# EPSG:9807

## Transverse Mercator

### Attributes

Data source: EPSG

Information source: EPSG guidance note #7-2, http://www.epsg.org

Revision date: 2022-01-06

### Formula

```Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided.

For the calculation of easting and northing from latitude and longitude, first calculate constants for the projection:

n = f / (2-f)
B = [a/(1+n)] (1 + n^2/4 + n^4/64)

h1 = n/2 – (2/3)n^2 + (5/16)n^3 + (41/180)n^4
h2 = (13/48)n^2 – (3/5)n^3 + (557/1440)n^4
h3 = (61/240)n^3 – (103/140)n^4
h4 = (49561/161280)n^4

Then the meridional arc distance from equator to the projection origin (Mo) is computed from:
If LatO = 0 then Mo = 0
else if LatO ≡ 90°N ≡ π/2 radians
Mo = B (π/2)
else if LatO ≡ 90°S ≡ -π/2 radians
Mo = B (-π/2)
else
Qo = asinh(tan LatO) – [e atanh(e sin LatO)]
βo = atan(sinh Qo)
ξO0 = asin (sin βo)
Note: The previous two steps are taken from the generic calculation flow given below for latitude Lat, but here for LatO may be simplified to ξO0 = βo = atan(sinh Qo).
ξO1 = h1 sin(2ξOo)
ξO2 = h2 sin(4ξOo)
ξO3 = h3 sin(6ξOo)
ξO4 = h4 sin(8ξOo)
ξO = ξO0+ ξO1+ ξO2+ ξO3+ ξO4
Mo = B ξO
end

Note: if the projection grid origin is very close to but not exactly at the pole (within 2" or 50m), in the equation above for Qo the tangent function is unstable and may fail. Rather than using Qo as above, Mo may instead be calculated from:

Mo = a[(1 – e^2/4 – 3e^4/64 – 5e^6/256 –....)LatO – (3e^2/8 + 3e^4/32 + 45e^6/1024+....)sin(2.LatO)
+ (15e^4/256 + 45e^6/1024 +.....)sin(4.LatO) – (35e^6/3072 + ....)sin(6.LatO)  + .....]

Then
Q = asinh(tan Lat) – [e atanh(e sin Lat)]
β = atan(sinh Q)
η0 = atanh [cos β sin(Lon – LonO)]
ξ0 = asin (sin β  cosh η0)
ξ1 = h1 sin(2ξ0) cosh(2η0)
η1 = h1 cos(2ξ0) sinh(2η0)
ξ2 = h2 sin(4ξ0) cosh(4η0)
η2 = h2 cos(4ξ0) sinh(4η0)
ξ3 = h3 sin(6ξ0) cosh(6η0)
η3 = h3 cos(6ξ0) sinh(6η0)
ξ4 = h4 sin(8ξ0) cosh(8η0)
η4 = h4 cos(8ξ0) sinh(8η0)
ξ = ξ0 + ξ1 + ξ2 + ξ3 + ξ4
η = η0 + η1 + η2 + η3 + η4
and
Easting, E =  FE + ko B η
Northing, N =  FN + ko (B ξ – Mo)

For the reverse formulas to convert Easting and Northing projected coordinates to latitude and longitude first calculate constants of the projection where n is as for the forward conversion, as are B and Mo:
h1' = n/2 – (2/3)n^2 + (37/96)n^3 – (1/360)n^4
h2' = (1/48)n^2 + (1/15)n^3 – (437/1440)n^4
h3' = (17/480)n^3 – (37/840)n^4
h4' = (4397/161280)n^4

Then
η' = (E –  FE) / (B ko)
ξ' = [(N – FN) + ko Mo] / (B ko)
ξ1' = h1' sin(2ξ') cosh(2η')
η1' = h1' cos(2ξ') sinh(2η')
ξ2' = h2' sin(4ξ') cosh(4η')
η2' = h2' cos(4ξ') sinh(4η')
ξ3' = h3' sin(6ξ') cosh(6η')
η3' = h3' cos(6ξ') sinh(6η')
ξ4' = h4' sin(8ξ') cosh(8η')
η4' = h4' cos(8ξ') sinh(8η')
ξ0' = ξ' – (ξ1' + ξ2' + ξ3' + ξ4')
η0' = η' – (η1' + η2' + η3' + η4')

β' = asin(sin ξ0' / cosh η0')
Q' = asinh(tan β')
Q" = Q' + [e atanh(e tanh Q')] = Q' + [e atanh(e tanh Q")] which should be iterated until the change in Q" is insignificant. Then
Lat = atan(sinh Q")
Lon = LonO + asin(tanh(η0') / cos β')```

### Example

```For Projected Coordinate System OSGB36 / British National Grid

Parameters:
Ellipsoid  Airy 1830  a = 6377563.396 m  1/f = 299.32496
then e'^2 = 0.00671534 and e^2 = 0.00667054

Latitude of natural origin (LatO) = 49°00'00"N = 0.85521133 rad
Longitude of natural origin (LonO) = 2°00'00"W = -0.03490659 rad
Scale factor (ko) = 0.9996013
False Eastings (FE) = 400000.00 m
False Northings (FN) = -100000.00 m

Forward calculation for:
Latitude = 50°30'00.00"N = 0.88139127 rad
Longitude = 00°30'00.00"E = 0.00872665 rad
Constants of the projection:

n = 0.00167322
B = 6366914.609
h1 = 0.0008347452
h2 = 0.0000007554
h3 = 1.18487E-09
h4 = 2.40864E-12
QO = 0.9787671618
ξO0 = 0.8518980373
ξO1 = 0.0008273732
ξO2 = -0.0000001986
ξO3 = -1.0918E-09
ξO4 = 1.2218E-12
Mo = 5429228.602
Q = 1.0191767215
β = 0.8781064142
η0 = 0.0278629616
ξ0 = 0.8785743280
η1 = -0.0000086229
ξ1 = 0.0008215669
η2 = -0.0000000786
ξ2 = -0.0000002768
η3 = 1.05551E-10
ξ3 = -1.01855E-09
η4 = 3.97791E-13
ξ4 = 1.67447E-12
η = 0.0278542603
ξ = 0.8793956171
Then
Easting	E = 577274.99 metres
Northing	N =  69740.50 metres

Reverse calculation for same easting and northing first gives:

h1' = 0.0008347455
h2' = 0.0000000586
h3' = 1.65563E-10
h4' = 2.13692E-13
Then
ξ' = 0.87939562
η' = 0.0278542603
ξ1' = 0.0008213109
η1' = -0.0000086953
ξ2' = -0.0000000217
η2' = -0.0000000061
ξ3' = -1.41881E-10
η3' = 1.486E-11
ξ4' = 1.49609E-13
η4' = 3.50657E-14
ξ0' = 0.8785743280
η0' = 0.0278629616
β' = 0.8781064142
Q' = 1.0191767215
Q" 1st iteration	= 1.0243166838
Q" 2nd iteration	= 1.0243306667
Q" 3rd iteration	= 1.0243307046
Q" 4th iteration	= 1.0243307047

Then
Latitude	(Lat) = 50°30'00.000"N
Longitude (Lon) =	00°30'00.000"E```