A spy plane must see the entire surface of the Earth in one flight; what is the shortest flight, starting...

[ Editor: no proof of a minimum was found in the collected postings;
the following are candidate paths and an argument about their shape. ]

David Wayne Ring <dwr2560@tamuts.tamu.edu> wrote:
> Assuming the earth is a unit sphere, I get 4pi + sqrt(2) - 1 ~ 12.980584

That should be 4pi + 2*(sqrt(2)-1) ~ 13.3948

> The plane goes straight up, then makes 4 semicircular arcs, each of which
> covers 1/4 of the surface (plus overlap).
> Hmmmm... If I adjust the first turn a bit, I get:
> 2*L + 2*S + 7*PI/2 ~ 12.888805
> where  L=sqrt(14-9*sqrt(2))/2   S=sqrt(2-sqrt(2))/2

I think I can improve that a little more, but it would require quite a bit
of effort to estimate the length. And I don't want to do that unless I
know I am on the right track.

Anyway, this is an interesting problem and there has been no discussion, so
I will go ahead and describe the path in more detail.

Zeroth order: The plane starts at (r,theta,phi)=(1,pi/4,0). It goes
straight up to (sqrt(2),pi/4,0). It follows (sqrt(2),pi/4,phi) to phi=pi/2.
then it makes a smooth transition to a semicircular arc which ends at
(sqrt(2),3pi/4,pi/2). It follows (sqrt(2),3pi/4,phi) to phi=0. The
second half of the trip is a mirror image of the first.

First order: Same starting position (x,y,z) = ( 1/sqrt(2) , 0 , 1/sqrt(2) ).
The plane heads straight for (x,y,z) = (1,tan(pi/8),1). Then it heads for
(r,th,phi)=(sqrt(2),pi/4,pi/4). The rest of the first half is the same. The
second half is the mirror image.

Second order: At start, theta is a bit smaller. The first corner has x a bit
less than 1, and y is different than before, but z=1. The second point also
has x<1, but it lies on the 'x' unit cylinder. The next section used to be an
arc, but is now a piece of a helix (still on the 'x' cylinder). It does
not extend all the way to z=0, the last small piece of it is replaced by
a piece of ellipse which ends at ( r=1 , th=pi/2 , phi=pi/4 ). The rest of
the first half is the same.

Note that the length must be calculated numerically because of the ellipse.

I am very anxious to know if there is a better strategy.

-- Dave Ring <Cdude@phys.tamu.edu>  (1994)

[ Editor: in 1998 Dean Inada <dmi@questrel.com> proposed to Chris Cole
"4pi+2sqrt(2)-2", describing the path as "up and back sqrt(2) to 4pi
baseball seam", and added "I'm pretty sure it can be improved on, but
it gets more tricky."  This is the same value as Ring's corrected
4pi + 2*(sqrt(2)-1).  Inada asked the puzzle's author, Paul Winner,
whether he knew the answer or a solution under 4pi; Winner replied: ]

I have a solution but no proof.  [...] my solution is > 4pi but < 13
[...] I conjecture > 4pi

-- Paul Winner <paul.winner@weburbia.com>

Mike Naylor <you.know.him@you.love.him> wrote:
> We will never see an exact solution to this here on rec.puzzles, and in
> fact I doubt that we will even see a believable approximation. There's
> just no way to prove that any specific tour that you plot is the
> shortest possible.

Really?

Any trajectory that is an answer (optimal or not) to this problem
has a convex hull (ie. a 'shrink-wrap' around its path, if I may)
that must fully enclose the Earth.  It may touch the surface of
the Earth, though, but only tangentially.  If the convex hull does
not touch the Earth in at least four points, the solution that it
represents can be made more optimal by shrinking the trajectory
linearly, until it DOES touch the Earth in four or more points.

  Hence, assume that the convex hull of the trajectory touches
  tangentially the surface of the Earth in at least four points.

