What would be used for “coordinates” on a large asteroid?












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Assuming we had a non-spherical asteroid that doesn't have a magnetic "north", how would the inhabitants define areas on the asteroid? How would they explain to a visitor to go to a very specific spot to retrieve or leave something besides "head over the hill sunward for 50 km"










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    Welcome to Worldbuilding! Great first question!
    – kingledion
    Nov 30 at 0:57






  • 5




    FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
    – Chris Cudmore
    Nov 30 at 17:16










  • There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
    – bobtato
    Dec 2 at 19:38










  • could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
    – bukwyrm
    Dec 3 at 15:07








  • 2




    This is what you're looking for: space.stackexchange.com/questions/21278/…
    – ben
    Dec 4 at 1:32
















50














Assuming we had a non-spherical asteroid that doesn't have a magnetic "north", how would the inhabitants define areas on the asteroid? How would they explain to a visitor to go to a very specific spot to retrieve or leave something besides "head over the hill sunward for 50 km"










share|improve this question




















  • 4




    Welcome to Worldbuilding! Great first question!
    – kingledion
    Nov 30 at 0:57






  • 5




    FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
    – Chris Cudmore
    Nov 30 at 17:16










  • There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
    – bobtato
    Dec 2 at 19:38










  • could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
    – bukwyrm
    Dec 3 at 15:07








  • 2




    This is what you're looking for: space.stackexchange.com/questions/21278/…
    – ben
    Dec 4 at 1:32














50












50








50


3





Assuming we had a non-spherical asteroid that doesn't have a magnetic "north", how would the inhabitants define areas on the asteroid? How would they explain to a visitor to go to a very specific spot to retrieve or leave something besides "head over the hill sunward for 50 km"










share|improve this question















Assuming we had a non-spherical asteroid that doesn't have a magnetic "north", how would the inhabitants define areas on the asteroid? How would they explain to a visitor to go to a very specific spot to retrieve or leave something besides "head over the hill sunward for 50 km"







map-making asteroids






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share|improve this question













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edited Nov 30 at 0:57









kingledion

72.7k26244431




72.7k26244431










asked Nov 30 at 0:33









TChris Gardner

25124




25124








  • 4




    Welcome to Worldbuilding! Great first question!
    – kingledion
    Nov 30 at 0:57






  • 5




    FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
    – Chris Cudmore
    Nov 30 at 17:16










  • There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
    – bobtato
    Dec 2 at 19:38










  • could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
    – bukwyrm
    Dec 3 at 15:07








  • 2




    This is what you're looking for: space.stackexchange.com/questions/21278/…
    – ben
    Dec 4 at 1:32














  • 4




    Welcome to Worldbuilding! Great first question!
    – kingledion
    Nov 30 at 0:57






  • 5




    FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
    – Chris Cudmore
    Nov 30 at 17:16










  • There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
    – bobtato
    Dec 2 at 19:38










  • could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
    – bukwyrm
    Dec 3 at 15:07








  • 2




    This is what you're looking for: space.stackexchange.com/questions/21278/…
    – ben
    Dec 4 at 1:32








4




4




Welcome to Worldbuilding! Great first question!
– kingledion
Nov 30 at 0:57




Welcome to Worldbuilding! Great first question!
– kingledion
Nov 30 at 0:57




5




5




FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
– Chris Cudmore
Nov 30 at 17:16




FYI, the largest asteroid in the solar system is Ceres, 578 km by 458 km. using the larger dimension, this gives a circumference of 1815 km, and a surface area of 262 388 km^2, which is bigger than the UK. So navigation actually becomes a significant problem.
– Chris Cudmore
Nov 30 at 17:16












There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
– bobtato
Dec 2 at 19:38




There is a convention for defining the poles of celestial bodies (en.wikipedia.org/wiki/Poles_of_astronomical_bodies), and you could define an arbitrary convention for defining their prime meridians, for example "the point nearest the parent body at its closest approach after 1 Jan 1970". That way, two independent sets of explorers from Earth wouldn't end up using different coordinate systems. But that is a separate question to what indigenous inhabitants might come up with, or what is the most useful system.
– bobtato
Dec 2 at 19:38












could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
– bukwyrm
Dec 3 at 15:07






could you be more precise about what you are asking? For instance, to meet my friend at the cinema, i usually say : 'meet me at the cinema'. I either use my knowledge of where it is, or google maps to find it. If i use google maps, i just follow the instructions ('left, right'). This would not change on an asteroid. Or is your question about what kind of map could represent a non-spherical object? (Then i'd recommend looking at maps of mountains) Or is your question about how to map said object? (Again, look at how mountains are mapped)
– bukwyrm
Dec 3 at 15:07






2




2




This is what you're looking for: space.stackexchange.com/questions/21278/…
– ben
Dec 4 at 1:32




This is what you're looking for: space.stackexchange.com/questions/21278/…
– ben
Dec 4 at 1:32










20 Answers
20






active

oldest

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33














My suggestion would be that you select a point on the asteroid to act as a pole. Perhaps the point of first landing? Then, using that point and asteroid's centre of gravity as references, you can map spherical coordinates.






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  • 2




    This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
    – JBH
    Nov 30 at 0:40








  • 4




    How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
    – Cadence
    Nov 30 at 2:26






  • 4




    Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
    – jamesqf
    Nov 30 at 6:56






  • 1




    It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
    – BillThePlatypus
    Nov 30 at 14:28






  • 2




    @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
    – anaximander
    Nov 30 at 15:12



















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Whoever is going to be on that asteroid will necessarily used radio communication to keep in contact with the rest of the crew.



To ensure communication a network of antennas has to be established, since a single antenna could at best serve half of the asteroid.



Each position can then be simply referred to the distance from the (closest) antennas.






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  • 2




    Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
    – Joe Bloggs
    Nov 30 at 16:08












  • True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
    – MttJocy
    Nov 30 at 16:32










  • But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
    – Alexis
    Dec 3 at 12:24










  • "Lumpal positioning unit" just doesn't have the same ring to it, does it?
    – The Nate
    Dec 7 at 16:06



















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I would suggest that they use a GeoHash which subdivides the asteroid into a hierarchical grid that can be navigated based on any desired granularity



https://en.wikipedia.org/wiki/Geohash



The origin point of the geohash should be the starting outpost location and this would provide a mostly sequential means of describing location where "most" objects that are physically close to each other, share similar geohash values.



NOTE: there are some minor cases where the hash of nearby locations will not be similar, but for most things it should be good enough.



Also NOTE: Geohash is a competing system of location to What3Words which is used here in Ireland and which produces non-sequential descriptors for location. This makes it impossible to know if two locations are close to each other just based on their 3 words which is why I would recommend using Geohash instead.






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  • I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
    – Ruadhan
    Nov 30 at 12:02






  • 3




    With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
    – Mazura
    Dec 1 at 0:05



















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I'd maybe consider using the axis of rotation - it would be a very rare asteroid that isn't rotating somehow. Imagine sticking a skewer through the asteroid along the axis. That would give you a top and bottom, and then you can use spinwise and counter-spinwise (or something similar).



Obviously only works if you have an asteroid that is rotating nicely, something that is rotating a bit more chaotically might be more of an issue. If it's not rotating at all, or is spinning chaotically then Arkenstein XII's answer is definitely the way to go.






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  • 2




    That works great for most asteroids, as they would rotate around a principal axis.
    – M. Stern
    Nov 30 at 6:47






  • 2




    @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
    – MttJocy
    Nov 30 at 15:48








  • 2




    I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
    – kasperd
    Nov 30 at 22:26






  • 3




    @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
    – el duderino
    Dec 1 at 17:29






  • 2




    Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
    – M. Stern
    Dec 2 at 7:49



















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Depending on the length of your characters' stay on the asteroid (is it a mining operation, or are we setting up a habitat?), I'd say just establish a series of beacons that your personal navigation system can triangulate with. If it's longer-term, you have to set up magnetic shielding against the worst bits of solar rays anyway, so you might as well use that system and have a magnetic north as your standard.



An alternative to either of those would be to use the rotation of the asteroid as your north/south, and set up from there. If it's not spinning, this obviously wouldn't work, but it's worth bearing in mind. There are lots of ways to do it, but if this is a corporate thing, then they'd be likely to do whatever is easiest. That would be the triangulation beacon system, which they'd need for comms on a large, dense body anyway, and it would work regardless of why the asteroid is being navigated. It could be standard procedure.






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  • 1




    The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
    – MttJocy
    Nov 30 at 15:51





















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You have to remember that even on Earth, apart from the poles and equator, the coordinate system is arbitrary. Greenwich isn't even a particularly important location in the grand scheme of things, apart from hosting the Royal Observatory that's used to define the 0 meridian.



The problem is the minimum number of points you need to define a coordinate system relative to a body. On a cube we conventionally select a vertex at one corner and three dimensions from that point. On a planet you take the axis of rotation to define the equator, and an arbitrary point on the surface to define a 0 meridian, everything is then defined relative to the equator, the 0 meridian, and the direction of rotation.



Asteroids are irregular. If it's tumbling then you could take the longest axis instead of the axis of rotation and the point of landing as 0 meridian, it could then be modelled as rotating around that axis, even if the axis itself is not stable. If it's spinning then you have poles and the landing point can be the 0 meridian. If you haven't landed on it yet then any arbitrary distinctive point can act as 0 meridian.






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  • I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
    – bobtato
    Dec 2 at 19:30










  • @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
    – Separatrix
    Dec 3 at 9:08



















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You can use the same coordinate system for celestial navigation (latitude and longitude) used by Terran mariners. What you need is : an almanac, a watch, a device for measuring the angle of the stars relative to some average horizon, and a map.



Almanac



The basic concept of celestial navigation is this : imagine several easily-recognizable stars. Next, imagine that you could draw a line from each of these stars that would pass through the center of whatever you are standing on. This line will touch the ground at one (only one) location. An almanac records these stars and the location (in latitude and longitude) of the point on the surface where the imaginary line from the star touches the ground.



Watch and Calendar



And, this spot will move as the object rotates around it's own axis (days); but will only move a little with the seasons.



Measurement Device (Sextant)



When you are standing on the spot where this imaginary line from your easily-recognizable star intersects the ground, that star will be directly overhead.



Map



Likely, you are not standing on one of these spots at any particular time. The angular measurement times the average radius of the asteroid provides you with the approximate circular (radial) distance between that point and where you are. Measure multiple stars to determine where these circles overlap on a map. That is your (approximate) location






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  • 1




    I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
    – Ruadhan
    Nov 30 at 12:00



















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Use Peaks and Valleys, and polar coordinates



In the old days, prior to GPS, cartography and mapping were major and necessary fields of study. Obviously without a magnetic pole you would be unable to create universal direction, but you could still create a 'map' by using surveying techniques found mainly in triangulation of mountain peaks.



Basically, you identify the highest point in a field of view, triangulate this with others to determine its distance and height, allowing you to build up an accurate map of peaks.



As it is an asteroid, it has a centre of mass, and should be relatively easy to determine which peak is the highest - this would be your reference point. The second tallest peak would be your reference base line, from which you could measure polar coordinates in a consistent anticlockwise direction from centre of gravity (z-axis) for all peaks thereon.



Often, when exploring Australia, Charles Sturt would have to 'artificially' create mini-piles of rocks to serve as references to determine coordinates. If your asteroid is flat, this could be used to create artificial reference points too.



Edit: To clarify the process:



As an example triangulation was used to map completely the Rocky Mountains in Colorado:



enter image description here



By using mountain peaks between 3 points, distances can be determined very accurately. Identify 2 peaks from one peak, measure the angle, then move to an alternate peak to measure the new angle between the first peaks. By plotting this on a map, you can use this to determine way finding, distance measures and reference points without necessarily relating to latitude / longitude. (It was also possible to use this technique without a compass).



Another example, giving each 'triangle' an identification in the Equadorian mountains:
enter image description here



By coding each triangulation, you can use this to describe locations. Pinpointing coordinates can be accomplished by mapping over a grid (polar or rectilinear) if necessary, perhaps based on the two tallest peaks as the origin.



This is a very flexible system, and was used in the 'old days' in particular when mapping difficult terrain, and is very accurate.






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  • For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
    – Willk
    Dec 4 at 2:32










  • @Willk Yes will do so. There are some examples which could assist the OP.
    – flox
    Dec 4 at 14:41



















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The asteroid would be mapped before anybody ever landed on it, rotation, procession etc would not, I think, be used due to the relative impact on these things from habitation and industry.



https://dawn.jpl.nasa.gov/multimedia/images/image-detail.html?id=PIA17480



It doesn't seem logical either to adopt magnetic polar attributes, as these are un-managed variables.



As with L Dutch & G.B. Robinson's answers, triangulation to communication antenna and 'relative stationary' satellites would be most useful, reliable & probably become ubiquitous. It's not like asteroid dwellers would be without electronic devices at any time, as their lives would depend on them. 3d mapping is not really any more complex, tho we would probably expect long term residents to develop abbreviations and colloquial terms as references






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  • 1




    The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
    – MttJocy
    Nov 30 at 16:01










  • True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
    – Giu Piete
    Dec 1 at 14:49





















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Maybe the easiest way would be to add a number of beacons and then just run bearings off, or between, them.



Possibly designate one, or a few, as Prime(s), analogous to magnetic poles, and relate the bearings of the others to them.



There are a number of navigation systems that use, or have used, fixed beacons to determine locations with varying degrees of accuracy depending on range from the beacons.






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    1














    The center of mass would be the origin, all other coordinates (x, y, and z ~ latitude, longitude, elevation) would be pulled off that. You would orient and find the location of the CoM by triangulation of the stars as it rotates, with repeated measurements over time and in different places on the surface of the rock.






    share|improve this answer





















    • CoM would be your first reference point, but what would you use for a second reference point?
      – Michael
      Nov 30 at 20:04










    • Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
      – kaas347
      Dec 1 at 22:59






    • 1




      I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
      – Michael
      Dec 1 at 23:26










    • There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
      – მამუკა ჯიბლაძე
      Dec 2 at 5:06










    • Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
      – Nij
      Dec 3 at 11:08



















    1














    TL;DR: Two spikes, a ruler and a protractor



    There is no perfect solution, just as there is no perfect solution for Earth. Witness differing data (as plural of datum, in which a datum is the basis of a specific system of reference). Here is a document describing conversion between two equally acceptable systems of datum: https://www.ngs.noaa.gov/CORS/Articles/WGS84NAD83.pdf



    Here's a short snippet from that paper which goes to the heart of this difference:




    First of all, one should understand that the 3-D Cartesian frames to
    which the coordinates of the NAD83 and WGS84 refer are not identical.
    Their origin, axes orientation in space, and the unit of scale
    differ. Why? Simply, because the definitions of these two frames are
    based on different sets of observations, processing algorithms, and
    perhaps, geodetic assumptions.




    So there is no one right answer for difficult surfaces -- we can't even get it settled rigorously for our relatively simple surface here on Earth. We simply agree which set of imperfections to try to work around. YET there must still be many more wrong answers than right answers.



