How far must one travel to reach the North Pole if one starts from the equator and always heads northwest?

Additional material

[ Editor: rec.puzzles discussion of January and February 1994, after
the puzzle was reposted as "Starting on the equator, you fly
continuously towards north-east ... assuming that the speed of the
plane is 1000 km/h, how soon will you be there (if ever)" (Risto
Lankinen).  It concerns the general formula and whether the traveller
really reaches the pole. ]

By the way, did anyone post the loxodrome solution?  If so, I didn't
see it, but my recollection is that last time that puzzle was posted
we agreed that the length of the path was sqrt(2) times the initial
distance from the north pole (measured along the earth's surface, of
course).  More generally, for a path that makes an angle x of less
than 90 degrees from geographic north, the length is sec(x) times the
initial distance from the pole.

-- David A. Karr <karr@cs.cornell.edu>

Well, I posted the puzzle, and all I asked was to determine if the
length of the path is finite or infinite.  However, your comments about
the natural next question to ask are correct.  In general, the length
of a loxodromic path between two points is l sec(x), where x is the
bearing and l is the difference in latitude between the two points (or
strictly speaking, the radius of the earth times the difference in
latitude).  This is quite easy to see directly, once you know the
answer, or it can be proved using calculus or spherical trig.  The
problem is also discussed in Heinrich Dorrie's _100_Great_Problems_of_
Elementary_Mathematics_.

-- Timothy Y. Chow <tycchow@math.mit.edu>

In article <2i3433$le2@zip.eecs.umich.edu> positron@quip.eecs.umich.edu
(Jonathan Haas) writes:
<You will *never* reach the north pole. If you could reach the north
<pole by travelling northeast, you could *leave* the north pole by
<travelling southwest, which you can't do. You'll continually spiral
<around the north pole, getting closer and closer but never reaching
<it.

I guess you're a fan of Zeno.  The path has finite length, so unless
you buy the argument that you can never get anywhere because you have
to get halfway there first, you will reach the North pole in finite
time.

Here's a suggested sketch of a new archive entry.

A path on the earth that cuts all meridians at a constant angle (i.e.,
a path where you maintain a constant bearing) is called a loxodrome.
If t is the difference in latitude between two points on a loxodromic
path of bearing a, then the length of the path between the points is
R*t*sec(a), where R is the radius of the earth.  This is intuitively
clear and is not hard to prove using calculus.  For example, if the
bearing is pi/4 radians and we take our endpoints to be the north pole
and a point on the equator, the length of the path will be about
14000 km (taking R = 6400 km).  Note that in this example the path will
circle the north pole infinitely many times but that its length is
nevertheless finite.

-- Timothy Y. Chow <tycchow@math.mit.edu>

I find I have to walk more and more slowly the faster I am turning
around, and I certainly cannot turn around or change the direction I'm
travelling at an infinite rate.  So I agree with Jonathan that I will
never reach the pole - not because of the (finite) length of the path,
but because of the infinite number of turns I will have to make.

This argument applies regardless of whether I keep facing the direction
I am travelling, or keep facing roughly the same direction (in
3-dimensional space, not compass direction) and take steps forward,
backward, sideways, etc., to follow the correct path.

-- David Seal <dseal@armltd.co.uk>

In article <16313@grayt> grayt@Software.Mitel.COM (Tom Gray) writes:
   Actually as one nears the north pole, the meridians come closer and
   closer together ... NE ... has degenerated into the same direction
   as N.

Eh?  The angle between NE and N is always 45 degrees.  If you're making
some other angle with the meridian, you're on a different heading.
(What you're saying is that the local coordinate system you're using to
measure NE is only valid over smaller and smaller areas as you approach
the pole.)

   Also the distance travelled along this path is finite. Therefore if
   you travel with a finite speed on this path your travel time will be
   finite and you will reach the destination which is the pole.

But your angular velocity will grow without bound.

   Isn't this just a variation of Zeno's paradox of Achilles and the
   Hare?

Yeah, but with a twist.

It's true that the path's length is finite.  As others have pointed
out, however, the number of circuits the path makes of the pole is
infinite.  No matter how many circuits you make, you still have an
infinite number left to go.  Unlike time and distance in Zeno's
original paradox, this quantity (circuits) never decreases.

If one accepts that an idealized plane can produce and tolerate
unbounded angular acceleration, then, perhaps, you'll be able to reach
the pole, in some sense.  I think that many of us refuse to accept that
premise as reasonable.  The plane will either be unable to maintain the
constant NE heading or will be unable to maintain a constant speed
along the spiral, eventually stalling and crashing a few hundred feet
from its destination.

