I have two points as below. I need to get the distance between them in meters.
POINT (80.99456 7.86795)
POINT (80.97454 7.872174)
How can this be done via GeoPandas?
Your points are in a lon, lat coordinate system (EPSG:4326 or WGS 84). To calculate a distance in meters, you would need to either use the Great-circle distance or project them in a local coordinate system to approximate the distance with a good precision.
For Sri Lanka, you can use EPSG:5234 and in GeoPandas, you can use the distance function between two GeoDataFrames.
from shapely.geometry import Point
import geopandas as gpd
pnt1 = Point(80.99456, 7.86795)
pnt2 = Point(80.97454, 7.872174)
points_df = gpd.GeoDataFrame({'geometry': [pnt1, pnt2]}, crs='EPSG:4326')
points_df = points_df.to_crs('EPSG:5234')
points_df2 = points_df.shift() #We shift the dataframe by 1 to align pnt1 with pnt2
points_df.distance(points_df2)
The result should be 2261.92843 m
Related
I would like to use pyephem or skyfield to find the degree of the local horizon (starting from South = 0°) at which is located the highest point of the ecliptic. Note that I am not trying to find the culmination of a planet. It could well be that, when looking South, there is no planet on the ecliptic by that time/latitude, but there will still be a highest point of the ecliptic. Can anyone help?
While it’s likely that someone will soon jump in with a closed-form solution that uses spherical trigonometry — in which case it’s likely that Skyfield or PyEphem will only be used to determine the Earth orientation — here’s a quick way to get an answer within about a degree:
Generate the ecliptic as 360 points one degree apart across the sky.
Compute the altitude and azimuth of each one.
Choose the highest.
The result agrees closely to what I see if I open Stellarium, turn on the ecliptic, and choose a field star right near the point where the ecliptic reaches the highest point in the sky.
import numpy as np
from skyfield import api, framelib
from skyfield.positionlib import Apparent
𝜏 = api.tau
ts = api.load.timescale()
eph = api.load('de421.bsp')
bluffton = api.Topos('40.74 N', '84.11 W')
t = ts.utc(2021, 2, 16, 22, 52)
angle = np.arange(360) / 360.0 * 𝜏
zero = angle * 0.0
f = framelib.ecliptic_frame
d = api.Distance([np.sin(angle), np.cos(angle), zero])
v = api.Velocity([zero, zero, zero])
p = Apparent.from_time_and_frame_vectors(t, f, d, v)
p.center = bluffton
alt, az, distance = p.altaz()
i = np.argmax(alt.degrees) # Which of the 360 points has highest altitude?
print('Altitude of highest point on ecliptic:', alt.degrees[i])
print('Azimuth of highest point on ecliptic:', az.degrees[i])
The result:
Altitude of highest point on ecliptic: 67.5477569215633
Azimuth of highest point on ecliptic: 163.42529398930515
This is probably a computationally expensive enough approach that it won’t interest you once you or someone else does the spherical trigonometry to find an equation for the azimuth; but at the very least this might provide numbers to check possible formulae against.
I am trying to calculate distance between two geographical coordinates and I want to convert geographical coordinates to the km. Therefore I used deg2km function. However, I realise that it is not convert points properly.
For instance, I used these two points.
p_x=[5; 10]; %degree
p_y=[8; 16]; %degree
pos_y=deg2km(p_y,6378);
pos_x=deg2km(p_x,6378);
It returns as:
pos_x= [556.58549846099 1113.17099692198]
pos_y= [890.536797537587 1781.07359507517]
When I calculate distance ( sqrt((556.5-1113.2)^2+(890.5368-1781.1)^2) ) between these points I obtained distance as : 1050.2464
However I checked it google map and also other websites it should be 1042 km.
Do you have any suggestion to calculate distance and also points as kilometers properly?
Thanks in advance!
edited as :
I've points(deg)and I need to convert them km and calculate distance between points.
LAT=[41.000173;41.010134]*pi/180;
LON=[28.995882;28.995584]*pi/180;
I used this code to calculate distance. It calculates properly.
But I can not convert my points to kilometers.
