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New Navigation Problem Solutions & Hint 1

I planned only to lurk, but this group seems to have prematurely dropped, or greatly transformed, the original navigation problem. (Go south one mile, then east one mile, then north one mile to return to your starting point.) In addition to the "North Pole" and "points near South Pole" solutions there are several other solutions. Two solutions are given in this posting, but one of them is subtlety flawed; however, a technologically impossible extension of the valid solution is also mentioned, so two new solutions are, in fact, given below.

I will soon post another valid, technically feasible, but extremely expensive solution, which permits starting from any point south of the Tropic of Cancer, if no one else posts it before I have sequentially posted four hints about it. (Hint 1 is given below.) Another distinct "set of solutions" also exists. Each member of the set differs from the others only in details as to how to implement the same irresponsible idea. None of my four hints relate to this idea, but calling it "irresponsible" is a hint. All solutions result from my attempt to "think outside the box." Some may think I went too far outside, especially with this solution set, despite it not being prohibited by the problem statement and the traveler traversing only the three specified legs. If no one guesses this set, I will post it after the "expensive solution" has been posted.

"Thinking outside the box" requires conscious examination of your mental logic to avoid self-imposed presumptions and to take full advantage of ambiguities in the problem statement. Thus I begin by discussing the problem’s travel instructions to expose some common presumptions, or at least to clearly state my interpretation of the problem’s actual requirements.

Each latitude circle defines a plane. For me, "going south" means crossing latitude circle planes, while remaining in only one longitude circle plane, and always going across latitude planes in the direction from north towards south. Thus, going south does not confine one to traveling on a great circle path on the surface of an assumed spherical Earth. (If you think it does, then I have no new solutions to offer you.) For example, I think an airplane can "go south" even if it is continuously changing altitude during the flight. Etc. for "go east" and "go north."

The "roll-a-coaster" solution given in the next paragraph is only a trivial extension of the South Pole solution. I give it mainly because it provides me an opportunity to illustrate my understanding of the north / south flexibility inherent in "go east" but it also has the merit of expanding the area of possible starting points from a set of measure zero to a region with finite area, which conceptually could be one continuous area surrounding the South Pole. An obvious further conceptual extension, using a circular track, non-existent tunneling technology, and remarks I make later about purely radial motion, can in principle permit one to depart from essentially any point on Earth. I mention this physically unrealizable further extension to preempt others from doing so and to technically validate the claim that this posing contains two solutions. (This further extension replaces the more interesting, but flawed, Tropical "solution," given later.)

"Roll-a-Coaster" Solution: Assume a monorail roll-a-coaster, with a total of one mile (or 1/n miles) of track, has been built around the South Pole. The rail support structure tilts strongly away from the local vertical so that the track, despite undulations, remains precisely in one latitude circle plane. (The wheels should grab the rail on top, bottom and one side because of this extreme tilt.) Designs with different undulation amplitudes require different latitudes for the departure point. Such design variations can be continuous. Thus the possible departure points occupy continuous bands around the Earth, and not just circles of measure zero. Unrealistically large undulations in some of the n=1 solutions can cause that band to overlap with the band of n=2 solutions, etc. to produce one continuous departure region surrounding the South Pole.

Now I want to discuss how the one mile distances are to be measured, because if you accepted my interpretation of "going south," the problem statement is ambiguous or at least not specific about how to do this. I note that going "straight up" is not traveling in any of the three specified directions and that the problem statement does not explicitly limit the total travel to three miles, as is commonly presumed; however, I will assume that the distance traveled is to be measured along the path, unless the motion is purely radial towards or from the center of the Earth. In my view, the problem statement allows any amount of pure radial motion despite most people’s self-imposed limitation of the total travel distance to three miles.

