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Suppose I have the geographic coordinates of "Saratoga, California, USA" as

Latitude:   37°15.8298′ N   
Longitude: 122° 1.3806′ W

I know from here that in the case of latitude 1° ≈ 69 miles and that longitude varies:

1° longitude = cosine (latitude) * length of degree (miles) at Equator.

How many miles is 1° longitude at longitude: 122°1.3806′ W?

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2 Answers 2

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It doesn't matter at what longitude you are. What matters is what latitude you are.

Length of 1 degree of Longitude = cosine (latitude in radians) * length of degree (miles) at equator.

Convert your latitude into decimal degrees ~ 37.26383

Convert your decimal degrees into radians ~ 0.65038

Take the cosine of the value in radians ~ 0.79585

1 degree of Longitude = ~0.79585 * 69.172 = ~ 55.051 miles

More useful information from the about.com website:

Degrees of latitude are parallel so the distance between each degree remains almost constant but since degrees of longitude are farthest apart at the equator and converge at the poles, their distance varies greatly.

Each degree of latitude is approximately 69 miles (111 kilometers) apart. The range varies (due to the earth's slightly ellipsoid shape) from 68.703 miles (110.567 km) at the equator to 69.407 (111.699 km) at the poles. This is convenient because each minute (1/60th of a degree) is approximately one [nautical] mile.

A degree of longitude is widest at the equator at 69.172 miles (111.321) and gradually shrinks to zero at the poles. At 40° north or south the distance between a degree of longitude is 53 miles (85 km)

Note that the original site (about.com) erroneously omitted the "nautical" qualifier.

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Equations

Suppose we are discussing latitude and longitude in WGS84 and we want to be accurate. Both meridian radius of curvature M (latitude length) and radius of curvature of parallels R (longitude length) in WGS84 depend only on latitude phi and are defined as:

M = a(1 - e^2) / (1 - e^2 sin(phi)^2)^(3/2) (1)

R = a cos(phi) / (1 - e^2 sin(phi)^2)^(1/2) (2)

where

  • e is the eccentricity of the referenced ellipsoid of WGS84
  • a is the semi-major axis of the ellipsoid.

WGS 84 defines the semi-major axis of the WGS 84 ellipsoid a and the flattening factor of the Earth f as:

a=6378137m

1/f=298.257223563

With these defining parameters in WGS 84, we can obtain the eccentricity:

e=sqrt(2f-f^2)=0.081819191

Verification of the above equations

To verify whether the above eq. (1) and eq. (2) are correct, we will check them against the results at USGS which said, at 38 degrees North latitude, one degree of latitude equals approximately 364,000 feet (69 miles) and one degree of longitude equals 288,200 feet (54.6 miles).

To find the meridian radius of curvature M (latitude length) and radius of curvature of parallels (longitude length) at 38 degrees North latitude, we plug in a=6378137, e=0.081819191 and phi=38° into eq. (1) and eq. (2):

M = 6359629.652 m/rad

R = 5032429.322 m/rad

Furthermore, to find length of 1° of latitude and longitude, we multiply the above equations by the radian of 1°:

Length of 1° latitude = M * 1°/180° * pi = 6359629.652*1*pi/180 = 110996.4766 m = 68.970013 miles

Length of 1° longitude= R * 1°/180° * pi = 5032429.322*1*pi/180 = 87832.46103 m = 54.57656101 miles

The results are in line with the results of USGS.

Answer to your question

Repeat the above steps but plug in phi=37°15.8298′ = 37+15.8298/60=37.26383, we can obtain:

Length of 1° latitude=68.96139 miles

Length of 1° longitude=55.11761 miles

So 1° longitude equals to 55.11761 miles. Apparently, your equation is a simplified equation assuming the Earth is a sphere and works best near the equator but not for your case.

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