Converting between grid eastings and northings and ellipsoidal latitude and longitude

All the formula here are coded into a spreadsheet available from the OS web site.

Converting latitude and longitude to eastings and northings

These formulae are used to convert the format of coordinates between grid eastings and northings and ellipsoidal coordinates (latitude and longitude) on the same datum. This is not a datum transformation. If you need to transform between GNSS coordinates and OSGB36 National Grid coordinates, you need to apply either the OSTN15 transformation or the approximate Helmert type transformation.

To convert (project) a position from latitude and longitude coordinates (φ, λ) to easting and northing coordinates (E, N) using a Transverse Mercator projection (for example, National Grid or UTM), compute the following formulae. Remember to express all angles in radians. You will need the ellipsoid constants a, b and and the following projection constants. Annexe A gives values of these constants for the ellipsoids and projections usually used in Britain.

N₀ — northing of true origin

E₀ — easting of true origin

F₀ — scale factor on central meridian

Φ₀ — latitude of true origin

λ₀ — longitude of true origin and central meridian.

n=aba+bv=aFo(1e2sin2ϕ)0.5ρ=aFo(1e2)(1e2sin2ϕ)1.5η=vρ1n = \frac{a-b}{a+b}\\ v = aF_o(1-e^2sin^2\phi)^{-0.5}\\ \rho = aF_o(1-e^2)(1-e^2sin^2\phi)^{-1.5}\\ \eta= \frac{v}{\rho}-1
M=bF0[(1+n+54n2+54n3)(ϕϕ0)(3n+3n2+218n3)sin(ϕphi0)cos(ϕ+ϕ0)+(158n2+158n3)sin(2(ϕϕ0))cos(2(ϕ+ϕ0))=3524n3sin(3(ϕϕ0))cos(3(ϕ+phi0))]M = bF_0[(1+n+\frac{5}{4}n^2+\frac{5}{4}n^3)(\phi-\phi_0)-(3n+3n^2+\frac{21}{8}n^3)sin(\phi-phi_0)cos(\phi+\phi_0)\\ + (\frac{15}{8}n^2+\frac{15}{8}n^3)sin(2(\phi-\phi_0))cos(2(\phi+\phi_0))=\frac{35}{24}n^3sin(3(\phi-\phi_0))cos(3(\phi+phi_0))]\\
I=M+N0II=v2sinϕcosϕIII=v24sinϕcos3ϕ(5tan2ϕ+9η2)IIIA=v720sinϕcos5ϕ(6158tan2ϕ4ϕ)I = M+N_0 II = \frac{v}{2}sin\phi cos\phi\\ III = \frac{v}{24}sin\phi cos^3\phi(5-tan^2 \phi + 9\eta^2)\\ IIIA = \frac{v}{720}sin\phi cos^5\phi(61-58tan^2\phi^4\phi)\\
IV=vcosϕV=v6cos3ϕ(vρtan2ϕ)VI=v120cos5ϕ(518tan2ϕ+tan4ϕ+14η258(tan2ϕ)η2) IV = vcos\phi\\ V = \frac{v}{6}cos^3\phi(\frac{v}{\rho}-tan^2\phi)\\ VI = \frac{v}{120}cos^5\phi(5-18tan^2\phi+tan^4\phi+14\eta^2-58(tan^2\phi)\eta^2)
N=I+II(λλo)2+III(λλ0)4+IIIA(λλ0)6E=e0+IV(λλ0)+V(λλ0)3+IV(λΛ0)5N = I + II(\lambda-\lambda_o)^2+III(\lambda-\lambda_0)^4+IIIA(\lambda-\lambda_0)^6\\ E = e_0+IV(\lambda-\lambda_0)+V(\lambda-\lambda_0)^3+IV(\lambda-\Lambda_0)^5\\

Here’s a worked example using the Airy 1830 ellipsoid and National Grid. Intermediate values are shown here to 10 decimal places. Compute all values using double-precision arithmetic.