    Here's my proposed less-wrong answer, which is similar to many answers provided here, but perhaps more complete:



    Select a north pole and drive a real or imaginary spike into it. Select an arbitrary prime meridien and drive another spike in where that intersects your chosen equator. We are unlikely to have a handy set of facts to support definitions like "equidistant at all longitude", so a naturally defined equator may not be available. Likewise, if the thing does not rotate appreciably, or if that rotation is perturbed or wholly inconsistent with an intuitional model of where poles and an equator "should" be, we can still get by with nothing more than our two chosen points -- a north pole, and a 0-0 point (corresponding to 0-0 in the Gulf of Guinea, check it out). With these two points we can always measure two facts:




    • angle of rotation about the north pole from the prime meridien

    • distance from the north pole as the space crow flies


    These are not perfect, but can be agreed upon and figured by independent observers. Any radio distancing scheme could also be based upon these two points to yield the same two facts about position. This would also support map projection such as we are familiar with, including all of the existing shortcomings.






    share|improve this answer





























      1














      Let me start by clearing one thing - poles are not defined by magnetic field. Poles are defined by axis of rotation. Magnetic poles are somewhat independent from geographic poles and are only used as handy approximation, especially when you can't get a hold of more accurate measurement e.g of sun and stars, when the sky is clouded.
      https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies



      As for your actual question - it depends on what you mean by large asteroid. Really big ones (bigger than 400km diameter) end up being nearly spherical due to their own gravity, we recently decided to call them dwarf planets. All dwarf planets rotate mostly regularly so you don't need to invent any new system, just choose a prime meridian, maybe a place of first landing, or the highest mountain, and you're good.
      https://en.wikipedia.org/wiki/Dwarf_planet



      Smaller, irregular asteroids, may not map nicely, especially if their rotation is very irregular, but you can still define the main axis by calculating mass distribution, choose the axis that gives maximum moment of intertia, which for regular bodies overlaps with axis of rotation, choose prime meridian (same as above), and you're good.
      https://en.wikipedia.org/wiki/Moment_of_inertia



      For very small asteroids, don't bother with Earth-like coordinates. If your total area is comparable with big cities like London, Moscow or Paris, do something similar. Define "districts" and "neighbourhoods" with memorable names. Place couple of beacons and plaques to make it clear which is where. Everybody will learn them after living there for couple of weeks.



      In conclusion.



      For any reasonably spherelike object you can define a
      North Pole, where the major axis crosses the surface, choose an arbitrary Prime Meridian and plot an Earth-like grid of coordinates by simply projecting an imagined sphere on the actual surface. The asteroid doesn't even have to be very regular. These basic rules will work for all kind of potato shapes, as long as it has mostly positive surface curvature, which will always be the case except for very small asteroids. For those, you don't need a grid because the very small area and irregularity allows you to name all sectors unambiguously and just use those names.






      share|improve this answer



















      • 1




        I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
        – Baldrickk
        Dec 3 at 15:35








      • 1




        @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
        – Milo Bem
        Dec 3 at 17:16



















      1















      1. enter image description here


      You can apply a graph to your asteroid and find the distances of each point. From one another with this equation.



      AB^2=(Bx-Ax)^2+(By-Ay)^2…………………^2=squared



      If you make the same graph on four sided sections of asteroid you will have a rough grid.
      Next




      1. enter image description here


      This is a better way to make a three dimensional coordinate site system labeling each point from the center as a distance with the equation:



      The root of [(x2-x1)+(y2-y1)+(z2-z1)] where each point is a labeled coordinate at the surface of the asteroid.
      Next




      1. enter image description here


      Use a circle grid to plot points from center with meters or kilometers or centimeters.



      enter image description here



      Then you find the angle degree of the point with this type of equation. Remember your circle grid always has a 90 degree angle already.



      enter image description here



      One can also find angle this way



      enter image description here



      Now apply the equation.
      X^2 + y^2=(r cos @)^2. + (r sin @)^2 Where @ is a Greek representational letter for degrees or radians.



      And this gives a unique coordinate for any point and it’s distamce from center of asteroid. You can put a stake or pin at any point and create a grid by stretching a bright colored string from one to the other. And use these shapes to find area.
      Next




      1. Another way is calculus if the asteroid is spinning you can find sections of surface area as zones of the asteroid.


      enter image description here



      And use integration.



      enter image description here



      For this you need the radius or diameter at each given integrated section. It if one can imagine a large amount of sections and each having a sub measure of length sections to each sliced section it also makes a grid.



      There are few more possibilities and I will add if asked.



      This is a rough idea and I am a little rusty and just woke up. Will edit later. I had a better presentation but the pictures would not copy here.






      share|improve this answer































        1














        To elaborate on and combine a few other answers, consider that the Earth's magnetic north is not in a constant position, so it is not the factor in how lat/long coordinates are determined. Rather it is the axis of rotation as has been pointed out by others.



        The equator is determined by being the great circle (circle which circumscribes the geoid of the planet), upon which twice per year (on the equinoxes) the sun travels exactly 180 degrees from horizon to horizon in a perpendicular fashion. Leap years are a thing because there is a slight difference in the time it takes for the earth to complete a single orbit of the sun vs. the period in which two equinoxes occur.



        In your case a similar equator could be determined by observing the relationship between the body which the astroid orbits vs any 'wobble' along its axis of rotation (although this may take longer than a single 'year' or orbital period about the body.) The 2 points at which any two lines that are perpendicular to that equator intersect are its 'poles'. Lines which then intersect at a 90 degree angle at the poles can be used as meridians, the points at which those meridians intersect the equator are effectively arbitrary (the prime meridian is Euro-centric due to the fact that European cartographers defined it, not for any special geometric reason.)



        The last step is then determined by math which describes projecting the surface of the asteroid to the abstract sphere that these lines define, surveying specific landmarks (natural or artificially placed) to act as well-known points which can be surveyed against as a reference system for any other point on the surface, allowing you to triangulate coordinates. These landmarks need to be offset along either the x or y axis from one another.



        On a personal note I am also a fan of the answer which mentions 'geohashing' for modern cartographical applications. I have worked on an abstract specification known as 'Discrete Global Grid Systems' with the Open Geospatial Consortium which is conceptually similar and I'd encourage you to look into it if this sort of thing interests you. Its basis divides the sphere defined by an equator and meridians into a grid of triangles which is then projected to the surface of the body. Ironically, this is shown in the illustration of the robot in the left sidebar of this site. However your question specifically mentions coordinates, which a hashed addressing system does not provide.






        share|improve this answer





























          1





          +500









          Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.



          enter image description here
          We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$

          Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.



          Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.






          share|improve this answer





















          • How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
            – NofP
            Dec 5 at 23:15






          • 1




            @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
            – Ross Millikan
            Dec 5 at 23:29



















          0














          Expanding a bit on LDutch's answer and MttJocy's comment on it. Take a page from pre-GPS aerial and nautical navigation systems.



          Aviation commonly uses (fading as GPS is taking over, but still present) the VOR system and ADFs to figure positions and navigate. Essentially the various beacons all broadcast at a different frequency, with a directional pulse that varies in phase with the base signal. By calculating the phase difference between the primary and secondary phases your ADF provides your bearing from the station. Getting a bearing from two different stations gives you a very precise location. There are also some things that can be done in terms of phasing and signal strength to give an approximate distance from a single station, allowing for a rough position fix off of a single signal, but I am not aware of it being done in practice as the two bearing option is much more reliable. The major drawback for this system as currently implemented, and your intended use is that it is line of site and relatively short range (~200 miles). If you are too far from a station, or a pesky mountain (or the horizon of your asteroid) happens to be between you and it, you are not going to be able to pick up the signals. Meaning this system is not commonly used for surface travel.



          The LORAN system was developed in WW2 and used until fairly recently for nautical navigation. It again works by using a series of fixed position transmitters, but in this case they come in paired units. Each member of the pairing is separated by a known distance, and they pulse out synchronized signals. The receiver compares the time difference in receiving the two pulses and from that difference you can plot a line of your relative distance from the two. Grab readings from an alternate pair to plot your location. The major advantage this system has over the other is range (~1500 miles) and being much less sensitive to LoS issues, though it is also somewhat less accurate. With the sensitivity and precision of modern electronics, it is conceivable that someone could build a 3+ point LORAN system and pinpoint their location pretty accurately off of one reading, the challenge being keeping the signals synchronized across the additional broadcast stations.



          Now the range and LOS issues are primarily a function of the frequencies that are used in broadcasting the signals. So you could theoretically implement a VOR system in the frequency bands used by LORAN and have better performance in that respect, but it will also change the timing and phase calculations for determining direction. These would be engineering problems and may impact how quickly and reliably you could find you position. (Way too long since I studies this stuff to remember that level of detail)



          Once the beacon network is established your coordinates become a range and bearing to a convenient beacon. "10 klicks from beacon XYZ bearing 175".



          Note, I would advise against trying to establish a GPS network around an asteroid. That system is dependent on known orbits & timing for the satellites, and I suspect it would be very difficult to maintain such with the irregular shape and weak gravity of an asteroid.






          share|improve this answer





















          • Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
            – MttJocy
            Dec 1 at 16:37



















          0














          Assuming a sufficiently advanced society, what about artificial poles, magnetic or otherwise?






          share|improve this answer





























            0














            A map with landmarks



            The asteroid would be mapped out and photographed to generate a 3d model with landmarks



            Like a pirate's map, directions would be given from the nearest landmark.






            share|improve this answer





























              0














              Mobile network sensors.
              The OP clearly states that there are inhabitants. I'm going to make a leap of faith assuming that all the areas of interest have at least some inhabitant, or have had some visitor.



              Local linear embedding has been used to reconstruct global maps from pairwise local distances. All you need is to equip the locals with peer-to-peer devices, measure the ping time, and resolve the local linear embedding problem, et voila: the map is ready. Bonus points, as long as the inhabitants don't move simultaneously in the same direction or according to some linear transformation, you can recompute their position over time and map unexplored areas of the asteroid.



              Note that, as long as you can track changes in the pairwise distances between inhabitants, then you can define an "absolute" map, i.e. One where the thing you called North remains in the same place relative to the asteroid.



              Any new inhabitant's location can be easily located on the map again by local linear embedding, or by trilateration from existing inhabitants.



              References: http://www.cs.toronto.edu/~fritz/absps/globalcoord.pdf
              http://www.pnas.org/content/early/2008/07/07/0709842104.abstract



              Etc...






              share|improve this answer





















              • P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                – NofP
                Dec 3 at 21:24










              protected by James Dec 3 at 17:15



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              20 Answers
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              33














              My suggestion would be that you select a point on the asteroid to act as a pole. Perhaps the point of first landing? Then, using that point and asteroid's centre of gravity as references, you can map spherical coordinates.






              share|improve this answer



















              • 2




                This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
                – JBH
                Nov 30 at 0:40








              • 4




                How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
                – Cadence
                Nov 30 at 2:26






              • 4




                Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
                – jamesqf
                Nov 30 at 6:56






              • 1




                It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
                – BillThePlatypus
                Nov 30 at 14:28






              • 2




                @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
                – anaximander
                Nov 30 at 15:12
















              33














              My suggestion would be that you select a point on the asteroid to act as a pole. Perhaps the point of first landing? Then, using that point and asteroid's centre of gravity as references, you can map spherical coordinates.






              share|improve this answer



















              • 2




                This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
                – JBH
                Nov 30 at 0:40








              • 4




                How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
                – Cadence
                Nov 30 at 2:26






              • 4




                Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
                – jamesqf
                Nov 30 at 6:56






              • 1




                It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
                – BillThePlatypus
                Nov 30 at 14:28






              • 2




                @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
                – anaximander
                Nov 30 at 15:12














              33












              33








              33






              My suggestion would be that you select a point on the asteroid to act as a pole. Perhaps the point of first landing? Then, using that point and asteroid's centre of gravity as references, you can map spherical coordinates.






              share|improve this answer














              My suggestion would be that you select a point on the asteroid to act as a pole. Perhaps the point of first landing? Then, using that point and asteroid's centre of gravity as references, you can map spherical coordinates.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 30 at 0:47

























              answered Nov 30 at 0:37









              Arkenstein XII

              2,106325




              2,106325








              • 2




                This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
                – JBH
                Nov 30 at 0:40








              • 4




                How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
                – Cadence
                Nov 30 at 2:26






              • 4




                Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
                – jamesqf
                Nov 30 at 6:56






              • 1




                It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
                – BillThePlatypus
                Nov 30 at 14:28






              • 2




                @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
                – anaximander
                Nov 30 at 15:12














              • 2




                This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
                – JBH
                Nov 30 at 0:40








              • 4




                How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
                – Cadence
                Nov 30 at 2:26






              • 4




                Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
                – jamesqf
                Nov 30 at 6:56






              • 1




                It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
                – BillThePlatypus
                Nov 30 at 14:28






              • 2




                @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
                – anaximander
                Nov 30 at 15:12








              2




              2




              This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
              – JBH
              Nov 30 at 0:40






              This is the only option. The details may vary: such as using geostationary satelites, but because astroids don't have poles and their arbitrary rotation makes external (independent of the asteroid) references almost meaningless, picking a point and pounding in the proverbial survey stake is all you can do to guarantee a predictable solution. Consider the Paris Meridian.
              – JBH
              Nov 30 at 0:40






              4




              4




              How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
              – Cadence
              Nov 30 at 2:26




              How well do spherical coordinates map to an asteroid that isn't necessarily spherical? Many of them are pretty substantially "squished" in one direction or another.
              – Cadence
              Nov 30 at 2:26




              4




              4




              Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
              – jamesqf
              Nov 30 at 6:56




              Who says asteroids don't have poles? Most do, especially the larger ones. There are only a few which tumble: sciencedirect.com/science/article/abs/pii/S0019103504002568 (Though if you want to be pedantic, the Earth does a bit, too, with a precession that takes about 26K years.) So you just pick a spot to define your prime meridian, and you're set. Here's Ceres, for an example: planetary.org/multimedia/space-images/small-bodies/…
              – jamesqf
              Nov 30 at 6:56




              1




              1




              It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
              – BillThePlatypus
              Nov 30 at 14:28




              It would probably be best for the initial landing point to be on the "equator" instead. If it was chosen as a pole,every direction would be worth from there
              – BillThePlatypus
              Nov 30 at 14:28




              2




              2




              @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
              – anaximander
              Nov 30 at 15:12




              @BillThePlatypus perhaps "pole" isn't the best term; it might be better to refer to it as an "origin", in the graph sense. So, your "northern hemisphere" would be +y, "southern hemisphere" would be -y, west would be +x, and east would be -x.
              – anaximander
              Nov 30 at 15:12











              25














              Whoever is going to be on that asteroid will necessarily used radio communication to keep in contact with the rest of the crew.



              To ensure communication a network of antennas has to be established, since a single antenna could at best serve half of the asteroid.



              Each position can then be simply referred to the distance from the (closest) antennas.






              share|improve this answer

















              • 2




                Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
                – Joe Bloggs
                Nov 30 at 16:08












              • True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
                – MttJocy
                Nov 30 at 16:32










              • But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
                – Alexis
                Dec 3 at 12:24










              • "Lumpal positioning unit" just doesn't have the same ring to it, does it?
                – The Nate
                Dec 7 at 16:06
















              25














              Whoever is going to be on that asteroid will necessarily used radio communication to keep in contact with the rest of the crew.



              To ensure communication a network of antennas has to be established, since a single antenna could at best serve half of the asteroid.



              Each position can then be simply referred to the distance from the (closest) antennas.






              share|improve this answer

















              • 2




                Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
                – Joe Bloggs
                Nov 30 at 16:08












              • True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
                – MttJocy
                Nov 30 at 16:32










              • But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
                – Alexis
                Dec 3 at 12:24










              • "Lumpal positioning unit" just doesn't have the same ring to it, does it?
                – The Nate
                Dec 7 at 16:06














              25












              25








              25






              Whoever is going to be on that asteroid will necessarily used radio communication to keep in contact with the rest of the crew.



              To ensure communication a network of antennas has to be established, since a single antenna could at best serve half of the asteroid.



              Each position can then be simply referred to the distance from the (closest) antennas.






              share|improve this answer












              Whoever is going to be on that asteroid will necessarily used radio communication to keep in contact with the rest of the crew.



