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Four ships at the ocean with the same distance
What's the more general case of this puzzler?Miss Danner and the Elmer BoysRobot Long and Robot HighHow can 64 = 65?The different twinsI like to travel. Can you tell where I've been?Help Billy retrieve his ball
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Four ships are located at the ocean, such that the distance between all four ships is the same.
How are the ships positioned?
lateral-thinking
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
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add a comment
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$begingroup$
Four ships are located at the ocean, such that the distance between all four ships is the same.
How are the ships positioned?
lateral-thinking
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
$endgroup$
1
$begingroup$
If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
$endgroup$
– Strawberry
Jul 8 at 14:57
add a comment
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$begingroup$
Four ships are located at the ocean, such that the distance between all four ships is the same.
How are the ships positioned?
lateral-thinking
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
$endgroup$
Four ships are located at the ocean, such that the distance between all four ships is the same.
How are the ships positioned?
lateral-thinking
lateral-thinking
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
asked Jul 5 at 16:41
ThomasLThomasL
8362 silver badges21 bronze badges
8362 silver badges21 bronze badges
1
$begingroup$
If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
$endgroup$
– Strawberry
Jul 8 at 14:57
add a comment
|
1
$begingroup$
If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
$endgroup$
– Strawberry
Jul 8 at 14:57
1
1
$begingroup$
If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
$endgroup$
– Strawberry
Jul 8 at 14:57
$begingroup$
If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
$endgroup$
– Strawberry
Jul 8 at 14:57
add a comment
|
8 Answers
8
active
oldest
votes
$begingroup$
Place them on the globe at the vertices of a (large) regular tetrahedron.
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
$endgroup$
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
|
show 3 more comments
$begingroup$
They could also be at the vertices of a tetrahedron, with three ships near each other on the surface of the ocean and the other being a shipwreck on the ocean floor.
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
$endgroup$
7
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
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$begingroup$
They could all be distance 0 apart, i.e. touching each other. For example, three small ships in a triangular formation (all touching each other) on board a larger shipping vessel.
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
$endgroup$
7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
2
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
2
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
add a comment
|
$begingroup$
If we take your phrasing very literally, "the ocean" shouldn't mean more than one ocean. My understanding of geography tells me ships on the surface of any single ocean could not be equidistant. But you never said they were floating on the surface. As @JonMarkPerry suggested, ships should be at the points of a regular tetrahedron, but I would position two ships floating on the ocean surface and two sunken on the floor. That way they don't have to be too far apart.
Inspired by @Jafe's answer... The Starship Enterprise is floating in the ocean, and inside the holodeck there is a blimp airship. Inside the balloon part of the airship there is typical sailboat, and inside the hold of the sailboat there is a "ship-in-a-bottle".
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
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2
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
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$begingroup$
One or more of the ships is a submarine. Either 1 below or 3 below, or any other form of a regular tetrahedron in 3D (ocean) space.
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
$endgroup$
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
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$begingroup$
One solution on the earth globe:
3 ships on a parallel at equal distances and the fourth ship on a pole, like a spherical tetrahedron. It must fulfill:
1) $D_1 = 2timespi times R_1/3$, where $D_1$ is the distance between the 3 boats on the parallel circle and $R_1$ the radius of this parallel whose length is $2pitimes R_1$.
2) $D_2 = Rtimesalpha$, where $D_2$ is the distance between any of the 3 boats and the boat on the pole, $R$ the radius of the earth and $alpha$ the angle in radians between the pole and the parallel (like a latitude but measured from the pole and not the equator).
3) $D_1=D_2$ as we want the same distance between all boats
Combining the previous equations with the fact that
the radius of the parallel is $R_1 = Rsin(alpha)$
we find:
$alpha/sin(alpha)=2pi/3$, so $alpha = 1.947 rm~radians = 111.5^o$ which is equivalent to a latitude of 21.5 degrees south
So if ...
the 3 boats are on the parallel latitude $21.5^orm S$ and the fourth is on the pole, the distances between them all are the same: $D_1 = Rtimesalpha= 6370rmkmtimes1.947= 12.402rmkm$
QED
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
$endgroup$
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
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Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
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sorry, my first time here
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– Xavier Gonzalez
Aug 4 at 23:39
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$begingroup$
Just to add to the number of possibilities regarding alternate solutions:
They are four spaceships arranged in the vertices of a tetrahedron orbiting above the ocean.
