Posted on 2014-09-15
Last Modified: 2014-09-15

How do I add, and substract 1 mile for a given latitude? And 1 mile to a given longitude?

Question by:SimpleDude
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Expert Comment

ID: 40323401
Latitude is easy.  Longitude is more complicated.
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Accepted Solution

ozo earned 400 total points
ID: 40323405
latitude + 0.014457
longitude + 0.014481/cos(latitude)
LVL 27

Assisted Solution

d-glitch earned 100 total points
ID: 40323417
And you need to be careful at the boundaries:  the N and S poles, the equator, the prime (0 deg) meridian and its opposite (180 deg).

Author Comment

ID: 40323426
Hi guys, I just need to select X number of records within a Lat and Long range.

I was using this formula, but it drastically decreases the server performance:

  (ACOS(SIN(47.988594 * PI() / 180) * SIN(dbo.ADDRESSES.latitude * PI() / 180) + COS(47.988594 * PI() / 180) * COS(dbo.ADDRESSES.latitude * PI()
                      / 180) * COS((- 122.201789 - dbo.ADDRESSES.longitude) * PI() / 180)) * 180 / PI() * 60 * 1.1515 BETWEEN 0 AND @distance
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LVL 27

Expert Comment

ID: 40323453
How big is the data base?  Is it sorted in any way?
What units are used in the database?  Degrees or radians?
You should search in the same units the data base uses.  Save a lot on conversions.

This   (SIN(47.988594 * PI() / 180)  (and several similar terms) are constants.  
You only have to calculate them once.

Is this related to the nearest k restaurants problem?

Author Comment

ID: 40323466
Something like that. The point of origin (lat-long) changes constantly, that's why I cannot calculate it just once.
LVL 84

Expert Comment

ID: 40323474
If the restaurants are in a small area, it may be preferable to convert the latitudes and longitudes into a flat grid to make distance calculations easier.
If the restaurants are in a large area where a flat approximation would be inaccurate, it may be preferable to convert into a 3-d orthogonal coordinate system to make distance calculations easier.

Author Closing Comment

ID: 40323557
Thanks guys

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