def test_Ex9_intersect():
# Two paths A and B are given by two pairs of positions:
# Enter elements as lat/long in deg:
n_EA1_E = lat_lon2n_E(rad(10), rad(20))
n_EA2_E = lat_lon2n_E(rad(30), rad(40))
n_EB1_E = lat_lon2n_E(rad(50), rad(60))
n_EB2_E = lat_lon2n_E(rad(70), rad(80))
# Find the intersection between the two paths, n_EC_E:
n_EC_E_tmp = unit(
np.cross(np.cross(n_EA1_E, n_EA2_E, axis=0),
np.cross(n_EB1_E, n_EB2_E, axis=0),
axis=0))
# n_EC_E_tmp is one of two solutions, the other is -n_EC_E_tmp. Select
# the one that is closet to n_EA1_E, by selecting sign from the dot
# product between n_EC_E_tmp and n_EA1_E:
n_EC_E = np.sign(np.dot(n_EC_E_tmp.T, n_EA1_E)) * n_EC_E_tmp
# When displaying the resulting position for humans, it is more
# convenient to see lat, long:
lat_EC, long_EC = n_E2lat_lon(n_EC_E)
msg = 'Ex9, Intersection: lat, long = {} {} deg'
print(msg.format(deg(lat_EC), deg(long_EC)))
assert_allclose(deg(lat_EC), 40.31864307)
assert_allclose(deg(long_EC), 55.90186788)
def test_Ex8_position_A_and_azimuth_and_distance_to_B():
# Position A is given as n_EA_E:
# Enter elements as lat/long in deg:
lat, lon = rad(80), rad(-90)
n_EA_E = lat_lon2n_E(lat, lon)
# The initial azimuth and great circle distance (s_AB), and Earth
# radius (r_Earth) are also given:
azimuth = rad(200)
s_AB = 1000 # m
r_Earth = 6371e3 # m, mean Earth radius
# Find the destination point B, as n_EB_E ("The direct/first geodetic
# problem" for a sphere)
# SOLUTION:
# Step1: Convert distance in meter into distance in [rad]:
distance_rad = s_AB / r_Earth
# Step2: Find n_EB_E:
n_EB_E = n_EA_E_distance_and_azimuth2n_EB_E(n_EA_E, distance_rad, azimuth)
# When displaying the resulting position for humans, it is more
# convenient to see lat, long:
lat_EB, long_EB = n_E2lat_lon(n_EB_E)
print('Ex8, Destination: lat, long = {0} {1} deg'.format(
deg(lat_EB), deg(long_EB)))
assert_allclose(deg(lat_EB), 79.99154867)
assert_allclose(deg(long_EB), -90.01769837)
azimuth1 = n_EA_E_and_n_EB_E2azimuth(n_EA_E, n_EB_E, a=r_Earth, f=0)
assert_allclose(azimuth, azimuth1 + 2 * np.pi)
def test_Ex6_interpolated_position():
# Position B at time t0 and t2 is given as n_EB_E_t0 and n_EB_E_t1:
# Enter elements as lat/long in deg:
n_EB_E_t0 = lat_lon2n_E(rad(89), rad(0))
n_EB_E_t1 = lat_lon2n_E(rad(89), rad(180))
# The times are given as:
t0 = 10
t1 = 20
ti = 16 # time of interpolation
# Find the interpolated position at time ti, n_EB_E_ti
# SOLUTION:
# Using standard interpolation:
n_EB_E_ti = unit(n_EB_E_t0 + (ti - t0) * (n_EB_E_t1 - n_EB_E_t0) /
(t1 - t0))
# When displaying the resulting position for humans, it is more
# convenient to see lat, long:
lat_EB_ti, long_EB_ti = n_E2lat_lon(n_EB_E_ti)
msg = 'Ex6, Interpolated position: lat, long = {} {} deg'
print(msg.format(deg(lat_EB_ti), deg(long_EB_ti)))
assert_allclose(deg(lat_EB_ti), 89.7999805)
assert_allclose(deg(long_EB_ti), 180.)
def test_n_EA_E_distance_and_azimuth2n_EB_E():
a = [[1.0, 1.0], [0.0, 0.0],
[0.0, 0.0]] # lat = [0, 0] and lon = [0, 0] degrees
distance_rad = np.pi / 2
azimuth = np.r_[np.pi / 2, 0]
b = n_EA_E_distance_and_azimuth2n_EB_E(a, distance_rad, azimuth)
print(deg(*n_E2lat_lon(a)))
print(deg(*n_E2lat_lon(b)))
print(b.tolist())
assert_allclose(b, [[0.0, 0.0], [1.0, 0.0], [0.0, 1.0]
]) # # lat = [0, 90] and lon = [90, 0] degrees
aa = n_EA_E_distance_and_azimuth2n_EB_E(b, distance_rad, azimuth + np.pi)
print(aa.tolist())
assert_allclose(aa, a)
def test_Ex1_A_and_B_to_delta_in_frame_N():