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
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$begingroup$
If the ships are specified to be on the surface of the ocean (one ocean, not multiple), and the ocean is shaped as a normal ocean (not a sphere), we can still cheat with physics:
Spatially, the ships would be equidistant if they are configured in a tetrahedron. But we need not deal only with space. If we consider distance in spacetime, we can make a similar tetrahedron where all ships are on the ocean's surface:
Three of the ships are currently in a triangle pattern, equidistant, at the same time. One ship was formerly in the middle of that triangle. Let's say it was a spaceship, and it warped out. The time it was at that position can be calculated so that the combination of time and spatial distance from that ship to one of the others is the same as the spacetime separating any of the other two ships (i.e., spatial distance). We can thus construct a tetrahedron without requiring that a ship be above or below, by using increased temporal distance to make up what lacks in spatial distance.
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Place them on the globe at the vertices of a (large) regular tetrahedron.
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
$endgroup$
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
|
show 3 more comments
$begingroup$
Place them on the globe at the vertices of a (large) regular tetrahedron.
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
$endgroup$
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
|
show 3 more comments
$begingroup$
Place them on the globe at the vertices of a (large) regular tetrahedron.
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
$endgroup$
Place them on the globe at the vertices of a (large) regular tetrahedron.
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
answered Jul 5 at 16:49
JMPJMP
27.2k6 gold badges54 silver badges116 bronze badges
27.2k6 gold badges54 silver badges116 bronze badges
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
|
show 3 more comments
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
3
3
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
$begingroup$
"positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid.
$endgroup$
– Mazura
Jul 6 at 2:36
4
4
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
$begingroup$
I'm curious if this would be possible with the real shape of the earth. I'd be interesting if someone tried to fit a tetrahedron into some geodetic model.
$endgroup$
– jinawee
Jul 6 at 20:35
8
8
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
There is so much water on Earth, it is easy. Have 2 ships on the 30ºW meridian, at 55ºN, North Atlantic, and 55ºS, South Atlantic, and 2 ships on the Equator, at 95ºE, Indian ocean, and 155ºW, Pacific.
$endgroup$
– Florian F
Jul 6 at 21:27
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
$begingroup$
Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth.
$endgroup$
– Michael Seifert
Jul 8 at 17:48
2
2
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
$begingroup$
@jinawee: Even on an ellipsoidal Earth, you could still place the ships at approx. 35.264°N and 35.264°S and at 90-degree intervals in longitude, with ships alternating Northern and Southern Hemispheres. For example: 35.264°N 20°W, 35.264°S 110°W, 35.264°S 70°E, and 35.264°N 160°W looks like it would work. Note that these latitudes would have to be geocentric rather than geodetic latitudes. If the geoid is in fact coincident with sea level, the rotational symmetry of the Earth then takes care of the rest.
$endgroup$
– Michael Seifert
Jul 8 at 18:11
|
show 3 more comments
$begingroup$
They could also be at the vertices of a tetrahedron, with three ships near each other on the surface of the ocean and the other being a shipwreck on the ocean floor.
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
$endgroup$
7
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
add a comment
|
$begingroup$
They could also be at the vertices of a tetrahedron, with three ships near each other on the surface of the ocean and the other being a shipwreck on the ocean floor.
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
$endgroup$
7
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
add a comment
|
$begingroup$
They could also be at the vertices of a tetrahedron, with three ships near each other on the surface of the ocean and the other being a shipwreck on the ocean floor.
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
$endgroup$
They could also be at the vertices of a tetrahedron, with three ships near each other on the surface of the ocean and the other being a shipwreck on the ocean floor.
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
answered Jul 6 at 9:24
Oscar CunninghamOscar Cunningham
8257 silver badges7 bronze badges
8257 silver badges7 bronze badges
7
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
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7
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I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
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– user1717828
Jul 7 at 1:10
7
7
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
$begingroup$
I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean...
