import math


def bng(input_lat, input_lon):
    """
    Convert lat and long to BNG
    Expects two floats as input
    Returns a tuple of Easting, Northing floats
    Accurate to ~5m

    For testing purposes:
    input: 51.44533267, -0.32824866
    output: 516275.97337480687, 173142.1442810801

    Entirely Based on code by Hannah Fry (CASA):
    hannahfry.co.uk/2012/02/01/converting-british-national-grid-to-latitude-and-longitude-ii/

    """
    # Simple bounds checking
    if not all([0 <= input_lat <= 90, -180 <= input_lon <= 180]):
        raise Exception(
            "input_lat should be between 0 and 90, input_lon should be between -180 and 80")
    # Convert input to degrees
    lat_1 = input_lat * math.pi / 180
    lon_1 = input_lon * math.pi / 180
    # The GSR80 semi-major and semi-minor axes used for WGS84 (m)
    a_1, b_1 = 6378137.000, 6356752.3141
    # The eccentricity (squared) of the GRS80 ellipsoid
    e2_1 = 1 - (b_1 * b_1) / (a_1 * a_1)
    # Transverse radius of curvature
    nu_1 = a_1 / math.sqrt(1 - e2_1 * math.sin(lat_1) ** 2)
    # Third spherical coordinate is 0, in this case
    H = 0
    x_1 = (nu_1 + H) * math.cos(lat_1) * math.cos(lon_1)
    y_1 = (nu_1 + H) * math.cos(lat_1) * math.sin(lon_1)
    z_1 = ((1 - e2_1) * nu_1 + H) * math.sin(lat_1)

    # Perform Helmert transform (to go between Airy 1830 (_1) and GRS80 (_2))
    s = 20.4894 * 10 ** -6
    # The translations along x,y,z axes respectively
    tx, ty, tz = -446.448, 125.157, -542.060
    # The rotations along x,y,z respectively, in seconds
    rxs, rys, rzs = -0.1502, -0.2470, -0.8421
    # In radians
    rx, ry, rz = \
        rxs * math.pi / (180 * 3600.),\
        rys * math.pi / (180 * 3600.),\
        rzs * math.pi / (180 * 3600.)
    x_2 = tx + (1 + s) * x_1 + (-rz) * y_1 + (ry) * z_1
    y_2 = ty + (rz) * x_1 + (1 + s) * y_1 + (-rx) * z_1
    z_2 = tz + (-ry) * x_1 + (rx) * y_1 + (1 + s) * z_1

    # The GSR80 semi-major and semi-minor axes used for WGS84 (m)
    a, b = 6377563.396, 6356256.909
    # The eccentricity of the Airy 1830 ellipsoid
    e2 = 1 - (b * b) / (a * a)
    p = math.sqrt(x_2 ** 2 + y_2 ** 2)
    # Initial value
    lat = math.atan2(z_2, (p * (1 - e2)))
    latold = 2 * math.pi

    # Latitude is obtained by an iterative procedure
    while abs(lat - latold) > 10 ** -16:
        lat, latold = latold, lat
        nu = a / math.sqrt(1 - e2 * math.sin(latold) ** 2)
        lat = math.atan2(z_2 + e2 * nu * math.sin(latold), p)

    lon = math.atan2(y_2, x_2)
    H = p / math.cos(lat) - nu
    # Scale factor on the central meridian
    F0 = 0.9996012717
    # Latitude of true origin (radians)
    lat0 = 49 * math.pi / 180
    # Longtitude of true origin and central meridian (radians)
    lon0 = -2 * math.pi / 180
    # Northing & easting of true origin (m)
    N0, E0 = -100000, 400000
    n = (a - b) / (a + b)
    # Meridional radius of curvature
    rho = a * F0 * (1-e2) * (1 - e2 * math.sin(lat) ** 2) ** (-1.5)
    eta2 = nu * F0 / rho - 1

    M1 = (1 + n + (5/4) * n ** 2 + (5 / 4) * n ** 3) \
        * (lat - lat0)
    M2 = (3 * n + 3 * n ** 2 + (21 / 8) * n ** 3) \
        * math.sin(lat - lat0) * math.cos(lat + lat0)
    M3 = ((15 / 8) * n ** 2 + (15 / 8) * n ** 3) \
        * math.sin(2 * (lat-lat0)) * math.cos(2 * (lat + lat0))
    M4 = (35 / 24) * n ** 3 * math.sin(3 * (lat - lat0)) \
        * math.cos(3 * (lat + lat0))

    M = b * F0 * (M1 - M2 + M3 - M4)

    I = M + N0
    II = nu * F0 * math.sin(lat) * \
        math.cos(lat) / 2
    III = nu * F0 * math.sin(lat) * \
        math.cos(lat) ** 3 * (5 - math.tan(lat) ** 2 + 9 * eta2) / 24
    IIIA = nu * F0 * math.sin(lat) * \
        math.cos(lat) ** 5 * (61 - 58 * math.tan(lat) ** 2 + math.tan(lat) ** 4) / 720
    IV = nu * F0 * math.cos(lat)
    V = nu * F0 * \
        math.cos(lat) ** 3 * (nu / rho - math.tan(lat) ** 2) / 6
    VI = nu * F0 * math.cos(lat) ** 5 * \
        (5 - 18 * math.tan(lat) ** 2 + math.tan(lat) ** 4 + 14 * eta2 - 58 * eta2 * math.tan(lat) ** 2) / 120

    N = I + II * (lon - lon0) \
        ** 2 + III * (lon - lon0) ** 4 + IIIA * (lon - lon0) ** 6
    E = E0 + IV * (lon - lon0) + V * (lon - lon0) \
        ** 3 + VI * (lon - lon0) ** 5
    return E, N

@@ -489,16 +489,18 @@
 ot all(%5B
+-9
 0 %3C= inp

Fix lon and lat bounds in Python implementation