# Four ships at the ocean with the same distance

Four ships are located at the ocean, such that the distance between all four ships is the same.
How are the ships positioned?

• 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. Jul 8, 2019 at 14:57

Place them on the globe at the vertices of a (large) regular tetrahedron.

• "positioned" with GPS, using their thrusters. The hard part is where, due to irregularity and 20% of the locations being invalid. Jul 6, 2019 at 2:36
• 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. Jul 6, 2019 at 20:35
• 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. Jul 6, 2019 at 21:27
• Having all four locations on land would be harder. IIRC this is a plot point in David Brin's novel Earth. Jul 8, 2019 at 17:48
• @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. Jul 8, 2019 at 18:11

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.

• I really thought this is what @ThomasL was going for, given his odd phrasing in the question: Four ships are located at the ocean... Jul 7, 2019 at 1:10

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.

• At the end of the day, this is probably the one that has actually occurred. Jul 5, 2019 at 17:17
• There is also the possibilty, that on ship is is positioned on a high wave building a tetrahedron with the other three ships. Jul 5, 2019 at 17:19
• @ThomasL that'd be quite the wave, but you could be right Jul 5, 2019 at 17:28
• Or only slightly more realistically a small circular ship jammed between 3 larger ships in a triangle. Jul 6, 2019 at 12:26
• shiplilly.com/blog/blue-marlin-ship-ships-shipping-ships
– armb
Jul 8, 2019 at 14:59

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".

• I like the second ariant Jul 8, 2019 at 0:18

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.

• or bar bs gurz vf fhaxra, this was my first idea as well. Jul 6, 2019 at 5:49
• Traditionally that would be considered a boat, not a ship.
– armb
Jul 8, 2019 at 15:03

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) $$D_1 = 2\times\pi \times R_1/3$$, 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 $$2\pi\times R_1$$.

2) $$D_2 = R\times\alpha$$, 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 $$R_1 = R\sin(\alpha)$$

we find:

$$\alpha/sin(\alpha)=2\pi/3$$, 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 $$21.5^o\rm S$$ and the fourth is on the pole, the distances between them all are the same: $$D_1 = R\times\alpha= 6370\rm{km}\times1.947= 12.402\rm{km}$$

QED

• 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.
– Rubio
Jul 6, 2019 at 20:58
• Also, pictures or it didn't happen. Jul 7, 2019 at 14:59
• sorry, my first time here Aug 4, 2019 at 23:39

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.

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.