5 ms·

> a knot is only approximately equal to 1 min of latitude per hour (there are numerous cases where it's unusably wrong) Whilst it isn't precisely equal it's ce

> a knot is only approximately equal to 1 min of latitude per hour (there are numerous cases where it's unusably wrong) Whilst it isn't precisely equal it's certainly precise enough for navigational purposes. What are the numerous cases of which you speak (i.e. give me a sea navigation scenario where the current definition of a nautical mile is unusuably wrong, may be easier to come up with an air one but again I think you'll struggle). Certainly as a sailor (who routinely navigates yachts off-shore) 1 nm = 1 knot = 1 minute of latitude is very useful, especially when combined with a mercator projection. Lines of constant bearing are straight which is a very useful property. For longer ocean passages where you may want to sail a great circle route to save distance other projections are available. Yes you could use kilometre lines but what's your positioning system going to be? It won't map well to lat/long so you'll need to come up with an entirely new coordinate system. Given basically all charts available (and the surveying data behind them) are done in a lat/long system (though many different datums exist) it's a huge job to switch to the new coordinate system for little benefit. The current system of units also works very nicely for celestial navigation. Measure angle (A) of body above the horizon. (90 - A) * 60 = distance in nm between you and the point on the earth that body is directly overhead. Have you actually got any experience of navigating by sea or air or are you just assuming the system of units used is ill thought out and useless and everyone's crying out to use metric instead?

logfromblammo 10y ago
> Yes you could use kilometre lines but what's your positioning system going to be? It won't map well to lat/long so you'll need to come up with an entirely new coordinate system. Will you really? Why not use UTM [0] + UPS [1] or MGRS [2]? [0] https://en.wikipedia.org/wiki/Universal_Transverse_Mercator_coordinate_system https://en.wikipedia.org/wiki/Universal_Transverse_Mercator_... [1] https://en.wikipedia.org/wiki/Universal_polar_stereographic_coordinate_system https://en.wikipedia.org/wiki/Universal_polar_stereographic_... [2] https://en.wikipedia.org/wiki/Military_grid_reference_system https://en.wikipedia.org/wiki/Military_grid_reference_system
- gchadwick 10y ago
  Well indeed such coordinate systems exist but you've still got the cost of translating a large body of cartography into them. Plus as pointed out by another commentor the number of KMs around the earth is not a round number, so you need to do some fudging to get a metricised spherical coordinate system (and you want a spherical coordinate system because the 60th of a degree = 1nm equivilance is very useful for simplfying the mathematics required for navigation).
  - logfromblammo 10y ago
    The cost of translating is a mouse click [after a software pro writes the code behind the control]. Modern, computerized charts don't need to lie flat or use only one projection visualization. Furthermore, you can add up-to-date ocean current, jet stream, wind, wave, and weather information to a computerized chart, along with a vehicle performance model, and automatically optimize a route for fastest travel time or lowest fuel consumption. Nowadays, the unit of distance on the printouts and reports is only needed for human readability and loss-of-power emergencies. Once you have planned out your route, you can scale your backup hard copy with whatever unit your navigator feels most comfortable with, but inside the computer, the software most definitely has a "native" measurement unit, and those are usually in the base SI unit. All other units are converted from that native representation as necessary. You can navigate your vessel in any coordinate system you like, as long as the conversion functions are coded correctly. And once they are, all charts are translated into all coordinate systems. If you like constant-bearing lines on Mercator projections, just print it out that way. If you like following currents and great circles, just make sure your navigation computer has a backup generator. The fudging you mention is 99.9% handled just by using the WGS84 ellipsoid. If you're doing any fancy rocket science, you also need a somewhat detailed gravitational model mapped on top of it. Those sorts of calculations are most easily done in an ECR cartesian coordinate system, which easily translates to ECEF cartesian, which somewhat less easily translates to geodetic coordinates--it is actually easier to translate that in the other direction. Unfortunately for that math, no one knows where their house is on the Earth in XYZ coordinates. The amount of error creeping into your double-precision floats is almost insignificant at the scale you need with one or two extra conversions. The math only needs one arbitrarily chosen unit. I have never seen nautical miles chosen as the unit for length.

logfromblammo 10y ago

> Yes you could use kilometre lines but what's your positioning system going to be? It won't map well to lat/long so you'll need to come up with an entirely new coordinate system. Will you really? Why not use UTM [0] + UPS [1] or MGRS [2]? [0] https://en.wikipedia.org/wiki/Universal_Transverse_Mercator_coordinate_system https://en.wikipedia.org/wiki/Universal_Transverse_Mercator_... [1] https://en.wikipedia.org/wiki/Universal_polar_stereographic_coordinate_system https://en.wikipedia.org/wiki/Universal_polar_stereographic_... [2] https://en.wikipedia.org/wiki/Military_grid_reference_system https://en.wikipedia.org/wiki/Military_grid_reference_system

gchadwick 10y ago
Well indeed such coordinate systems exist but you've still got the cost of translating a large body of cartography into them. Plus as pointed out by another commentor the number of KMs around the earth is not a round number, so you need to do some fudging to get a metricised spherical coordinate system (and you want a spherical coordinate system because the 60th of a degree = 1nm equivilance is very useful for simplfying the mathematics required for navigation).
- logfromblammo 10y ago
  The cost of translating is a mouse click [after a software pro writes the code behind the control]. Modern, computerized charts don't need to lie flat or use only one projection visualization. Furthermore, you can add up-to-date ocean current, jet stream, wind, wave, and weather information to a computerized chart, along with a vehicle performance model, and automatically optimize a route for fastest travel time or lowest fuel consumption. Nowadays, the unit of distance on the printouts and reports is only needed for human readability and loss-of-power emergencies. Once you have planned out your route, you can scale your backup hard copy with whatever unit your navigator feels most comfortable with, but inside the computer, the software most definitely has a "native" measurement unit, and those are usually in the base SI unit. All other units are converted from that native representation as necessary. You can navigate your vessel in any coordinate system you like, as long as the conversion functions are coded correctly. And once they are, all charts are translated into all coordinate systems. If you like constant-bearing lines on Mercator projections, just print it out that way. If you like following currents and great circles, just make sure your navigation computer has a backup generator. The fudging you mention is 99.9% handled just by using the WGS84 ellipsoid. If you're doing any fancy rocket science, you also need a somewhat detailed gravitational model mapped on top of it. Those sorts of calculations are most easily done in an ECR cartesian coordinate system, which easily translates to ECEF cartesian, which somewhat less easily translates to geodetic coordinates--it is actually easier to translate that in the other direction. Unfortunately for that math, no one knows where their house is on the Earth in XYZ coordinates. The amount of error creeping into your double-precision floats is almost insignificant at the scale you need with one or two extra conversions. The math only needs one arbitrarily chosen unit. I have never seen nautical miles chosen as the unit for length.