$endgroup$
– user1717828
Jul 7 at 1:10
add a comment
|
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They could all be distance 0 apart, i.e. touching each other. For example, three small ships in a triangular formation (all touching each other) on board a larger shipping vessel.
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
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7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
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– Dark Thunder
Jul 5 at 17:17
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There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
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– ThomasL
Jul 5 at 17:19
2
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@ThomasL that'd be quite the wave, but you could be right
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– Dark Thunder
Jul 5 at 17:28
2
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Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
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– armb
Jul 8 at 14:59
add a comment
|
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They could all be distance 0 apart, i.e. touching each other. For example, three small ships in a triangular formation (all touching each other) on board a larger shipping vessel.
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
$endgroup$
7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
2
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
2
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
add a comment
|
$begingroup$
They could all be distance 0 apart, i.e. touching each other. For example, three small ships in a triangular formation (all touching each other) on board a larger shipping vessel.
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
$endgroup$
They could all be distance 0 apart, i.e. touching each other. For example, three small ships in a triangular formation (all touching each other) on board a larger shipping vessel.
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
answered Jul 5 at 17:05
jafejafe
37.5k6 gold badges104 silver badges372 bronze badges
37.5k6 gold badges104 silver badges372 bronze badges
7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
2
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
2
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
add a comment
|
7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
2
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
2
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
7
7
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
At the end of the day, this is probably the one that has actually occurred.
$endgroup$
– Dark Thunder
Jul 5 at 17:17
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
$begingroup$
There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships.
$endgroup$
– ThomasL
Jul 5 at 17:19
2
2
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
$begingroup$
@ThomasL that'd be quite the wave, but you could be right
$endgroup$
– Dark Thunder
Jul 5 at 17:28
2
2
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
$begingroup$
Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle.
$endgroup$
– boboquack
Jul 6 at 12:26
2
2
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
$begingroup$
shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
$endgroup$
– armb
Jul 8 at 14:59
add a comment
|
$begingroup$
If we take your phrasing very literally, "the ocean" shouldn't mean more than one ocean. My understanding of geography tells me ships on the surface of any single ocean could not be equidistant. But you never said they were floating on the surface. As @JonMarkPerry suggested, ships should be at the points of a regular tetrahedron, but I would position two ships floating on the ocean surface and two sunken on the floor. That way they don't have to be too far apart.
Inspired by @Jafe's answer... The Starship Enterprise is floating in the ocean, and inside the holodeck there is a blimp airship. Inside the balloon part of the airship there is typical sailboat, and inside the hold of the sailboat there is a "ship-in-a-bottle".
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
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2
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I like the second ariant
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– Hagen von Eitzen
Jul 8 at 0:18
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$begingroup$
If we take your phrasing very literally, "the ocean" shouldn't mean more than one ocean. My understanding of geography tells me ships on the surface of any single ocean could not be equidistant. But you never said they were floating on the surface. As @JonMarkPerry suggested, ships should be at the points of a regular tetrahedron, but I would position two ships floating on the ocean surface and two sunken on the floor. That way they don't have to be too far apart.
Inspired by @Jafe's answer... The Starship Enterprise is floating in the ocean, and inside the holodeck there is a blimp airship. Inside the balloon part of the airship there is typical sailboat, and inside the hold of the sailboat there is a "ship-in-a-bottle".
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
$endgroup$
2
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
add a comment
|
$begingroup$
If we take your phrasing very literally, "the ocean" shouldn't mean more than one ocean. My understanding of geography tells me ships on the surface of any single ocean could not be equidistant. But you never said they were floating on the surface. As @JonMarkPerry suggested, ships should be at the points of a regular tetrahedron, but I would position two ships floating on the ocean surface and two sunken on the floor. That way they don't have to be too far apart.
Inspired by @Jafe's answer... The Starship Enterprise is floating in the ocean, and inside the holodeck there is a blimp airship. Inside the balloon part of the airship there is typical sailboat, and inside the hold of the sailboat there is a "ship-in-a-bottle".
