Calculate distance between two lat/lon/alt points in Python

distancegeographiclibpython

I'm trying to find out how to calculate the distance between two points with latitude, longitude, and elevation data.

I found that geographiclib.geodesic.Geodesic.WGS84.Inverse() outputs a s12 parameter that has the distance data according to the geographiclib documentation. I can also see a Wikipedia entry on altitude correction.

Is there any pre-built function that can take all three dimentions and output the correct distance or would the solution be to run a correction on the 2d method output?

I'm very new to this, so I'm also trying to learn and am open to using a different library

Best Answer

As outlined in Calculating distance between two points using latitude longitude and altitude (elevation), Geopy: Calculating Distance, [GPS] Use pyproj to calculate distance, azimuth, and elevation from GPS latitude and longitude or Haversine formula that includes an altitude parameter?, you need to calculate first the 2D great-circle distance or the geodesic distance between the 2 points (distance between flat coordinates) and combine it with the difference in altitudes inside the Euclidean Formula for the 3D distance calculation. The solution assumes that the altitude is in meters and thus the great_circle's or geodesic's distance needs to be in meters

You can use GeoPy, pyproj or geographiclib for example.

With GeoPy

import numpy as np
# points with latitude,longitude units in degrees ,altitude unit in meters
pt1 =[35.3524,135.0302, 100]
pt2 =[35.3532,135.0305,500]
# 2D geodesic distance in meters
from geopy import distance 
distance_2d= distance.distance(pt1[:2], pt2[:2]).m
print(distance_2d)
92.85194331754518
# 3D euclidean distance
distance_3d = np.sqrt(distance_2d**2 + (pt2[2] - pt1[2])**2)
print(distance_3d)
410.6354628838632

With pyproj

from pyproj import Geod
g = Geod(ellps='WGS84')
# 2D distance in meters with longitude, latitude of the points
azimuth1, azimuth2, distance_2d = g.inv(pt1[1], pt1[0], pt2[1], pt2[0]) 
print(distance_2d)
92.85194331754519
# 3D euclidean distance
distance_3d = np.hypot(distance,pt2[2]-pt1[2])
print(distance_3d)
410.63546288386

With geographiclib

from geographiclib.geodesic import Geodesic
geod = Geodesic.WGS84 
g = geod.Inverse(pt1[0], pt1[1],pt2[0],pt2[1])
#2D geodesic distance in meters
print(g['s12'])
92.85194331754519
# 3D euclidean distance
distance_3d = np.hypot(g['s12'],pt2[2]-pt1[2])
print(distance_3d)
410.63546288386

Related Solutions

[GIS] Finding the minimum and maximum distance between an observer and a lat/lon bounding box

The "bounding box" can have several interpretations. The two most natural are (a) the sides are meridians and the top and bottom are circles of constant latitude and (b) the sides are still meridians, but now the top and bottom are geodesics (which therefore do not necessarily have constant latitude). Fortunately, the solution method is the same in both interpretations, because in all cases the sides are portions of circles. This reduces the problem to optimizing the distance between a fixed point x and an arc of a circle (in three dimensions). Do this for each of the four sides and retain the best results.

A simple way to represent the arc of a circle is to choose coordinates for x in which the circle's center is at the origin, the circle is determined by two vectors e and f of equal length (say, r) and at right angles to each other, and the arc is given by all points of the form cos(q) * e + sin(q) * f for angles q ranging from q0 through q1. Critical points for the optima occur either at the endpoints of the arc or where the tangent vector is perpendicular to the direction vector. The latter tells us the inner product of x - (cos(q) * e + sin(q) * f) with (-sin(q) * e + cos(q) * f) is zero. This easily reduces to the equation

tan(q) = -xe / xf

where xe is the inner product of x with e and xf is the inner product of x with f. There will usually be two solutions (differing by 180 degrees). Use ATan (or its equivalent) to find them. Any of these solutions that lies in the interval [q0, q1] is a critical point.

In this fashion you collect (in most cases) up to two critical points per side plus the four vertices of the bounding box: that gives between 4 and 12 distance calculations to make in order to find the largest and smallest distance. These are 3D (Pythagorean) distances: compute them using the Pythagorean formula, but don't bother to take square roots when comparing; just take square roots for the largest and smallest distances.

^{Portions of the four circles determined by the sides of this bounding box are shown in gray. The arcs of latitude are in red and the arcs of the meridians are in blue.}

There is a notable exception to this nice situation: x might lie on, or extremely close to, the axis of the circle. In this case, every point along the circle (and therefore every point along the arc) has the same distance to x. Make sure to check for this situation. It's easy to detect: both xf and xe are zero (or extremely small compared to x itself).

It remains, then, to describe how to find e, f, q0, and q1. That's easy:

For a meridian at longitude lambda, the circle's center is already at (0,0,0), so don't change the coordinates of x (which presumably is referenced to a geocentric system). Take e = (Rcos(lambda), Rsin(lambda), 0) and f = (0,0,R) where R is the earth's radius (in a spherical model). Here, q0 will be the minimum latitude among the bounding box coordinates and q1 will be the maximum latitude.
For a circle of latitude phi, take the center to be (0, 0, Rsin(phi)). Subtract these coordinates from x. In these new coordinates, we can let e = (Rcos(phi), 0, 0) and f = (0, R*cos(phi), 0). Now q0 is the westernmost longitude and q1 is the easternmost longitude in the bounding box.
Equations for e, f, q0, and q1 are more complicated for arbitrary geodesics (case (b) in the first paragraph). Because it is not apparent these are needed, I won't derive them here.

In summary,

Only simple calculations are needed to establish the 3D coordinate representation of x, e, f and to find q0 and q1. (This is all done four times, once for each side of the bounding box.)
Four arctangent calculations will determine between 0 and 8 critical points. At this time, you can inspect xe and xf (four times) to check whether any of the sides of the bounding box are equidistant from x. (You will have to decide how such a situation is to be handled).
Squared pythagorean distances (sums of squares) have to be determined between x and the four corners of the bounding box plus these other critical points. Take the square roots of the min and max of these distances.

[GIS] Measuring distance between two coordinate pairs taken from NMEA file

If your data is in a cartesian/rectangular coordinate system you could simply do 2 pythagorean calculations... where the first one would calculate the distance in XY plane, where a would be x1-x2 and b would be y1-y2. The second calculation would use the result from the first calculation as a and b would be z1-z2.

perhaps something like this

sqrt((sqrt((deltaX)^2+(deltaY)^2)))^2+(deltaZ)^2)

Best Answer

Related Solutions

[GIS] Finding the minimum and maximum distance between an observer and a lat/lon bounding box

[GIS] Measuring distance between two coordinate pairs taken from NMEA file

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