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 think). 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 http://mathforum.org/library/drmath/view/61135.html 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 http://mathforum.org/dr.math/
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