Calculating the Distance Between Two Points

Calculate distance between two latitude-longitude points? (Haversine formula)

This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance.

Excerpt:

This script [in Javascript] calculates great-circle distances between the two points –
that is, the shortest distance over the earth’s surface – using the
‘Haversine’ formula.

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
  var R = 6371; // Radius of the earth in km
  var dLat = deg2rad(lat2-lat1);  // deg2rad below
  var dLon = deg2rad(lon2-lon1); 
  var a = 
    Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * 
    Math.sin(dLon/2) * Math.sin(dLon/2)
    ; 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  var d = R * c; // Distance in km
  return d;
}

function deg2rad(deg) {
  return deg * (Math.PI/180)
}

Calculate distance between 2 GPS coordinates

Calculate the distance between two coordinates by latitude and longitude, including a Javascript implementation.

West and South locations are negative.
Remember minutes and seconds are out of 60 so S31 30' is -31.50 degrees.

Don't forget to convert degrees to radians. Many languages have this function. Or its a simple calculation: radians = degrees * PI / 180.

function degreesToRadians(degrees) {
  return degrees * Math.PI / 180;
}

function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
  var earthRadiusKm = 6371;

  var dLat = degreesToRadians(lat2-lat1);
  var dLon = degreesToRadians(lon2-lon1);

  lat1 = degreesToRadians(lat1);
  lat2 = degreesToRadians(lat2);

  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
          Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  return earthRadiusKm * c;
}

Here are some examples of usage:

distanceInKmBetweenEarthCoordinates(0,0,0,0)  // Distance between same 
                                              // points should be 0
0

distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
                                                          // to Arlington
5918.185064088764

Rough computation of distance between 2 points

As long as you're getting the absolute value (like you stated |X|) and not using the modulus function then that will give you the manhattan distance between the two points

If that is what you want, then you've not missed anything

If you want the straight line distance use the pythagorean theorem. This is sqrt((x1 - x2) ^ 2 + (y1 - y2) ^ 2)

Calculating the distance between 2 points

measure the square distance from one point to the other:

((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)) < d*d

where d is the distance, (x1,y1) are the coordinates of the 'base point' and (x2,y2) the coordinates of the point you want to check.

or if you prefer:

(Math.Pow(x1-x2,2)+Math.Pow(y1-y2,2)) < (d*d);

Noticed that the preferred one does not call Pow at all for speed reasons, and the second one, probably slower, as well does not call Math.Sqrt, always for performance reasons. Maybe such optimization are premature in your case, but they are useful if that code has to be executed a lot of times.

Of course you are talking in meters and I supposed point coordinates are expressed in meters too.

Getting distance between two points based on latitude/longitude

Edit: 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 below instead of re-implementing Haversine -- see his post for rationale.

This answer focuses just on answering the specific bug OP ran into.

---

It's because in Python, all the trig 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.

Calculate the distance between two CGPoints

Well, with stuff your refering too where is the full code:

CGPoint p2; //[1]
CGPoint p1;
//Assign the coord of p2 and p1...
//End Assign...
CGFloat xDist = (p2.x - p1.x); //[2]
CGFloat yDist = (p2.y - p1.y); //[3]
CGFloat distance = sqrt((xDist * xDist) + (yDist * yDist)); //[4]

The distance is the variable distance.

What is going on here:

1. So first off we make two points...
2. Then we find the distance between x coordinates of the points.
3. Now we find the distance between the y coordinates.
4. These lengths are two sides of the triangle, infact they are the legs, time to find the hypotenuse which means after doing some math to rearragne c^2 = a^2 + b^2 we get the hypotenuse to equal sqrt((xDist^2) + (yDist^2)). xDist^2 = (xDist * xDist). And likewise: yDist^2 = (yDist * yDist)

You can't really make a CGPoint be the distance, distance doesn't have an x and y component. It is just 1 number.

If you think CGPoint is a unit of measurement (for example feet is a unit of measurement) it is not.

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