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.

Figure 1:
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:

  1. What is the minimum speed needed to fly from the North Pole to the South Pole (on the equinox) to remain entirely in daylight?
  2. 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 absent-minded 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:

  1. The earth is a perfect sphere and rotates about a fixed axis passing through the North and South Poles.
  2. The day–night boundary is discrete (no refraction of light).
  3. 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 Challenge: Reach the South Pole from the North Pole in daylight. The difficulty level corresponds to the your plane’s speed. The speed is constant, but appears to change due to the distortion of the map. Use your arrow keys to turn the plane.

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 high-speed military jet that can be refueled mid-flight, 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 day-night 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 vv and the latitude at which your plane’s speed equals the westward motion of the day–night boundary ϕ(v)\phi_*(v). The real challenge is to get from ϕ(v)\phi_*(v) degrees north to ϕ(v)\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.

Speed of Plane (mph)2505007501000Longitude-90-60-300306090North PoleSouth Polefaster than dawnfaster than dawn
Speed of Dawn: On this graph, the y-axis represents latitude and the x-axis represents velocity. The bold red curve is the speed of dawn at a given a latitude. The blue vertical is the speed of the plane and the horizontal purple lines are $\phi_*$. The shaded regions are the latitudes at which the plane is faster than dawn.

This leads to an alternative proposal: for a plane with a max speed vv, fly south from the North Pole to the latitude ϕ(v)\phi_*(v) along some path that avoids flying into the night and ensures that you reach latitude ϕ(v)\phi_*(v) at the dawn boundary. Then fly due south until you reach ϕ(v)\phi_*(v) degrees south. As long as you arrive at ϕ(v)\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.

Figure 2: Swimmers attempt to cross a river. All swimmers (colored dots) swim at the same speed but at different angles against the current. The arrow leaving each dot is the direction they swim. The lines behind the swimmers mark their path. Blue swimmers swim mostly upstream (10 degrees across from up), while red swimmers swim mostly across (70 degrees across from up). The rivers get faster from left to right. The rightmost river is more than twice as fast as the leftmost river. Notice that the slower the river the more successful the blue swimmer and the faster the river the more successful the red swimmer. In order from fast to slow, the best swimmer is the swimmer who swims at angle 10 degrees, 30 degrees, 50 degrees, and 70 degrees from upstream.

Let VV denote the speed of the river and v<Vv < 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 body-length 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 xx denote the direction across the river and yy denote downstream direction. Let θ\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][0,-V], and the velocity at which you swim, v[sin(θ),cos(θ)]v [\sin(\theta), -\cos(\theta)]. Vectors add tip-to-tail, so we can represent your drift velocity as a triangle with a downstream leg of length VV plus a shorter leg of length vv oriented in the direction you swim. These are the blue and red vectors shown in Figure 3.

Figure 3: The drift velocity of a swimmer swimming across a river. The red arrow represents the swimmer’s velocity, the blue arrow represents the river’s velocity, and the purple arrow is the swimmer’s drift velocity, which is the sum of the river’s velocity $V$ and swimmer’s velocity $v$. The direction and length of the drift velocity depends on the declination of the swimmer, marked in red as the angle $\theta$. If the swimmer swims in a different direction, the resulting drift velocity will change. The purple circle represents all possible drift velocities. The dashed red and purple lines show the resulting drift velocity if the swimmer swam at a different declination. The declination shown is optimal since it minimizes the slope of the purple line.

The purple circle in Figure 3 represents all possible drift velocities given all possible declinations θ\theta. Our goal is to minimize the angle between the drift velocity and the horizontal direction xx—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 VV, swimmer velocity vv, and drift velocity is a right triangle. It follows that cos(θ)=v/V\cos(\theta) = v/V, so the optimal declination is given by:

θ(v,V)=cos1(vV).\theta(v,V) = \cos^{-1} \left( \frac{v}{V} \right).

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/Vv/V, so the smaller the angle. More simply, the faster we are, the more we should swim upstream.

