def grav(lat, p=0):
r"""Calculates acceleration due to gravity as a function of latitude and as
a function of pressure in the ocean.
Parameters
----------
lat : array_like
latitude in decimal degrees north [-90...+90]
p : number or array_like. Default p = 0
pressure [dbar]
Returns
-------
g : array_like
gravity [m s :sup:`2`]
See Also
--------
TODO
Notes
-----
In the ocean z is negative.
Examples
--------
>>> import gsw
>>> lat = [-90, -60, -30, 0]
>>> p = 0
>>> gsw.grav(lat, p)
array([ 9.83218621, 9.81917886, 9.79324926, 9.780327 ])
>>> gsw.grav(45)
9.8061998770458008
References
----------
.. [1] IOC, SCOR and IAPSO, 2010: The international thermodynamic equation
of seawater - 2010: Calculation and use of thermodynamic properties.
Intergovernmental Oceanographic Commission, Manuals and Guides No. 56,
UNESCO (English), 196 pp.
.. [2] Moritz (2000) Goedetic reference system 1980. J. Geodesy, 74,
128-133.
.. [3] Saunders, P.M., and N.P. Fofonoff (1976) Conversion of pressure to
depth in the ocean. Deep-Sea Res.,pp. 109 - 111.
Modifications:
2011-03-29. Trevor McDougall & Paul Barker
"""
X = np.sin(lat * DEG2RAD)
sin2 = X ** 2
gs = 9.780327 * (1.0 + (5.2792e-3 + (2.32e-5 * sin2)) * sin2)
z = z_from_p(p, lat)
# z is the height corresponding to p.
grav = gs * (1 - gamma * z)
return grav
def distance(lon, lat, p=0):
r"""Calculates the distance in met res between successive points in the
vectors lon and lat, computed using the Haversine formula on a spherical
earth of radius 6,371 km, being the radius of a sphere having the same
volume as Earth. For a spherical Earth of radius 6,371,000 m, one nautical
mile is 1,853.2488 m, thus one degree of latitude is 111,194.93 m.
Haversine formula:
R = earth's radius (mean radius = 6,371 km)
.. math::
a = \sin^2(\delta \text{lat}/2) +
\cos(\text{lat}_1) \cos(\text{lat}_2) \sin^2(\delta \text{lon}/2)
c = 2 \times \text{atan2}(\sqrt{a}, \sqrt{(1-a)})
d = R \times c
Parameters
----------
lon : array_like
decimal degrees east [0..+360] or [-180 ... +180]
lat : array_like
latitude in decimal degrees north [-90..+90]
p : number or array_like. Default p = 0
pressure [dbar]
Returns
-------
dist: array_like
distance between points on a spherical Earth at pressure (p) [m]
See Also
--------
TODO
Notes
-----
z is height and is negative in the oceanographic.
Distances are probably good to better than 1\% of the "true" distance on
the ellipsoidal earth.
Examples
--------
>>> import gsw
>>> lon = [159, 220]
>>> lat = [-35, 35]
>>> gsw.distance(lon, lat)
array([[ 10030974.652916]])
>>> p = [200, 1000]
>>> gsw.distance(lon, lat, p)
array([[ 10030661.63878009]])
>>> p = [[200], [1000]]
>>> gsw.distance(lon, lat, p)
array([[ 10030661.63878009],
[ 10029412.58776001]])
References
----------
.. [1] http://www.eos.ubc.ca/~rich/map.html
Modifications:
2000-11-06. Rich Pawlowicz
2011-04-04. Paul Barker and Trevor McDougall
"""
# FIXME? The argument handling seems much too complicated.
# Maybe we can come up with some simple specifications of
# what argument combinations are permitted, and handle everything
# with broadcasting. - EF
# FIXME: Eric what do you think? This assume p(stations, depth)
lon, lat, = np.atleast_2d(lon), np.atleast_2d(lat)
if (lon.size == 1) & (lat.size == 1):
raise ValueError('more than one point is needed to compute distance')
elif lon.ndim != lat.ndim:
raise ValueError('lon, lat must have the same dimension')
lon, lat, p = np.broadcast_arrays(lon, lat, p)
dlon = np.diff(lon * DEG2RAD)
dlat = np.diff(lat * DEG2RAD)
a = ((np.sin(dlat / 2)) ** 2 + np.cos(lat[:, :-1] * DEG2RAD) *
np.cos(lat[:, 1:] * DEG2RAD) * (np.sin(dlon / 2)) ** 2)
angles = 2 * np.arctan2(np.sqrt(a), np.sqrt(1 - a))
p_mid = 0.5 * (p[:, 0:-1] + p[:, 0:-1])
lat_mid = 0.5 * (lat[:, :-1] + lat[:, 1:])
z = z_from_p(p_mid, lat_mid)
distance = (earth_radius + z) * angles
return distance
