Given a list of coordinates, how can I find the closest pair of coordinates using only ONE SQL statement? I have been stuck on this question for a while now and haven't managed to come up with any ideas yet. Since I'm only allowed to use one sql statement I assume variables are not allowed too.
for visualization:
id lat long
1 -12 35
2 -1 100
3 -9 3
You can use a self join:
select t1.*, t2.*
from t t1 join
t t2
on t1.id < t2.id
order by power(t1.lat - t2.lat, 2) + power(t1.long - t2.long, 2)
limit 1
Longitude and latitude are measures associated with the surface of a sphere (our Earth) and the shortest distances are great circles, not straight lines. This is one reason airplanes flying from New York to Rome fly over Newfoundland. Your Euclidean measure would only be shortest if the world looked like this:
This is an equi-rectangular projection of the spherical earth onto a flat map. Note the distortion at the poles. The calculation of great circles is WAY more complex and I doubt you could calculate it in a SQL statement. I suspect, although I have not tried to find any cases, that there are points that calculate as further away but are actually closer as you travel on the earth's surface.
Related
I have MySQL table with 500 location records that has a similar structure to:
id, name, lat, long
Lat & long are decimal (float) location values.
My need is to return a random 100 record set that are a minimum 200 meters and a maximum 500 meters away from each other. I'm familiar with using the great circle formula to get the distance between two points. However, I have no idea how to write a select statement to compare all locations against each other to ensure the distance requirements for the random 100 selected? Any thoughts or help would be greatly appreciated. My only tools are a MySQL database so the solution needs to be written in MySQL SQL. Thank you in advance.
SELECT *
FROM (
SELECT p.latitude1 AS latitude1, p.longitude1 AS longitude1, p.latitude2 AS latitude2, p.longitude2 AS longitude2,
(((ACOS(SIN((latitude2*PI()/180)) * SIN((latitude1*PI()/180))+COS((latitude2*PI()/180)) * COS((latitude1*PI()/180)) * COS(((longitude2- longitude1)* PI()/180))))*180/PI())*60*1.1515) AS distance
FROM places p
)
WHERE distance > 200 AND distance < 500
ORDER BY RAND()
LIMIT 100
If you are only looking at 500 metres then take some short cuts and pretend this is an xy space and use simple trigonometry. Can you even get away with a square instead of donut ?
You could use that as the first cut and then to proper maths on the remainder.
Your second cut will now be small enough. Assign a random no to each and take top X. Voila
edit For the first part, cheat and use Google maps. Find out what fraction of a degree equates to 500 metres at the equator. Use x +/- that value and the same for y. Quick and dirty first cut. **ok it won't work at the poles! **
I have a SQL database set of places to which I am assigned coordinates (lat, long). I would like to ask those points that lie within a radius of 5km from my point inside. I wonder how to construct a query in a way that does not collect unnecessary records?
Since you are talking about small distances of about 5 km and we are probably not in the direct vicinity of the north or south pole we can work with an approximated grid system of longitude and latitude values. Each degree in latidude is equivalent to a distance of km_per_lat=6371km*2*pi/360degrees = 111.195km. The distance between two longitudinal lines that are 1 degree apart depends on the actual latitude:
km_per_long=km_per_lat * cos(lat)
For areas here in North Germany (51 degrees north) this value would be around 69.98km.
So, assuming we are interested in small distances around lat0 and long0 we can safely assume that the translation factors for longitudinal and latitudinal angles will stay the same and we can simply apply the formula
SELECT 111.195*sqrt(power(lat-#lat0,2)
+power(cos(pi()/180*#lat0)*(long-#long0),2)) dist_in_km FROM tbl
Since you want to use the formula in the WHERE clause of your select you could use the following:
SELECT * FROM tbl
WHERE 111.195*sqrt(power(lat-#lat0,2)
+power(cos(pi()/180*#lat0)*(long-#long0),2)) < 5
The select statement will work for latitude and longitude values given in degree (in a decimal notation). Because of that we have to convert the value inside the cos() function to radians by multiplying it with pi()/180.
If you have to work with larger distances (>500km) then it is probably better to apply the appropriate distance formula used in navigation like
cos(delta)=cos(lat0)*cos(lat)*cos(long-long0) + sin(lat0)*sin(lat)
After calculating the actual angle delta by applying acos() you simply multiply that value by the earth's radius R = 6371km = 180/pi()*111.195km and you have your desired distance (see here: Wiki: great circle distance)
Update (reply to comment):
Not sure what you intend to do. If there is only one reference position you want to compare against then you can of course precompile your distance calculation a bit like
SELECT #lat0:=51,#long0:=-9; -- assuming a base position of: 51°N 9°E
SELECT #rad:=PI()/180,#fx:=#rad*6371,#fy:=#fx*cos(#rad*#lat0);
Your distance calculation will then simplify to just
SELECT #dist:=sqrt(power(#fx*(lat-#lat0),2)+power(#fy*(long-#long0),2))
with current positions in lat and long (no more cosine functions necessary). It is up to you whether you want to store all incoming positions in the database first or whether you want to do the calculations somewhere outside in Spring, Java or whatever language you are using. The equations are there and easy to use.