Swimmer Declination 2
Figure 4:
The optimal declination of the swimmer relative to the upstream direction as a function of the ratio of the swimmer’s speed vv to the speed of the river VV. 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 ϕ\phi. Let RR be the radius of the earth. Let r(ϕ)r(\phi) be the distance between us and the earth’s axis of rotation. Then:

r(ϕ)=Rcos(ϕ).r(\phi) = R \cos(\phi).

The earth spins at ω=2π24rad/hr\omega = \frac{2 \pi}{24} \text{rad/hr}. Therefore the speed of the day–night boundary at latitude ϕ\phi is:

V(ϕ)=ωRcos(ϕ).V(\phi) = \omega R \cos(\phi).

Recall that there are regions near the poles where the plane is faster than the day-night boundary. The bounding latitude, ϕ\phi_*, is:

ϕ(v)=cos1(vωR).\phi_*(v) = \cos^{-1}\left( \frac{v}{\omega R}\right).

Let ρ\rho be the ratio v/(ωR)v/(\omega R). Then ρ\rho is the ratio of our velocity to the velocity we would need to fly to keep up with daylight at the equator and ϕ=cos1(ρ)\phi_* = \cos^{-1}(\rho).

Figure 5: The speed at which daylight moves west as a function of latitude $\phi$. The speed is given in miles per hours on the horizontal axis and is represented by the blue curve. The orange line is the cruising speed of a Boeing 747. The orange shaded regions show where the plane travels faster than the day–night boundary.

Because we can outrun night indefinitely anywhere within ϕ\phi_* of either pole, it is enough to be able to fly from ϕ\phi_* to ϕ- \phi_* in daylight in order to fly from the North Pole to the South Pole in daylight.

Between ϕ\phi_* and ϕ-\phi_*, we are slower than the day-night 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 vv we seek the trajectory that minimizes the elapsed solar time in traveling from latitude ϕ\phi_* to ϕ-\phi_*.

Let θ\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 θ\theta as a function of latitude ϕ\phi uniquely specifies a trajectory. The declination θ(ϕ)\theta(\phi) defines this trajectory implicitly and can be thought of as a bearing given to the pilot at each latitude (fly θ\theta degrees south of west when at latitude ϕ\phi).

The solar time elapsed constantly increases while we are between ϕ(v)\phi_*(v) and ϕ(v)-\phi_*(v) since we are slower than the dawn-dusk boundaries in that range. Let S(t)S(t) denote the solar time elapsed at the actual time tt since we passed ϕ(v)\phi_*(v).

Let ϕ(t)\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 ϕ(t)\phi(t) and S(t)S(t) are both monotonic (one always decreases, the other always increases), we can define a function S(ϕ)S(\phi), which is the solar time elapsed in going from latitude ϕ(v)\phi_*(v) to latitude ϕ\phi over the chosen trajectory.

Then, using the Fundamental Theorem of Calculus, we can express the total solar time elapsed over the trip from ϕ(v)\phi_*(v) to ϕ(v)-\phi_*(v) as an integral of the rate at which solar time elapses per change in latitude:

S(ϕ(v))=ϕ(v)ϕ(v)ddϕS(ϕ)dϕ.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:

ddϕS(ϕ)=ddtS(t)ddtϕ(t).\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 θ\theta and how quickly solar time elapses.

If we fly at a declination θ\theta south from due west, our southward velocity is vsin(θ)v \sin(\theta). Then our rate of change in latitude is:

ddtϕ=vsin(θ)R.\frac{d}{dt} \phi = -\frac{v \sin(\theta)}{R}.

If we fly at a declination θ\theta south from due west, our westward velocity is vcos(θ)v \cos(\theta). The dawn-dusk boundary moves west at speed V(ϕ)=ωRcos(ϕ)V(\phi) = \omega R \cos(\phi), so solar time thus elapses at rate:

ddtS(t)=V(ϕ)vcos(θ)V(ϕ).\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:

ddϕS(ϕ)=RV(ϕ)V(ϕ)vcos(θ)vsin(θ)\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(ϕ(v))=ϕ(v)ϕ(v)RV(ϕ)V(ϕ)/vcos(θ(ϕ))sin(θ(ϕ))dϕS(-\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 θ\theta is expressed as a function of our latitude, since our optimal declination may change depending on our latitude.