You are of course free to disagree, perhaps claiming either (1) that no radial motion component is permitted, just as no westward movement is allowed, even though these prohibitions are not explicitly stated or (2) that the distance traveled radially is irrelevant. I.e. that one must make a one mile displacement in each of the three directions, regardless of distance traveled, and never travel westward. Position (1) means that you also reject my interpretation of "going south." I.e. you restrict travel to great circles on the surface of an idealized Earth. Thus, for people taking this position, airplanes can taxi south on the runway but never fly south. IMHO position (2) also is not reasonable as the problem speaks of "traveling" or "going", not of "displacements." Perhaps you disagree for some other reason, but fortunately, all of my solutions can avoid pure radial motion; however, I think I must make one additional idealization, similar to assuming a spherical Earth, for my
"expensive" solution with four hints. (If I make this idealization and am allowed to have uncounted purely radial motion, then that solution can start anywhere, not just from points south of the Tropic of Cancer.)

To help others find this physically-possible but "expensive" solution, I will give four, two-word, hints about it. – The first is below and others will follow, one at a time, at 3 to 6 day intervals, or all be given in one post if someone else posts this solution before I do. When combined with the others, the last hint should give the solution away to most physicists. (This will save me the effort of describing it. The "irresponsible" physically-questionable solution set can be described in a short paragraph after the "expensive" solution has been described, so I am not giving any more hints about it.)

Hint 1 = "space ship"

To help myself remember my planned future hints, I give the first letters of the remaining hints now: Hint 2 = CC, Hint 3 = YL, Hint 4 = TF. I plan to repost this Email for the benefit of those who missed it the first time, but will list all hints already disclosed at the top, so it need not be read more than once.

If you find the solution before I post the fourth hint, please privately tell it to me or only publicly post what TF stands for. (It is something essential to the solution and every English speaking physicist refers to it this way.) Doing this will allow others to continue thinking about the problem until all hints are given. Now I give the subtlety-flawed solution in detail and invite you to find the flaw.

"Tropical Solution": I will be traveling by rocket ship. My starting point is just north of the Tropic of Capricorn, perhaps only by meters. I will launch in the dark but very near the day/night terminator line shortly before the moment of the equinox of 21 March. Although this completely defines the launch conditions, it may be helpful to redundantly give more information. Thus, I note that essentially ninety degrees to my East, the sun will soon be passing over the equator. In case you have forgotten, I remind you that one, and only one, point on the Tropic of Capricorn circle is always in the plane of the ecliptic. (All others are always south of that plane.) Because my launch point is only slightly north of the point in the Tropic of Capricorn circle which is currently in the plane of the ecliptic, my launch site is just slightly north of the plane of the ecliptic. Twelve hours after launch, my launch point will be far south of the plane of the ecliptic, enjoying the last minutes
of daylight, before rotating into the 12 hour darkness period. I also note that my launch point is very close to that point on a spherical Earth, which is currently the most advanced (leading all others) in the Earth’s journey around the sun. This is convenient because it means that my rocket is essentially vertical but with a very slight inclination towards the south, if I launch with initial velocity vector parallel to the plane of the ecliptic. (I want to be careful to be traveling slightly towards the south even as I break contact with the launch pad. I also want to describe a completely realistic, easily achievable trip. Thus, I note that my planned trajectory is very similar to the trajectory of many of the earliest sub-orbital, up/down rocket launches.)

This "solution" is more general than the very detailed description of my journey which follows. The starting point could be anywhere in the tropics and still have a launch that is almost vertical, but very slightly inclined to the south, if made from a point, which is slightly north of that point which is currently leading all others in the Earth’s journey around the sun. The longitude of the start point is arbitrary, but only a narrow range of "vertical-launch, critically-timed" longitudes is available each year. E.g. if you want to start from near Havana Cuba, you may need to wait many years, but eventually you will be able to launch from its suburbs. (The problem statement does not require the journey to begin in this or any other millenium. This is another tacit, self-imposed, limitation many needlessly make.)

At the end of the initial southbound leg, I will be almost one mile above the Earth’s surface and exactly in the plane of the Tropic of Capricorn at point I’ll call "P1." I of course never deviated from the longitude plane of the departure point during my assent and did continuously make some small but steady progress southward while traveling one mile along a slightly curved, nearly vertical, path to P1. My south bound velocity component was always very small during this first mile of the journey and just as I reach P1, it has been reduced to zero by my small thruster control rockets. I’ll call P1’s altitude above the Earth’s surface "A1."

