| Re: Calculate the distance between 2 points [message #171330 is a reply to message #171327] |
Thu, 30 December 2010 12:31   |
me
Messages: 192
Registered: September 2010
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Senior Member |
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On 12/30/2010 7:15 AM, Denis McMahon wrote:
> On 30/12/10 00:43, Sarah wrote:
>> I know this
>>
>> $distanceValue =
>> rad2deg(acos(sin(deg2rad($LATITUDINE1))*sin(deg2rad($LATITUDINE2))
>> +cos(deg2rad($LATITUDINE1))*cos(deg2rad($LATITUDINE2))*cos(deg2rad($LONGITU DINE1-
>> $LONGITUDINE2))))*69.09;
>>
>> or something like this.... but I don't want to have all records of DB
>> and than analize them one by one .... I want a query the return only
>> nearest values
>
> Then you have to do the calculation as part of the query.
>
> What you can do is pre-calculate some latitude and longitude limits, for
> example 1 degree of latitude always equates to the same distance north
> to south.
>
> Polar circumference is 40,008 km, half that is 20,004 km, divide by 180
> degrees, and you get 111.13333 km per degree of latitude. So, to make it
> easy, for every 100 km of distance that you want to limit to, you only
> need to check +- 1 degree of latitude of your given point (a more
> precise figure is 0.9 degrees).
>
> Therefore, if you're checking for places within 300km, you can
> immediately dismiss anywhere that's more than 3 (or 2.7) degrees of
> latitude away from your specified location.
>
> Longitude is trickier, because the distance per degree of longitude
> varies from 111.3222222 miles at the equator to 0 miles at the pole.
It may be possible to develop a non-linearity table with would produce a
correction factor for every X degrees of longitude, and store that in
the database as LONGADJ or the like. It would be as accurate as it is
detailed. A preliminary calculation would take the declining distance
per degree of longitude into account, and the result of that calculation
would be carried forward for the rest of the equation.
Bill B
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