              To ensure communication a network of antennas has to be established, since a single antenna could at best serve half of the asteroid.



              Each position can then be simply referred to the distance from the (closest) antennas.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 0:51









              L.Dutch

              74.8k24180362




              74.8k24180362








              • 2




                Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
                – Joe Bloggs
                Nov 30 at 16:08












              • True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
                – MttJocy
                Nov 30 at 16:32










              • But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
                – Alexis
                Dec 3 at 12:24










              • "Lumpal positioning unit" just doesn't have the same ring to it, does it?
                – The Nate
                Dec 7 at 16:06














              • 2




                Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
                – Joe Bloggs
                Nov 30 at 16:08












              • True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
                – MttJocy
                Nov 30 at 16:32










              • But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
                – Alexis
                Dec 3 at 12:24










              • "Lumpal positioning unit" just doesn't have the same ring to it, does it?
                – The Nate
                Dec 7 at 16:06








              2




              2




              Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
              – Joe Bloggs
              Nov 30 at 16:08






              Triangulation off know antennae makes perfect sense. If you’re occupying for long enough it would even make sense to set up a local gps network (yes, I know the acronym doesn’t make sense, but you get the drift).
              – Joe Bloggs
              Nov 30 at 16:08














              True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
              – MttJocy
              Nov 30 at 16:32




              True enough you could simply do that and make a network of what would be NDB's (Non Directional Beacons). That said perhaps better would be to pulse this as a second rotating directional signal passes through the 0 degree bearing from the transmitter, that is to say in other words upgrade your NDB's to VOR's that allow receivers to calculate their relative bearing to the transmitter easily with no moving parts like a directional receiver antenna (A fixed base station on the ground can accommodate a directional transmitter and machinery to move it more easily than a portable device).
              – MttJocy
              Nov 30 at 16:32












              But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
              – Alexis
              Dec 3 at 12:24




              But then if you want to compute distances or optimal path between two points, you still need to know the position of their nearby antennas in some reference coordinate system applicable to the entire asteroid.
              – Alexis
              Dec 3 at 12:24












              "Lumpal positioning unit" just doesn't have the same ring to it, does it?
              – The Nate
              Dec 7 at 16:06




              "Lumpal positioning unit" just doesn't have the same ring to it, does it?
              – The Nate
              Dec 7 at 16:06











              22














              I would suggest that they use a GeoHash which subdivides the asteroid into a hierarchical grid that can be navigated based on any desired granularity



              https://en.wikipedia.org/wiki/Geohash



              The origin point of the geohash should be the starting outpost location and this would provide a mostly sequential means of describing location where "most" objects that are physically close to each other, share similar geohash values.



              NOTE: there are some minor cases where the hash of nearby locations will not be similar, but for most things it should be good enough.



              Also NOTE: Geohash is a competing system of location to What3Words which is used here in Ireland and which produces non-sequential descriptors for location. This makes it impossible to know if two locations are close to each other just based on their 3 words which is why I would recommend using Geohash instead.






              share|improve this answer





















              • I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
                – Ruadhan
                Nov 30 at 12:02






              • 3




                With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
                – Mazura
                Dec 1 at 0:05
















              22














              I would suggest that they use a GeoHash which subdivides the asteroid into a hierarchical grid that can be navigated based on any desired granularity



              https://en.wikipedia.org/wiki/Geohash



              The origin point of the geohash should be the starting outpost location and this would provide a mostly sequential means of describing location where "most" objects that are physically close to each other, share similar geohash values.



              NOTE: there are some minor cases where the hash of nearby locations will not be similar, but for most things it should be good enough.



              Also NOTE: Geohash is a competing system of location to What3Words which is used here in Ireland and which produces non-sequential descriptors for location. This makes it impossible to know if two locations are close to each other just based on their 3 words which is why I would recommend using Geohash instead.






              share|improve this answer





















              • I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
                – Ruadhan
                Nov 30 at 12:02






              • 3




                With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
                – Mazura
                Dec 1 at 0:05














              22












              22








              22






              I would suggest that they use a GeoHash which subdivides the asteroid into a hierarchical grid that can be navigated based on any desired granularity



              https://en.wikipedia.org/wiki/Geohash



              The origin point of the geohash should be the starting outpost location and this would provide a mostly sequential means of describing location where "most" objects that are physically close to each other, share similar geohash values.



              NOTE: there are some minor cases where the hash of nearby locations will not be similar, but for most things it should be good enough.



              Also NOTE: Geohash is a competing system of location to What3Words which is used here in Ireland and which produces non-sequential descriptors for location. This makes it impossible to know if two locations are close to each other just based on their 3 words which is why I would recommend using Geohash instead.






              share|improve this answer












              I would suggest that they use a GeoHash which subdivides the asteroid into a hierarchical grid that can be navigated based on any desired granularity



              https://en.wikipedia.org/wiki/Geohash



              The origin point of the geohash should be the starting outpost location and this would provide a mostly sequential means of describing location where "most" objects that are physically close to each other, share similar geohash values.



              NOTE: there are some minor cases where the hash of nearby locations will not be similar, but for most things it should be good enough.



              Also NOTE: Geohash is a competing system of location to What3Words which is used here in Ireland and which produces non-sequential descriptors for location. This makes it impossible to know if two locations are close to each other just based on their 3 words which is why I would recommend using Geohash instead.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 11:07









              bicarbon8

              3213




              3213












              • I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
                – Ruadhan
                Nov 30 at 12:02






              • 3




                With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
                – Mazura
                Dec 1 at 0:05


















              • I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
                – Ruadhan
                Nov 30 at 12:02






              • 3




                With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
                – Mazura
                Dec 1 at 0:05
















              I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
              – Ruadhan
              Nov 30 at 12:02




              I'd say this is the answer. Use a series of GPS beacons around the asteroid for navigation and geo-hash specific locations. Good answer.
              – Ruadhan
              Nov 30 at 12:02




              3




              3




              With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
              – Mazura
              Dec 1 at 0:05




              With a computer program that someone wrote to map asteroids... go figure. People who don't like that idea are probably the same people against using MechJeb in KSP.
              – Mazura
              Dec 1 at 0:05











              20














              I'd maybe consider using the axis of rotation - it would be a very rare asteroid that isn't rotating somehow. Imagine sticking a skewer through the asteroid along the axis. That would give you a top and bottom, and then you can use spinwise and counter-spinwise (or something similar).



              Obviously only works if you have an asteroid that is rotating nicely, something that is rotating a bit more chaotically might be more of an issue. If it's not rotating at all, or is spinning chaotically then Arkenstein XII's answer is definitely the way to go.






              share|improve this answer

















              • 2




                That works great for most asteroids, as they would rotate around a principal axis.
                – M. Stern
                Nov 30 at 6:47






              • 2




                @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
                – MttJocy
                Nov 30 at 15:48








              • 2




                I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
                – kasperd
                Nov 30 at 22:26






              • 3




                @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
                – el duderino
                Dec 1 at 17:29






              • 2




                Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
                – M. Stern
                Dec 2 at 7:49
















              20














              I'd maybe consider using the axis of rotation - it would be a very rare asteroid that isn't rotating somehow. Imagine sticking a skewer through the asteroid along the axis. That would give you a top and bottom, and then you can use spinwise and counter-spinwise (or something similar).



              Obviously only works if you have an asteroid that is rotating nicely, something that is rotating a bit more chaotically might be more of an issue. If it's not rotating at all, or is spinning chaotically then Arkenstein XII's answer is definitely the way to go.






              share|improve this answer

















              • 2




                That works great for most asteroids, as they would rotate around a principal axis.
                – M. Stern
                Nov 30 at 6:47






              • 2




                @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
                – MttJocy
                Nov 30 at 15:48








              • 2




                I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
                – kasperd
                Nov 30 at 22:26






              • 3




                @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
                – el duderino
                Dec 1 at 17:29






              • 2




                Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
                – M. Stern
                Dec 2 at 7:49














              20












              20








              20






              I'd maybe consider using the axis of rotation - it would be a very rare asteroid that isn't rotating somehow. Imagine sticking a skewer through the asteroid along the axis. That would give you a top and bottom, and then you can use spinwise and counter-spinwise (or something similar).



              Obviously only works if you have an asteroid that is rotating nicely, something that is rotating a bit more chaotically might be more of an issue. If it's not rotating at all, or is spinning chaotically then Arkenstein XII's answer is definitely the way to go.






              share|improve this answer












              I'd maybe consider using the axis of rotation - it would be a very rare asteroid that isn't rotating somehow. Imagine sticking a skewer through the asteroid along the axis. That would give you a top and bottom, and then you can use spinwise and counter-spinwise (or something similar).



              Obviously only works if you have an asteroid that is rotating nicely, something that is rotating a bit more chaotically might be more of an issue. If it's not rotating at all, or is spinning chaotically then Arkenstein XII's answer is definitely the way to go.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 2:50









              PainlessDocJ

              3513




              3513








              • 2




                That works great for most asteroids, as they would rotate around a principal axis.
                – M. Stern
                Nov 30 at 6:47






              • 2




                @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
                – MttJocy
                Nov 30 at 15:48








              • 2




                I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
                – kasperd
                Nov 30 at 22:26






              • 3




                @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
                – el duderino
                Dec 1 at 17:29






              • 2




                Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
                – M. Stern
                Dec 2 at 7:49














              • 2




                That works great for most asteroids, as they would rotate around a principal axis.
                – M. Stern
                Nov 30 at 6:47






              • 2




                @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
                – MttJocy
                Nov 30 at 15:48








              • 2




                I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
                – kasperd
                Nov 30 at 22:26






              • 3




                @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
                – el duderino
                Dec 1 at 17:29






              • 2




                Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
                – M. Stern
                Dec 2 at 7:49








              2




              2




              That works great for most asteroids, as they would rotate around a principal axis.
              – M. Stern
              Nov 30 at 6:47




              That works great for most asteroids, as they would rotate around a principal axis.
              – M. Stern
              Nov 30 at 6:47




              2




              2




              @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
              – MttJocy
              Nov 30 at 15:48






              @M.Stern Probably for the largest ones like Ceres and such at least the smaller objects tend to tumble more than rotate in an orderly fashion and are of course heavily perturbed by gravitational interactions or even radiation pressure from the Sun over time especially given the highly non uniform distribution of mass and surface area common on the smaller objects. Course that said the smaller ones kinda have less need of a co-ordinate system as they have surface areas more like a large retail store than anything you need a map to navigate usefully.
              – MttJocy
              Nov 30 at 15:48






              2




              2




              I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
              – kasperd
              Nov 30 at 22:26




              I think preservation of angular momentum is going to guarantee that there will always be a well defined axis of rotation.
              – kasperd
              Nov 30 at 22:26




              3




              3




              @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
              – el duderino
              Dec 1 at 17:29




              @kasperd The relationship between angular momentum and rotation gets a lot wonkier when you introduce 3D objects that don't have symmetry about an axis. Most importantly, angular momentum does not have to lie on the same axis that the object rotates about. So while angular momentum is always constant, the axis of rotation will wobble around and precess for lumpy objects like asteroids.
              – el duderino
              Dec 1 at 17:29




              2




              2




              Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
              – M. Stern
              Dec 2 at 7:49




              Only rotation around a principal axis is stable. An unstable rotation leads to deformations of the body, such that rotational energy is converted to heat. Thus any rotation will decay to a rotation around a principal axis. I would add to this nice answer that even if the rotation is not stable yet, you could use the principal axes as a reference.
              – M. Stern
              Dec 2 at 7:49











              6














              Depending on the length of your characters' stay on the asteroid (is it a mining operation, or are we setting up a habitat?), I'd say just establish a series of beacons that your personal navigation system can triangulate with. If it's longer-term, you have to set up magnetic shielding against the worst bits of solar rays anyway, so you might as well use that system and have a magnetic north as your standard.



              An alternative to either of those would be to use the rotation of the asteroid as your north/south, and set up from there. If it's not spinning, this obviously wouldn't work, but it's worth bearing in mind. There are lots of ways to do it, but if this is a corporate thing, then they'd be likely to do whatever is easiest. That would be the triangulation beacon system, which they'd need for comms on a large, dense body anyway, and it would work regardless of why the asteroid is being navigated. It could be standard procedure.






              share|improve this answer

















              • 1




                The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
                – MttJocy
                Nov 30 at 15:51


















              6














              Depending on the length of your characters' stay on the asteroid (is it a mining operation, or are we setting up a habitat?), I'd say just establish a series of beacons that your personal navigation system can triangulate with. If it's longer-term, you have to set up magnetic shielding against the worst bits of solar rays anyway, so you might as well use that system and have a magnetic north as your standard.



              An alternative to either of those would be to use the rotation of the asteroid as your north/south, and set up from there. If it's not spinning, this obviously wouldn't work, but it's worth bearing in mind. There are lots of ways to do it, but if this is a corporate thing, then they'd be likely to do whatever is easiest. That would be the triangulation beacon system, which they'd need for comms on a large, dense body anyway, and it would work regardless of why the asteroid is being navigated. It could be standard procedure.






              share|improve this answer

















              • 1




                The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
                – MttJocy
                Nov 30 at 15:51
















              6












              6








              6






              Depending on the length of your characters' stay on the asteroid (is it a mining operation, or are we setting up a habitat?), I'd say just establish a series of beacons that your personal navigation system can triangulate with. If it's longer-term, you have to set up magnetic shielding against the worst bits of solar rays anyway, so you might as well use that system and have a magnetic north as your standard.



              An alternative to either of those would be to use the rotation of the asteroid as your north/south, and set up from there. If it's not spinning, this obviously wouldn't work, but it's worth bearing in mind. There are lots of ways to do it, but if this is a corporate thing, then they'd be likely to do whatever is easiest. That would be the triangulation beacon system, which they'd need for comms on a large, dense body anyway, and it would work regardless of why the asteroid is being navigated. It could be standard procedure.






              share|improve this answer












              Depending on the length of your characters' stay on the asteroid (is it a mining operation, or are we setting up a habitat?), I'd say just establish a series of beacons that your personal navigation system can triangulate with. If it's longer-term, you have to set up magnetic shielding against the worst bits of solar rays anyway, so you might as well use that system and have a magnetic north as your standard.



              An alternative to either of those would be to use the rotation of the asteroid as your north/south, and set up from there. If it's not spinning, this obviously wouldn't work, but it's worth bearing in mind. There are lots of ways to do it, but if this is a corporate thing, then they'd be likely to do whatever is easiest. That would be the triangulation beacon system, which they'd need for comms on a large, dense body anyway, and it would work regardless of why the asteroid is being navigated. It could be standard procedure.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 5:53









              G. B. Robinson

              1517




              1517








              • 1




                The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
                – MttJocy
                Nov 30 at 15:51
















              • 1




                The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
                – MttJocy
                Nov 30 at 15:51










              1




              1




              The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
              – MttJocy
              Nov 30 at 15:51






              The rotation also works less well when the object rotates on more than one axis (Common on minor bodies like asteroids due to their low mass and non uniform shapes making them very sensitive to gravitational or even solar radiation perturbing their orbits and rotations). For short stays using the principal axis of rotation might work not so much on longer time scales as this tends to change over time for the reasons stated above.
              – MttJocy
              Nov 30 at 15:51













              3














              You have to remember that even on Earth, apart from the poles and equator, the coordinate system is arbitrary. Greenwich isn't even a particularly important location in the grand scheme of things, apart from hosting the Royal Observatory that's used to define the 0 meridian.



              The problem is the minimum number of points you need to define a coordinate system relative to a body. On a cube we conventionally select a vertex at one corner and three dimensions from that point. On a planet you take the axis of rotation to define the equator, and an arbitrary point on the surface to define a 0 meridian, everything is then defined relative to the equator, the 0 meridian, and the direction of rotation.



