3d distance calculations with GeoDjango

Question

I am using

• python 2.7.12
• django 1.10.6
• postgreSQL 9.5.6
• postGIS 2.2.2

First question

I ne

Answer 1

** python 3.7, Django 2.2.10

some useful functions ive written to help me with lat/lon coordinates in applications

from django.contrib.gis.geos import Point
from django.contrib.gis.measure import Distance


def get_point(lat, lon):
    try:
        lat = float(lat)
        lon = float(lon)
        if lat and lon:
            return Point(x=float(round(lon, 6)), y=float(round(lat, 6)), srid=4326)
        else:
            return None
    except Exception:
        return None


def get_point_to_point_distance(point_one, point_two, measurement_unit="mile"):
    return Distance(**{measurement_unit: point_one.distance(point_two)})

where measurement_unit can be any of the following:

    STANDARD_UNIT = "m"
    UNITS = {
        'chain': 20.1168,
        'chain_benoit': 20.116782,
        'chain_sears': 20.1167645,
        'british_chain_benoit': 20.1167824944,
        'british_chain_sears': 20.1167651216,
        'british_chain_sears_truncated': 20.116756,
        'cm': 0.01,
        'british_ft': 0.304799471539,
        'british_yd': 0.914398414616,
        'clarke_ft': 0.3047972654,
        'clarke_link': 0.201166195164,
        'fathom': 1.8288,
        'ft': 0.3048,
        'german_m': 1.0000135965,
        'gold_coast_ft': 0.304799710181508,
        'indian_yd': 0.914398530744,
        'inch': 0.0254,
        'km': 1000.0,
        'link': 0.201168,
        'link_benoit': 0.20116782,
        'link_sears': 0.20116765,
        'm': 1.0,
        'mi': 1609.344,
        'mm': 0.001,
        'nm': 1852.0,
        'nm_uk': 1853.184,
        'rod': 5.0292,
        'sears_yd': 0.91439841,
        'survey_ft': 0.304800609601,
        'um': 0.000001,
        'yd': 0.9144,
    }

    # Unit aliases for `UNIT` terms encountered in Spatial Reference WKT.
    ALIAS = {
        'centimeter': 'cm',
        'foot': 'ft',
        'inches': 'inch',
        'kilometer': 'km',
        'kilometre': 'km',
        'meter': 'm',
        'metre': 'm',
        'micrometer': 'um',
        'micrometre': 'um',
        'millimeter': 'mm',
        'millimetre': 'mm',
        'mile': 'mi',
        'yard': 'yd',
        'British chain (Benoit 1895 B)': 'british_chain_benoit',
        'British chain (Sears 1922)': 'british_chain_sears',
        'British chain (Sears 1922 truncated)': 'british_chain_sears_truncated',
        'British foot (Sears 1922)': 'british_ft',
        'British foot': 'british_ft',
        'British yard (Sears 1922)': 'british_yd',
        'British yard': 'british_yd',
        "Clarke's Foot": 'clarke_ft',
        "Clarke's link": 'clarke_link',
        'Chain (Benoit)': 'chain_benoit',
        'Chain (Sears)': 'chain_sears',
        'Foot (International)': 'ft',
        'German legal metre': 'german_m',
        'Gold Coast foot': 'gold_coast_ft',
        'Indian yard': 'indian_yd',
        'Link (Benoit)': 'link_benoit',
        'Link (Sears)': 'link_sears',
        'Nautical Mile': 'nm',
        'Nautical Mile (UK)': 'nm_uk',
        'US survey foot': 'survey_ft',
        'U.S. Foot': 'survey_ft',
        'Yard (Indian)': 'indian_yd',
        'Yard (Sears)': 'sears_yd'
    }

Answer 2

Let's break the problem down:

1. In the Distance class documentation, we can read the following:

   Accepts two geographic fields or expressions and returns the distance between them, as a Distance object.

   So the Distance(p1, p2) returns a Distance object.
   If you do:

   p1 = Instrument.objects.get(pk=151071000).coordinates
   p2 = Instrument.objects.get(pk=151071008).coordinates
   d = Distance(m=p1.distance(p2))
   print d.m
   

   You will get the measurement in meters.
   

   I would stick with the annotate solution, which seems more solid! (opinionated response)

---

2. Distance calculates the 2D distance between two points. In order to get a 3D calculation, you need to create one yourself.
   You can have a look at my method from this question: Calculating distance between two points using latitude longitude and altitude (elevation)

   EDIT 2019: Since the initial answer I have composed a Q&A style example here: How to calculate 3D distance (including altitude) between two points in GeoDjango that uses a far better (and less calculation error-prone) distance calculation between 2 points with altitude.

   In sort:

   We need to calculate the 2D great-circle distance between 2 points using either the Haversine formula or the Vicenty formula and then we can combine it with the difference (delta) in altitude between the 2 points to calculate the Euclidean distance between them as follows:

   dist = sqrt(great_circle((lat_1, lon_1), (lat-2, lon_2).m**2, (alt_1 - alt_2)**2)
   

   The solution assumes that the altitude is in meters and thus converts the great_circle's result into meters as well.

---

Leaving this here for comment continuation purposes.

2. Distance calculates the 2D distance between two points. In order to get a 3D calculation, you need to create one yourself.
You can have a look at my method from this question: Calculating distance between two points using latitude longitude and altitude (elevation)

• Let polar_point_1 = (long_1, lat_1, alt_1) and polar_point_2 = (long_2, lat_2, alt_2)

• Translate each point to it's Cartesian equivalent by utilizing this formula:

  x = alt * cos(lat) * sin(long)
  y = alt * sin(lat)
  z = alt * cos(lat) * cos(long)
  

  and you will have p_1 = (x_1, y_1, z_1) and p_2 = (x_2, y_2, z_2) points respectively.

• Finally use the Euclidean formula:

  dist = sqrt((x_2-x_1)**2 + (y_2-y_1)**2 + (z_2-z_1)**2)