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Dead Reckoning

Date: 08/12/2003 at 09:15:30
From: Chris
Subject: Navigation

I am looking to do a form of 'dead reckoning' using a fixed lat/lon
position, velocity components for north and west, and a time delay to
compute an extrapolated lat/lon position. I am having trouble finding 
specific references to do this. The WGS84 ellipsoid is used for the 
lat/lon, and the assumption is that only small distances will be 
traveled between fixes, i.e. Great-Circle distances don't apply (I

For example, if you are at 34 degrees latitude 56 degrees longitude
traveling in a vehicle that has a northward vector component of 
velocity of 100 meters/sec and a westward vector component of velocity
of 75 meters/sec, and it has been 500 milliseconds since the last
known fixed position, what would be the extrapolated position?

Getting the expected results from a extrapolation can be done by
getting the distance between two points using arc length formula and
assuming a perfect sphere. Using a given time delay, determine both
velocity components that would be needed to get to the new lat/lon, 
but actually determining the new position only by velocity is the 
tricky part.

Date: 08/13/2003 at 08:18:00
From: Doctor Rick
Subject: Re: Navigation

Hi, Chris.

If it's true that "great circle distances don't apply," that is, that 
the earth can be treated as locally flat, then all you need is to 
convert latitude and longitude to x, y coordinates in a plane tangent 
to the earth at the initial point. See this item in the Dr. Math 
Archives for the transformation:

   Circle of Radius 1 km around Given Latitude/Longitude 

Given north (y) and east (x) components of velocity, and time of 
travel, you can compute x and y components of the vehicle location 
after that time. In your example,

  x = -75 m/s * 0.500 s = -37.5 m
  y = 100 m/s * 0.500 s =  50.0 m

Use the transformation in the page above to convert (x,y) to latitude 
and longitude (a2, b2) given the latitude and longitude of the 
initial point, (a1, b1).

Your example is definitely a short enough distance that this 
approximation applies. Just use the new latitude and longitude, 
(a2,b2), as the starting point (a1, b1) for the calculation of the 
next leg.

- Doctor Rick, The Math Forum 
Associated Topics:
College Higher-Dimensional Geometry

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