LAT=[41.000173;41.010134]*pi/180;
LON=[28.995882;28.995584]*pi/180;
R=6378; %km
for i=1:length(LAT)-1
psi(i,1) = atan2( sin (LON(i+1)-LON(i)) * cos (LAT(i+1)) , cos (LAT(i)) *sin (LAT(i+1)) - sin (LAT(i)) * cos (LAT(i+1)) * cos (LON(i+1)-LON(i)) );
a=(sin((LAT(i+1)-LAT(i))/2))^2+cos(LAT(i))*cos(LAT(i+1))*(sin((LON(i+1)-LON(i))/2))^2;
c=2*atan2(sqrt(a),sqrt(1-a));
d(i,1)=R*c;
end
I want to calculate the distance between two points on surface of earth in meteres
I have tried with both basemap and cartopy but both result in different numbers.
Basemap:
import mpl_toolkits.basemap.pyproj as pyproj
k = pyproj.Geod(ellps="WGS84")
distance = k.inv(c0[1], c0[0], c1[1], c1[0])[-1]/1000.
Cartopy:
import cartopy.geodesic as gd
k = gd.Geodesic() // defaults to WGS84
distance = k.inverse(c0, c1).base[0,0]/1000
where both coord0 and coord1 are numpy arrays of size 2 having lat and lon of a coordinate.
c0 = numpy.array([77.343750, 22.593726])
c1 = numpy.array([86.945801, 23.684774])
Cartopy Output: 990.6094719605074
Basemap Output: 1072.3456344712142
With Basemap, you must use proper order of (long, lat):
distance = k.inv(c0[0], c0[1], c1[0], c1[1])[-1]/1000.
and the result will agree with Cartopy's, which is the correct result:
990.6094719605074
I am using geopy to simply calculate the distance between two long,lat co-ordinates.
However I am confused on what scale the longitude should be on.
The latitude co-ordinated go from -90 to +90, and currently I've put my Longitude on a scale from 0-360 degrees - should this be -180 to 180 to satisfy :
great_circle(NYC, test).miles
where NYC and test are the co-ord pairs.
Thanks,
Izzy
Geopy Point latitude must be in range [-90; 90], longitude should be in range [-180; 180].
Longitude is automatically normalized (e.g. 185 is normalized to -175), while out-of-band latitudes would result in a ValueError being thrown. See the Point normalization testcases for getting the better idea on how normalization works: test/test_point.py
An example from docs for calculating distance between two points:
>>> from geopy import distance
>>> newport_ri = (41.49008, -71.312796)
>>> cleveland_oh = (41.499498, -81.695391)
>>> print(distance.distance(newport_ri, cleveland_oh).miles)
538.39044536
I want to plot a latitude and longitude using matlab. Using that latitude and longitude as center of the circle, I want to plot a circle of radius 5 Nm.
r = 5/60;
nseg = 100;
x = 25.01;
y = 55.01;
theta = 0 : (2 * pi / nseg) : (2 * pi);
pline_x = r * cos(theta) + x;
pline_y = r * sin(theta) + y;
hold all
geoshow(pline_x, pline_y)
geoshow(x, y)
The circle does not look of what I expected.
Drawing a circle on earth is more complex that it looks like.
Drawing a line or a poly line is simple, because the vertices are defined.
Not so on circle.
a circle is defined by all points having the same distance from center (in meters! not in degrees!!!)
Unfortuantley lat and lon coordinates have not the same scale.
(The distance between two degrees of latidtude is always approx. 111.3 km, while for longitude this is only true at the equator. At the poles the distance between two longitudes approach zero. In Europe the factor is about 0.6. (cos(48deg))
There are two solution, the first is more universal, usefull for nearly all problems.
convert spherical coordinate (of circle center) to cartesian plane with unit = 1m, using a transformation (e.g equidistant transformation, also called equirectangular transf., this transformation works with the cos(centerLat) compensation factor)
calculate points (e.g circle points) in x,y plane using school mathematics.
transform all (x,y) points back to spherical (lat, lon) coordinates, using the inverse transformation of point 1.
Other solution
1. write a function which draws an ellipse in defined rectangle (all cartesian x,y)
2. define bounding of the circle to draw:
2a: calculate north-south diameter of circle/ in degrees: this a bit tricky: the distance is define in meters, you need a transformation to get the latitudeSpan: one degrees of lat is approx 111.3 km (eart circumence / 360.0): With this meters_per_degree value calc the N-S disatcne in degrees.
2b: calculate E-W span in degrees: now more tricky: calculate like 2a, but now divide by cos(centerLatitude) to compensate that E-W distances need more degrees when moving north to have the same meters.
Now draw ellipseInRectangle using N-S and E_W span for heigh and width.
But a circle on a sphere looks on the projected monitor display (or paper) only like a circle in the center of the projection. This shows:
Tissot's Error Ellipse