I want to be very clear, so I note P1 is always a point that is in both the plane of the Tropic of Capricorn and the plane of the longitude circle passing through the launch point and always A1 above the surface of the Earth. There are two such points, and even though it is redundant to do so, I explicitly state that when I arrive at P1, P1 is the one that is on the "advancing" or "leading" side of the Earth in its orbital motion about the sun. Twelve hours later, P1 will be a trailing point making the journey around the sun, but it is not close to the most trailing point because it is then far south of the ecliptic. (The leading and trailing points are always on the ecliptic.)

I arrive at P1 going fast and it is approximately one minute before the exact moment of the equinox. Although P1 is sunlit when I arrive because of its altitude, it is still slightly (perhaps only meters) on the dark side of the plane defined by the "terminator line" (really a circle) and north of the plane of the ecliptic. It is north of the ecliptic because P1 is on a larger circle in the Tropic of Capricorn plane than the Tropic of Capricorn circle. I’ll call this larger circle "C" and note that some points on C are always north of the ecliptic. When I arrive at P1, it’s distance from the plane of the ecliptic is still slightly increasing due to the Earth’s rotation. I will call P1’s distance from ecliptic plane, at the moment I arrive there, "D." Although P1 is still rising above the plane of the ecliptic when I arrive there, it’s distance from the ecliptic will soon begin to decrease again. (Recall that because of the Earth’s rotation, 12 hours later P1 will be far south of the
plane of the ecliptic.) The time interval between my arrival at P1 and the moment when P1’s distance from the ecliptic plane has been reduced by the Earth’s rotation enough to again be D, I will call "T."

Just as I pass through P1, I initiate a west-pointing, thruster-rocket burn of short duration. This impulse gives me some motion to the east, relative to the launch plane, but for half a minute or so after passing through point P1 my primary velocity component is still "up" (away from the center of the Earth). The eastward motion velocity component achieved with the impulse burn of my west pointing thruster is, however, sufficient to carry me through the terminator plane into the daylight side of that plane. This eastward velocity component is reduced to exactly zero precisely T after passing through P1 by a brief compensating "impulse burn" of an east pointing thruster. This thruster burn begins very near the end of the time interval T and terminates exactly at T at a point, still in the Tropic of Capricorn plane, I’ll call "P2." This completes the eastbound leg of my journey, but I want to describe in more detail how I made sure that the distance traveled eastward, as measured
along the path traveled, was made to be exactly one mile.

As I passed through P1, I fired retro rockets vertically and took advantage of the Earth’s gravitational attraction after passing through the point P1 so that my trajectory between the points P1 and P2 resembles a narrow inverter parabola. I carefully regulated my speed along this quasi parabolic path to return me to a distance D above the ecliptic at precisely T after passing through the point P1. Because of my careful navigation through space, the length traveled along the "parabolic arc" is precisely one mile. (I controlled the strength and duration of my retro rocket burn, to adjust the highest altitude point, or peak of the "parabola," to make this true.) That is, I controlled my speed along the path so that traveling this east bound leg required exactly a time interval T and the height of the quasi parabola so that the total path length was exactly one mile.

During all this "parabolic arc" and eastward motion, I have been very careful to never leave the plane of the Tropic of Capricorn. Thus both I and the point P1, which has also been traveling eastward through space as the Earth rotates, and is now on its downward descent towards the ecliptic, arrive at a point that is A1 above the Earth’s surface, D from the plane of the ecliptic, and in both the Tropic of Capricorn plane and the plane of the original launch point, which is now rotated from its position at the time I passed P1 by (T/24)x360 degrees, where T is measured in hours. That is to say, the points P1 and P2 are the same point!

