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???
01/17/08 12:54
Modified:
  01/17/08 13:09

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#149640 - Adjusting for latitude
Responding to: ???'s previous message
Ap said:
Can you please explain me why Adjust east-west distance for latitude is required.

Sure. At the equator, one degree of longitude corresponds to an east-west distance of about 69 miles. (Circumference of the earth / 360 degrees = 25,000 miles / 360 ~= 69 miles.) However, as you move north or south away from the equator, each degree of longitude corresponds to a smaller and smaller distance as the longitude lines get closer and closer together. Right at the north and south poles, where all the longitude lines come together, there is no space between them. That's why the adjustment is necessary.

Ap said:
Also is the formula equally well

d = acos(sin(Lat1) × sin(Lat2) + cos(Lat1) × cos(Lat2) × cos(Lon1 – Lon2))

where d= distance and result is in radians.

That formula gives the exact distance, and takes into account the curvature of the earth. In other words, it does not make the approximation that a small chunk of the earth's surface is essentially flat. There are two problems with that formula. One is that it is time consuming to calculate. (I count six trig functions!) The other is that for small distances, you're going to be bothered by roundoff errors in the calculations. Look here for more discussion.

For what you're doing, stick with something simple.

-- Russ


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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