I have about 100 cities all over the USA and want to calculate some distances. According to that I want use one coordinate projection and need to know which one is the best so I can get the most accurate results. I have compared the results using NAD83 for Texas and NAD 83 for Washington for a same data and the results are not the same.


If you want to calculate distances then you should use an Equidistant projection that best covers the USA. USA Contiguous Equidistant Conic would likely be an appropriate choice.

  • Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice? – user122678 Nov 3 '18 at 10:25
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    Given your goal to calculate distances, no. Most projections preserve one of three properties: distance, area or angles, at the expense of the other two. The one you just suggested would ensure that areas of a feature anywhere on the map are conserved, but not distance. I suggest you read about map projections, how they work and how they distort some properties of features to maintain accuracy of another: desktop.arcgis.com/en/arcmap/10.3/guide-books/map-projections/… – wfgeo Nov 3 '18 at 10:29
  • Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances? – user122678 Nov 3 '18 at 10:34
  • To be more specific: It is a good choice because you want to calculate distances in the USA. – wfgeo Nov 3 '18 at 10:35
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    Be careful with asserting the fitness of an equidistant projection for random distance calculations. The equal distance property only holds for one or two points to any other point. For many applications, an equal area projection may be better suited for general "as the bird flies" distance, and UTM for distance within 4-6 degrees. Calculating spheroidal distance is more expensive but far more accurate. – Vince Nov 3 '18 at 12:17

Your best workflow may depend on whether you want "as the crow flies" distances or "via road" distances. For the latter, you would need to have a line data layer. For the "as the crow flies", you should look at tools that support a "geodesic" option. If it supports a geodesic option, that will give the shortest distances on the ellipsoid surface. The calculations won't take elevation into account though.

As Vince points out in comments, equidistant projections have very limited equidistant lines. For a conic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.


A simpler option might be not to use a projection and compute distances directly from geographical coordinates using the Haversine formula.

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    Single sentence answers are automatically flagged for poor quality, so please add some details to flesh out this answer some more. Haversine is spherical, not spheroidal, so this answer needs a caveat about accuracy. – Vince Nov 4 '18 at 12:36

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