Now I leave the Tropic of Capricorn plane, but I am always careful to stay in the longitude plane of the launch site and to never have any southern velocity component. I could simply exactly reverse my assent path, but there are infinitely many other return paths which also remain in this longitude plane and have a total length of one mile with only northbound velocity components and terminate at the original launch point. For example, I could head straight toward launch site when I leave the Tropic of Capricorn plane and later compensate for the shorter length of that direct part of the path by a final part of the path that is more curved than the assent path.

PS – This "solution" illustrates an "Easter Egg" – a plausible, but false, statements (or other error) hidden in text. I hope most of you will at least think there must be an Easter Egg here even if you can’t find it. If you find it, wait midweek before posting exactly where it is so others who do not read their PHY-1 Emails every day have a chance to hunt for it. I have also hidden "Easter Eggs" in my book, Dark Visitor. I give hints about some of them in the book’s postscript. If you like hunting for Easter Eggs, take a look at Dark Visitor. The site given below tells how to read it for free.

I put the Easter Eggs in Dark Visitor because my target reader is intelligent and likes intellectual challenges, games, etc. but is not currently very interested in physics. Dark Visitor is a science student recruiting tool. Wall Street is actively recruiting / stealing science students with cash grants. (More details privately by request.) Dark Visitor is my attempt to counter punch.

Despite integrating all the physics in Dark Visitor into the text as a natural part of a physically possible, scary story (Ice age beginning in 2008 due to a 2.2 solar mass "dark visitor" now approaching from deep space), parts of Dark Visitor will bore my target reader with science (astrophysics, cosmology, climatology, orbital mechanics, etc. – a complete list given at web site below). Thus, the text recommends that several large sections, and all of Chapter 10, be skipped on the first reading. I included Easter Eggs in the text to encourage subsequent reading of these sections in a search for these hidden eggs. (Please privately tell me if you know of any prior book that has Easter Eggs for the reader to find. -I know they are quite common in computer software.) If you want to know more about my book, or hunt for more of my eggs, visit, where I have just put up a new page that graphically gives both the trajectory of the dark visitor, and how it perturbs the
Earth’s orbit – all the results of a finite-time-step computation. Alternatively, contact me privately.

I could be wrong, but I don’t believe there is any "egg" in the "roll-a-coaster" solutions given above or in the "expensive solution" for which I will be giving hint 2 in a few days.

Bill Powell

John Denker <jsd@AV8N.COM> wrote:Here is my solution to the great-circle navigation problem.

This problem is a veritable poster-child for illustrating
the power and elegance of the Clifford Algebra approach
... which could also be called the quaternion approach,
since quaternions are isomorphic to unit bivectors in
Clifford Algebra. So if you can do bivectors, you get
quaternions for free.

Statement of the problem: We are given two points on the
sphere, a and b. The task is to find the heading that we
should use when departing a enroute to b.

WLoG we scale things so that our sphere is the unit sphere.
Consequently a and b are unit vectors.

Assumption #1: We assume a is not a multiple of b, because
otherwise the problem is trivial ... *any* direction will
equally well get you from a to b. To say the same thing
more formally, we assume a/\b is nonzero.

Assumption #2: We assume a is not the north or south pole.
Otherwise expressing the answer as a heading is not useful.
(Some useful alternatives will be given below.)

By way of preparation, it will be convenient to expand b
as follows:
b := c a + s w
where w is some vector that lies in the a/\b plane and
is perpendicular to a, i.e. w.a = 0. Here c and s
are scalars. (They are the cosine and sine of some
angle, but that's actually more than we need to know.)

You can easily verify that a/\b = s a/\w. That makes
sense; the a/\b plane is the same as the a/\w plane,
except for a scalar scale-factor.

Thence you can easily verify that a(a/\b) = s w, or
equivalently a.(a/\b) = s w. That gives us a
convenient way of constructing w:
w = a.(a/\b) / ||a/\b|| [eq 1]

... where the denominator is well-behaved by virtue
of assumption #1 (above).

As usual the norm is defined as
||a/\b|| := sqrt((a/\b).(b/\a)) [eq 2]

So much for preparation. Let's actually attack the
problem now.

Let x(theta) be a point that roves along the great
circle from a to b, starting with x=a when theta=0.