              Asteroids are irregular. If it's tumbling then you could take the longest axis instead of the axis of rotation and the point of landing as 0 meridian, it could then be modelled as rotating around that axis, even if the axis itself is not stable. If it's spinning then you have poles and the landing point can be the 0 meridian. If you haven't landed on it yet then any arbitrary distinctive point can act as 0 meridian.






              share|improve this answer























              • I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
                – bobtato
                Dec 2 at 19:30










              • @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
                – Separatrix
                Dec 3 at 9:08
















              3














              You have to remember that even on Earth, apart from the poles and equator, the coordinate system is arbitrary. Greenwich isn't even a particularly important location in the grand scheme of things, apart from hosting the Royal Observatory that's used to define the 0 meridian.



              The problem is the minimum number of points you need to define a coordinate system relative to a body. On a cube we conventionally select a vertex at one corner and three dimensions from that point. On a planet you take the axis of rotation to define the equator, and an arbitrary point on the surface to define a 0 meridian, everything is then defined relative to the equator, the 0 meridian, and the direction of rotation.



              Asteroids are irregular. If it's tumbling then you could take the longest axis instead of the axis of rotation and the point of landing as 0 meridian, it could then be modelled as rotating around that axis, even if the axis itself is not stable. If it's spinning then you have poles and the landing point can be the 0 meridian. If you haven't landed on it yet then any arbitrary distinctive point can act as 0 meridian.






              share|improve this answer























              • I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
                – bobtato
                Dec 2 at 19:30










              • @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
                – Separatrix
                Dec 3 at 9:08














              3












              3








              3






              You have to remember that even on Earth, apart from the poles and equator, the coordinate system is arbitrary. Greenwich isn't even a particularly important location in the grand scheme of things, apart from hosting the Royal Observatory that's used to define the 0 meridian.



              The problem is the minimum number of points you need to define a coordinate system relative to a body. On a cube we conventionally select a vertex at one corner and three dimensions from that point. On a planet you take the axis of rotation to define the equator, and an arbitrary point on the surface to define a 0 meridian, everything is then defined relative to the equator, the 0 meridian, and the direction of rotation.



              Asteroids are irregular. If it's tumbling then you could take the longest axis instead of the axis of rotation and the point of landing as 0 meridian, it could then be modelled as rotating around that axis, even if the axis itself is not stable. If it's spinning then you have poles and the landing point can be the 0 meridian. If you haven't landed on it yet then any arbitrary distinctive point can act as 0 meridian.






              share|improve this answer














              You have to remember that even on Earth, apart from the poles and equator, the coordinate system is arbitrary. Greenwich isn't even a particularly important location in the grand scheme of things, apart from hosting the Royal Observatory that's used to define the 0 meridian.



              The problem is the minimum number of points you need to define a coordinate system relative to a body. On a cube we conventionally select a vertex at one corner and three dimensions from that point. On a planet you take the axis of rotation to define the equator, and an arbitrary point on the surface to define a 0 meridian, everything is then defined relative to the equator, the 0 meridian, and the direction of rotation.



              Asteroids are irregular. If it's tumbling then you could take the longest axis instead of the axis of rotation and the point of landing as 0 meridian, it could then be modelled as rotating around that axis, even if the axis itself is not stable. If it's spinning then you have poles and the landing point can be the 0 meridian. If you haven't landed on it yet then any arbitrary distinctive point can act as 0 meridian.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Dec 3 at 12:05

























              answered Nov 30 at 16:16









              Separatrix

              75.4k30176297




              75.4k30176297












              • I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
                – bobtato
                Dec 2 at 19:30










              • @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
                – Separatrix
                Dec 3 at 9:08


















              • I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
                – bobtato
                Dec 2 at 19:30










              • @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
                – Separatrix
                Dec 3 at 9:08
















              I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
              – bobtato
              Dec 2 at 19:30




              I believe the meridian is / was based on the Royal Observatory, which is on the hill above the Naval College (though it's also more or less due North of it so it doesn't make much difference)...
              – bobtato
              Dec 2 at 19:30












              @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
              – Separatrix
              Dec 3 at 9:08




              @bobtato, the two aren't strictly independent, the Observatory came under the Admiralty as its funding was tied to work on navigation.
              – Separatrix
              Dec 3 at 9:08











              2














              You can use the same coordinate system for celestial navigation (latitude and longitude) used by Terran mariners. What you need is : an almanac, a watch, a device for measuring the angle of the stars relative to some average horizon, and a map.



              Almanac



              The basic concept of celestial navigation is this : imagine several easily-recognizable stars. Next, imagine that you could draw a line from each of these stars that would pass through the center of whatever you are standing on. This line will touch the ground at one (only one) location. An almanac records these stars and the location (in latitude and longitude) of the point on the surface where the imaginary line from the star touches the ground.



              Watch and Calendar



              And, this spot will move as the object rotates around it's own axis (days); but will only move a little with the seasons.



              Measurement Device (Sextant)



              When you are standing on the spot where this imaginary line from your easily-recognizable star intersects the ground, that star will be directly overhead.



              Map



              Likely, you are not standing on one of these spots at any particular time. The angular measurement times the average radius of the asteroid provides you with the approximate circular (radial) distance between that point and where you are. Measure multiple stars to determine where these circles overlap on a map. That is your (approximate) location






              share|improve this answer

















              • 1




                I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
                – Ruadhan
                Nov 30 at 12:00
















              2














              You can use the same coordinate system for celestial navigation (latitude and longitude) used by Terran mariners. What you need is : an almanac, a watch, a device for measuring the angle of the stars relative to some average horizon, and a map.



              Almanac



              The basic concept of celestial navigation is this : imagine several easily-recognizable stars. Next, imagine that you could draw a line from each of these stars that would pass through the center of whatever you are standing on. This line will touch the ground at one (only one) location. An almanac records these stars and the location (in latitude and longitude) of the point on the surface where the imaginary line from the star touches the ground.



              Watch and Calendar



              And, this spot will move as the object rotates around it's own axis (days); but will only move a little with the seasons.



              Measurement Device (Sextant)



              When you are standing on the spot where this imaginary line from your easily-recognizable star intersects the ground, that star will be directly overhead.



              Map



              Likely, you are not standing on one of these spots at any particular time. The angular measurement times the average radius of the asteroid provides you with the approximate circular (radial) distance between that point and where you are. Measure multiple stars to determine where these circles overlap on a map. That is your (approximate) location






              share|improve this answer

















              • 1




                I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
                – Ruadhan
                Nov 30 at 12:00














              2












              2








              2






              You can use the same coordinate system for celestial navigation (latitude and longitude) used by Terran mariners. What you need is : an almanac, a watch, a device for measuring the angle of the stars relative to some average horizon, and a map.



              Almanac



              The basic concept of celestial navigation is this : imagine several easily-recognizable stars. Next, imagine that you could draw a line from each of these stars that would pass through the center of whatever you are standing on. This line will touch the ground at one (only one) location. An almanac records these stars and the location (in latitude and longitude) of the point on the surface where the imaginary line from the star touches the ground.



              Watch and Calendar



              And, this spot will move as the object rotates around it's own axis (days); but will only move a little with the seasons.



              Measurement Device (Sextant)



              When you are standing on the spot where this imaginary line from your easily-recognizable star intersects the ground, that star will be directly overhead.



              Map



              Likely, you are not standing on one of these spots at any particular time. The angular measurement times the average radius of the asteroid provides you with the approximate circular (radial) distance between that point and where you are. Measure multiple stars to determine where these circles overlap on a map. That is your (approximate) location






              share|improve this answer












              You can use the same coordinate system for celestial navigation (latitude and longitude) used by Terran mariners. What you need is : an almanac, a watch, a device for measuring the angle of the stars relative to some average horizon, and a map.



              Almanac



              The basic concept of celestial navigation is this : imagine several easily-recognizable stars. Next, imagine that you could draw a line from each of these stars that would pass through the center of whatever you are standing on. This line will touch the ground at one (only one) location. An almanac records these stars and the location (in latitude and longitude) of the point on the surface where the imaginary line from the star touches the ground.



              Watch and Calendar



              And, this spot will move as the object rotates around it's own axis (days); but will only move a little with the seasons.



              Measurement Device (Sextant)



              When you are standing on the spot where this imaginary line from your easily-recognizable star intersects the ground, that star will be directly overhead.



              Map



              Likely, you are not standing on one of these spots at any particular time. The angular measurement times the average radius of the asteroid provides you with the approximate circular (radial) distance between that point and where you are. Measure multiple stars to determine where these circles overlap on a map. That is your (approximate) location







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 3:11









              James McLellan

              5,6311633




              5,6311633








              • 1




                I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
                – Ruadhan
                Nov 30 at 12:00














              • 1




                I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
                – Ruadhan
                Nov 30 at 12:00








              1




              1




              I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
              – Ruadhan
              Nov 30 at 12:00




              I suspect for regular navigation over the surface of a large asteroid that celestial navigation isn't going to be effective. Either the asteroid is spinning too fast to get decent measurements, or the granularity of the result isn't accurate enough for more than the most basic navigation.
              – Ruadhan
              Nov 30 at 12:00











              2














              Use Peaks and Valleys, and polar coordinates



              In the old days, prior to GPS, cartography and mapping were major and necessary fields of study. Obviously without a magnetic pole you would be unable to create universal direction, but you could still create a 'map' by using surveying techniques found mainly in triangulation of mountain peaks.



              Basically, you identify the highest point in a field of view, triangulate this with others to determine its distance and height, allowing you to build up an accurate map of peaks.



              As it is an asteroid, it has a centre of mass, and should be relatively easy to determine which peak is the highest - this would be your reference point. The second tallest peak would be your reference base line, from which you could measure polar coordinates in a consistent anticlockwise direction from centre of gravity (z-axis) for all peaks thereon.



              Often, when exploring Australia, Charles Sturt would have to 'artificially' create mini-piles of rocks to serve as references to determine coordinates. If your asteroid is flat, this could be used to create artificial reference points too.



              Edit: To clarify the process:



              As an example triangulation was used to map completely the Rocky Mountains in Colorado:



              enter image description here



              By using mountain peaks between 3 points, distances can be determined very accurately. Identify 2 peaks from one peak, measure the angle, then move to an alternate peak to measure the new angle between the first peaks. By plotting this on a map, you can use this to determine way finding, distance measures and reference points without necessarily relating to latitude / longitude. (It was also possible to use this technique without a compass).



              Another example, giving each 'triangle' an identification in the Equadorian mountains:
              enter image description here



              By coding each triangulation, you can use this to describe locations. Pinpointing coordinates can be accomplished by mapping over a grid (polar or rectilinear) if necessary, perhaps based on the two tallest peaks as the origin.



              This is a very flexible system, and was used in the 'old days' in particular when mapping difficult terrain, and is very accurate.






              share|improve this answer























              • For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
                – Willk
                Dec 4 at 2:32










              • @Willk Yes will do so. There are some examples which could assist the OP.
                – flox
                Dec 4 at 14:41
















              2














              Use Peaks and Valleys, and polar coordinates



              In the old days, prior to GPS, cartography and mapping were major and necessary fields of study. Obviously without a magnetic pole you would be unable to create universal direction, but you could still create a 'map' by using surveying techniques found mainly in triangulation of mountain peaks.



              Basically, you identify the highest point in a field of view, triangulate this with others to determine its distance and height, allowing you to build up an accurate map of peaks.



              As it is an asteroid, it has a centre of mass, and should be relatively easy to determine which peak is the highest - this would be your reference point. The second tallest peak would be your reference base line, from which you could measure polar coordinates in a consistent anticlockwise direction from centre of gravity (z-axis) for all peaks thereon.



              Often, when exploring Australia, Charles Sturt would have to 'artificially' create mini-piles of rocks to serve as references to determine coordinates. If your asteroid is flat, this could be used to create artificial reference points too.



              Edit: To clarify the process:



              As an example triangulation was used to map completely the Rocky Mountains in Colorado:



              enter image description here



              By using mountain peaks between 3 points, distances can be determined very accurately. Identify 2 peaks from one peak, measure the angle, then move to an alternate peak to measure the new angle between the first peaks. By plotting this on a map, you can use this to determine way finding, distance measures and reference points without necessarily relating to latitude / longitude. (It was also possible to use this technique without a compass).



              Another example, giving each 'triangle' an identification in the Equadorian mountains:
              enter image description here



              By coding each triangulation, you can use this to describe locations. Pinpointing coordinates can be accomplished by mapping over a grid (polar or rectilinear) if necessary, perhaps based on the two tallest peaks as the origin.



              This is a very flexible system, and was used in the 'old days' in particular when mapping difficult terrain, and is very accurate.






              share|improve this answer























              • For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
                – Willk
                Dec 4 at 2:32










              • @Willk Yes will do so. There are some examples which could assist the OP.
                – flox
                Dec 4 at 14:41














              2












              2








              2






              Use Peaks and Valleys, and polar coordinates



              In the old days, prior to GPS, cartography and mapping were major and necessary fields of study. Obviously without a magnetic pole you would be unable to create universal direction, but you could still create a 'map' by using surveying techniques found mainly in triangulation of mountain peaks.



              Basically, you identify the highest point in a field of view, triangulate this with others to determine its distance and height, allowing you to build up an accurate map of peaks.



              As it is an asteroid, it has a centre of mass, and should be relatively easy to determine which peak is the highest - this would be your reference point. The second tallest peak would be your reference base line, from which you could measure polar coordinates in a consistent anticlockwise direction from centre of gravity (z-axis) for all peaks thereon.



              Often, when exploring Australia, Charles Sturt would have to 'artificially' create mini-piles of rocks to serve as references to determine coordinates. If your asteroid is flat, this could be used to create artificial reference points too.



              Edit: To clarify the process:



              As an example triangulation was used to map completely the Rocky Mountains in Colorado:



              enter image description here



              By using mountain peaks between 3 points, distances can be determined very accurately. Identify 2 peaks from one peak, measure the angle, then move to an alternate peak to measure the new angle between the first peaks. By plotting this on a map, you can use this to determine way finding, distance measures and reference points without necessarily relating to latitude / longitude. (It was also possible to use this technique without a compass).



              Another example, giving each 'triangle' an identification in the Equadorian mountains:
              enter image description here



              By coding each triangulation, you can use this to describe locations. Pinpointing coordinates can be accomplished by mapping over a grid (polar or rectilinear) if necessary, perhaps based on the two tallest peaks as the origin.



              This is a very flexible system, and was used in the 'old days' in particular when mapping difficult terrain, and is very accurate.






              share|improve this answer














              Use Peaks and Valleys, and polar coordinates



              In the old days, prior to GPS, cartography and mapping were major and necessary fields of study. Obviously without a magnetic pole you would be unable to create universal direction, but you could still create a 'map' by using surveying techniques found mainly in triangulation of mountain peaks.



              Basically, you identify the highest point in a field of view, triangulate this with others to determine its distance and height, allowing you to build up an accurate map of peaks.



              As it is an asteroid, it has a centre of mass, and should be relatively easy to determine which peak is the highest - this would be your reference point. The second tallest peak would be your reference base line, from which you could measure polar coordinates in a consistent anticlockwise direction from centre of gravity (z-axis) for all peaks thereon.



              Often, when exploring Australia, Charles Sturt would have to 'artificially' create mini-piles of rocks to serve as references to determine coordinates. If your asteroid is flat, this could be used to create artificial reference points too.