P.S.  While I'm here:

In article <CK96LF.75s@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Daniel
A. Asimov) writes:
   [quoting Jonathan Haas:]
   >You will *never* reach the north pole. If you could reach the north
   >pole by travelling northeast, you could *leave* the north pole by
   >travelling southwest, which you can't do.

   (The kind of reasoning highlighted above could lead one to think
   that it is possible to parachute from the ground up to an airplane.)

Eh?  I don't think Mr. Haas was suggesting that a plane can fly
backwards.  Let's not get extreme in trying to deflate another's
arguments.

-- Patrick A. O'Donnell <pao@ai.mit.edu>

[ Editor: Asimov highlighted the second sentence of the quotation. ]

In article <1994Jan28.192023.15035@galois.mit.edu> I wrote:
>Are you claiming that if we fix the speed of travel along the path,
>then the magnitude of the acceleration vector approaches infinity?
>This looks a little messy to calculate; have you crunched it out
>already?  I'd be interested in seeing the results.

Well, I got impatient and decided to crunch it out myself.  It turns
out that the magnitude of the acceleration *does* approach infinity (in
fact it is proportional to sec^2(t) + 1 where t is the time and the
speed of the plane has been normalized so that the latitude of the
plane increases by 1 radian per unit time interval).  I had been
hoping that the distance to the pole would decrease fast enough that
the magnitude of the acceleration would remain bounded even though the
angular acceleration wouldn't, but no such luck.

This might be worth noting in the archive too.

-- Timothy Y. Chow <tycchow@math.mit.edu>

[ Editor: Chow sent Chris Cole his derivation: ]

Replace the earth with a sphere of radius R centered at the origin.
For simplicity let us consider only the case of travel in the northern
hemisphere.  Using spherical coordinates T and P (for "theta" and
"phi," where T is the angle from the positive z-axis), we see that the
velocity of the plane at the point (R sin(T)cos(P), R sin(T)sin(P),
R cos(T)) can be decomposed into latitudinal and longitudinal
components.  These components are

  v sin(B) * (-sin(P), cos(P), 0)   and
  v cos(B) * (-cos(T)cos(P), -cos(T)sin(P), sin(T))

respectively, where v is the (constant) speed of the plane and B is the
bearing (measured clockwise from north, of course).  Now velocity is
just the derivative of position with respect to time t.  This fact
yields the following system of simultaneous differential equations for
T and P (one for each of the x, y, and z coordinates).

 RT'cos(T)cos(P) - RP'sin(T)sin(P) = -v sin(B)sin(P) - v cos(B)cos(T)cos(P)
 RT'cos(T)sin(P) + RP'sin(T)cos(P) =  v sin(B)cos(P) - v cos(B)cos(T)sin(P)
                        -RT'sin(T) =  v cos(B)sin(T)

Eliminating T' from the first two equations yields RP'sin(T) =
v sin(B).  Anywhere other than the poles, sin(T) is nonzero, so we can
cancel sin(T) from the last equation and solve to obtain
T = -vt cos(B)/R + c for some constant c.  Using this we can solve for
P:

  P = tan(B) log(csc[vt cos(B)/R - c] + cot[vt cos(B)/R - c]) + k,

where k is some constant.  From this we see that the plane's latitude
increases with time at a rate proportional to v cos(B).  This allows us
to find the length of the loxodrome immediately given the latitudes of
the endpoints.

The magnitude of the acceleration can be calculated by differentiating
the velocity with respect to time and taking the square root of the
sum of the squares of the components.  The acceleration is

  / -vP'sin(B)cos(P) + vT'cos(B)sin(T)cos(P) + vP'cos(B)cos(T)sin(P) \
 |  -vP'sin(B)sin(P) + vT'cos(B)sin(T)sin(P) - vP'cos(B)cos(T)cos(P)  |
  \                          vT'cos(B)cos(T)                         /

The sum of the squares of the components is -2v^2 P'T'sin(B)cos(B)sin(T)
+ (v sin(B))^2 P'^2 + (v cos(B))^2 T'^2 + (v cos(B))^2 P'^2 cos^2(T).
Substituting P' = v sin(B)/(R sin(T)) and T'= -v cos(B), we see that we
obtain some constant terms, a term in csc^2(T) and a term in cot^2(T).
The latter terms clearly diverge as T -> 0.