φ: 52° 39′ 27.2531″ N

III: 1.5606875430E+05

λ: 001° 43′ 04.5177″ E

IIIA: -2.0671123013E+04

IV: 3.8751205752E+06

ν: 6.3885023339E+06

V: -1.7000078207E+05

ρ: 6.3727564398E+06

VI: -1.0134470437E+05

η²: 2.4708137334E-03

M: 4.0668829595E+05

E: 651409.903 m

I: 3.0668829595E+05

N: 313177.270 m

II: 1.5404079094E+06

Converting eastings and northings to latitude and longitude

Obtaining (φ, λ) from (E, N) is an iterative procedure. You need values for the ellipsoid and projection constants a,b, e², N, E₀, F₀, Φ₀ and λ₀ as in the previous section. Remember to express all angles in radians. First compute:

ϕ=(NN0aF0)+ϕ0\phi' = (\frac{N-N_0}{aF_0})+\phi_0

and M from equation (C3), substituting Φ' for Φ. If the absolute value of N-N0-M ≥ 0.01mm obtain a new value for Φ' using

ϕnew=(NN0MaF0)+ϕ\phi_{new}=(\frac{N-N_0-M}{aF_0})+\phi'

and recompute M substituting Φ' for Φ . Iterate until the absolute value of N-N0-M < 0.01mm then compute , and using equation (C2) and compute

VIII=tanϕ2ρvVIII=tanϕ24ρv3(5+3tan2ϕ+η29(tan2ϕ)η2)IX=tanϕ720ρv5(61+90tan2ϕ+45tan4ϕ)VIII = \frac{tan\phi'}{2\rho v}\\ VIII = \frac{tan\phi '}{24\rho v^3}(5+3tan^2\phi'+\eta^2-9(tan^2\phi')\eta^2)\\ IX = \frac{tan\phi'}{720\rho v^5}(61+90tan^2\phi'+45tan^4\phi')\\
X=secϕVXI=secϕ6v3(vρ+2tan2ϕ)XII=secϕ5040v7(61+662tan2ϕ+1320tan4ϕ+720tan5ϕ)X = \frac{sec\phi'}{V}\\ XI = \frac{sec\phi'}{6v^3}(\frac{v}{\rho}+2tan^2\phi')\\ XII = \frac{sec\phi'}{5040v^7}(61+662tan^2\phi'+1320tan^4\phi'+720tan^5\phi')\\
ϕ=ϕVII(EEO)2+VIII(EE0)4IX(EE0)5λ=λ0+X(EE0)XI(EE0)3+XII(EE0)5XIIIA(EE0)7\phi = \phi'-VII(E-E_O)^2+VIII(E-E_0)^4-IX(E-E_0)^5\\ \lambda = \lambda_0 + X(E-E_0)-XI(E-E_0)^3+XII(E-E_0)^5-XIIIA(E-E_0)^7

Here’s a worked example using the Airy 1830 ellipsoid and National Grid. Intermediate values are shown here to 10 decimal places. Compute all values using double-precision arithmetic.

E: 651409.903 m

final φ′: 9.2006620954E-01 rad

N: 313177.270 m

ν: 6.3885233415E+06

ρ: 6.3728193094E+06

φ′ #1: 9.2002324604E-01 rad

η2: 2.4642206357E-03

M #1: 4.1290347143E+05

VII: 1.6130562489E-14

N-N0-M#1: 2.7379857228E+02

VIII: 3.3395547427E-28

IX: 9.4198561675E-42

φ′ #2: 9.2006619470E-01 rad

X: 2.5840062507E-07

M #2: 4.1317717541E+05

XI: 4.6985969956E-21

N-N0-M#2: 9.4594338385E-02

XII: 1.6124316614E-34

XIIA: 6.6577316285E-48

φ′ #3: 9.2006620954E-01 rad

M #3: 4.1317726997E+05

N-N0-M#3: 3.2661366276E-05

φ: 52° 39′ 27.2531″ N

λ: 001° 43′ 04.5177″ E

φ′ #4: 9.2006620954E-01 rad

M #4: 4.1317727000E+05

N-N0-M#4: 1.1350493878E-08

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