Which coordinate system is the best for a distances from whole USA?

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up vote
1
down vote

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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.

coordinate-system united-states epsg

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edited 10 hours ago

PolyGeo♦

52.3k1779236

asked 10 hours ago

user122678

345

add a comment | 

up vote
1
down vote

favorite

1

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.

coordinate-system united-states epsg

share|improve this question

edited 10 hours ago

PolyGeo♦

52.3k1779236

asked 10 hours ago

user122678

345

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

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.

coordinate-system united-states epsg

share|improve this question

edited 10 hours ago

PolyGeo♦

52.3k1779236

asked 10 hours ago

user122678

345

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.

coordinate-system united-states epsg

coordinate-system united-states epsg

share|improve this question

edited 10 hours ago

PolyGeo♦

52.3k1779236

asked 10 hours ago

user122678

345

share|improve this question

edited 10 hours ago

PolyGeo♦

52.3k1779236

asked 10 hours ago

user122678

345

share|improve this question

share|improve this question

edited 10 hours ago

PolyGeo♦

52.3k1779236

edited 10 hours ago

PolyGeo♦

52.3k1779236

edited 10 hours ago

PolyGeo♦

52.3k1779236

52.3k1779236

asked 10 hours ago

user122678

345

asked 10 hours ago

user122678

345

asked 10 hours ago

user122678

345

345

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

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.

share|improve this answer

answered 10 hours ago

1saac

1,055315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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/…
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be more specific: It is a good choice because you want to calculate distances in the USA.
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
4
down vote

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 comic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 2 hours ago

julien

7,92534380

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

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.

share|improve this answer

answered 10 hours ago

1saac

1,055315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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/…
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be more specific: It is a good choice because you want to calculate distances in the USA.
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

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.

share|improve this answer

answered 10 hours ago

1saac

1,055315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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/…
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be more specific: It is a good choice because you want to calculate distances in the USA.
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

up vote
1
down vote



accepted

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.

share|improve this answer

answered 10 hours ago

1saac

1,055315

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.

share|improve this answer

answered 10 hours ago

1saac

1,055315

share|improve this answer

share|improve this answer

answered 10 hours ago

1saac

1,055315

answered 10 hours ago

1saac

1,055315

answered 10 hours ago

1saac

1,055315

1,055315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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/…
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be more specific: It is a good choice because you want to calculate distances in the USA.
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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/…
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
  – user122678
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be more specific: It is a good choice because you want to calculate distances in the USA.
  – 1saac
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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
  8 hours ago
  

  
  
  
  

Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
– user122678
9 hours ago

Is "USA_Contigous_Alberts_Equal_Area_Conic" a good choice?
– user122678
9 hours ago

1

1

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/…
– 1saac
9 hours ago

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/…
– 1saac
9 hours ago

Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
– user122678
9 hours ago

Good. So, according to this my best choice is a ESRI: 102005 because I want to calculate distances?
– user122678
9 hours ago

To be more specific: It is a good choice because you want to calculate distances in the USA.
– 1saac
9 hours ago

To be more specific: It is a good choice because you want to calculate distances in the USA.
– 1saac
9 hours ago

3

3

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
8 hours ago

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
8 hours ago

 | 
show 1 more comment

up vote
4
down vote

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 comic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

add a comment | 

up vote
4
down vote

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 comic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

add a comment | 

up vote
4
down vote

up vote
4
down vote

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 comic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

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 comic or cylindrical equidistant projection, the standard parallels and all meridians only are equidistant.

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

share|improve this answer

share|improve this answer

answered 2 hours ago

mkennedy

15.3k13156

answered 2 hours ago

mkennedy

15.3k13156

answered 2 hours ago

mkennedy

15.3k13156

15.3k13156

add a comment | 

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 2 hours ago

julien

7,92534380

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 2 hours ago

julien

7,92534380

add a comment | 

up vote
0
down vote

up vote
0
down vote

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

share|improve this answer

answered 2 hours ago

julien

7,92534380

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

share|improve this answer

answered 2 hours ago

julien

7,92534380

share|improve this answer

share|improve this answer

answered 2 hours ago

julien

7,92534380

answered 2 hours ago

julien

7,92534380

answered 2 hours ago

julien

7,92534380

7,92534380

add a comment | 

add a comment | 

 

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