-- Timothy Y. Chow

[ Editor: from the equation for T above, T' = -v cos(B)/R; the
substitution drops the factor 1/R. ]

In article <2i2ihn$3jd@louhi.icl.fi>, ristol@xerver.icl.fi (Risto
Lankinen) says:
>Starting on the equator, you fly continuously towards north-east.
>It's no big surprise, that you eventually reach the north pole, but,
>assuming that the speed of the plane is 1000 km/h, how soon will you
>be there (if ever).

This touched off a long thread.  The path described spirals inward
toward the pole.  The distance from the pole falls with linearity,
reaching 0 at a computable time T.  The bearing from the pole changes
more and more rapidly, becoming discontinuous at time T.

Some argued that you "never" reach the pole.  This can mean one of two
things:
(a) A mathematical point moving along the loxodromic path never
    reaches the pole.  This is false, because if you look at it at time
    T, there it is!
(b) An airplane would in fact be unable to follow the path in
    question, because it would have to do impossible things - rotate
    with infinite torque, ultimately - and would come apart, or at any
    rate deviate from the path.

    This depends on whether you are talking about a theoretical
    "puzzle world" airplane or a real-world nuts-and-bolts airplane.
    There are plenty of puzzles that involve flies, dogs, and other
    creatures doing impossible things, reversing direction
    "instantaneously" and so on, so why not an airplane?

    However, I think it MIGHT be possible to meet the conditions of the
    puzzle even with a REAL airplane.  It depends on what you mean by
    things like "travel on the path" and "reach the pole".

    I am going to assume that when we say the plane "travels on the
    path" we mean that the point representing the center of mass of the
    plane travels on the path.  As for "reaching the pole", there are
    three possibilities:

    a. (least stringent) Some part of the plane passes over the pole.

    b. (letter-of-the-puzzle) "You" reach the pole - that is, some part
       of the body of some occupant of the plane passes over the pole.

    c. (most stringent) The center of mass of the plane has to pass
       over the pole, following the loxodromic path.

  In cases a and b, the center of mass of the plane just has to come
  "close enough" to the pole and does not have to do an infinite
  pirouette.

  Now if you actually look at the shape of the spiral - which
  approximates a logarithmic spiral on a flat surface at regions close
  to the pole - one realizes that almost all of the twirling takes
  place VERY close to the pole and VERY late in the trip, really in the
  last second.  In fact - (these calculations are for the logarithmic
  spiral, but it's a close enough approximation) -

       Let the distance from the pole be r
       For a small northward shift dr, the circumferential shift is
           also = dr
       The change in the bearing from the pole dtheta / dr (in
           degrees) = 360 dr / 2*pi*r
       Integrating both sides,
           Change in theta [r1, r2] = (360 / 2*pi) ln r [r1, r2]
       And for a full circle, change in theta = 360 degrees
           360 = (360 / 2*pi) ln r [r1, r2]
           ln r [r1, r2] = 2*pi
       That is to say that for each full circuit of the pole, coming
       back to the same meridian, the distance from the pole decreases
       by a factor of 1 / e-to-the-2*pi, that is, about 1/530.  If you
       start from 1 km away, then by the time you've made ONE circuit
       and are back to the meridian you started on, you are only about
       2 m away; by the time you've made another circuit (won't take
       long!) you'd be less than .4 cm away.  You get awful close to the
       pole - practically there - without much spinning.
     Furthermore, at any given time the turning radius of the path is
     1.414 times the distance from the pole.

     Implications:

  a) For type a solutions (some part of the plane passes over the
     pole) any plane is capable of doing this whose "special wingspan"
     is at least .707 of its horizontal turning radius.  By "special
     wingspan" I mean the greatest extension of the left-side wing in
     the direction of "10:30 o'clock" from the center of mass.

     I believe some of the low-speed ultralights are capable of this,
     especially those things designed for human-powered flight with a
     hundred-foot wingspan on the albatross model.  Granted this sort
     of thing is not likely to fly from the equator (or at 1000 km/hr).

     A fighter plane would also be able to do it, if you allow it to
     climb or dive the last half-km.  That way the projection of the
     center of mass on the ground could trace out a spiral down to a
     turning radius of 10 m. or less, without the plane itself having to
     turn in such a tight curve.  It might even meet the speed and range
     requirements, if it refuels in midair.

  b) "You" reach the pole.  If "you" have a 3-foot reach, then the
     center of mass will be moving on a path with about a 4.2 turning
     radius when you reach it.  This is probably still possible either
     for the aerobatic fighter plane (in a dive or climb) or for a real
     ultralight, basically a converted hang-glider.

  c) The mathematical center of mass has to reach the pole.  Granted, I
     don't see this as possible with a real plane.  But why use a real
     plane and then pose tasks that have to be accomplished with
     impossible exactitude?