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
$endgroup$
If we take your phrasing very literally, "the ocean" shouldn't mean more than one ocean. My understanding of geography tells me ships on the surface of any single ocean could not be equidistant. But you never said they were floating on the surface. As @JonMarkPerry suggested, ships should be at the points of a regular tetrahedron, but I would position two ships floating on the ocean surface and two sunken on the floor. That way they don't have to be too far apart.
Inspired by @Jafe's answer... The Starship Enterprise is floating in the ocean, and inside the holodeck there is a blimp airship. Inside the balloon part of the airship there is typical sailboat, and inside the hold of the sailboat there is a "ship-in-a-bottle".
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
edited Jul 5 at 17:15
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
answered Jul 5 at 17:00
Dark ThunderDark Thunder
2,4795 silver badges28 bronze badges
2,4795 silver badges28 bronze badges
2
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
add a comment
|
2
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
2
2
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
$begingroup$
I like the second ariant
$endgroup$
– Hagen von Eitzen
Jul 8 at 0:18
add a comment
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One or more of the ships is a submarine. Either 1 below or 3 below, or any other form of a regular tetrahedron in 3D (ocean) space.
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
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or bar bs gurz vf fhaxra, this was my first idea as well.
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– Guntram Blohm
Jul 6 at 5:49
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Traditionally that would be considered a boat, not a ship.
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– armb
Jul 8 at 15:03
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One or more of the ships is a submarine. Either 1 below or 3 below, or any other form of a regular tetrahedron in 3D (ocean) space.
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
$endgroup$
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
add a comment
|
$begingroup$
One or more of the ships is a submarine. Either 1 below or 3 below, or any other form of a regular tetrahedron in 3D (ocean) space.
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
$endgroup$
One or more of the ships is a submarine. Either 1 below or 3 below, or any other form of a regular tetrahedron in 3D (ocean) space.
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
answered Jul 6 at 4:50
John ChurchillJohn Churchill
1212 bronze badges
1212 bronze badges
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
add a comment
|
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
or bar bs gurz vf fhaxra, this was my first idea as well.
$endgroup$
– Guntram Blohm
Jul 6 at 5:49
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
$begingroup$
Traditionally that would be considered a boat, not a ship.
$endgroup$
– armb
Jul 8 at 15:03
add a comment
|
$begingroup$
One solution on the earth globe:
3 ships on a parallel at equal distances and the fourth ship on a pole, like a spherical tetrahedron. It must fulfill:
1) $D_1 = 2timespi times R_1/3$, where $D_1$ is the distance between the 3 boats on the parallel circle and $R_1$ the radius of this parallel whose length is $2pitimes R_1$.
2) $D_2 = Rtimesalpha$, where $D_2$ is the distance between any of the 3 boats and the boat on the pole, $R$ the radius of the earth and $alpha$ the angle in radians between the pole and the parallel (like a latitude but measured from the pole and not the equator).
3) $D_1=D_2$ as we want the same distance between all boats
Combining the previous equations with the fact that
the radius of the parallel is $R_1 = Rsin(alpha)$
we find:
$alpha/sin(alpha)=2pi/3$, so $alpha = 1.947 rm~radians = 111.5^o$ which is equivalent to a latitude of 21.5 degrees south
So if ...
the 3 boats are on the parallel latitude $21.5^orm S$ and the fourth is on the pole, the distances between them all are the same: $D_1 = Rtimesalpha= 6370rmkmtimes1.947= 12.402rmkm$
QED
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
$endgroup$
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
add a comment
|
$begingroup$
One solution on the earth globe:
3 ships on a parallel at equal distances and the fourth ship on a pole, like a spherical tetrahedron. It must fulfill:
1) $D_1 = 2timespi times R_1/3$, where $D_1$ is the distance between the 3 boats on the parallel circle and $R_1$ the radius of this parallel whose length is $2pitimes R_1$.
2) $D_2 = Rtimesalpha$, where $D_2$ is the distance between any of the 3 boats and the boat on the pole, $R$ the radius of the earth and $alpha$ the angle in radians between the pole and the parallel (like a latitude but measured from the pole and not the equator).