Our goal is to minimize this integral by choosing an optimal declination function θ(ϕ)\theta(\phi). Finding the optimal declination is easy since the only part of the integral we change by changing the declination θ\theta is the integrand (i.e., the rate at which solar time elapses per unit change in latitude). Thus, if we can choose θ(ϕ)\theta(\phi) to minimize the integrand at each latitude ϕ\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 body-length traveled across the river. Moreover, minimizing the integrand is easy because the part of the integrand that depends on θ\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:

θ(ϕ)=cos1(vV(ϕ))=cos1(vωRsec(ϕ))=cos1(ρsec(ϕ)).\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.

Figure 6: The optimal declination (degrees south from west) as a function of the latitude of the plane for planes with velocities ranging from one-tenth the speed of the day–night boundary at the equator to nine-tenths the speed of the day–night boundary at the equator.

Now we know what direction to point the plane given our velocity and latitude. Near ϕ(v)\phi_*(v) or ϕ(v)-\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:

S=1ωcos1(ρ)cos1(ρ)ρ2cos(ϕ)2dϕ.\begin{aligned} S = \frac{1}{\omega} \int_{\cos^{-1}(\rho)}^{-\cos^{-1}(\rho)} \sqrt{ \rho^{-2} - \cos(\phi)^{-2}} d \phi. \end{aligned}

The angular velocity ω=2π/T\omega = 2 \pi/T where TT is the period of the earth’s rotation (24 hours). Hence the total solar time elapsed is:

S=T2πcos1(ρ)cos1(ρ)ρ2cos(ϕ)2dϕ.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/2T/2, which requires finding the smallest ρ\rho such that:

cos1(ρ)cos1(ρ)ρ2cos(ϕ)2dϕπ.\int_{-\cos^{-1}(\rho)}^{\cos^{-1}(\rho)} \sqrt{ \rho^{-2} - \cos(\phi)^{-2}} d \phi \leq \pi.

We have now arrived at the non-dimensionalized version of the problem. If we can solve for the ratio ρ\rho that sets this integral equal to π\pi, we can solve for the minimum speed vv needed to go from the North to South Pole entirely in daylight for any spherical planet (any radius RR and period TT). Specifically, we are looking for ρ\rho such that:

cos1(ρ)cos1(ρ)ρ2cos(ϕ)2dϕ=π.\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 ρ\rho is at least zero and at most one, so we can perform a bisection search over ratios to find the ratio ρ\rho such that the numerical approximation to this integral is close to π\pi. This search gives the estimate ρ=1/2\rho = 1/2.

Time Lost Given Ratio
Figure 7:
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 ρ=1/2\rho = 1/2 the integral equals π\pi, then we have proved that the smallest value of the ratio ρ\rho for which it is possible to circumnavigate the globe in daylight is 1/21/2. Therefore we evaluate the integral at ρ=1/2\rho = 1/2, where, as predicted, it equals π\pi. The details are provided in the expandable below.

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=12ωR.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:

  1. 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=12ωRv = \frac{1}{2} \omega R.
  2. The optimal trajectory for a plane traveling at speed v=12ωRv = \frac{1}{2} \omega R is uniquely specified by settings its declination south from west at latitude ϕ\phi to: θ(ϕ)=cos1((1/2)sec(ϕ))\theta(\phi) = \cos^{-1}((1/2) \sec(\phi)).
  3. The latitude ϕ(v)\phi_*(v) at which the plane is faster than the day–night boundary is cos1(1/2)=π/3=60\cos^{-1}(1/2) = \pi/3 = 60 degrees.

In Numbers

What is v=12ωRv = \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!

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 vv 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.

Time Lost on Earth
Figure 8: Solar time elapsed (in hours) on earth as a function of velocity for planes following optimal trajectories. Any plane that travels faster than 518.2 mph takes less than 12 hours to complete its trip, and can fly from the North to South Pole entirely in daylight.