              Edit: To clarify the process:



              As an example triangulation was used to map completely the Rocky Mountains in Colorado:



              enter image description here



              By using mountain peaks between 3 points, distances can be determined very accurately. Identify 2 peaks from one peak, measure the angle, then move to an alternate peak to measure the new angle between the first peaks. By plotting this on a map, you can use this to determine way finding, distance measures and reference points without necessarily relating to latitude / longitude. (It was also possible to use this technique without a compass).



              Another example, giving each 'triangle' an identification in the Equadorian mountains:
              enter image description here



              By coding each triangulation, you can use this to describe locations. Pinpointing coordinates can be accomplished by mapping over a grid (polar or rectilinear) if necessary, perhaps based on the two tallest peaks as the origin.



              This is a very flexible system, and was used in the 'old days' in particular when mapping difficult terrain, and is very accurate.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Dec 4 at 14:46

























              answered Dec 3 at 17:23









              flox

              7,469626




              7,469626












              • For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
                – Willk
                Dec 4 at 2:32










              • @Willk Yes will do so. There are some examples which could assist the OP.
                – flox
                Dec 4 at 14:41


















              • For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
                – Willk
                Dec 4 at 2:32










              • @Willk Yes will do so. There are some examples which could assist the OP.
                – flox
                Dec 4 at 14:41
















              For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
              – Willk
              Dec 4 at 2:32




              For the bounty, lay out an example triangulation using the highest point in a field of view and artificial reference points.
              – Willk
              Dec 4 at 2:32












              @Willk Yes will do so. There are some examples which could assist the OP.
              – flox
              Dec 4 at 14:41




              @Willk Yes will do so. There are some examples which could assist the OP.
              – flox
              Dec 4 at 14:41











              1














              The asteroid would be mapped before anybody ever landed on it, rotation, procession etc would not, I think, be used due to the relative impact on these things from habitation and industry.



              https://dawn.jpl.nasa.gov/multimedia/images/image-detail.html?id=PIA17480



              It doesn't seem logical either to adopt magnetic polar attributes, as these are un-managed variables.



              As with L Dutch & G.B. Robinson's answers, triangulation to communication antenna and 'relative stationary' satellites would be most useful, reliable & probably become ubiquitous. It's not like asteroid dwellers would be without electronic devices at any time, as their lives would depend on them. 3d mapping is not really any more complex, tho we would probably expect long term residents to develop abbreviations and colloquial terms as references






              share|improve this answer

















              • 1




                The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
                – MttJocy
                Nov 30 at 16:01










              • True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
                – Giu Piete
                Dec 1 at 14:49


















              1














              The asteroid would be mapped before anybody ever landed on it, rotation, procession etc would not, I think, be used due to the relative impact on these things from habitation and industry.



              https://dawn.jpl.nasa.gov/multimedia/images/image-detail.html?id=PIA17480



              It doesn't seem logical either to adopt magnetic polar attributes, as these are un-managed variables.



              As with L Dutch & G.B. Robinson's answers, triangulation to communication antenna and 'relative stationary' satellites would be most useful, reliable & probably become ubiquitous. It's not like asteroid dwellers would be without electronic devices at any time, as their lives would depend on them. 3d mapping is not really any more complex, tho we would probably expect long term residents to develop abbreviations and colloquial terms as references






              share|improve this answer

















              • 1




                The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
                – MttJocy
                Nov 30 at 16:01










              • True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
                – Giu Piete
                Dec 1 at 14:49
















              1












              1








              1






              The asteroid would be mapped before anybody ever landed on it, rotation, procession etc would not, I think, be used due to the relative impact on these things from habitation and industry.



              https://dawn.jpl.nasa.gov/multimedia/images/image-detail.html?id=PIA17480



              It doesn't seem logical either to adopt magnetic polar attributes, as these are un-managed variables.



              As with L Dutch & G.B. Robinson's answers, triangulation to communication antenna and 'relative stationary' satellites would be most useful, reliable & probably become ubiquitous. It's not like asteroid dwellers would be without electronic devices at any time, as their lives would depend on them. 3d mapping is not really any more complex, tho we would probably expect long term residents to develop abbreviations and colloquial terms as references






              share|improve this answer












              The asteroid would be mapped before anybody ever landed on it, rotation, procession etc would not, I think, be used due to the relative impact on these things from habitation and industry.



              https://dawn.jpl.nasa.gov/multimedia/images/image-detail.html?id=PIA17480



              It doesn't seem logical either to adopt magnetic polar attributes, as these are un-managed variables.



              As with L Dutch & G.B. Robinson's answers, triangulation to communication antenna and 'relative stationary' satellites would be most useful, reliable & probably become ubiquitous. It's not like asteroid dwellers would be without electronic devices at any time, as their lives would depend on them. 3d mapping is not really any more complex, tho we would probably expect long term residents to develop abbreviations and colloquial terms as references







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 30 at 11:23









              Giu Piete

              1676




              1676








              • 1




                The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
                – MttJocy
                Nov 30 at 16:01










              • True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
                – Giu Piete
                Dec 1 at 14:49
















              • 1




                The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
                – MttJocy
                Nov 30 at 16:01










              • True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
                – Giu Piete
                Dec 1 at 14:49










              1




              1




              The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
              – MttJocy
              Nov 30 at 16:01




              The last part is a good point most people rarely describe things by co-ordinates especially in settled areas where there are landmarks or pieces of infrastructure to use as reference points instead. This of course makes sense when you consider that our primate brains evolved to navigate the world using our eyes and things that we can physically see within the world and orient around rather than numbers assigned to virtual lines that exist only on paper.
              – MttJocy
              Nov 30 at 16:01












              True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
              – Giu Piete
              Dec 1 at 14:49






              True, but we have plenty of examples of people adapting to organised cartographical/reference systems, travelling to major cities across the world that have different fundamental concepts for their mass transit systems or no city block systems can make for a headache to those used to a different system. Most of them take a great deal of getting used to, but they are fundamentally arbitrary but organised systems...and then you have London =)
              – Giu Piete
              Dec 1 at 14:49













              1














              Maybe the easiest way would be to add a number of beacons and then just run bearings off, or between, them.



              Possibly designate one, or a few, as Prime(s), analogous to magnetic poles, and relate the bearings of the others to them.



              There are a number of navigation systems that use, or have used, fixed beacons to determine locations with varying degrees of accuracy depending on range from the beacons.






              share|improve this answer


























                1














                Maybe the easiest way would be to add a number of beacons and then just run bearings off, or between, them.



                Possibly designate one, or a few, as Prime(s), analogous to magnetic poles, and relate the bearings of the others to them.



                There are a number of navigation systems that use, or have used, fixed beacons to determine locations with varying degrees of accuracy depending on range from the beacons.






                share|improve this answer
























                  1












                  1








                  1






                  Maybe the easiest way would be to add a number of beacons and then just run bearings off, or between, them.



                  Possibly designate one, or a few, as Prime(s), analogous to magnetic poles, and relate the bearings of the others to them.



                  There are a number of navigation systems that use, or have used, fixed beacons to determine locations with varying degrees of accuracy depending on range from the beacons.






                  share|improve this answer












                  Maybe the easiest way would be to add a number of beacons and then just run bearings off, or between, them.



                  Possibly designate one, or a few, as Prime(s), analogous to magnetic poles, and relate the bearings of the others to them.



                  There are a number of navigation systems that use, or have used, fixed beacons to determine locations with varying degrees of accuracy depending on range from the beacons.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 30 at 15:33









                  GeeTee

                  812




                  812























                      1














                      The center of mass would be the origin, all other coordinates (x, y, and z ~ latitude, longitude, elevation) would be pulled off that. You would orient and find the location of the CoM by triangulation of the stars as it rotates, with repeated measurements over time and in different places on the surface of the rock.






                      share|improve this answer





















                      • CoM would be your first reference point, but what would you use for a second reference point?
                        – Michael
                        Nov 30 at 20:04










                      • Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                        – kaas347
                        Dec 1 at 22:59






                      • 1




                        I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                        – Michael
                        Dec 1 at 23:26










                      • There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                        – მამუკა ჯიბლაძე
                        Dec 2 at 5:06










                      • Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                        – Nij
                        Dec 3 at 11:08
















                      1














                      The center of mass would be the origin, all other coordinates (x, y, and z ~ latitude, longitude, elevation) would be pulled off that. You would orient and find the location of the CoM by triangulation of the stars as it rotates, with repeated measurements over time and in different places on the surface of the rock.






                      share|improve this answer





















                      • CoM would be your first reference point, but what would you use for a second reference point?
                        – Michael
                        Nov 30 at 20:04










                      • Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                        – kaas347
                        Dec 1 at 22:59






                      • 1




                        I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                        – Michael
                        Dec 1 at 23:26










                      • There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                        – მამუკა ჯიბლაძე
                        Dec 2 at 5:06










                      • Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                        – Nij
                        Dec 3 at 11:08














                      1












                      1








                      1






                      The center of mass would be the origin, all other coordinates (x, y, and z ~ latitude, longitude, elevation) would be pulled off that. You would orient and find the location of the CoM by triangulation of the stars as it rotates, with repeated measurements over time and in different places on the surface of the rock.






                      share|improve this answer












                      The center of mass would be the origin, all other coordinates (x, y, and z ~ latitude, longitude, elevation) would be pulled off that. You would orient and find the location of the CoM by triangulation of the stars as it rotates, with repeated measurements over time and in different places on the surface of the rock.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Nov 30 at 19:48









                      kaas347

                      111




                      111












                      • CoM would be your first reference point, but what would you use for a second reference point?
                        – Michael
                        Nov 30 at 20:04










                      • Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                        – kaas347
                        Dec 1 at 22:59






                      • 1




                        I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                        – Michael
                        Dec 1 at 23:26










                      • There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                        – მამუკა ჯიბლაძე
                        Dec 2 at 5:06










                      • Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                        – Nij
                        Dec 3 at 11:08


















                      • CoM would be your first reference point, but what would you use for a second reference point?
                        – Michael
                        Nov 30 at 20:04










                      • Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                        – kaas347
                        Dec 1 at 22:59






                      • 1




                        I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                        – Michael
                        Dec 1 at 23:26










                      • There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                        – მამუკა ჯიბლაძე
                        Dec 2 at 5:06










                      • Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                        – Nij
                        Dec 3 at 11:08
















                      CoM would be your first reference point, but what would you use for a second reference point?
                      – Michael
                      Nov 30 at 20:04




                      CoM would be your first reference point, but what would you use for a second reference point?
                      – Michael
                      Nov 30 at 20:04












                      Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                      – kaas347
                      Dec 1 at 22:59




                      Literally every other point you define is the second point. The coordinates of that point would be with respect to the origin, i.e. the CoM.
                      – kaas347
                      Dec 1 at 22:59




                      1




                      1




                      I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                      – Michael
                      Dec 1 at 23:26




                      I'm saying one point isn't sufficient. If you use x,y,z then you need another point to define one of your axes; if you use lat/lon you need to define where your origin is on the surface.
                      – Michael
                      Dec 1 at 23:26












                      There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                      – მამუკა ჯიბლაძე
                      Dec 2 at 5:06




                      There actually exist nonconvex asteroids. For such CoM might be easily located in outer space :)
                      – მამუკა ჯიბლაძე
                      Dec 2 at 5:06












                      Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                      – Nij
                      Dec 3 at 11:08




                      Three dimensions requires three points to define a unique coordinate basis. This won't work without a means of defining, absolutely or arbitrarily, two more points.
                      – Nij
                      Dec 3 at 11:08











                      1














                      TL;DR: Two spikes, a ruler and a protractor



                      There is no perfect solution, just as there is no perfect solution for Earth. Witness differing data (as plural of datum, in which a datum is the basis of a specific system of reference). Here is a document describing conversion between two equally acceptable systems of datum: https://www.ngs.noaa.gov/CORS/Articles/WGS84NAD83.pdf



                      Here's a short snippet from that paper which goes to the heart of this difference:




                      First of all, one should understand that the 3-D Cartesian frames to
                      which the coordinates of the NAD83 and WGS84 refer are not identical.
                      Their origin, axes orientation in space, and the unit of scale
                      differ. Why? Simply, because the definitions of these two frames are
                      based on different sets of observations, processing algorithms, and
                      perhaps, geodetic assumptions.




                      So there is no one right answer for difficult surfaces -- we can't even get it settled rigorously for our relatively simple surface here on Earth. We simply agree which set of imperfections to try to work around. YET there must still be many more wrong answers than right answers.



                      Here's my proposed less-wrong answer, which is similar to many answers provided here, but perhaps more complete:



                      Select a north pole and drive a real or imaginary spike into it. Select an arbitrary prime meridien and drive another spike in where that intersects your chosen equator. We are unlikely to have a handy set of facts to support definitions like "equidistant at all longitude", so a naturally defined equator may not be available. Likewise, if the thing does not rotate appreciably, or if that rotation is perturbed or wholly inconsistent with an intuitional model of where poles and an equator "should" be, we can still get by with nothing more than our two chosen points -- a north pole, and a 0-0 point (corresponding to 0-0 in the Gulf of Guinea, check it out). With these two points we can always measure two facts:




                      • angle of rotation about the north pole from the prime meridien

                      • distance from the north pole as the space crow flies


                      These are not perfect, but can be agreed upon and figured by independent observers. Any radio distancing scheme could also be based upon these two points to yield the same two facts about position. This would also support map projection such as we are familiar with, including all of the existing shortcomings.






                      share|improve this answer


























                        1














                        TL;DR: Two spikes, a ruler and a protractor



                        There is no perfect solution, just as there is no perfect solution for Earth. Witness differing data (as plural of datum, in which a datum is the basis of a specific system of reference). Here is a document describing conversion between two equally acceptable systems of datum: https://www.ngs.noaa.gov/CORS/Articles/WGS84NAD83.pdf



                        Here's a short snippet from that paper which goes to the heart of this difference:




                        First of all, one should understand that the 3-D Cartesian frames to
                        which the coordinates of the NAD83 and WGS84 refer are not identical.
                        Their origin, axes orientation in space, and the unit of scale
                        differ. Why? Simply, because the definitions of these two frames are
                        based on different sets of observations, processing algorithms, and
                        perhaps, geodetic assumptions.




                        So there is no one right answer for difficult surfaces -- we can't even get it settled rigorously for our relatively simple surface here on Earth. We simply agree which set of imperfections to try to work around. YET there must still be many more wrong answers than right answers.



                        Here's my proposed less-wrong answer, which is similar to many answers provided here, but perhaps more complete:



                        Select a north pole and drive a real or imaginary spike into it. Select an arbitrary prime meridien and drive another spike in where that intersects your chosen equator. We are unlikely to have a handy set of facts to support definitions like "equidistant at all longitude", so a naturally defined equator may not be available. Likewise, if the thing does not rotate appreciably, or if that rotation is perturbed or wholly inconsistent with an intuitional model of where poles and an equator "should" be, we can still get by with nothing more than our two chosen points -- a north pole, and a 0-0 point (corresponding to 0-0 in the Gulf of Guinea, check it out). With these two points we can always measure two facts:




                        • angle of rotation about the north pole from the prime meridien

                        • distance from the north pole as the space crow flies


                        These are not perfect, but can be agreed upon and figured by independent observers. Any radio distancing scheme could also be based upon these two points to yield the same two facts about position. This would also support map projection such as we are familiar with, including all of the existing shortcomings.






