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???
01/16/08 14:33
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#149561 - Not entirely true
Responding to: ???'s previous message
Russ Cooper said:

On the other hand, if you only need to find the closest object, and don't care how far you actually are from it, then you don't have to worry about any of that. Just pretend that a degree of latitude is the same as a degree of longitude and calculate the "distances" in terms of degrees.


This could be a problem as you get away from the equator. As you said, degrees of longitude decrease in distance as you move away from the equator. At the equator a the distance a degree of longitude is (~69 miles) about the same as the distance a degree of latitude is, but this is not true at a longitude of 45º. Even if the distance is short, you run into a problem of making Longitude Distance = Latitude Distance.

For example, if you were around 45º of longitude and you had 2 objects, each 0 degrees and 1 minute away from your current location. Object 1 was at the same latitude and only differed in longitude while Object 2 was at the same logitude and only differed in latitude. Object 1 is actually about 0.82 miles away while Object 2 is about 1.15 miles away.

If you are talking about a short distance what you can do is adjust for your longitude by multiplying it by a constant.




List of 32 messages in thread
TopicAuthorDate
Global positioning algorithm            01/01/70 00:00      
   Global?            01/01/70 00:00      
   Ehm what do you receive?            01/01/70 00:00      
      coordinates            01/01/70 00:00      
         Length/absolute value of a vector ?            01/01/70 00:00      
         How do they know?            01/01/70 00:00      
         An old greek...            01/01/70 00:00      
            What about doppler            01/01/70 00:00      
               not doppler & math            01/01/70 00:00      
                  Haversine formula is it accurate for short distanc            01/01/70 00:00      
               Are you thinking of D-VOR            01/01/70 00:00      
            An old greek... not accurate enough for            01/01/70 00:00      
               Well, you're stating that it's a 2D space.            01/01/70 00:00      
                  Well, you're stating that it's a 2D space.            01/01/70 00:00      
                     So how wide are your roads?!            01/01/70 00:00      
               You said approximate            01/01/70 00:00      
               Some questions            01/01/70 00:00      
                  Answers to Cooper            01/01/70 00:00      
                     Plane should be fine, then            01/01/70 00:00      
                        Plane should be fine, then            01/01/70 00:00      
                           If you have cartesian coordinates ...            01/01/70 00:00      
                           complex math trough LUT            01/01/70 00:00      
                              Fixed point            01/01/70 00:00      
                                 Scaling            01/01/70 00:00      
                     You have wide roads, then!            01/01/70 00:00      
                        That's still wide!            01/01/70 00:00      
                     Much better            01/01/70 00:00      
                         Much better            01/01/70 00:00      
                           Adjusting for latitude            01/01/70 00:00      
                              Thanks            01/01/70 00:00      
                  Not entirely true            01/01/70 00:00      
                     You\'re right            01/01/70 00:00      

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