Sphere, and light source
Clinton Pierce
supposed to place sattelites in orbits. These satelites need to be able to "see"
100% of the planet's surface at all times. What is the minumum number of satellites
that are needed? My high school geometry instructor never was able to satisfy me
or anyone else with an explanation.
As you move the satellite further and further away, the area "visible" _approaches_
50% of the planets surface area. So the answer is not two. Would 3 satellites leave
unwatched "wedges" in the corners? Four? Four seems like it most certainly would
cover it, wouldn't it?
---
-------------------------------------------------------------------------------
Clinton A. Pierce | "If you rush a Miracle Man, you get rotten miracles..."
c...@epoch.com | -- Miracle Max, The Princess Bride.
-------------------------------------------------------------------------------
David Karr
>Let's say you have a planet that's a perfect sphere. Above the
>planet, you are supposed to place sattelites in orbits. These
>satelites need to be able to "see" 100% of the planet's surface at
>corners?
The three points (locations of the satellites) determine a plane.
"Draw" a line perpendicular to this plane through the center of the
planet. Then one or both of the points where this line passes
through the sphere will be invisible from every point on the plane.
> Four? Four seems like it most certainly would
>cover it, wouldn't it?
Sure, when they're at the vertices of a large tetrahedron enclosing
the planet. The trick is getting them to stay in position. (A neat
trick if you can accomplish it!)
-- David A. Karr (ka...@cs.cornell.edu)
Andrew C. Plotkin
Clinton Pie...@tigger.co (958)
> Let's say you have a planet that's a perfect sphere. Above the planet,
> you are supposed to place sattelites in orbits. These satelites need
> to be able to "see" 100% of the planet's surface at all times. What is
> the minumum number of satellites that are needed? My high school
> geometry instructor never was able to satisfy me or anyone else with an
explanation.
> As you move the satellite further and further away, the area "visible"
> _approaches_ 50% of the planets surface area. So the answer is not two.
> Would 3 satellites leave unwatched "wedges" in the corners? Four?
Four seems like it most certainly would cover it, wouldn't it?
Nice question.
Three satellites cannot cover the globe. The proof is, um... Well, put
the first two satellites anywhere. Draw a great circle that passes
underneath both of them. Say (for convenience) that this is the equator.
Now neither satellite can see the north pole, or a circle of radius
epsilon around it, since we are assuming that they're a finite distance
from the planet. Similarly, neither satellite can see the south pole.
And you can't put a third satellite to watch both the north and south
poles.
You *can* position four satellites to cover the globe. (Put them at the
corners of a tetrahedron which circumscribes the globe. I forget how big
this is.) However, I don't think they can *orbit* in such a way as to
remain in a tetrahedron! This gets tricky.
I can arrange six orbiting satellites to do it. Put three in an
equatorial orbit (equally spaced), far enough away that they constantly
watch all the ground between 45 north and 45 south latitude. Then put
another three in (the same) polar orbit (also equally spaced) at the
same distance. That does it. (You might want to move the second trio a
little farther out, so they don't collide with the first trio. :-)
But there may be a way to do it with four or five satellites.
--Z
"And Aholibamah bare Jeush, and Jaalam, and Korah: these were the borogoves..."
Isaac Kuo
Andrew C. Plotkin <ap...@andrew.cmu.edu> wrote:
>Clinton Pie...@tigger.co (958)
>> Let's say you have a planet that's a perfect sphere. Above the planet,
>> you are supposed to place sattelites in orbits. These satelites need
>> to be able to "see" 100% of the planet's surface at all times. What is
>> the minumum number of satellites that are needed? My high school
>I can arrange six orbiting satellites to do it. Put three in an
>equatorial orbit (equally spaced), far enough away that they constantly
>watch all the ground between 45 north and 45 south latitude. Then put
>another three in (the same) polar orbit (also equally spaced) at the
>same distance. That does it. (You might want to move the second trio a
I haven't figured out any way with 4 satellites, nor a proof of its
impossibility, but here's a way to do it with 5 satellites:
Position 4 satellites equally spaced 90 degrees away from each
other in a far away equatorial orbit. Call them A, B, C, and D,
so that A is opposite C and B is opposite D. Call the period
of their orbits a "day", divided into 24 hours in the obvious
way. Make the orbit radius far away enough such that they
have a 178.9999 degree field of view.