-- (PK) (Univ. of Ill. at Chicago) <U31850@uicvm.uic.edu>

In article <94031.020403U31850@uicvm.uic.edu>, <U31850@uicvm.uic.edu>
wrote:
>      c) The mathematical center of mass has to reach the pole.
>     Granted, I don't see this as possible with a real plane.  But
>     why use a real plane and then pose tasks that have to be
>     accomplished with impossible exactitude?

It should be noted that with physics as we understand it, the center of
mass of an object is a position which can't be determined precisely.

If one assumes Newtonian mechanics as well as rigid bodies (a common
idealization), the assertion that infinite accelerations are impossible
is simply false.  When one rigid body collides with another, infinite
accelerations necessarily occur.

If one makes the airplane a rigid body and collide it with an
appropriately shaped track (also infinitely rigid), a logarithmic
spiral to the pole is conceivable.

-- Isaac Kuo <isaackuo@math.berkeley.edu>

Tom Gray wrote:

   After much discussion, I have not seen a consensus form on the issue
   of the Constant NE Flight Puzzle. [...] The FAQ states that a
   constant NE flight cannot reach the North Pole since a time reversed
   version of this flight would mean leaving the pole in the direction
   SW. According to this view only the direction S has any meaning at
   the pole.

Vaughan R. Pratt replies with the famous "fly flying back and forth
between two locomotives" puzzle, where the locomotives certainly
collide, and the fly covers a finite distance (with infinitely many
reversals of direction) before being crushed.

Nonetheless, the FAQ's argument has some merit.  Whether you regard its
answer as correct is, I claim, a question of semantics.

The Constant NE Flight defines a smooth (indeed, analytic) mapping from
the half open interval [0,1) to the sphere.  There is no smooth
extension to a mapping from [0,1] to the sphere (though there is a
unique continuous extension).  The FAQ's argument is an informal proof
of this fact.

Does the journey reach the pole?  That depends on how fussy you are
with your definitions of "journey" and "reach".

-- Michael Weiss <columbus@strident.think.com> (rec.puzzles and
   sci.math, February 1994)

Well, *I* wouldn't go on any "journey" where my acceleration would be
unbounded.  I'd be afraid of what my condition would be once I "reach"
my destination (probably a gooey puddle of crushed bones, blood,
mangled guts, and, unless this cold clears up, phlegm).

Consider the acceleration of a NE flight of constant speed (that is,
the velocity vector has constant magnitude, but not direction, since NE
is not a direction in the sense of physics).  As the flight approaches
the pole, the acceleration goes to infinity...

-- Bob "better get a mop" Beals <beals@gargoyle.uchicago.edu>

In article <CL2KLL.9Ez@freenet.carleton.ca>, Ross Cowie
<ao383@FreeNet.Carleton.CA> wrote:
>It seems in this puzzle, many people are focusing on the inevitable
>infinite acceleration, but I wonder if this only happens infinitely
>close to the pole?

Poles.  If you start from the South pole it takes infinite acceleration
to leave it in such a way that at all points of your trajectory (save
at the South pole itself where it's undefined) you are traveling NE.

That the two poles are the only two places where infinite acceleration
is encountered is easily seen from my earlier observation that a NE
course is a straight line on the Mercator projection.  The poles are
the only singularities on the Mercator projection.

>On a slightly different tack, and to help put some practical
>visualisation on this thing, the following question:
>
>From a given start position on the equator, where exactly are the
>first few locations at which one re-crosses the starting longitude?

This follows once you have a rule positioning latitude L (in radians)
on the Mercator projection.  This is given by the inverse gudermannian
function,

                invgd(L) = ln(tan(L/2 + pi/4))

(So at the equator, L=0, tan(pi/4) = 1, and ln(1) = 0.  At the North
pole, L = pi/2, tan(pi/4 + pi/4) = oo, and ln(oo) = oo, i.e. the north
pole is infinitely high on the Mercator projection.)  Its Taylor
expansion about L=0 starts out L + (L^3)/6 + (L^5)/24 + (L^7)/5040 + ...