Choose four points on the surface of the Earth so that whichever
way you divide the Earth into two hemispheres (just don't do it
for real! :-) at least one point is in either side.  Extend the
tangential planes of each point to infinity to get a solid that
is isomorphical to a tetrahedron (the platonic solid with four
triangular faces).  A trajectory that produces this tetrahedron
for its convex hull must trace through all four corners of it,
and it can be proved that such trajectory is minimized when the
tetrahedron is a regular one (ie. all of its edges are equally
long).  A hamiltonian cycle through its corners traces four of
its edges; replace one edge with a trip from and to the nearest
point on the surface, and we have a "believable approximation".

(You can further truncate one or two corners of the tetrahedron
by removing a smaller tetrahedron to get better choice of edges
for even shorter surface trip.)

I guess all I'm trying to say is that, in my experience, those
who say "We will never..." are often the ones who indeed won't.

-- Risto Lankinen <rlankine@us.oracle.com>

[ Editor: Lankinen gives no length for his tetrahedral path.  For a
regular tetrahedron circumscribed about the unit sphere the edge is
2*sqrt(6) ~ 4.90 and the vertices are at distance 3 from the center,
so three edges plus a trip from a vertex to the surface and back come
to about 3*4.90 + 2*2 ~ 18.7 before any truncation, longer than the
paths above. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 06); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
A spy plane is sent on a mission to observe the entire Earth in one
flight. What is the shortest distance it must travel to be able to see
the entire surface of the Earth, starting and ending at the same ground
level point?

Assume the Earth is a perfect sphere of radius 1.

Some variations:

2 - the plane does not have to start on the planets surface but must
return to its starting point,

3 - the plane must start on the ground but does not have to return to
the starting point.

4 - it must start and end on the ground but in different places

5 - it does not have to start on the ground and does not have to return
to the starting point.

In all cases the plane can go to any height and distances are measured
in a frame in which the planet is stationary and non-rotating.

-- Paul Winner <paul.winner@weburbia.com>

A related problem (Flying Saucer):

You are the captain of a flying saucer. You are meeting with the leader
of a spherical planet at her palace on the equator.  She wishes to take
a trip to see the entire equator of her planet. Find the minimum length
of a flight path relative to the palace which will accomplish this.
Work in units of the planet radius.

NonObPuzzle: Find extensions, complications, misinterpretations and
assumptions to produce problems with the following answers:
1. 0 (or very small)
2. 2 PI
3. No solution
4. A large solution
5. A very very large solution
6. A solution a little smaller than the Spoiler
7. A solution a very very little smaller than the Spoiler

-- lflynn@sbuv.wwb.noaa.gov


The spy plane:

I never did see anyone post the answer so here is a clue: There is a
very simple route [Editor: "root" in the original] with a length of

12.88576587631673249403176198...

-- Paul Winner <paul.winner@weburbia.com>

In mail to Dean Inada, who had sent an answer "assuming there isn't
some better minimum based on a completely different path":

Good effort, nobody else has offered any answers. There is a much
simpler path with length 12.8857658763167324... It is possible that
your method may be able to beat that.

-- Phil Gibbs <philip.gibbs@weburbia.com>

> Maybe he is referring to the edges of a polyhedron, suitably rounded.
-- Chris Cole

The "baseball seam" can be considered the edges of a cube, suitably
rounded.  I think I have a characterization of (a) minimal path (but
not its length) in terms of elementary functions, I will followup mail
when I have it written down.

-- Dean Inada <dmi@questrel.com>

[ Editor: neither the path giving 12.8857... nor Dean Inada's
follow-up was found in the collected material. ]

The flying saucer:

rot13 Spoiler
Onpxjneqf:rreugsbgbberenhdfrugfhycvcfugkvfarirf:fverjfanruG

-- lflynn@sbuv.wwb.noaa.gov

[ Editor: decoded, the spoiler reads "Backwards:" followed by the
reversed text of "The answer is: seven sixths pi plus the square root
of three", i.e. 7*pi/6 + sqrt(3) = 5.3972... planet radii. ]