Modify the orbits (rotate their axes by 1 degree) such that
the satellites oscillate north and south of the equator.
Note that this oscillation will naturally have a period of
a day. Note that no matter how these modifactions occur,
the planet's surface between the 88 degrees S and 88 degrees
N lattitude will always be covered. Call the surface
south of 88 degrees S and north of 88 degrees N the polar
caps.
Make it such that A and C are northmost at 12 AM, and such
that B and D are northmost at 6 AM.
A and C will cover the north polar cap for more than 7 PM
until 5 AM. B and D will cover the north polar cap for
more than 1 AM until 11 AM. Thus, the north polar cap
is covered for more than 7 PM until 11 AM.
A and C will cover the south polar cap for more than 7 AM
until 5 PM. B and D will cover the south polar cap for
more than 1 PM until 11 PM. Thus, the south polar cap
is covered for more than 7 AM until 11 PM.
The only gaps left are the north polar cap from 11 AM
until 7 PM and the south polar cap from 11 PM until 7 AM.
This is covered by a fith satellite, E. This satellite
is at the same orbital radius as the others, but is in
a polar orbit. It is northmost at 3 PM. It covers the
north polar cap for more than 10 AM until 8 PM. It
covers the south polar cap for more than 10 PM until 8 AM.
These 5 satellites will cover the whole surface all the time.
--
_____ Isaac Kuo (isaa...@math.berkeley.edu)
__|_>o<_|__
/___________\ "Yo! Shampoo!"
\=\>-----</=/ --Saotome Ranma, or Ranma Saotome
Ed Foster
(Isaac Kuo) wrote:
> In article <siI0Wv200...@andrew.cmu.edu>,
> Andrew C. Plotkin <ap...@andrew.cmu.edu> wrote:
> >Clinton Pie...@tigger.co (958)
> >> Let's say you have a planet that's a perfect sphere. Above the planet,
> >> you are supposed to place sattelites in orbits. These satelites need
> >> to be able to "see" 100% of the planet's surface at all times. What is
> >> the minumum number of satellites that are needed? My high school
>
> >I can arrange six orbiting satellites to do it. Put three in an
> >equatorial orbit (equally spaced), far enough away that they constantly
> >watch all the ground between 45 north and 45 south latitude. Then put
> >another three in (the same) polar orbit (also equally spaced) at the
> >same distance. That does it. (You might want to move the second trio a
>
> I haven't figured out any way with 4 satellites, nor a proof of its
> impossibility
>
How about putting 4 satellites at the corners of a regular tetrahedron
which completely encloses the earth?
Timothy E Vaughan
be the edges of a tetrahedron (whose volume encompasses the earth), and
there could be cameras at the vertices.
8-)
Tim
Seth Breidbart
Ed Foster <efo...@draper.com> wrote:
>> >Clinton Pie...@tigger.co (958)
>> >> Let's say you have a planet that's a perfect sphere. Above the planet,
>> >> you are supposed to place sattelites in orbits. These satelites need
>> >> to be able to "see" 100% of the planet's surface at all times. What is
>> >> the minumum number of satellites that are needed?
>How about putting 4 satellites at the corners of a regular tetrahedron
>which completely encloses the earth?
Specify their orbits.
Seth
Andrew C. Plotkin
source "Andrew C. Plotkin"@andr (2106)
> You *can* position four satellites to cover the globe. (Put them at the
> corners of a tetrahedron which circumscribes the globe. I forget how big
> this is.) However, I don't think they can *orbit* in such a way as to
> remain in a tetrahedron! This gets tricky.
A side note I thought of after I left work yesterday:
Instead of having (all of) the satellites orbit, have them remain
stationary (relative to the planet and its star). Hang a big solar sail
off each one, angled and sized so that the pressure of the reflecting
light counterbalances the gravitational pull of the planet. You can do
this for any position farther from the star than the planet is. (The
farther "back" it hangs, the less sail it needs, because it can reflect
the light at a sharper angle and thus get more momentum out of it. Draw
some pictures, you'll get it.)
