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

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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 transformationarrow-up-right or the approximate Helmert type transformationarrow-up-right.

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+bn = \frac{a-b}{a+b}\\
v=aFo(1e2sin2ϕ)0.5v = aF_o(1-e^2sin^2\phi)^{-0.5}\\
ρ=aFo(1e2)(1e2sin2ϕ)1.5\rho = aF_o(1-e^2)(1-e^2sin^2\phi)^{-1.5}\\
η=vρ1\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)6N = I + II(\lambda-\lambda_o)^2+III(\lambda-\lambda_0)^4+IIIA(\lambda-\lambda_0)^6\\
E=e0+IV(λλ0)+V(λλ0)3+IV(λΛ0)5E = 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=secϕνXI=secϕ6v3(vρ+2tan2ϕ)XII=secϕ5040v7(61+662tan2ϕ+1320tan4ϕ+720tan5ϕ)X = \frac{sec\phi'}{\nu}\\ 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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