示例#1
0
def airborne_position(msg0, msg1, t0, t1):
    """Decode airborn position from a pair of even and odd position message

    Args:
        msg0 (string): even message (28 bytes hexadecimal string)
        msg1 (string): odd message (28 bytes hexadecimal string)
        t0 (int): timestamps for the even message
        t1 (int): timestamps for the odd message

    Returns:
        (float, float): (latitude, longitude) of the aircraft
    """

    mb0 = common.hex2bin(msg0)[32:]
    mb1 = common.hex2bin(msg1)[32:]

    # 131072 is 2^17, since CPR lat and lon are 17 bits each.
    cprlat_even = common.bin2int(mb0[22:39]) / 131072.0
    cprlon_even = common.bin2int(mb0[39:56]) / 131072.0
    cprlat_odd = common.bin2int(mb1[22:39]) / 131072.0
    cprlon_odd = common.bin2int(mb1[39:56]) / 131072.0

    air_d_lat_even = 360.0 / 60
    air_d_lat_odd = 360.0 / 59

    # compute latitude index 'j'
    j = common.floor(59 * cprlat_even - 60 * cprlat_odd + 0.5)

    lat_even = float(air_d_lat_even * (j % 60 + cprlat_even))
    lat_odd = float(air_d_lat_odd * (j % 59 + cprlat_odd))

    if lat_even >= 270:
        lat_even = lat_even - 360

    if lat_odd >= 270:
        lat_odd = lat_odd - 360

    # check if both are in the same latidude zone, exit if not
    if common.cprNL(lat_even) != common.cprNL(lat_odd):
        return None

    # compute ni, longitude index m, and longitude
    if (t0 > t1):
        lat = lat_even
        nl = common.cprNL(lat)
        ni = max(common.cprNL(lat) - 0, 1)
        m = common.floor(cprlon_even * (nl - 1) - cprlon_odd * nl + 0.5)
        lon = (360.0 / ni) * (m % ni + cprlon_even)
    else:
        lat = lat_odd
        nl = common.cprNL(lat)
        ni = max(common.cprNL(lat) - 1, 1)
        m = common.floor(cprlon_even * (nl - 1) - cprlon_odd * nl + 0.5)
        lon = (360.0 / ni) * (m % ni + cprlon_odd)

    if lon > 180:
        lon = lon - 360

    return round(lat, 5), round(lon, 5)
示例#2
0
def surface_position_with_ref(msg, lat_ref, lon_ref):
    """Decode surface position with only one message,
    knowing reference nearby location, such as previously calculated location,
    ground station, or airport location, etc. The reference position shall
    be with in 45NM of the true position.

    Args:
        msg (string): even message (28 bytes hexadecimal string)
        lat_ref: previous known latitude
        lon_ref: previous known longitude

    Returns:
        (float, float): (latitude, longitude) of the aircraft
    """

    mb = common.hex2bin(msg)[32:]

    cprlat = common.bin2int(mb[22:39]) / 131072.0
    cprlon = common.bin2int(mb[39:56]) / 131072.0

    i = int(mb[21])
    d_lat = 90.0 / 59 if i else 90.0 / 60

    j = common.floor(lat_ref / d_lat) + common.floor(
        0.5 + ((lat_ref % d_lat) / d_lat) - cprlat
    )

    lat = d_lat * (j + cprlat)

    ni = common.cprNL(lat) - i

    if ni > 0:
        d_lon = 90.0 / ni
    else:
        d_lon = 90.0

    m = common.floor(lon_ref / d_lon) + common.floor(
        0.5 + ((lon_ref % d_lon) / d_lon) - cprlon
    )

    lon = d_lon * (m + cprlon)

    return round(lat, 5), round(lon, 5)
示例#3
0
def surface_position(msg0, msg1, t0, t1, lat_ref, lon_ref):
    """Decode surface position from a pair of even and odd position message,
    the lat/lon of receiver must be provided to yield the correct solution.

    Args:
        msg0 (string): even message (28 bytes hexadecimal string)
        msg1 (string): odd message (28 bytes hexadecimal string)
        t0 (int): timestamps for the even message
        t1 (int): timestamps for the odd message
        lat_ref (float): latitude of the receiver
        lon_ref (float): longitude of the receiver

    Returns:
        (float, float): (latitude, longitude) of the aircraft
    """

    msgbin0 = common.hex2bin(msg0)
    msgbin1 = common.hex2bin(msg1)

    # 131072 is 2^17, since CPR lat and lon are 17 bits each.
    cprlat_even = common.bin2int(msgbin0[54:71]) / 131072.0
    cprlon_even = common.bin2int(msgbin0[71:88]) / 131072.0
    cprlat_odd = common.bin2int(msgbin1[54:71]) / 131072.0
    cprlon_odd = common.bin2int(msgbin1[71:88]) / 131072.0

    air_d_lat_even = 90.0 / 60
    air_d_lat_odd = 90.0 / 59

    # compute latitude index 'j'
    j = common.floor(59 * cprlat_even - 60 * cprlat_odd + 0.5)

    # solution for north hemisphere
    lat_even_n = float(air_d_lat_even * (j % 60 + cprlat_even))
    lat_odd_n = float(air_d_lat_odd * (j % 59 + cprlat_odd))

    # solution for north hemisphere
    lat_even_s = lat_even_n - 90.0
    lat_odd_s = lat_odd_n - 90.0

    # chose which solution corrispondes to receiver location
    lat_even = lat_even_n if lat_ref > 0 else lat_even_s
    lat_odd = lat_odd_n if lat_ref > 0 else lat_odd_s

    # check if both are in the same latidude zone, rare but possible
    if common.cprNL(lat_even) != common.cprNL(lat_odd):
        return None

    # compute ni, longitude index m, and longitude
    if t0 > t1:
        lat = lat_even
        nl = common.cprNL(lat_even)
        ni = max(common.cprNL(lat_even) - 0, 1)
        m = common.floor(cprlon_even * (nl - 1) - cprlon_odd * nl + 0.5)
        lon = (90.0 / ni) * (m % ni + cprlon_even)
    else:
        lat = lat_odd
        nl = common.cprNL(lat_odd)
        ni = max(common.cprNL(lat_odd) - 1, 1)
        m = common.floor(cprlon_even * (nl - 1) - cprlon_odd * nl + 0.5)
        lon = (90.0 / ni) * (m % ni + cprlon_odd)

    # four possible longitude solutions
    lons = [lon, lon + 90.0, lon + 180.0, lon + 270.0]

    # make sure lons are between -180 and 180
    lons = [(l + 180) % 360 - 180 for l in lons]

    # the closest solution to receiver is the correct one
    dls = [abs(lon_ref - l) for l in lons]
    imin = min(range(4), key=dls.__getitem__)
    lon = lons[imin]

    return round(lat, 5), round(lon, 5)