One recrosses the starting longitude where the inverse gudermannian is
an integer multiple of 2 pi.  To see this, use the fact that the
Mercator projection is conformal (infinitesimal squares on the Earth
project to squares).  Hence the NE trajectory is plotted as a 45-degree
line.  The starting longitude reappears 2 pi to the right, hence we are
2 pi up, so we want the latitude whose inverse gudermannian is 2 pi.
That is, we want the values of the gudermannian at multiples of 2 pi.
The gudermannian is given by

                gd(x) = 2 arctan(exp(x)) - pi/2.

(A really cute thing about the gudermannian is that its Taylor
expansion about x=0 starts out x - (x^3)/6 + (x^5)/24 - (x^7)/5040 + ...
identical to its inverse except for the alternation of signs!  So
gd(x) = i invgd(i x).)

[ Editor: the coefficient of the seventh-power term in both series is
61/5040, not 1/5040. ]

The following program calculates gd(x) for x = 0, 2pi, 4pi, 6pi, ...,
subtracts it from pi/2 to give the distance in radians to the North
pole, then scales it by 2e7/pi meters/radian to convert to meters.  It
prints this distance and the remaining flying time assuming a NE
velocity of 1000 km/hr.

#include <math.h>

main()
{
        double cros, dist;
        for (cros = 0; cros < 6; cros += 1) {
                dist = 2e7/M_PI*(M_PI/2-2*atan(tanh(M_PI*cros)));
                printf("%g\t%8g\t%g\n", cros, dist, .0036*sqrt(2.)*dist);
        }
}

The first few crossings occur at the following distances in meters and
seconds from the pole.

crossing  meters to go        seconds to go
0          1e+07        50911.7           (1e7 = 18th cent. defn. of meter)
1          23777        121.053
2        44.4022        0.226059
3       0.0829186       0.000422153
4       0.000154845     7.88342e-07
5       2.8837e-07      1.46814e-09

So 14 hours to the first crossing, 2 minutes to the next, then a
fraction of a second later your very quarks part company.

>Or similarly, how many spirals before one is within a meter of the
>pole?

Just under three.  Acceleration should be around 8000 G at that point
(v = 277 m/s, r ~ 1 m, so a = v^2/r ~ 80000 m/s/s ~ 8000 G).

-- Vaughan Pratt <pratt@cs.stanford.edu> (rec.puzzles and sci.math,
   February 1994)

[ Editor: two more replies from the January 1994 thread. ]

In article <2i3433$le2@zip.eecs.umich.edu> positron@quip.eecs.umich.edu (Jonathan Haas) writes:
>Risto Lankinen <ristol@xerver.icl.fi> wrote:
>>
>>Starting on the equator, you fly continuously towards north-east.  It's no
>>big surprise, that you eventually reach the north pole, but, assuming that
>>the speed of the plane is 1000 km/h, how soon will you be there (if ever).
[...]
>You will *never* reach the north pole. If you could reach the north pole by
>travelling northeast, you could *leave* the north pole by travelling southwest,
>which you can't do. You'll continually spiral around the north pole, getting
>closer and closer but never reaching it.
I know this has been discussed before, but:

The reasoning above somehow smacks of Zeno's paradox.  Consider the
following slight modification of that paradox:

Achilles and the Tortoise are constrained to a Euclidean plane, with
Achilles at position A(t) and the Tortoise at position T(t) at time t.
Given that ||A(0) - T(0)|| (the distance between A and T at time 0) is
100, and that Achilles moves at speed 10 in the direction of ray AT
while the tortoise moves at speed 1 in the same direction, does
Achilles ever catch the Tortoise?

Most people answer "yes," but since the rules of motion for Achilles
and the Tortoise do not specify how to reverse the motion from the
configuration A(t) = T(t) (in what direction should they go?), it
is clearly impossible to reach that configuration ... NOT!



On the other hand, suppose we assume that the position of the airplane
is a continuous function of time, and so we conclude that since the
position approaches the North pole as t approaches t1, then
position(t1) = North pole.  Now we ask, what is the *direction* of
the airplane at t1?  It turns out that this cannot be defined as
a continuous function.  That's a problem.

A more reasonable model might be to assume that the airplane carries
on board a clock, a location finder (reading latitude and longitude),
and a direction finder (reading compass direction, except at latitude
90 where it indicates a line of longitude).  Each of these instruments
has some maximum error in its reading: e1, e2, e3 respectively.
Rephrase the problem to read, "The airplane always flies so that the
direction finder reads northeast.  What are the possible readings of
the clock when the location finder reads within e2 of the North Pole
(i.e., when it is possible we have arrived), if that ever occurs?"
Then take the limit as e1, e2, and e3 all approach 0, i.e., as the
instruments become more and more accurate.  (At some point you might
want to worry about Heisenberg uncertainty, but perhaps you can
eliminate that by making the airplane bigger ... )


-- David A. Karr <karr@cs.cornell.edu>

Either this isn't in the archive or I don't believe the archive. The
northward component of velocity is always 707 km / hr. So the time to reach
the north pole is 10000/707 = 14.14 hr.