3) $D_1=D_2$ as we want the same distance between all boats
Combining the previous equations with the fact that
the radius of the parallel is $R_1 = Rsin(alpha)$
we find:
$alpha/sin(alpha)=2pi/3$, so $alpha = 1.947 rm~radians = 111.5^o$ which is equivalent to a latitude of 21.5 degrees south
So if ...
the 3 boats are on the parallel latitude $21.5^orm S$ and the fourth is on the pole, the distances between them all are the same: $D_1 = Rtimesalpha= 6370rmkmtimes1.947= 12.402rmkm$
QED
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
$endgroup$
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
add a comment
|
$begingroup$
One solution on the earth globe:
3 ships on a parallel at equal distances and the fourth ship on a pole, like a spherical tetrahedron. It must fulfill:
1) $D_1 = 2timespi times R_1/3$, where $D_1$ is the distance between the 3 boats on the parallel circle and $R_1$ the radius of this parallel whose length is $2pitimes R_1$.
2) $D_2 = Rtimesalpha$, where $D_2$ is the distance between any of the 3 boats and the boat on the pole, $R$ the radius of the earth and $alpha$ the angle in radians between the pole and the parallel (like a latitude but measured from the pole and not the equator).
3) $D_1=D_2$ as we want the same distance between all boats
Combining the previous equations with the fact that
the radius of the parallel is $R_1 = Rsin(alpha)$
we find:
$alpha/sin(alpha)=2pi/3$, so $alpha = 1.947 rm~radians = 111.5^o$ which is equivalent to a latitude of 21.5 degrees south
So if ...
the 3 boats are on the parallel latitude $21.5^orm S$ and the fourth is on the pole, the distances between them all are the same: $D_1 = Rtimesalpha= 6370rmkmtimes1.947= 12.402rmkm$
QED
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
$endgroup$
One solution on the earth globe:
3 ships on a parallel at equal distances and the fourth ship on a pole, like a spherical tetrahedron. It must fulfill:
1) $D_1 = 2timespi times R_1/3$, where $D_1$ is the distance between the 3 boats on the parallel circle and $R_1$ the radius of this parallel whose length is $2pitimes R_1$.
2) $D_2 = Rtimesalpha$, where $D_2$ is the distance between any of the 3 boats and the boat on the pole, $R$ the radius of the earth and $alpha$ the angle in radians between the pole and the parallel (like a latitude but measured from the pole and not the equator).
3) $D_1=D_2$ as we want the same distance between all boats
Combining the previous equations with the fact that
the radius of the parallel is $R_1 = Rsin(alpha)$
we find:
$alpha/sin(alpha)=2pi/3$, so $alpha = 1.947 rm~radians = 111.5^o$ which is equivalent to a latitude of 21.5 degrees south
So if ...
the 3 boats are on the parallel latitude $21.5^orm S$ and the fourth is on the pole, the distances between them all are the same: $D_1 = Rtimesalpha= 6370rmkmtimes1.947= 12.402rmkm$
QED
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
edited Jul 6 at 21:14
Rubio♦
33.1k6 gold badges78 silver badges205 bronze badges
33.1k6 gold badges78 silver badges205 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
answered Jul 6 at 20:23
Xavier GonzalezXavier Gonzalez
312 bronze badges
312 bronze badges
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
add a comment
|
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
2
2
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Please use >! spoilers, like every other answer here. Speaking of other answers, yours is a lot of extra specifics that are nevertheless just dressing up a duplicate of the accepted answer. You should read the other answers to ensure you're not adding a duplicate.
$endgroup$
– Rubio♦
Jul 6 at 20:58
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
Also, pictures or it didn't happen.
$endgroup$
– Mr Lister
Jul 7 at 14:59
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
$begingroup$
sorry, my first time here
$endgroup$
– Xavier Gonzalez
Aug 4 at 23:39
add a comment
|
$begingroup$
Just to add to the number of possibilities regarding alternate solutions:
They are four spaceships arranged in the vertices of a tetrahedron orbiting above the ocean.
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
$endgroup$
add a comment
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$begingroup$
Just to add to the number of possibilities regarding alternate solutions:
They are four spaceships arranged in the vertices of a tetrahedron orbiting above the ocean.
