle $\phi_*(v)$ for an arbitrary speed $v$. If you fly at speed $v$ you cannot tilt past $\phi_*(v)$ without starting out by flying into night. If we are looking for solutions where the middle leg consists of a half great circle, the farthest north the great circle leg can start is $\phi_*(v)$.
Seven flight paths (marked in blue) starting from 0, 15, 30, 45, 60, 75, and 90 degrees north. All of the flight paths are halves of great circles, and traverse half the globe. All paths start at dawn and all planes (blue circles) fly at onehalf the speed of dawn at the equator. The red circles mark the equator and 60 degrees south and north. The black circle marks the boundary between day and night. Note that planes starting north of 60 degrees start by flying into the night, and are caught by the night again before finishing their flight paths.
Flying at half earth speed, the most we can tilt the great circle is 60 degrees, so we start the great circle leg from 60 degrees north, reach a maximum declination of 60 degrees, and finish the great circle leg at 60 degrees south.
Why can’t we go any slower than half equatorial speed?
If we consider all possible trajectories, we would need to go back to our calculusbased approach. If, however, we limit our possible flight paths, we can answer this question with simple geometry.
Let’s restrict our attention to flight paths that consist of:
 flying south from the North Pole along the dawn boundary until we reach $\phi_*(v)$.
 taking a great circle west from $\phi_*(v)$ to the opposite point on the globe.
 flying south along the dusk boundary to the South Pole.
For what range of speeds $v$ is this plan viable? We already know that it works if $v$ is set to half equatorial speed, and it is obvious that it will work for faster speeds since if we fly faster we will complete the great circle leg in less than 12 solar hours. However, if we fly slower than half equatorial speed we will take more than 24 hours to complete the great circle leg, so we will take more than 12 solar hours to get to $\phi_*(v)$. This means that, even if we leave at dawn, night will catch us before we get far enough south. For this limited class of paths, half equatorial speed is the slowest we could possibly fly^{2}.
Afterword
There are a couple takeways here. First, while it took some patience to reach the answers, the answers are surprisingly simple and strikingly elegant. Second, inspiration is usually bought with perseverance. Use familiar tools to work at a problem until you get enough clues to see it from a new perspective. It might take all day, but if you fly fast enough, you can make that day last as long as you want.

It gets even worse. No land on earth exists at 60 degrees south. Even if we wanted to land and call our journey a success, we would need a seaplane.
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We could consider other more general classes of flight paths. For example, we could change the middle leg to a great circle, leaving $\phi_*(v)$ north and arriving at $\phi_*(v)$ south but without necessarily going 180 degrees west. We could then search for the optimum number of degrees west to fly for a given speed $v$ and subsequently for the minimum $v$ needed to complete the journey. This requires a numericallyinspired guessandcheck method much like what we used before, so it doesn’t provide much new insight.
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