How can I calculate the distance between two points using Latitude and Longitude?
python - Getting distance between two points based on latitude/longitude - Stack Overflow
Distance between latitude and longitude points
python - How can I quickly estimate the distance between two (latitude, longitude) points? - Stack Overflow
What is the distance per degree of longitude?
What is longitude?
What is the distance between two consecutive latitudes?
The distance between any two adjacent latitudes is approximately 69 miles or 111 km. Latitude lines run parallel to each other. That's why the distance between them remains constant from the South to the North pole. On the other hand, longitude lines are furthest apart at the equator and meet at the poles.
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I have a list of locations and their latitude and longitudes. I want to compare them to a specific point and use a formula to output their distances from that point in terms of how many miles East and North of that point they are, not in a straight line beween them.
So I'd have my origin point and point A. Lines 2 and 3 would look something like this:
Name of Origin point | latitude O | longitude O | 0 | 0
Name of Point A | latitude A | longitude A | X (number of miles) | Y (number of miles)
What formulas would I use in columns D and E to calculate these distances?
The Vincenty distance is now deprecated since GeoPy version 1.13 - you should use geopy.distance.distance() instead!
Some previous answers were based on the haversine formula, which assumes the earth is a sphere, which results in errors of up to about 0.5% (according to help(geopy.distance)). The Vincenty distance uses more accurate ellipsoidal models, such as WGS-84, and is implemented in geopy. For example,
import geopy.distance
coords_1 = (52.2296756, 21.0122287)
coords_2 = (52.406374, 16.9251681)
print(geopy.distance.geodesic(coords_1, coords_2).km)
will print the distance of 279.352901604 kilometers using the default ellipsoid WGS-84. (You can also choose .miles or one of several other distance units.)
Just as a note, if you just need a quick and easy way of finding the distance between two points, I strongly recommend using the approach described in Kurt's answer instead of reimplementing Haversine—see his post for rationale.
This answer focuses just on answering the specific bug the OP ran into.
It's because in Python, all the trigonometry functions use radians, not degrees.
You can either convert the numbers manually to radians, or use the radians function from the math module:
from math import sin, cos, sqrt, atan2, radians
# Approximate radius of earth in km
R = 6373.0
lat1 = radians(52.2296756)
lon1 = radians(21.0122287)
lat2 = radians(52.406374)
lon2 = radians(16.9251681)
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
distance = R * c
print("Result: ", distance)
print("Should be: ", 278.546, "km")
The distance is now returning the correct value of 278.545589351 km.
The answers to Haversine Formula in Python (Bearing and Distance between two GPS points) provide Python implementations that answer your question.
Using the implementation below I performed 100,000 iterations in less than 1 second on an older laptop. I think for your purposes this should be sufficient. However, you should profile anything before you optimize for performance.
from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
"""
Calculate the great circle distance between two points
on the earth (specified in decimal degrees)
"""
# convert decimal degrees to radians
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
# Radius of earth in kilometers is 6371
km = 6371* c
return km
To underestimate haversine(lat1, long1, lat2, long2) * 0.90 or whatever factor you want. I don't see how introducing error to your underestimation is useful.
Since the distance is relatively small, you can use the equirectangular distance approximation. This approximation is faster than using the Haversine formula. So, to get the distance from your reference point (lat1, lon1) to the point you're testing (lat2, lon2) use the formula below:
from math import sqrt, cos, radians
R = 6371 # radius of the earth in km
x = (radians(lon2) - radians(lon1)) * cos(0.5 * (radians(lat2) + radians(lat1)))
y = radians(lat2) - radians(lat1)
d = R * sqrt(x*x + y*y)
Since R is in km, the distance d will be in km.
Reference: http://www.movable-type.co.uk/scripts/latlong.html