Then the tangent vector will be d(x)/d(theta).

My physical intuition tells me that the tangent vector,
tangent to the point a, will be just w (or perhaps a
multiple thereof, depending on how you scale theta).
(If that's not obvious to you, see the appendix, below.)

So here's the recipe: If the two vectors a and b are
already expressed in rectangular coordinates, that's
great. Otherwise, e.g. if they are in polar coordinates
(latitude and longitude), convert them to rectangular
coordinates ... which will be the last (or almost the
last) use of trig functions in the whole problem!

Find the vector w using equation [1] as given above.

Next we need to know what w looks like to the guy on
the ground, i.e. in the tangent plane, tangent to the
sphere at point a. To get our bearings, let's determine
the vector (call it v) that points from a to the north
pole. We can find this immediately by using equation
[1] again ... just imagine traveling from a to the north

Now we project out the component of w parallel to v (call it
w1). The remainder of w (call it w2) will be perpendicular
to v. All of these vectors (w, v, w1, and w2) lie in the
tangent plane. We use the usual formula for the projection
w1 = v (v.w) / v.v
w2 = w - w1

At this point we know enough to draw the vector that points
from a to b. Lay a piece of graph paper on the floor
(aligned NS/EW of course), and draw a vector from the origin
to the point (w2,w1).

Note that starting from the rectangular components of a
and b, we have gotten to this point without using any trig
functions or any other transcendental functions. We have
just performed algebraic computations on the components.
It's just add, subtract, multiply, and divide. We can
even leave out the square roots, since we only care about
the direction of w, not its magnitude. This is arguably
important if you need to do this calculation a zillion
times per second, and you don't want to spend all your CPU
time evaluating transcendental functions.

In general there are two great-circle routes from a to b:
a short one and a long one. The formalism given here is
not guaranteed to give you the short one. There is an
ambiguity associated (conceptually, at least) with the
choice of positive versus negative square root in
equation [2]. Or you could just put an explicit +-
sign in the definition of w in equation [1]. (I suspect
it's possible to set things up so that the formalism
always coughs up the short route, but I haven't pursued
this idea.)

Finally: If you insist on expressing the answer as a
heading angle, calculate atan2(w1,w2).


For slight extra credit, we now consider the special case
where a is at the north pole. The desired departure heading
is south (duh!) but that probably isn't what you wanted to
know. To determine which meridian you should follow, you
can skip the entire calculation; the answer is just the
longitude of b, by inspection.

However, you can appreciate the power and beauty of the
bivector formulation by observing that you *can*
calculate the vector w just fine, even when a is at the
north pole. We cannot calculate the projections w1 and
w2, because v doesn't exist ... but w exists as a
perfectly well-behaved vector in real space.

This is important if you're building something like an
autopilot or video game. The flight director can display
the vector w as a pointer that points that-a-way ... and
it will work just fine at the poles and everywhere else.

To say the same thing in fancier terms:
-- expressions involving polar coordinates are singular
at the poles and numerically unstable near the poles, but
-- the actual physics and geometry of the sphere have
no singularities anywhere, and
-- the bivector representation has no singularities



You can easily prove that the tangent to the great
circle at point a is just w, a vector in the a/\b
plane perpendicular to a.

This should be obvious from the geometry of the situation,
but it is also possible to prove it by directly calculating
the tangent vector. It's an amusing exercise in Clifford
Algebra. We express the roving point x(theta) by applying
the rotation operator R(theta) to the vector a, and then
differentiating w.r.t theta. Of course we are talking about
the particular rotation operator R() that causes rotations
in the a/\b plane.

Specifically, for an infinitesimal angle theta, we can
use the standard formula
for producing a rotation, which gives us:
R(theta) a = [1 + (theta/2)w/\a] a [1 + (theta/2)a/\w]
and if you just turn the crank, expanding the RHS and
simplifying using the axiomatic properties of vectors
a and w, you get
... = a + theta w + irrelevant terms involving theta^2

This can be immediately differentiated w.r.t theta,
whereupon we find that the tangent vector is w, QED.

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