I think this trick was first suggested by Bob Forward in an Analog
science article; he called the things "statites" as opposed to
"satellites".
You can now blanket the globe with three satellites (in a triangle in
equatorial orbit) and two statites (one watching the north polar area,
one south.)
Of course, this is a different puzzle than the satellites-only one.
Girish Babu Maraliga
In article <32puea$4...@agate.berkeley.edu>, isaa...@lhasa.berkeley.edu (Isaac Kuo) writes...
>In article <siI0Wv200...@andrew.cmu.edu>,
>Andrew C. Plotkin <ap...@andrew.cmu.edu> wrote:
>>Clinton Pie...@tigger.co (958)
>>> Let's say you have a planet that's a perfect sphere. Above the planet,
>>> you are supposed to place sattelites in orbits. These satelites need
>>> to be able to "see" 100% of the planet's surface at all times. What is
>>> the minumum number of satellites that are needed? My high school
>
>
>impossibility, but here's a way to do it with 5 satellites:
>
>
>Position 4 satellites equally spaced 90 degrees away from each
>other in a far away equatorial orbit. Call them A, B, C, and D,
>so that A is opposite C and B is opposite D. Call the period
Pardon me, if my solution seem to absurdly obvious, but the more I think about
it, the more convicing it appears to me.
As someone suggested earlier, the optimal solution is to place 4 satellites at
the vertices of the tetra-hedron enclosing the spherical planet.
The conditions are :
1. The planets should be in a geo-stationary orbit.
This means that they should be placed at a distance 'd' from the centre of the
earth, but not along the polar axis.
'd' is a function of the radius of the planet and it's gravity. The actual
formula escapes me now, but it is there for all to see in the textbooks.
(For earth, it is about 36,000 Kilometres)
In the following figure, if P-P is the polar axis and assuming a unique
origin on the equator, the satellites should be placed as follows
a. At a distance 'd' above 0, 0 (origin)
b. At a distance 'd' above 120 E, 60 N
c. At a distance 'd' above 120 W, 60 N
d. At a distance 'd' above 180 (E/W), 60 S
.... assuming, of course, that the polar axis is N-S.
+
P
+ * +
P
+
A better way to visualize this is to image a solid tetrahedron with a
spehirical void within it. This tetrahedron is rotating along an axis through
the centre of the sphere, running parallel to any plane formed by any three
vertices. This forms the polar axis of the spherical planet inside.
Any comments ??
/G/
--
Girish B Maraliga | 'Comedy is funnier than Science,
OB Port Consultant | .. but, science is more scientific !'
DEC, Marlboro, Mass. | mara...@gnpike.enet.dec.com
Mark Brader
> Let's say you have a planet that's a perfect sphere.
I'll also assume that its mass distribution is spherically symmetrical,
otherwise the question becomes exceedingly hard. And similarly, I'll
ignore questions of stability and assume the satellites go to, and stay
in, exactly in the orbit they're intended for. I'm also ignoring
relativistic effects and the price of corn. :-)
> Above the
> planet, you are supposed to place sattelites in orbits. These
> satelites need to be able to "see" 100% of the planet's surface
> at all times. What is the minumum number of satellites
> that are needed?
Interesting question. I also like the technique of trying several
different spellings of "satellite", to be sure that at least one
was right. :-)
> As you move the satellite further and further away, the area "visible"
> _approaches_ 50% of the planets surface area. So the answer is not two.
Right.
> Would 3 satellites leave unwatched "wedges" in the corners?
Yes.
The important thing to realize is that you can't solve the problem with
one set of satellites that are in a *constant* geometrical relationship
to each other. The reason is that this is only possible if the satellites
are all following the *same circular orbit*. But if they're in the same
orbit, then they're all in a common plane which also goes through the
center of the planets. And then the bits of the planet's surface that
are parallel to that plane can't be "seen" by any of the satellites.
For example, if the orbits are in the equatorial plane, the north and
south poles can't be "seen". (Again, as you move the satellites farther
and farther out, the points *approach* being visible.)
In fact, considering any 2 satellites, whether they're in the same orbit
at any time they area they leave unseen will include at any time, for
similar reasons to the above, some pair of diametrically opposite points.
