# EPSG:9815

## Hotine Oblique Mercator (variant B)

### Attributes

Data source: EPSG

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

Revision date: 2018-08-29

### 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.

The following constants for the projection may be calculated :

B = {1 + [e^2 * cos^4(latc) / (1 - e^2 )]}^0.5
A = a * B * kc *(1 - e^2 )^0.5 / ( 1 - e^2 * sin^2(latc))
to = tan(pi/4 - latc/2) / ((1 - e*sin(latc)) / (1 + e*sin(latc)))^(e/2)
D = B (1 - e^2)^0.5  / (cos(latc) * ( 1 - e^2*sin^2(latc))^0.5)
if D < 1 to avoid problems with computation of F make D^2  = 1
F = D + (D^2 - 1)^0.5  * SIGN(latc)
H = F*(to)^B
G = (F - 1/F) / 2
gammao = asin(sin(alphac) / D)
lonO = lonc - (asin(G*tan(gamma0))) / B
vc = 0
In general: uc = (A / B) atan2((Dsq - 1)^0.5 , cos (alphac) ) * SIGN(latc)  (see Gn7-2 implementation notes in preface for atan2 convention)
but for the special cases where alphac = 90 degrees (e.g. Hungary, Switzerland) then
uc = A*(lonc - lonO)

Forward case: To compute (E,N) from a given (lat,lon) :

t = tan(pi/4 - lat/2) / ((1 - e sin (lat)) / (1 + e sin (lat)))^(e/2)
Q = H / t^B
S = (Q - 1 / Q) / 2
T = (Q + 1 / Q) / 2
V = sin(B (lon - lonO))
U = (- V cos(gammao) + S sin(gammao)) / T
v = A ln((1 - U) / (1 + U)) / 2 B

In general:
u = (A atan2((S cos(gammao) + V sin(gammao)) , cos(B (lon - lonO))) / B) - (ABS(uc) . SIGN(latc))

but when alphac = pi/2 rad
if lon = lonc, u = 0
else u = (A atan((S cos(gammao) + V sin(gammao)) / cos(B (lon - lonO))) / B) - (ABS(uc) . SIGN(latc) . SIGN(lonc – lon))

The rectified skew co-ordinates are then derived from:
E = v cos(gammac) + u sin(gammac) + Ec
N = u cos(gammac) - v sin(gammac) + Nc

Reverse case: Compute (lat,lon)  from a given (E,N)  :

v’ = (E - Ec) cos(gammac) - (N - Nc) sin(gammac)
u’ = (N - Nc) cos(gammac) + (E - Ec) sin(gammac) + (ABS(uc) . SIGN(latc))

Q’ = e- (B v ‘/ A)  where e is the base of natural logarithms.
S' = (Q’ - 1 / Q’) / 2
T’ = (Q’ + 1 / Q’) / 2
V’ = sin (B u’ / A)
U’ = (V’ cos(gammac) + S’ sin(gammac)) / T’
t’ = (H / ((1 + U’) / (1 - U’))^0.5)^(1 / B)

chi = pi / 2 - 2 atan(t’)

lat = chi + sin(2chi).( e^2 / 2 + 5*e^4 / 24 + e^6 / 12 + 13*e^8 / 360) +  sin(4*chi).( 7*e^4 /48 + 29*e^6 / 240 + 811*e8 / 11520) +  sin(6chi).( 7*e^6 / 120 + 81*e8 / 1120) +  sin(8chi).(4279 e^8 / 161280)

lon=  lonO - atan2 ((S’ cos(gammao) , V’ sin(gammao)) / cos(B*u’ / A)) / B```

### Example

```For Projected Coordinate System  Timbalai 1948 / R.S.O. Borneo (m)

Parameters:
Ellipsoid:  Everest 1830 (1967 Definition)
a = 6377298.556 metres 1/f = 300.8017
then e = 0.081472981and e^2 = 0.006637847

Latitude Projection Centre fc = 4°00'00"N = 0.069813170 rad
Longitude Projection Centre lc = 115°00'00"E = 2.007128640 rad
Azimuth of central line ac = 53°18'56.9537" = 0.930536611 rad
Rectified to skew gc= 53°07'48.3685" = 0.927295218 rad
Scale factor ko= 0.99984
Easting at projection centre Ec = 590476.87 m
Northing at projection centre Nc = 442857.65 m

Forward calculation for:
Latitude lat = 5°23'14.1129"N = 0.094025313 rad
Longitude lon = 115°48'19.8196"E = 2.021187362 rad

B = 1.003303209            F = 1.072121256
A =6376278.686            H = 1.000002991
to = 0.932946976          g0 = 0.927295218
D = 1.002425787           lon0 = 1.914373469
D2 =1.004857458
uc =738096.09              vc =0.00

t =0.910700729             Q =1.098398182
S =0.093990763            T = 1.004407419
V =0.106961709            U = 0.010967247
v =-69702.787                u = 163238.163

Then Easting E =      679245.73 m
Northing N =     596562.78 m

Reverse calculations for same easting and northing first gives :
v’ =   -69702.787              u’ = 901334.257
Q’ = 1.011028053
S’  = 0.010967907          T’ = 1.000060146
V’ = 0.141349378           U’ = 0.093578324
t’ = 0.910700729             c = 0.093404829

Then Latitude = 5°23'14.113"N
Longitude = 115°48'19.820"E```