how to get accurate or (near accurate) distance between two places? Mysql/PHP

Question

SELECT postcode, lat, lng, truncate( (degrees(acos (sin(radians(lat)) * sin( radians('.$latitude.')) + cos(radians(lat)) * cos( radians('.$latitude.')) * cos( radians(lng - ('.$longitude.'))) ) ) * 69.172), 2) as distance FROM myData
This query calculates distance (in miles). But when I check distance for same lat and longitude at google maps my result doesnt match. If the distance is around 10 miles then my result is a bit accurate, but over that it gets wrong (for example, my result showed 13 miles and google showed 22 miles for same post code values)

I got this query from http://forums.mysql.com/read.php?23,3868,3868#msg-3868

How can I get it accurate. Any ideas? Thanks for help.

UPDATE

I tried @Blixt code in PHP. Picked up 2 sample postcodes and their lats longs

//B28 9ET
$lat1 = 52.418819;
$long1 = -1.8481053;

//CV5 8BX
$lat2 = 52.4125573;
$long2 = -1.5407743;

$dtr = M_PI / 180;
$latA = $lat1 * $dtr;
$lonA = $long1 * $dtr;
$latB = $lat2 * $dtr;
$lonB = $long2 * $dtr;
$EarthRadius = 3958.76; //miles
echo $distance = $EarthRadius * acos(cos($latA) * cos($latB) * cos($lonB - $lonA) + sin($latA) * sin($latB));

Results:

My app - 12.95 miles

Google - 17.8 miles

Any ideas how to get it right?

Are you sure Google doesn't take stuff like roads into account when calculating distances? — 0scar
hmm .. may be because of this. I checked using Get directions on google maps. Is there any formula to get something similiar to Googles? — TigerTiger
There is no formula to take roads into account, other than some kind of average bird distance/drive distance ratio that will be wrong 99% of the time. You'll need to use an API that lets you get the road distance in that case. Have a look at the Google Maps API: code.google.com/apis/maps — Blixt

Blixt Blixt · Accepted Answer · 2009-06-12T11:47:50

Have a look at this source code. When I tested it against various other measurement services it seemed to get the same results. It's C# but the math should be easy enough to convert.

Here are the relevant pieces:

public const double EarthRadius = 6371.0072; // Kilometers
//public const double EarthRadius = 3958.76; // Miles

/* ... */

const double dtr = Math.PI / 180;
double latA = this.Latitude * dtr;
double lonA = this.Longitude * dtr;
double latB = other.Latitude * dtr;
double lonB = other.Longitude * dtr;

return GpsLocation.EarthRadius * Math.Acos(Math.Cos(latA) * Math.Cos(latB) * Math.Cos(lonB - lonA) + Math.Sin(latA) * Math.Sin(latB));

Note: The Earth is not perfectly spherical, so the constants used may differ. There is no easy way to make the measurements truly exact. See Wikipedia.

how to get accurate or (near accurate) distance between two places? Mysql/PHP

1 Answers