I would go with Euklid. dist=sqrt(power(x1-x2,2)+power(y1-y2,2)) . It works everywhere. Maybe you have to add a conversion to the x/y-coordinates, if degrees can't be translated in km that easy.
Than you can go and select everything you like WHERE x IS BETWEEN (x-5) AND (x+5) AND y IS BETWEEN (y-5) AND (y+5) . Now you can check the results with Euklid.
With an optimisation of the result order, you can get better results at first. Maybe there's a way to take Euklid to SQL, too.
In my MySQL database I have three fields, x,y,z representing a position.
I would like to transform these coordinates into polar coordinates az,el,r, and based on these, select the rows where (e.g.) az are within some region.
How would I go about doing this in MySQL?
EDIT:
This in not a question of how to actually do the coordinate transformation, but rather, if MySQL is capable of transforming the data based on some method, and then selecting data once it is transformed with a criterion based on a comparison of the transformed data.
Solve the Triangle ...
Cartesian = How far along and how far up
Polar = How far away and what angle
In order to convert you need to solve the right triangle for the two known sides
you need to use Pythagoras theorem to find the long side (hypotenuse)
you need the Tangent Function to find the angle
r = √ ( x2 + y2 ) = Pythagoras
θ = tan-1 ( y / x ) = Tangent Function
assuming there's no negative values - then you would have to take the inverse of tan function, or convert them to their positive counterpart
Mysql Pythagorus
SQRT((POWER(242-'EAST',2)) + (POWER(463-'NORT',2))) < 50
assuming your coordinates look like this.... here is an example
http://www.tek-tips.com/viewthread.cfm?qid=1397712
Tangent Function here
http://dev.mysql.com/doc/refman/5.0/en/mathematical-functions.html#function_tan
IMHO this is really a spherical coordinate system maths problem, not a MySQL-specific question.
MySQL just happens to be the data container in this instance.
For any solution you need to work out the maths first, then it becomes a matter of applying the equations to the data.
I can help with MySQL, but I'd have to Google solving these equations and my fingers are tired =)
I'm trying to query any locations within a specified distance from another location. The query is not the problem, but the distance returned by geography.STDistance is.
It seems STDistance makes fairly accurate calculations on locations close to the equator, but I need this to work with locations in the nordic countries. Norway, Sweden, Finland and so on...
According to my calculations, made on locations in northern Sweden, the distance is wrong by a factor of around 2.38?!
Expected result is 1070 meters and returned distance is 2537,28850694302 meters
My query looks like this:
DECLARE #g geography = geography::STGeomFromText('POINT(65.580254 22.179428)', 4326)
SELECT name, [pos].STSrid as srdi, [pos].STDistance(#g) as d
FROM [GPSCHAT].[dbo].[USERS]
and the "other location" has coordinates (65,578541 22,202286) (stored with SRID 4326)
I'm guessing this has to do with the distance from the equator (close to the polar circle), but there has to be a way to calculate this more accurately based on the Latitude or am i wrong?
It looks like you're creating your point using 'X, Y'.
When creating a point from text, use 'Y, X' instead.
Check out this MSDN Article for some more info.
Why don't you make use of another spatial reference identifier which fits better the earth curvature around your position. SRID 4326 might not been measured as accurate as other local referential systems
I am trying to calculate distance between two locations using spatial functions in both Mysql and PostgresSQL. I have taken the latitude and longitude from Google. The details are below
Location one - Lat: 42.260223; Lon: -71.800010
Location two - Lat: 42.245647; Lon: -71.802521
SQL Query used:
SELECT DISTANCE(GEOMFROMTEXT('Point(42.260223 -71.800010)'),GEOMFROMTEXT('Point(42.245647 -71.802521)'))
The both databases are giving the same result 0.014790703059697. But when I calculate distance in other systems the results are different. Please refer the below links
http://www.zip-codes.com/distance_calculator.asp?zip1=01601&zip2=01610&Submit=Search = 1.44 miles
http://www.distancecheck.com/zipcode-distance.php?start=01601&end=01610 = 1.53 miles
So I want to know whether my calculation method/query is right or not. And if it is wrong, then what is the right way of querying the db for the distance.
The simple answer is to use the Haversine formula. This assumes the earth is a sphere, which it isn't, but it's not a bad approximation. This, with lots of other details are described in this presentation:
http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL
In the case above, MySql is simply applying the pythagorean theorem: c2 = a^2 + b^2. In this specific case SQRT((42.245647 - 42.260223)^2 + (-71.802521^2 - -71.800010)^2) = 0.014790703.
There are actually two problems with using the MySql distance functon for distance with coordinates on a sphere. (1) MySql is caclulating distance on a plane, not a sphere. (2) The result is coming back in a form of degrees, not miles. To get a true, spherical distance in miles, km, ft etc, you need to convert your lat and long degrees into the units you want to measure by determining the radius from a line through the center of the earth for the latitude(s) you are measuring.
To get a true measure is quite complicated, many individuals and companies have made careers out of this.