                        share|improve this answer
























                          1












                          1








                          1






                          TL;DR: Two spikes, a ruler and a protractor



                          There is no perfect solution, just as there is no perfect solution for Earth. Witness differing data (as plural of datum, in which a datum is the basis of a specific system of reference). Here is a document describing conversion between two equally acceptable systems of datum: https://www.ngs.noaa.gov/CORS/Articles/WGS84NAD83.pdf



                          Here's a short snippet from that paper which goes to the heart of this difference:




                          First of all, one should understand that the 3-D Cartesian frames to
                          which the coordinates of the NAD83 and WGS84 refer are not identical.
                          Their origin, axes orientation in space, and the unit of scale
                          differ. Why? Simply, because the definitions of these two frames are
                          based on different sets of observations, processing algorithms, and
                          perhaps, geodetic assumptions.




                          So there is no one right answer for difficult surfaces -- we can't even get it settled rigorously for our relatively simple surface here on Earth. We simply agree which set of imperfections to try to work around. YET there must still be many more wrong answers than right answers.



                          Here's my proposed less-wrong answer, which is similar to many answers provided here, but perhaps more complete:



                          Select a north pole and drive a real or imaginary spike into it. Select an arbitrary prime meridien and drive another spike in where that intersects your chosen equator. We are unlikely to have a handy set of facts to support definitions like "equidistant at all longitude", so a naturally defined equator may not be available. Likewise, if the thing does not rotate appreciably, or if that rotation is perturbed or wholly inconsistent with an intuitional model of where poles and an equator "should" be, we can still get by with nothing more than our two chosen points -- a north pole, and a 0-0 point (corresponding to 0-0 in the Gulf of Guinea, check it out). With these two points we can always measure two facts:




                          • angle of rotation about the north pole from the prime meridien

                          • distance from the north pole as the space crow flies


                          These are not perfect, but can be agreed upon and figured by independent observers. Any radio distancing scheme could also be based upon these two points to yield the same two facts about position. This would also support map projection such as we are familiar with, including all of the existing shortcomings.






                          share|improve this answer












                          TL;DR: Two spikes, a ruler and a protractor



                          There is no perfect solution, just as there is no perfect solution for Earth. Witness differing data (as plural of datum, in which a datum is the basis of a specific system of reference). Here is a document describing conversion between two equally acceptable systems of datum: https://www.ngs.noaa.gov/CORS/Articles/WGS84NAD83.pdf



                          Here's a short snippet from that paper which goes to the heart of this difference:




                          First of all, one should understand that the 3-D Cartesian frames to
                          which the coordinates of the NAD83 and WGS84 refer are not identical.
                          Their origin, axes orientation in space, and the unit of scale
                          differ. Why? Simply, because the definitions of these two frames are
                          based on different sets of observations, processing algorithms, and
                          perhaps, geodetic assumptions.




                          So there is no one right answer for difficult surfaces -- we can't even get it settled rigorously for our relatively simple surface here on Earth. We simply agree which set of imperfections to try to work around. YET there must still be many more wrong answers than right answers.



                          Here's my proposed less-wrong answer, which is similar to many answers provided here, but perhaps more complete:



                          Select a north pole and drive a real or imaginary spike into it. Select an arbitrary prime meridien and drive another spike in where that intersects your chosen equator. We are unlikely to have a handy set of facts to support definitions like "equidistant at all longitude", so a naturally defined equator may not be available. Likewise, if the thing does not rotate appreciably, or if that rotation is perturbed or wholly inconsistent with an intuitional model of where poles and an equator "should" be, we can still get by with nothing more than our two chosen points -- a north pole, and a 0-0 point (corresponding to 0-0 in the Gulf of Guinea, check it out). With these two points we can always measure two facts:




                          • angle of rotation about the north pole from the prime meridien

                          • distance from the north pole as the space crow flies


                          These are not perfect, but can be agreed upon and figured by independent observers. Any radio distancing scheme could also be based upon these two points to yield the same two facts about position. This would also support map projection such as we are familiar with, including all of the existing shortcomings.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Dec 1 at 9:36









                          Haakon Dahl

                          1195




                          1195























                              1














                              Let me start by clearing one thing - poles are not defined by magnetic field. Poles are defined by axis of rotation. Magnetic poles are somewhat independent from geographic poles and are only used as handy approximation, especially when you can't get a hold of more accurate measurement e.g of sun and stars, when the sky is clouded.
                              https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies



                              As for your actual question - it depends on what you mean by large asteroid. Really big ones (bigger than 400km diameter) end up being nearly spherical due to their own gravity, we recently decided to call them dwarf planets. All dwarf planets rotate mostly regularly so you don't need to invent any new system, just choose a prime meridian, maybe a place of first landing, or the highest mountain, and you're good.
                              https://en.wikipedia.org/wiki/Dwarf_planet



                              Smaller, irregular asteroids, may not map nicely, especially if their rotation is very irregular, but you can still define the main axis by calculating mass distribution, choose the axis that gives maximum moment of intertia, which for regular bodies overlaps with axis of rotation, choose prime meridian (same as above), and you're good.
                              https://en.wikipedia.org/wiki/Moment_of_inertia



                              For very small asteroids, don't bother with Earth-like coordinates. If your total area is comparable with big cities like London, Moscow or Paris, do something similar. Define "districts" and "neighbourhoods" with memorable names. Place couple of beacons and plaques to make it clear which is where. Everybody will learn them after living there for couple of weeks.



                              In conclusion.



                              For any reasonably spherelike object you can define a
                              North Pole, where the major axis crosses the surface, choose an arbitrary Prime Meridian and plot an Earth-like grid of coordinates by simply projecting an imagined sphere on the actual surface. The asteroid doesn't even have to be very regular. These basic rules will work for all kind of potato shapes, as long as it has mostly positive surface curvature, which will always be the case except for very small asteroids. For those, you don't need a grid because the very small area and irregularity allows you to name all sectors unambiguously and just use those names.






                              share|improve this answer



















                              • 1




                                I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                                – Baldrickk
                                Dec 3 at 15:35








                              • 1




                                @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                                – Milo Bem
                                Dec 3 at 17:16
















                              1














                              Let me start by clearing one thing - poles are not defined by magnetic field. Poles are defined by axis of rotation. Magnetic poles are somewhat independent from geographic poles and are only used as handy approximation, especially when you can't get a hold of more accurate measurement e.g of sun and stars, when the sky is clouded.
                              https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies



                              As for your actual question - it depends on what you mean by large asteroid. Really big ones (bigger than 400km diameter) end up being nearly spherical due to their own gravity, we recently decided to call them dwarf planets. All dwarf planets rotate mostly regularly so you don't need to invent any new system, just choose a prime meridian, maybe a place of first landing, or the highest mountain, and you're good.
                              https://en.wikipedia.org/wiki/Dwarf_planet



                              Smaller, irregular asteroids, may not map nicely, especially if their rotation is very irregular, but you can still define the main axis by calculating mass distribution, choose the axis that gives maximum moment of intertia, which for regular bodies overlaps with axis of rotation, choose prime meridian (same as above), and you're good.
                              https://en.wikipedia.org/wiki/Moment_of_inertia



                              For very small asteroids, don't bother with Earth-like coordinates. If your total area is comparable with big cities like London, Moscow or Paris, do something similar. Define "districts" and "neighbourhoods" with memorable names. Place couple of beacons and plaques to make it clear which is where. Everybody will learn them after living there for couple of weeks.



                              In conclusion.



                              For any reasonably spherelike object you can define a
                              North Pole, where the major axis crosses the surface, choose an arbitrary Prime Meridian and plot an Earth-like grid of coordinates by simply projecting an imagined sphere on the actual surface. The asteroid doesn't even have to be very regular. These basic rules will work for all kind of potato shapes, as long as it has mostly positive surface curvature, which will always be the case except for very small asteroids. For those, you don't need a grid because the very small area and irregularity allows you to name all sectors unambiguously and just use those names.






                              share|improve this answer



















                              • 1




                                I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                                – Baldrickk
                                Dec 3 at 15:35








                              • 1




                                @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                                – Milo Bem
                                Dec 3 at 17:16














                              1












                              1








                              1






                              Let me start by clearing one thing - poles are not defined by magnetic field. Poles are defined by axis of rotation. Magnetic poles are somewhat independent from geographic poles and are only used as handy approximation, especially when you can't get a hold of more accurate measurement e.g of sun and stars, when the sky is clouded.
                              https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies



                              As for your actual question - it depends on what you mean by large asteroid. Really big ones (bigger than 400km diameter) end up being nearly spherical due to their own gravity, we recently decided to call them dwarf planets. All dwarf planets rotate mostly regularly so you don't need to invent any new system, just choose a prime meridian, maybe a place of first landing, or the highest mountain, and you're good.
                              https://en.wikipedia.org/wiki/Dwarf_planet



                              Smaller, irregular asteroids, may not map nicely, especially if their rotation is very irregular, but you can still define the main axis by calculating mass distribution, choose the axis that gives maximum moment of intertia, which for regular bodies overlaps with axis of rotation, choose prime meridian (same as above), and you're good.
                              https://en.wikipedia.org/wiki/Moment_of_inertia



                              For very small asteroids, don't bother with Earth-like coordinates. If your total area is comparable with big cities like London, Moscow or Paris, do something similar. Define "districts" and "neighbourhoods" with memorable names. Place couple of beacons and plaques to make it clear which is where. Everybody will learn them after living there for couple of weeks.



                              In conclusion.



                              For any reasonably spherelike object you can define a
                              North Pole, where the major axis crosses the surface, choose an arbitrary Prime Meridian and plot an Earth-like grid of coordinates by simply projecting an imagined sphere on the actual surface. The asteroid doesn't even have to be very regular. These basic rules will work for all kind of potato shapes, as long as it has mostly positive surface curvature, which will always be the case except for very small asteroids. For those, you don't need a grid because the very small area and irregularity allows you to name all sectors unambiguously and just use those names.






                              share|improve this answer














                              Let me start by clearing one thing - poles are not defined by magnetic field. Poles are defined by axis of rotation. Magnetic poles are somewhat independent from geographic poles and are only used as handy approximation, especially when you can't get a hold of more accurate measurement e.g of sun and stars, when the sky is clouded.
                              https://en.wikipedia.org/wiki/Poles_of_astronomical_bodies



                              As for your actual question - it depends on what you mean by large asteroid. Really big ones (bigger than 400km diameter) end up being nearly spherical due to their own gravity, we recently decided to call them dwarf planets. All dwarf planets rotate mostly regularly so you don't need to invent any new system, just choose a prime meridian, maybe a place of first landing, or the highest mountain, and you're good.
                              https://en.wikipedia.org/wiki/Dwarf_planet



                              Smaller, irregular asteroids, may not map nicely, especially if their rotation is very irregular, but you can still define the main axis by calculating mass distribution, choose the axis that gives maximum moment of intertia, which for regular bodies overlaps with axis of rotation, choose prime meridian (same as above), and you're good.
                              https://en.wikipedia.org/wiki/Moment_of_inertia



                              For very small asteroids, don't bother with Earth-like coordinates. If your total area is comparable with big cities like London, Moscow or Paris, do something similar. Define "districts" and "neighbourhoods" with memorable names. Place couple of beacons and plaques to make it clear which is where. Everybody will learn them after living there for couple of weeks.



                              In conclusion.



                              For any reasonably spherelike object you can define a
                              North Pole, where the major axis crosses the surface, choose an arbitrary Prime Meridian and plot an Earth-like grid of coordinates by simply projecting an imagined sphere on the actual surface. The asteroid doesn't even have to be very regular. These basic rules will work for all kind of potato shapes, as long as it has mostly positive surface curvature, which will always be the case except for very small asteroids. For those, you don't need a grid because the very small area and irregularity allows you to name all sectors unambiguously and just use those names.







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited Dec 3 at 17:15

























                              answered Dec 3 at 14:57









                              Milo Bem

                              1,153111




                              1,153111








                              • 1




                                I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                                – Baldrickk
                                Dec 3 at 15:35








                              • 1




                                @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                                – Milo Bem
                                Dec 3 at 17:16














                              • 1




                                I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                                – Baldrickk
                                Dec 3 at 15:35








                              • 1




                                @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                                – Milo Bem
                                Dec 3 at 17:16








                              1




                              1




                              I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                              – Baldrickk
                              Dec 3 at 15:35






                              I'd probably go so far as to say that a spherical "globe" map of an asteroid is still both possible and possibly the best solution. You would still use longditude and lattitude to define position, even with irregular bodies - it's a measure of the direction from the centre of mass after all. A 2D map would end up a little distorted though, but I don't know how any projection could prevent that.
                              – Baldrickk
                              Dec 3 at 15:35






                              1




                              1




                              @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                              – Milo Bem
                              Dec 3 at 17:16




                              @Baldrickk I agree, I added a conclusion that hopefully makes this clearer
                              – Milo Bem
                              Dec 3 at 17:16











                              1















                              1. enter image description here


                              You can apply a graph to your asteroid and find the distances of each point. From one another with this equation.



                              AB^2=(Bx-Ax)^2+(By-Ay)^2…………………^2=squared



                              If you make the same graph on four sided sections of asteroid you will have a rough grid.
                              Next




                              1. enter image description here


                              This is a better way to make a three dimensional coordinate site system labeling each point from the center as a distance with the equation:



                              The root of [(x2-x1)+(y2-y1)+(z2-z1)] where each point is a labeled coordinate at the surface of the asteroid.
                              Next




                              1. enter image description here


                              Use a circle grid to plot points from center with meters or kilometers or centimeters.



                              enter image description here



                              Then you find the angle degree of the point with this type of equation. Remember your circle grid always has a 90 degree angle already.



                              enter image description here



                              One can also find angle this way



                              enter image description here



                              Now apply the equation.
                              X^2 + y^2=(r cos @)^2. + (r sin @)^2 Where @ is a Greek representational letter for degrees or radians.



                              And this gives a unique coordinate for any point and it’s distamce from center of asteroid. You can put a stake or pin at any point and create a grid by stretching a bright colored string from one to the other. And use these shapes to find area.
                              Next




                              1. Another way is calculus if the asteroid is spinning you can find sections of surface area as zones of the asteroid.


                              enter image description here



                              And use integration.



                              enter image description here



                              For this you need the radius or diameter at each given integrated section. It if one can imagine a large amount of sections and each having a sub measure of length sections to each sliced section it also makes a grid.



                              There are few more possibilities and I will add if asked.



                              This is a rough idea and I am a little rusty and just woke up. Will edit later. I had a better presentation but the pictures would not copy here.






                              share|improve this answer




























                                1















                                1. enter image description here


                                You can apply a graph to your asteroid and find the distances of each point. From one another with this equation.



                                AB^2=(Bx-Ax)^2+(By-Ay)^2…………………^2=squared



                                If you make the same graph on four sided sections of asteroid you will have a rough grid.
                                Next




                                1. enter image description here


                                This is a better way to make a three dimensional coordinate site system labeling each point from the center as a distance with the equation:



                                The root of [(x2-x1)+(y2-y1)+(z2-z1)] where each point is a labeled coordinate at the surface of the asteroid.
                                Next




                                1. enter image description here


                                Use a circle grid to plot points from center with meters or kilometers or centimeters.



                                enter image description here



                                Then you find the angle degree of the point with this type of equation. Remember your circle grid always has a 90 degree angle already.



                                enter image description here



                                One can also find angle this way



                                enter image description here



                                Now apply the equation.
                                X^2 + y^2=(r cos @)^2. + (r sin @)^2 Where @ is a Greek representational letter for degrees or radians.



                                And this gives a unique coordinate for any point and it’s distamce from center of asteroid. You can put a stake or pin at any point and create a grid by stretching a bright colored string from one to the other. And use these shapes to find area.
                                Next




                                1. Another way is calculus if the asteroid is spinning you can find sections of surface area as zones of the asteroid.


                                enter image description here



                                And use integration.