The plane will be spinning pretty fast when it gets there - you might argue
that if there is a specific point in the airplane that is approaching the
north pole as specified the plane would be torn apart by inertial forces
in the instant before you reached your goal. Still, it would take 14.14
hours. 

Though the spiral does go round the pole an infinite number of times, the 
length of the spiral is finite (14142.1... km) and the loops are very small.

What's the formula for the latitudes at which you will cross the original
meridian?


-- Michael Tobis <tobis@skool.ssec.wisc.edu>

[ Editor: a January 1995 posting, "Heading northwest", about the
archive solution as it then stood; the solution above was later
rewritten. ]

***BEGIN ARCHIVE QUOTE***
==> geometry/spiral.p <==
How far must one travel to reach the North Pole if one starts from the
equator and always heads northwest?

==> geometry/spiral.s <==
One can't reach the North Pole by going northwest.  If one could, then
one could leave the North Pole by going southeast.  But from the Pole
all directions are due south.

If one heads northwest continuously, one will spiral closer and closer
to the North Pole, until finally one can't turn that sharply.
***END ARCHIVE QUOTE***

This "solution" missed the point to the puzzle. What we have here is just
some quibbling over the wording of the puzzle - we've seen a lot of that
recently in this newsgroup. It is certainly true that you will spiral
arbitrarily often around the North Pole, and you never actually reach the
Pole. However, the path you follow "converges" to the Pole, and the complete
path happens to have a finite length. You are asked to find that length.

Note that we are simplifying things by assuming the Earth is a perfect
sphere. A solution by calculus would determine the path that you follow
along the surface of that sphere, and then integrate to find the arc-length
of that path.

However, there is a short-cut. The correct way of looking at this puzzle
lets you determine the answer without using calculus. I don't think it is
too hard to find this short-cut, so I will leave it as a problem for others
to solve. The whole point to this puzzle is that this clever answer exists.

After some answers are posted, the entry in the archive for this puzzle
should be changed. Tom? Chris?


-- Barry Wolk <wolk@ccm.UManitoba.CA>

In article <HOEY.95Jan6101709@sun13.aic.nrl.navy.mil> hoey@aic.nrl.navy.mil (Dan Hoey) writes:
>[...]  I do not support the archives' statement that
>there is any positive limit to how _sharply_ a body can turn, but the
>_number_ of times a body can turn is certainly finite.  Physical
>objects do not perform infinite changes of direction, so those
>problems that rely on infinite change of direction are inherently
>unrealistic.  I include also the puzzle of the fly in a train wreck
>and the puzzle of the the dog running back and forth between a jogger
>and a hiker.  They all lead to the question of what direction the body
>is going where the infinite turns occur, and there is no answer.

But the other two questions you mention are part of the established
canon of puzzles, and they have canonical answers that assume the path
is possible.  This can perhaps be justified by taking the question as
a shorthand for a certain theoretical limiting case of the motions of
the objects mentioned.  Just as we don't worry about the fact that the
fly's finite diameter implies he'll be crushed a finite time before
the time the trains contact each other, we also don't worry about the
infinite trips he was unable to successfully complete in the brief
instant before that.

>I do think, though, that the fact that the path length is a continuous
>curve of finite length (pi r / sqrt 2) should be mentioned.  This can
>easily be seen by noting that locally, regions bounded by meridians
>and parallels approximate rectangles, and that northwest is 45 degrees
>from due north.

I would prefer the archive to give the above paragraph (or something
equivalent) as the *first* paragraph of the solution.  Then, in a
footnote, it can point out that this puzzle belongs to that class in
which the limiting case of the path is not physically realizable, or
in any case leaves certain other properties (in this case, the
final direction) ill-defined.

In other words, the point of the puzzle is to compute the path length;
the problem of having a physical object follow the path is secondary.



-- David A. Karr <karr@cs.cornell.edu>

[ Editor: forwarded to Chris Cole by Mark Brader, who added: ]

   I agree with this.  I think the problem statement should be changed
to refer to a moving *point*, and then in the *answer*, you point out
that it doesn't work for a physical object.


-- Mark Brader <msb@sq.com>