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
$endgroup$
add a comment
|
$begingroup$
Just to add to the number of possibilities regarding alternate solutions:
They are four spaceships arranged in the vertices of a tetrahedron orbiting above the ocean.
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
$endgroup$
Just to add to the number of possibilities regarding alternate solutions:
They are four spaceships arranged in the vertices of a tetrahedron orbiting above the ocean.
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
answered Jul 8 at 8:42
zovitszovits
1501 silver badge4 bronze badges
1501 silver badge4 bronze badges
add a comment
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add a comment
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If the ships are specified to be on the surface of the ocean (one ocean, not multiple), and the ocean is shaped as a normal ocean (not a sphere), we can still cheat with physics:
Spatially, the ships would be equidistant if they are configured in a tetrahedron. But we need not deal only with space. If we consider distance in spacetime, we can make a similar tetrahedron where all ships are on the ocean's surface:
Three of the ships are currently in a triangle pattern, equidistant, at the same time. One ship was formerly in the middle of that triangle. Let's say it was a spaceship, and it warped out. The time it was at that position can be calculated so that the combination of time and spatial distance from that ship to one of the others is the same as the spacetime separating any of the other two ships (i.e., spatial distance). We can thus construct a tetrahedron without requiring that a ship be above or below, by using increased temporal distance to make up what lacks in spatial distance.
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
$endgroup$
add a comment
|
$begingroup$
If the ships are specified to be on the surface of the ocean (one ocean, not multiple), and the ocean is shaped as a normal ocean (not a sphere), we can still cheat with physics:
Spatially, the ships would be equidistant if they are configured in a tetrahedron. But we need not deal only with space. If we consider distance in spacetime, we can make a similar tetrahedron where all ships are on the ocean's surface:
Three of the ships are currently in a triangle pattern, equidistant, at the same time. One ship was formerly in the middle of that triangle. Let's say it was a spaceship, and it warped out. The time it was at that position can be calculated so that the combination of time and spatial distance from that ship to one of the others is the same as the spacetime separating any of the other two ships (i.e., spatial distance). We can thus construct a tetrahedron without requiring that a ship be above or below, by using increased temporal distance to make up what lacks in spatial distance.
answered Jul 8 at 10:11
piojopiojo
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If the ships are specified to be on the surface of the ocean (one ocean, not multiple), and the ocean is shaped as a normal ocean (not a sphere), we can still cheat with physics:
Spatially, the ships would be equidistant if they are configured in a tetrahedron. But we need not deal only with space. If we consider distance in spacetime, we can make a similar tetrahedron where all ships are on the ocean's surface:
Three of the ships are currently in a triangle pattern, equidistant, at the same time. One ship was formerly in the middle of that triangle. Let's say it was a spaceship, and it warped out. The time it was at that position can be calculated so that the combination of time and spatial distance from that ship to one of the others is the same as the spacetime separating any of the other two ships (i.e., spatial distance). We can thus construct a tetrahedron without requiring that a ship be above or below, by using increased temporal distance to make up what lacks in spatial distance.
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
$endgroup$
If the ships are specified to be on the surface of the ocean (one ocean, not multiple), and the ocean is shaped as a normal ocean (not a sphere), we can still cheat with physics:
Spatially, the ships would be equidistant if they are configured in a tetrahedron. But we need not deal only with space. If we consider distance in spacetime, we can make a similar tetrahedron where all ships are on the ocean's surface:
Three of the ships are currently in a triangle pattern, equidistant, at the same time. One ship was formerly in the middle of that triangle. Let's say it was a spaceship, and it warped out. The time it was at that position can be calculated so that the combination of time and spatial distance from that ship to one of the others is the same as the spacetime separating any of the other two ships (i.e., spatial distance). We can thus construct a tetrahedron without requiring that a ship be above or below, by using increased temporal distance to make up what lacks in spatial distance.
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
answered Jul 8 at 10:11
piojopiojo
1814 bronze badges
1814 bronze badges
add a comment
|
add a comment
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If I was deploying nuclear subs, this is how many I'd have (ok 5, not 4), and this is where I'd stick 'em.
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– Strawberry
Jul 8 at 14:57