So a third satellite can't possibly cover both at once. With 3 satellites
in the *same orbit*, you can't even cover either one of the two points.
> Four? Four seems like it most certainly would cover it, wouldn't it?
With 4 you would be able to "cover it" momentarily -- place them at
the vertices of a tetrahedron, far enough out. But as explained above,
you can't maintain the satellites in that position, so this still isn't
a solution.
The first solution I came up with involves 6 satellites. Choose any or-
bital plane you like, and then a second orbital plane that's perpendicular
to it. In each plane, choose a circular orbit sufficiently far out, and
put 3 satellites at 120 degree intervals along that orbit. So you have
2 sets of 3 satellites each, with constant geometrical relationships
within each set; each set covers a thick band around the planet, and its
coverage area includes both of the holes left by the other band.
There's plenty of overlap here, so the 90 and 120 degree angles aren't
critical in this solution. Even the orbits of the satellites in a set
don't need to be exactly circular or exactly identical, though they DO
have to have the same length of major axis so that the period is the same.
Oh, and you'd better make sure the satellites aren't on a collision course--
you can make the circular orbit in one of the planes a bit higher than
the one in the other plane, for instance.
The overlap made it conceivable that a solution with 5 satellites would
work, and just as I was writing the assertion that I couldn't see how
to work on that problem, I came up with this answer.
The idea is that you have 3 satellites in distinct orbital planes each
inclined at say 25 degrees to the equator. The planes are symmetrically
placed and the satellites placed in the same fashion in each plane, so
that they always form an equilateral triangle. (This is not, of course,
a triangle of constant size; for that they'd have to be in the same orbit.)
For instance, on a particular orbit, one satellite might cross from north
to south of the equatorial plane at longitude 0, another at 120 W, and the
third at 120 E, all of these occurring simultaneously.
These satellites will cover all of the planet except for roughly triangular
areas near the poles -- but in this plan, these uncovered areas will
oscillate in size, and each one will disappear for a time while the other
one is near its maximum size. Place the satellites at a great distance
out, so they see almost half of the planet each. The orbits are circular.
Now choose a circular orbit around the poles, i.e. inclined 90 degrees to
the equator, and place the other satellites in it, one following the other
maybe 170 degrees apart along the orbit. Put the satellites at the same
distance out as the previous set. Then the area that they don't cover
will be a thin crescent, extending slightly more than halfway around the
planet by connecting two diametrically opposite areas that are both on
the equator. The end areas of the crescent will be fixed, while the arc
will sweep around the planet, like the moving bar in some ice cream scoops,
only continuously in one direction -- north to south on one side of the
planet and then south to north on the other.
Since the satellites are all at the same distance out, their orbital
periods are all the same. So the idea is that you synchronize the
sweeping bar so that when it sweeps past the north pole, the uncovered
area is at the south pole only, and vice versa. Thus the whole planet
is covered -- each pole being covered either by at least one of the two
polar satellites, or by all three of the others, at any particular time.
I think intuitively that it works, but I'm not going to do the math to
see whether the polar uncovered areas at the north and south poles can
really be made to disappear long enough for the uncovered crescent to
sweep past. I also think intuitively that no solution with 4 satellites
is going to be possible, but I wouldn't know how to begin proving it.
--
Mark Brader, m...@sq.com "He seems unable to win without the added
SoftQuad Inc., Toronto thrill of changing sides." -- Chess
This article is in the public domain.
David Karr
>1. The planets should be in a geo-stationary orbit.
>
>This means that they should be placed at a distance 'd' from the centre of the
>earth, but not along the polar axis.
"Not along the polar axis" isn't strong enough. The satellites have to be
placed exactly over the Equator, otherwise they can't be geostationary.
Any satellite in a circular orbit at the right distance 'd' will be
_geosynchronous_, but that's a different thing.
>a. At a distance 'd' above 0, 0 (origin)
>b. At a distance 'd' above 120 E, 60 N
>c. At a distance 'd' above 120 W, 60 N
>d. At a distance 'd' above 180 (E/W), 60 S
A satellite has to orbit the planet's center. If it has a circular
orbit, it must spend half its time in the northern hemisphere, half in
the south (unless it stays on the equator all the time). So your
satellites at 60 N can't possibly stay there, though if they're in
geosynchronous orbit they will show up at that spot every 24 hours.