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 θ(ϕ)=cos1((1/2)sec(ϕ))\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 implicitly-defined 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:

ddtϕ(t)=vsin(θ(ϕ))R=ω44sec(ϕ)2.\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:

ddtlongitude(t)=vcos(θ(ϕ))Rcos(ϕ)=ω4sec(ϕ)2.\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 ϕ(v)\phi_*(v) and ϕ(v)-\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 ϕ(v)=60\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.

A simulated trajectory is illustrated below.

Around the Earth

Figure 9: The optimal trajectory for a plane traveling at the minimum speed needed to complete the trip from the North to South Pole entirely in daylight. The trajectory is shown in cyan, starting from 60 degrees north and finishing at 60 degrees south. The day and night boundaries are in yellow. Notice that the optimal trajectory starts parallel to dawn and finishes parallel to dusk. The blue shaded region is night. The panel in the upper right shows the declination of the plane as it flies south. The panel in the lower right shows the solar time elapsed. Notice that the plane arrives at 60 degrees south just at 12 hours of elapsed solar time.

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 777-200LR 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 south1. 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 ϕ(v)\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 ϕ(v)\phi_*(v) and below ϕ(v)-\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 ϕ(v)\phi_*(v) and ϕ(v)-\phi_*(v) where we can’t help but lose solar time and is a natural solution under other objectives.

The resulting trajectory is animated below.

Figure 10: The complete optimal trajectory for a plane traveling at the minimum speed needed to complete the trip from the North to South Pole entirely in daylight. The trajectory is red, the plane a blue dot, and the shaded region night. The dashed line marks the 60th parallels. Once the plane reaches the South Pole it turns around and uses the same technique to fly back to the North Pole without entering night.

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 ϕ(v)=60\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 cos1(1/2)=π/3\cos^{-1}(1/2) = \pi/3 or 60 degrees. As we pass the equator we orient the plane west, until, at ϕ(v)=60-\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.

Figure 11: The complete optimal trajectory for a plane traveling at the minimum speed needed to complete the trip from the North to South Pole entirely in daylight. The trajectory is red, the plane is a blue dot, and the shaded region night. The solid circles in the Northern and Southern Hemispheres mark the 60th parallels above and below which the plane can fly west faster than night moves. Once the plane reaches the South Pole it turns around and uses the same technique to fly back to the North Pole without entering night.

A (Suspiciously) Beautiful Solution

This was where I originally stopped. After all, the questions we asked at the start are answered:

  1. The minimum speed is half the speed of the equatorial day–night boundary.
  2. The special latitudes ϕ(v)\phi_*(v) are at 60 degrees north and south.
  3. The maximum declination of the plane occurs at the equator and equals 60 degrees south from west.
  4. It takes exactly one day (24 hours in real time) to fly the optimal trajectory from 60 degrees north to 60 degrees south.
  5. 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.
  6. 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 implicitly-defined 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 calculus-based 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 ϕ\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.

Figure 12:
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).

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 (ϕ\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 half-circumference 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 ϕ\phi degrees off the equator so that it starts at latitude ϕ\phi, and arrives at ϕ-\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 half-equatorial 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.

Figure 13:
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 ϕ(v)\phi_*(v) for an arbitrary speed vv. If you fly at speed vv you cannot tilt past ϕ(v)\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 ϕ(v)\phi_*(v).

Figure 14:
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 one-half 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 calculus-based 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:

  1. flying south from the North Pole along the dawn boundary until we reach ϕ(v)\phi_*(v).
  2. taking a great circle west from ϕ(v)\phi_*(v) to the opposite point on the globe.
  3. flying south along the dusk boundary to the South Pole.

For what range of speeds vv is this plan viable? We already know that it works if vv 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 ϕ(v)-\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 fly2.


There are a couple take-ways 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.

  1. 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.

  2. We could consider other more general classes of flight paths. For example, we could change the middle leg to a great circle, leaving ϕ(v)\phi_*(v) north and arriving at ϕ(v)\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 vv and subsequently for the minimum vv needed to complete the journey. This requires a numerically-inspired guess-and-check method much like what we used before, so it doesn’t provide much new insight.

© 2021, Built by Alex and Jack Strang