# Positions A and B are given in (decimal) degrees and depths:
lat_EA, lon_EA, z_EA = rad(1), rad(2), 3
lat_EB, lon_EB, z_EB = rad(4), rad(5), 6
# Find the exact vector between the two positions, given in meters
# north, east, and down, i.e. find p_AB_N.
# SOLUTION:
# Step1: Convert to n-vectors (rad() converts to radians):
n_EA_E = lat_lon2n_E(lat_EA, lon_EA)
n_EB_E = lat_lon2n_E(lat_EB, lon_EB)
# Step2: Find p_AB_E (delta decomposed in E).
# WGS-84 ellipsoid is default:
p_AB_E = n_EA_E_and_n_EB_E2p_AB_E(n_EA_E, n_EB_E, z_EA, z_EB)
# Step3: Find R_EN for position A:
R_EN = n_E2R_EN(n_EA_E)
# Step4: Find p_AB_N
p_AB_N = np.dot(R_EN.T, p_AB_E)
# (Note the transpose of R_EN: The "closest-rule" says that when
# decomposing, the frame in the subscript of the rotation matrix that
# is closest to the vector, should equal the frame where the vector is
# decomposed. Thus the calculation np.dot(R_NE, p_AB_E) is correct,
# since the vector is decomposed in E, and E is closest to the vector.
# In the example we only had R_EN, and thus we must transpose it:
# R_EN'=R_NE)
# Step5: Also find the direction (azimuth) to B, relative to north:
azimuth = np.arctan2(p_AB_N[1], p_AB_N[0])
# positive angle about down-axis
print('Ex1, delta north, east, down = {0}, {1}, {2}'.format(
p_AB_N[0], p_AB_N[1], p_AB_N[2]))
print('Ex1, azimuth = {0} deg'.format(deg(azimuth)))
assert_allclose(p_AB_N[0], 331730.23478089)
assert_allclose(p_AB_N[1], 332997.87498927)
assert_allclose(p_AB_N[2], 17404.27136194)
assert_allclose(deg(azimuth), 45.10926324)
def test_Ex3_ECEF_vector_to_geodetic_latitude():
# Position B is given as p_EB_E ("ECEF-vector")
p_EB_E = 6371e3 * np.vstack((0.9, -1, 1.1)) # m
# Find position B as geodetic latitude, longitude and height
# SOLUTION:
# Find n-vector from the p-vector:
n_EB_E, z_EB = p_EB_E2n_EB_E(p_EB_E)
# Convert to lat, long and height:
lat_EB, long_EB = n_E2lat_lon(n_EB_E)
h_EB = -z_EB
msg = 'Ex3, Pos B: lat, long = {} {} deg, height = {} m'
print(msg.format(deg(lat_EB), deg(long_EB), h_EB))
assert_allclose(deg(lat_EB), 39.37874867)
assert_allclose(deg(long_EB), -48.0127875)
assert_allclose(h_EB, 4702059.83429485)
def test_Ex2_B_and_delta_in_frame_B_to_C_in_frame_E():
# delta vector from B to C, decomposed in B is given:
p_BC_B = np.r_[3000, 2000, 100].reshape((-1, 1)) # pylint: disable=too-many-function-args
# Position and orientation of B is given:
n_EB_E = unit([[1], [2], [3]]) # unit to get unit length of vector
z_EB = -400
R_NB = zyx2R(rad(10), rad(20), rad(30))
# the three angles are yaw, pitch, and roll
# A custom reference ellipsoid is given (replacing WGS-84):
a, f = 6378135, 1.0 / 298.26 # (WGS-72)
# Find the position of C.
# SOLUTION:
# Step1: Find R_EN:
R_EN = n_E2R_EN(n_EB_E)
# Step2: Find R_EB, from R_EN and R_NB:
R_EB = np.dot(R_EN, R_NB) # Note: closest frames cancel
# Step3: Decompose the delta vector in E:
p_BC_E = np.dot(R_EB, p_BC_B)
# no transpose of R_EB, since the vector is in B
# Step4: Find the position of C, using the functions that goes from one
# position and a delta, to a new position:
n_EC_E, z_EC = n_EA_E_and_p_AB_E2n_EB_E(n_EB_E, p_BC_E, z_EB, a, f)
# When displaying the resulting position for humans, it is more
# convenient to see lat, long:
lat_EC, long_EC = n_E2lat_lon(n_EC_E)
# Here we also assume that the user wants output height (= - depth):
msg = 'Ex2, Pos C: lat, long = {},{} deg, height = {} m'
print(msg.format(deg(lat_EC), deg(long_EC), -z_EC))
assert_allclose(deg(lat_EC), 53.32637826)
assert_allclose(deg(long_EC), 63.46812344)
assert_allclose(z_EC, -406.00719607)
def test_rad_and_deg(values):
radians = rad(values)
assert_allclose(deg(radians), values)