                                enter image description here



                                For this you need the radius or diameter at each given integrated section. It if one can imagine a large amount of sections and each having a sub measure of length sections to each sliced section it also makes a grid.



                                There are few more possibilities and I will add if asked.



                                This is a rough idea and I am a little rusty and just woke up. Will edit later. I had a better presentation but the pictures would not copy here.






                                share|improve this answer


























                                  1












                                  1








                                  1







                                  1. enter image description here


                                  You can apply a graph to your asteroid and find the distances of each point. From one another with this equation.



                                  AB^2=(Bx-Ax)^2+(By-Ay)^2…………………^2=squared



                                  If you make the same graph on four sided sections of asteroid you will have a rough grid.
                                  Next




                                  1. enter image description here


                                  This is a better way to make a three dimensional coordinate site system labeling each point from the center as a distance with the equation:



                                  The root of [(x2-x1)+(y2-y1)+(z2-z1)] where each point is a labeled coordinate at the surface of the asteroid.
                                  Next




                                  1. enter image description here


                                  Use a circle grid to plot points from center with meters or kilometers or centimeters.



                                  enter image description here



                                  Then you find the angle degree of the point with this type of equation. Remember your circle grid always has a 90 degree angle already.



                                  enter image description here



                                  One can also find angle this way



                                  enter image description here



                                  Now apply the equation.
                                  X^2 + y^2=(r cos @)^2. + (r sin @)^2 Where @ is a Greek representational letter for degrees or radians.



                                  And this gives a unique coordinate for any point and it’s distamce from center of asteroid. You can put a stake or pin at any point and create a grid by stretching a bright colored string from one to the other. And use these shapes to find area.
                                  Next




                                  1. Another way is calculus if the asteroid is spinning you can find sections of surface area as zones of the asteroid.


                                  enter image description here



                                  And use integration.



                                  enter image description here



                                  For this you need the radius or diameter at each given integrated section. It if one can imagine a large amount of sections and each having a sub measure of length sections to each sliced section it also makes a grid.



                                  There are few more possibilities and I will add if asked.



                                  This is a rough idea and I am a little rusty and just woke up. Will edit later. I had a better presentation but the pictures would not copy here.






                                  share|improve this answer















                                  1. enter image description here


                                  You can apply a graph to your asteroid and find the distances of each point. From one another with this equation.



                                  AB^2=(Bx-Ax)^2+(By-Ay)^2…………………^2=squared



                                  If you make the same graph on four sided sections of asteroid you will have a rough grid.
                                  Next




                                  1. enter image description here


                                  This is a better way to make a three dimensional coordinate site system labeling each point from the center as a distance with the equation:



                                  The root of [(x2-x1)+(y2-y1)+(z2-z1)] where each point is a labeled coordinate at the surface of the asteroid.
                                  Next




                                  1. enter image description here


                                  Use a circle grid to plot points from center with meters or kilometers or centimeters.



                                  enter image description here



                                  Then you find the angle degree of the point with this type of equation. Remember your circle grid always has a 90 degree angle already.



                                  enter image description here



                                  One can also find angle this way



                                  enter image description here



                                  Now apply the equation.
                                  X^2 + y^2=(r cos @)^2. + (r sin @)^2 Where @ is a Greek representational letter for degrees or radians.



                                  And this gives a unique coordinate for any point and it’s distamce from center of asteroid. You can put a stake or pin at any point and create a grid by stretching a bright colored string from one to the other. And use these shapes to find area.
                                  Next




                                  1. Another way is calculus if the asteroid is spinning you can find sections of surface area as zones of the asteroid.


                                  enter image description here



                                  And use integration.



                                  enter image description here



                                  For this you need the radius or diameter at each given integrated section. It if one can imagine a large amount of sections and each having a sub measure of length sections to each sliced section it also makes a grid.



                                  There are few more possibilities and I will add if asked.



                                  This is a rough idea and I am a little rusty and just woke up. Will edit later. I had a better presentation but the pictures would not copy here.







                                  share|improve this answer














                                  share|improve this answer



                                  share|improve this answer








                                  edited Dec 3 at 17:27

























                                  answered Dec 3 at 17:01









                                  Robus

                                  1527




                                  1527























                                      1














                                      To elaborate on and combine a few other answers, consider that the Earth's magnetic north is not in a constant position, so it is not the factor in how lat/long coordinates are determined. Rather it is the axis of rotation as has been pointed out by others.



                                      The equator is determined by being the great circle (circle which circumscribes the geoid of the planet), upon which twice per year (on the equinoxes) the sun travels exactly 180 degrees from horizon to horizon in a perpendicular fashion. Leap years are a thing because there is a slight difference in the time it takes for the earth to complete a single orbit of the sun vs. the period in which two equinoxes occur.



                                      In your case a similar equator could be determined by observing the relationship between the body which the astroid orbits vs any 'wobble' along its axis of rotation (although this may take longer than a single 'year' or orbital period about the body.) The 2 points at which any two lines that are perpendicular to that equator intersect are its 'poles'. Lines which then intersect at a 90 degree angle at the poles can be used as meridians, the points at which those meridians intersect the equator are effectively arbitrary (the prime meridian is Euro-centric due to the fact that European cartographers defined it, not for any special geometric reason.)



                                      The last step is then determined by math which describes projecting the surface of the asteroid to the abstract sphere that these lines define, surveying specific landmarks (natural or artificially placed) to act as well-known points which can be surveyed against as a reference system for any other point on the surface, allowing you to triangulate coordinates. These landmarks need to be offset along either the x or y axis from one another.



                                      On a personal note I am also a fan of the answer which mentions 'geohashing' for modern cartographical applications. I have worked on an abstract specification known as 'Discrete Global Grid Systems' with the Open Geospatial Consortium which is conceptually similar and I'd encourage you to look into it if this sort of thing interests you. Its basis divides the sphere defined by an equator and meridians into a grid of triangles which is then projected to the surface of the body. Ironically, this is shown in the illustration of the robot in the left sidebar of this site. However your question specifically mentions coordinates, which a hashed addressing system does not provide.






                                      share|improve this answer


























                                        1














                                        To elaborate on and combine a few other answers, consider that the Earth's magnetic north is not in a constant position, so it is not the factor in how lat/long coordinates are determined. Rather it is the axis of rotation as has been pointed out by others.



                                        The equator is determined by being the great circle (circle which circumscribes the geoid of the planet), upon which twice per year (on the equinoxes) the sun travels exactly 180 degrees from horizon to horizon in a perpendicular fashion. Leap years are a thing because there is a slight difference in the time it takes for the earth to complete a single orbit of the sun vs. the period in which two equinoxes occur.



                                        In your case a similar equator could be determined by observing the relationship between the body which the astroid orbits vs any 'wobble' along its axis of rotation (although this may take longer than a single 'year' or orbital period about the body.) The 2 points at which any two lines that are perpendicular to that equator intersect are its 'poles'. Lines which then intersect at a 90 degree angle at the poles can be used as meridians, the points at which those meridians intersect the equator are effectively arbitrary (the prime meridian is Euro-centric due to the fact that European cartographers defined it, not for any special geometric reason.)



                                        The last step is then determined by math which describes projecting the surface of the asteroid to the abstract sphere that these lines define, surveying specific landmarks (natural or artificially placed) to act as well-known points which can be surveyed against as a reference system for any other point on the surface, allowing you to triangulate coordinates. These landmarks need to be offset along either the x or y axis from one another.



                                        On a personal note I am also a fan of the answer which mentions 'geohashing' for modern cartographical applications. I have worked on an abstract specification known as 'Discrete Global Grid Systems' with the Open Geospatial Consortium which is conceptually similar and I'd encourage you to look into it if this sort of thing interests you. Its basis divides the sphere defined by an equator and meridians into a grid of triangles which is then projected to the surface of the body. Ironically, this is shown in the illustration of the robot in the left sidebar of this site. However your question specifically mentions coordinates, which a hashed addressing system does not provide.






                                        share|improve this answer
























                                          1












                                          1








                                          1






                                          To elaborate on and combine a few other answers, consider that the Earth's magnetic north is not in a constant position, so it is not the factor in how lat/long coordinates are determined. Rather it is the axis of rotation as has been pointed out by others.



                                          The equator is determined by being the great circle (circle which circumscribes the geoid of the planet), upon which twice per year (on the equinoxes) the sun travels exactly 180 degrees from horizon to horizon in a perpendicular fashion. Leap years are a thing because there is a slight difference in the time it takes for the earth to complete a single orbit of the sun vs. the period in which two equinoxes occur.



                                          In your case a similar equator could be determined by observing the relationship between the body which the astroid orbits vs any 'wobble' along its axis of rotation (although this may take longer than a single 'year' or orbital period about the body.) The 2 points at which any two lines that are perpendicular to that equator intersect are its 'poles'. Lines which then intersect at a 90 degree angle at the poles can be used as meridians, the points at which those meridians intersect the equator are effectively arbitrary (the prime meridian is Euro-centric due to the fact that European cartographers defined it, not for any special geometric reason.)



                                          The last step is then determined by math which describes projecting the surface of the asteroid to the abstract sphere that these lines define, surveying specific landmarks (natural or artificially placed) to act as well-known points which can be surveyed against as a reference system for any other point on the surface, allowing you to triangulate coordinates. These landmarks need to be offset along either the x or y axis from one another.



                                          On a personal note I am also a fan of the answer which mentions 'geohashing' for modern cartographical applications. I have worked on an abstract specification known as 'Discrete Global Grid Systems' with the Open Geospatial Consortium which is conceptually similar and I'd encourage you to look into it if this sort of thing interests you. Its basis divides the sphere defined by an equator and meridians into a grid of triangles which is then projected to the surface of the body. Ironically, this is shown in the illustration of the robot in the left sidebar of this site. However your question specifically mentions coordinates, which a hashed addressing system does not provide.






                                          share|improve this answer












                                          To elaborate on and combine a few other answers, consider that the Earth's magnetic north is not in a constant position, so it is not the factor in how lat/long coordinates are determined. Rather it is the axis of rotation as has been pointed out by others.



                                          The equator is determined by being the great circle (circle which circumscribes the geoid of the planet), upon which twice per year (on the equinoxes) the sun travels exactly 180 degrees from horizon to horizon in a perpendicular fashion. Leap years are a thing because there is a slight difference in the time it takes for the earth to complete a single orbit of the sun vs. the period in which two equinoxes occur.



                                          In your case a similar equator could be determined by observing the relationship between the body which the astroid orbits vs any 'wobble' along its axis of rotation (although this may take longer than a single 'year' or orbital period about the body.) The 2 points at which any two lines that are perpendicular to that equator intersect are its 'poles'. Lines which then intersect at a 90 degree angle at the poles can be used as meridians, the points at which those meridians intersect the equator are effectively arbitrary (the prime meridian is Euro-centric due to the fact that European cartographers defined it, not for any special geometric reason.)



                                          The last step is then determined by math which describes projecting the surface of the asteroid to the abstract sphere that these lines define, surveying specific landmarks (natural or artificially placed) to act as well-known points which can be surveyed against as a reference system for any other point on the surface, allowing you to triangulate coordinates. These landmarks need to be offset along either the x or y axis from one another.



                                          On a personal note I am also a fan of the answer which mentions 'geohashing' for modern cartographical applications. I have worked on an abstract specification known as 'Discrete Global Grid Systems' with the Open Geospatial Consortium which is conceptually similar and I'd encourage you to look into it if this sort of thing interests you. Its basis divides the sphere defined by an equator and meridians into a grid of triangles which is then projected to the surface of the body. Ironically, this is shown in the illustration of the robot in the left sidebar of this site. However your question specifically mentions coordinates, which a hashed addressing system does not provide.







                                          share|improve this answer












                                          share|improve this answer



                                          share|improve this answer










                                          answered Dec 4 at 0:59









                                          navigator_

                                          42134




                                          42134























                                              1





                                              +500









                                              Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.



                                              enter image description here
                                              We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$

                                              Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.



                                              Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.






                                              share|improve this answer





















                                              • How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                                – NofP
                                                Dec 5 at 23:15






                                              • 1




                                                @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                                – Ross Millikan
                                                Dec 5 at 23:29
















                                              1





                                              +500









                                              Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.



                                              enter image description here
                                              We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$

                                              Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.



                                              Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.






                                              share|improve this answer





















                                              • How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                                – NofP
                                                Dec 5 at 23:15






                                              • 1




                                                @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                                – Ross Millikan
                                                Dec 5 at 23:29














                                              1





                                              +500







                                              1





                                              +500



                                              1




                                              +500




                                              Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.



                                              enter image description here
                                              We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$

                                              Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.



                                              Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.






                                              share|improve this answer












                                              Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.



                                              enter image description here
                                              We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$

                                              Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.



                                              Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.







                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered Dec 5 at 16:45









                                              Ross Millikan

                                              81558




                                              81558












                                              • How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                                – NofP
                                                Dec 5 at 23:15






                                              • 1




                                                @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                                – Ross Millikan
                                                Dec 5 at 23:29


















                                              • How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                                – NofP
                                                Dec 5 at 23:15






                                              • 1




                                                @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                                – Ross Millikan
                                                Dec 5 at 23:29
















                                              How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                              – NofP
                                              Dec 5 at 23:15




                                              How many such markers are needed for a large asteroid? [reference for sizes en.wikipedia.org/wiki/Asteroid#Size_distribution]
                                              – NofP
                                              Dec 5 at 23:15




                                              1




                                              1




                                              @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                              – Ross Millikan
                                              Dec 5 at 23:29




                                              @NofP: It depends how tall a pole you put at each point and how tall your person is. If the poles are 2m high and the person is also 2m tall (to make the calcs easy) and the radius is R, you get a right triangle with hypotenuse $R+2$ and one leg $R$. The other leg, which is half the distance you can see a pole is $sqrt{(R+2)^2-R^2}=sqrt{4R+4}$. For a $5 km$ radius asteroid the sight distance is $280$ m or a dozen around the equator.
                                              – Ross Millikan
                                              Dec 5 at 23:29











                                              0














                                              Expanding a bit on LDutch's answer and MttJocy's comment on it. Take a page from pre-GPS aerial and nautical navigation systems.



                                              Aviation commonly uses (fading as GPS is taking over, but still present) the VOR system and ADFs to figure positions and navigate. Essentially the various beacons all broadcast at a different frequency, with a directional pulse that varies in phase with the base signal. By calculating the phase difference between the primary and secondary phases your ADF provides your bearing from the station. Getting a bearing from two different stations gives you a very precise location. There are also some things that can be done in terms of phasing and signal strength to give an approximate distance from a single station, allowing for a rough position fix off of a single signal, but I am not aware of it being done in practice as the two bearing option is much more reliable. The major drawback for this system as currently implemented, and your intended use is that it is line of site and relatively short range (~200 miles). If you are too far from a station, or a pesky mountain (or the horizon of your asteroid) happens to be between you and it, you are not going to be able to pick up the signals. Meaning this system is not commonly used for surface travel.



                                              The LORAN system was developed in WW2 and used until fairly recently for nautical navigation. It again works by using a series of fixed position transmitters, but in this case they come in paired units. Each member of the pairing is separated by a known distance, and they pulse out synchronized signals. The receiver compares the time difference in receiving the two pulses and from that difference you can plot a line of your relative distance from the two. Grab readings from an alternate pair to plot your location. The major advantage this system has over the other is range (~1500 miles) and being much less sensitive to LoS issues, though it is also somewhat less accurate. With the sensitivity and precision of modern electronics, it is conceivable that someone could build a 3+ point LORAN system and pinpoint their location pretty accurately off of one reading, the challenge being keeping the signals synchronized across the additional broadcast stations.