I'd really like to see a proof that total continous coverage *can't*
be accomplished with fewer than 5 satellites, though. None of us
seems to have come up with such a proof yet.
Stein Kulseth
|> I'd really like to see a proof that total continous coverage *can't*
|> be accomplished with fewer than 5 satellites, though. None of us
|> seems to have come up with such a proof yet.
It think it might be possible to accomplish it with 4:
Let all the satellites be distant enough to see a near half-sphere.
All satellites have same period in nearly circular, nearly equatorial
orbits, with :
A farthest away at 0 EW, at a N latitude
B closest at 0 EW at a S latitude
C farthest away at 90 E at a S latitude
D closest at 90 E at a N latitude
A and B is approximately opposite C and D, as in this polar view
(+ means northern latitude, - means southern, no sign is approx. 0 lat):
view: 1 2 3 4
A+
. . . . . . .C D. . . .
. B- . . . . . . .
. . . . . . . .
. . . A . . A .
. o . C-D+ o . . o . . o C+D-
. . . B . . B .
. . . . . . . .
. . . . . A- . . .
.C D. . . . . . . . . .
B+
(Due to the elliptical orbit B will be ahead of A when they cross the
equator in view 2, etc.)
Because the orbits are far away we treat the earth as a point o, and
the entire surface is visible if this point is within the
irregular tetrahedron defined by ABCD, this is obviously the case
in all the four snapshots above, inbetween these we have the situation
like this (symmetrical for other inbetween situations):
. . .
. .A+
. B-
. .
. o .
. .
C- .
D+ .
. . .
Line AD is obviously 'above' o, and BC obviously below o, now
if ABCD can be shown to always enclose o in this projection then
we're done.
Let A' be the A mirrored around o into C's orbit in the projection.
Initially at view 1 C will be ahead of A', and as A is at its slowest
orbital speed, C will gain on A' to start with. Later, A's speed
increases and C's speed increases, but due to symmetry (both A and C
are in the slow far half of their orbits) they will travel the
same distance between view1 and view 2. Thus AC will stay to the
left/above of o throughout this quarterperiod.
Similarily let B' be the mirror image of B. Initially B' is ahead
of D, and as B's speed is at its greatest it will gain on D to
start with. Similar symmetry applies, and line BD will stay to the
right/below o throughout.
Thus ABCD encloses o at all times, and thus these orbits should
allow the entire planet to be seen at all times.
I'll let someone else figure out exactly how far out these orbits
must be, and if they are physically realisable.
(... or, perhaps more likely, I'll let someone else figure out
just where I goofed :-)
--
stein....@tf.tele.no [X.400] stein....@nta.no [internet]
'When murders are committed by mathematics, they can be solved by
mathematics. Most of them aren't, and this one wasn't'
- Nick Charles (Dashiell Hammett's "The Thin Man")
Girish B Maraliga
geo-stationary orbits ???
--
Girish B Maraliga | "Money talks ... all mine says is G'bye !!"
OB Port Consultant | >> eMail: mara...@gnpike.enet.dec.com <<
Digital Equipment Corporation| "All opinions are mine..none others.."
Seth Breidbart
Girish B Maraliga <gir...@schism.mro.dec.com, mara...@gnpike.enet.dec.com> wrote:
>
>In article <32quvf$2...@panix3.panix.com>, se...@panix.com (Seth Breidbart) writes:
>|>In article <efoster-16...@ebmc08.draper.com>,
>|>Ed Foster <efo...@draper.com> wrote:
>|>>> >Clinton Pie...@tigger.co (958)
>|>>> >> Let's say you have a planet that's a perfect sphere. Above the planet,
>|>>> >> you are supposed to place sattelites in orbits. These satelites need
>|>>> >> to be able to "see" 100% of the planet's surface at all times. What is
>|>>> >> the minumum number of satellites that are needed?
>|>
>|>>How about putting 4 satellites at the corners of a regular tetrahedron
>|>>which completely encloses the earth?
