Just recently I came across the following statement

I am trying firstly understand the equations and secondly evaluate them for accuracy. In using Office 365 for Mac, Excel refuses to recognise the formulas for M & R. Could this be because the formulas might be missing a Math Operator? Simply replacing "PHI" with the Latitude (in Decimal Degrees) still results in an error message " are you trying to enter a formula?". Next I run into an error when I tried to check the validity of the value of "e". The formula as written in the statement above returns a "Num error because we haven't define the value of "f". When I replaced "f" in the equation to determine "e", it returned the #Num error message. What am I doing wrong?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)

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

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

If I understand the 1/f =298.257223563 statement correctly, by transposing f= 298.257223563*1, or is this assumption incorrect?

Eventually I'd like to test the formulas to determine the lengths of 1° of Latitude and or Longitude. Any assistance would be greatly appreciated. On a side note. Supporters of Office 365 often say that the improvements under 365, significantly reduce the need to resort to VBA. Given that we are dealing with values with 9 decimals places can this overcome the limitation of floating point error, or is Office 365 still a fair way off this accuracy required?