                                              Now the range and LOS issues are primarily a function of the frequencies that are used in broadcasting the signals. So you could theoretically implement a VOR system in the frequency bands used by LORAN and have better performance in that respect, but it will also change the timing and phase calculations for determining direction. These would be engineering problems and may impact how quickly and reliably you could find you position. (Way too long since I studies this stuff to remember that level of detail)



                                              Once the beacon network is established your coordinates become a range and bearing to a convenient beacon. "10 klicks from beacon XYZ bearing 175".



                                              Note, I would advise against trying to establish a GPS network around an asteroid. That system is dependent on known orbits & timing for the satellites, and I suspect it would be very difficult to maintain such with the irregular shape and weak gravity of an asteroid.






                                              share|improve this answer





















                                              • Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                                – MttJocy
                                                Dec 1 at 16:37
















                                              0














                                              Expanding a bit on LDutch's answer and MttJocy's comment on it. Take a page from pre-GPS aerial and nautical navigation systems.



                                              Aviation commonly uses (fading as GPS is taking over, but still present) the VOR system and ADFs to figure positions and navigate. Essentially the various beacons all broadcast at a different frequency, with a directional pulse that varies in phase with the base signal. By calculating the phase difference between the primary and secondary phases your ADF provides your bearing from the station. Getting a bearing from two different stations gives you a very precise location. There are also some things that can be done in terms of phasing and signal strength to give an approximate distance from a single station, allowing for a rough position fix off of a single signal, but I am not aware of it being done in practice as the two bearing option is much more reliable. The major drawback for this system as currently implemented, and your intended use is that it is line of site and relatively short range (~200 miles). If you are too far from a station, or a pesky mountain (or the horizon of your asteroid) happens to be between you and it, you are not going to be able to pick up the signals. Meaning this system is not commonly used for surface travel.



                                              The LORAN system was developed in WW2 and used until fairly recently for nautical navigation. It again works by using a series of fixed position transmitters, but in this case they come in paired units. Each member of the pairing is separated by a known distance, and they pulse out synchronized signals. The receiver compares the time difference in receiving the two pulses and from that difference you can plot a line of your relative distance from the two. Grab readings from an alternate pair to plot your location. The major advantage this system has over the other is range (~1500 miles) and being much less sensitive to LoS issues, though it is also somewhat less accurate. With the sensitivity and precision of modern electronics, it is conceivable that someone could build a 3+ point LORAN system and pinpoint their location pretty accurately off of one reading, the challenge being keeping the signals synchronized across the additional broadcast stations.



                                              Now the range and LOS issues are primarily a function of the frequencies that are used in broadcasting the signals. So you could theoretically implement a VOR system in the frequency bands used by LORAN and have better performance in that respect, but it will also change the timing and phase calculations for determining direction. These would be engineering problems and may impact how quickly and reliably you could find you position. (Way too long since I studies this stuff to remember that level of detail)



                                              Once the beacon network is established your coordinates become a range and bearing to a convenient beacon. "10 klicks from beacon XYZ bearing 175".



                                              Note, I would advise against trying to establish a GPS network around an asteroid. That system is dependent on known orbits & timing for the satellites, and I suspect it would be very difficult to maintain such with the irregular shape and weak gravity of an asteroid.






                                              share|improve this answer





















                                              • Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                                – MttJocy
                                                Dec 1 at 16:37














                                              0












                                              0








                                              0






                                              Expanding a bit on LDutch's answer and MttJocy's comment on it. Take a page from pre-GPS aerial and nautical navigation systems.



                                              Aviation commonly uses (fading as GPS is taking over, but still present) the VOR system and ADFs to figure positions and navigate. Essentially the various beacons all broadcast at a different frequency, with a directional pulse that varies in phase with the base signal. By calculating the phase difference between the primary and secondary phases your ADF provides your bearing from the station. Getting a bearing from two different stations gives you a very precise location. There are also some things that can be done in terms of phasing and signal strength to give an approximate distance from a single station, allowing for a rough position fix off of a single signal, but I am not aware of it being done in practice as the two bearing option is much more reliable. The major drawback for this system as currently implemented, and your intended use is that it is line of site and relatively short range (~200 miles). If you are too far from a station, or a pesky mountain (or the horizon of your asteroid) happens to be between you and it, you are not going to be able to pick up the signals. Meaning this system is not commonly used for surface travel.



                                              The LORAN system was developed in WW2 and used until fairly recently for nautical navigation. It again works by using a series of fixed position transmitters, but in this case they come in paired units. Each member of the pairing is separated by a known distance, and they pulse out synchronized signals. The receiver compares the time difference in receiving the two pulses and from that difference you can plot a line of your relative distance from the two. Grab readings from an alternate pair to plot your location. The major advantage this system has over the other is range (~1500 miles) and being much less sensitive to LoS issues, though it is also somewhat less accurate. With the sensitivity and precision of modern electronics, it is conceivable that someone could build a 3+ point LORAN system and pinpoint their location pretty accurately off of one reading, the challenge being keeping the signals synchronized across the additional broadcast stations.



                                              Now the range and LOS issues are primarily a function of the frequencies that are used in broadcasting the signals. So you could theoretically implement a VOR system in the frequency bands used by LORAN and have better performance in that respect, but it will also change the timing and phase calculations for determining direction. These would be engineering problems and may impact how quickly and reliably you could find you position. (Way too long since I studies this stuff to remember that level of detail)



                                              Once the beacon network is established your coordinates become a range and bearing to a convenient beacon. "10 klicks from beacon XYZ bearing 175".



                                              Note, I would advise against trying to establish a GPS network around an asteroid. That system is dependent on known orbits & timing for the satellites, and I suspect it would be very difficult to maintain such with the irregular shape and weak gravity of an asteroid.






                                              share|improve this answer












                                              Expanding a bit on LDutch's answer and MttJocy's comment on it. Take a page from pre-GPS aerial and nautical navigation systems.



                                              Aviation commonly uses (fading as GPS is taking over, but still present) the VOR system and ADFs to figure positions and navigate. Essentially the various beacons all broadcast at a different frequency, with a directional pulse that varies in phase with the base signal. By calculating the phase difference between the primary and secondary phases your ADF provides your bearing from the station. Getting a bearing from two different stations gives you a very precise location. There are also some things that can be done in terms of phasing and signal strength to give an approximate distance from a single station, allowing for a rough position fix off of a single signal, but I am not aware of it being done in practice as the two bearing option is much more reliable. The major drawback for this system as currently implemented, and your intended use is that it is line of site and relatively short range (~200 miles). If you are too far from a station, or a pesky mountain (or the horizon of your asteroid) happens to be between you and it, you are not going to be able to pick up the signals. Meaning this system is not commonly used for surface travel.



                                              The LORAN system was developed in WW2 and used until fairly recently for nautical navigation. It again works by using a series of fixed position transmitters, but in this case they come in paired units. Each member of the pairing is separated by a known distance, and they pulse out synchronized signals. The receiver compares the time difference in receiving the two pulses and from that difference you can plot a line of your relative distance from the two. Grab readings from an alternate pair to plot your location. The major advantage this system has over the other is range (~1500 miles) and being much less sensitive to LoS issues, though it is also somewhat less accurate. With the sensitivity and precision of modern electronics, it is conceivable that someone could build a 3+ point LORAN system and pinpoint their location pretty accurately off of one reading, the challenge being keeping the signals synchronized across the additional broadcast stations.



                                              Now the range and LOS issues are primarily a function of the frequencies that are used in broadcasting the signals. So you could theoretically implement a VOR system in the frequency bands used by LORAN and have better performance in that respect, but it will also change the timing and phase calculations for determining direction. These would be engineering problems and may impact how quickly and reliably you could find you position. (Way too long since I studies this stuff to remember that level of detail)



                                              Once the beacon network is established your coordinates become a range and bearing to a convenient beacon. "10 klicks from beacon XYZ bearing 175".



                                              Note, I would advise against trying to establish a GPS network around an asteroid. That system is dependent on known orbits & timing for the satellites, and I suspect it would be very difficult to maintain such with the irregular shape and weak gravity of an asteroid.







                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered Nov 30 at 21:18









                                              Rozwel

                                              1,385511




                                              1,385511












                                              • Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                                – MttJocy
                                                Dec 1 at 16:37


















                                              • Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                                – MttJocy
                                                Dec 1 at 16:37
















                                              Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                              – MttJocy
                                              Dec 1 at 16:37




                                              Distance measurement using a single station these days is generally done using DME instead which basically works like secondary radar but with the roles reversed as the interrogator is on the aircraft and the transponder is on the ground. Those are more often found co-located with VOR/NDB/LNAV stations used in non precision approach procedures at airports though but there is no real reason why you couldn't have this on all your beacons on the asteroid if you wanted it. Could be useful especially if the asteroid is a very irregular shape where LoS to two or more stations might be an issue.
                                              – MttJocy
                                              Dec 1 at 16:37











                                              0














                                              Assuming a sufficiently advanced society, what about artificial poles, magnetic or otherwise?






                                              share|improve this answer


























                                                0














                                                Assuming a sufficiently advanced society, what about artificial poles, magnetic or otherwise?






                                                share|improve this answer
























                                                  0












                                                  0








                                                  0






                                                  Assuming a sufficiently advanced society, what about artificial poles, magnetic or otherwise?






                                                  share|improve this answer












                                                  Assuming a sufficiently advanced society, what about artificial poles, magnetic or otherwise?







                                                  share|improve this answer












                                                  share|improve this answer



                                                  share|improve this answer










                                                  answered Dec 2 at 18:39









                                                  Buns Glazing

                                                  1012




                                                  1012























                                                      0














                                                      A map with landmarks



                                                      The asteroid would be mapped out and photographed to generate a 3d model with landmarks



                                                      Like a pirate's map, directions would be given from the nearest landmark.






                                                      share|improve this answer


























                                                        0














                                                        A map with landmarks



                                                        The asteroid would be mapped out and photographed to generate a 3d model with landmarks



                                                        Like a pirate's map, directions would be given from the nearest landmark.






                                                        share|improve this answer
























                                                          0












                                                          0








                                                          0






                                                          A map with landmarks



                                                          The asteroid would be mapped out and photographed to generate a 3d model with landmarks



                                                          Like a pirate's map, directions would be given from the nearest landmark.






                                                          share|improve this answer












                                                          A map with landmarks



                                                          The asteroid would be mapped out and photographed to generate a 3d model with landmarks



                                                          Like a pirate's map, directions would be given from the nearest landmark.







                                                          share|improve this answer












                                                          share|improve this answer



                                                          share|improve this answer










                                                          answered Dec 2 at 22:25









                                                          Thorne

                                                          14.1k42040




                                                          14.1k42040























                                                              0














                                                              Mobile network sensors.
                                                              The OP clearly states that there are inhabitants. I'm going to make a leap of faith assuming that all the areas of interest have at least some inhabitant, or have had some visitor.



                                                              Local linear embedding has been used to reconstruct global maps from pairwise local distances. All you need is to equip the locals with peer-to-peer devices, measure the ping time, and resolve the local linear embedding problem, et voila: the map is ready. Bonus points, as long as the inhabitants don't move simultaneously in the same direction or according to some linear transformation, you can recompute their position over time and map unexplored areas of the asteroid.



                                                              Note that, as long as you can track changes in the pairwise distances between inhabitants, then you can define an "absolute" map, i.e. One where the thing you called North remains in the same place relative to the asteroid.



                                                              Any new inhabitant's location can be easily located on the map again by local linear embedding, or by trilateration from existing inhabitants.



                                                              References: http://www.cs.toronto.edu/~fritz/absps/globalcoord.pdf
                                                              http://www.pnas.org/content/early/2008/07/07/0709842104.abstract



                                                              Etc...






                                                              share|improve this answer





















                                                              • P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                                – NofP
                                                                Dec 3 at 21:24
















                                                              0














                                                              Mobile network sensors.
                                                              The OP clearly states that there are inhabitants. I'm going to make a leap of faith assuming that all the areas of interest have at least some inhabitant, or have had some visitor.



                                                              Local linear embedding has been used to reconstruct global maps from pairwise local distances. All you need is to equip the locals with peer-to-peer devices, measure the ping time, and resolve the local linear embedding problem, et voila: the map is ready. Bonus points, as long as the inhabitants don't move simultaneously in the same direction or according to some linear transformation, you can recompute their position over time and map unexplored areas of the asteroid.



                                                              Note that, as long as you can track changes in the pairwise distances between inhabitants, then you can define an "absolute" map, i.e. One where the thing you called North remains in the same place relative to the asteroid.



                                                              Any new inhabitant's location can be easily located on the map again by local linear embedding, or by trilateration from existing inhabitants.



                                                              References: http://www.cs.toronto.edu/~fritz/absps/globalcoord.pdf
                                                              http://www.pnas.org/content/early/2008/07/07/0709842104.abstract



                                                              Etc...






                                                              share|improve this answer





















                                                              • P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                                – NofP
                                                                Dec 3 at 21:24














                                                              0












                                                              0








                                                              0






                                                              Mobile network sensors.
                                                              The OP clearly states that there are inhabitants. I'm going to make a leap of faith assuming that all the areas of interest have at least some inhabitant, or have had some visitor.



                                                              Local linear embedding has been used to reconstruct global maps from pairwise local distances. All you need is to equip the locals with peer-to-peer devices, measure the ping time, and resolve the local linear embedding problem, et voila: the map is ready. Bonus points, as long as the inhabitants don't move simultaneously in the same direction or according to some linear transformation, you can recompute their position over time and map unexplored areas of the asteroid.



                                                              Note that, as long as you can track changes in the pairwise distances between inhabitants, then you can define an "absolute" map, i.e. One where the thing you called North remains in the same place relative to the asteroid.



                                                              Any new inhabitant's location can be easily located on the map again by local linear embedding, or by trilateration from existing inhabitants.



                                                              References: http://www.cs.toronto.edu/~fritz/absps/globalcoord.pdf
                                                              http://www.pnas.org/content/early/2008/07/07/0709842104.abstract



                                                              Etc...






                                                              share|improve this answer












                                                              Mobile network sensors.
                                                              The OP clearly states that there are inhabitants. I'm going to make a leap of faith assuming that all the areas of interest have at least some inhabitant, or have had some visitor.



                                                              Local linear embedding has been used to reconstruct global maps from pairwise local distances. All you need is to equip the locals with peer-to-peer devices, measure the ping time, and resolve the local linear embedding problem, et voila: the map is ready. Bonus points, as long as the inhabitants don't move simultaneously in the same direction or according to some linear transformation, you can recompute their position over time and map unexplored areas of the asteroid.



                                                              Note that, as long as you can track changes in the pairwise distances between inhabitants, then you can define an "absolute" map, i.e. One where the thing you called North remains in the same place relative to the asteroid.



                                                              Any new inhabitant's location can be easily located on the map again by local linear embedding, or by trilateration from existing inhabitants.



                                                              References: http://www.cs.toronto.edu/~fritz/absps/globalcoord.pdf
                                                              http://www.pnas.org/content/early/2008/07/07/0709842104.abstract



                                                              Etc...







                                                              share|improve this answer












                                                              share|improve this answer



                                                              share|improve this answer










                                                              answered Dec 3 at 21:23









                                                              NofP

                                                              2,893421




                                                              2,893421












                                                              • P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                                – NofP
                                                                Dec 3 at 21:24


















                                                              • P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                                – NofP
                                                                Dec 3 at 21:24
















                                                              P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                              – NofP
                                                              Dec 3 at 21:24




                                                              P.S. The math applies to higher dimensions, in case your asteroid gets eaten by Interstellar's tesseract :)
                                                              – NofP
                                                              Dec 3 at 21:24





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