>|>
>|>Specify their orbits.
>geo-stationary orbits ???
You can only have a geo-stationary orbit over the equator.
Seth
Isaac Kuo
>Let all the satellites be distant enough to see a near half-sphere.
>All satellites have same period in nearly circular, nearly equatorial
>orbits, with :
[solution deleted]
>I'll let someone else figure out exactly how far out these orbits
>must be, and if they are physically realisable.
>(... or, perhaps more likely, I'll let someone else figure out
>just where I goofed :-)
Your solution works, as is clear if one looks at the plane defined
by A, B, and O. The area viewed by A and B is everything except
for a narrow crescent opposite them, with the two ends of that
crescent covering just beyond the points "normal" to that plane.
It is clear that C and D cover the middle part of that crescent
simply because they are nearly opposite A and B. All that
remains are the ends of the crescent, which are covered if and
only if C stays enough on one side of the plane while D stays
enough on the other.
One can visualize the motion of the plane ABO. At all times, C
stays on one side while D stays on the other. The question of
whether thay are "far enough" from the plane at all times is
necessarily solvable merely by scaling up the distances until
"far enough" is small enough.
Of course, this solution depends on using elliptical orbits.
I wonder if it is possible with circular orbits (I still haven't
figured out a way or a proof of its impossibility).
Girish B Maraliga
You're right (..as I painfully realised later).
If I indulge in nitpicking, we _can_ use propulsion to maintain the satellites
in position (there is nothing in the original puzzle that rules it out !!)
<Idea originated from an e.mail discussion>
Of course, that would take the fun out of the puzzle !!
/G/
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
Girish B Maraliga | Internet : mara...@gnpike.enet.dec.com
OB Port Consultant, DEC | "Comedy, they say, is funnier than science
Marlboro, MA. Ph:508.467.6745| ... but science is more 'Scientific' "
David Karr
>|>>|>>How about putting 4 satellites at the corners of a regular tetrahedron
>|>>|>>which completely encloses the earth?
>
>in position (there is nothing in the original puzzle that rules it out !!)
Or as someone else suggested, you could build a framework between the
satellites to keep them all at the proper distance from each other.
How that framework "orbits" is inconsequential, as long as it stays
around the planet.
>Of course, that would take the fun out of the puzzle !!
Actually, not entirely. I thought it was worthwhile pointing out that
under no circumstances can three satellites cover the surface, but four
can easily do it if they don't move. It then *adds* to the fun to
wonder what happens when the satellites go into free-fall orbit.
This is a good example of how one puzzle becomes multiple puzzles
depending on your assumptions.
So the score for this puzzle so far, including all significant variants, is:
Four satellites suffice if you employ exotic means of maintaining
distance, for example continuous rocket propulsion. In fact it can be
argued that a tetrahedral "framework" can constitute one huge
satellite with four cameras.
Four satellites also suffice if you require free-fall orbits of small
satellites, but allow elliptical orbits. (Credit Stein Kulseth for a
very clever construction.)
Five satellites suffice if the orbits must be circular free-fall
orbits and the satellites are small. It is not known if there is a
solution for four satellites in this case.
Seth Breidbart
David Karr <ka...@cs.cornell.edu> wrote:
>So the score for this puzzle so far, including all significant variants, is:
>
>Four satellites suffice if you employ exotic means of maintaining
>distance, for example continuous rocket propulsion. In fact it can be
>argued that a tetrahedral "framework" can constitute one huge
>satellite with four cameras.
>
>Four satellites also suffice if you require free-fall orbits of small
>satellites, but allow elliptical orbits. (Credit Stein Kulseth for a
>very clever construction.)
>
>Five satellites suffice if the orbits must be circular free-fall
>orbits and the satellites are small. It is not known if there is a
>solution for four satellites in this case.
If the satellites can be large, but are required to be convex, then
two are necessary and sufficient.
Seth
David Karr
>
>If the satellites can be large, but are required to be convex, then
>two are necessary and sufficient.
OK, OK, this is also a variant of the problem.
While we're bringing up new variants, if the universe has an
appropriate curvature, one small satellite is sufficient to cover the
entire planet.
-- David