Finding Points on the EarthDate: 06/08/2001 at 12:06:26 From: Ryan McGarty Subject: Impossible Equation? Hi, I'm using an equation to determine the distance between two points on the surface of the earth, and want to get the reverse of this: i.e. Find the point who has latitude and longitude five miles north of the given point, and the other three points to the south, east, and west. Given d = distance, lat1 = latitude 1, lat2 = latitude 2, lon1 = longitude 1, and lon2 = longitude 2. d=sin^2( (lat2-lat1)/2 ) + cos(lat1)*cos(lat2)*sin^2( (lon2-lon1)/2 ) How would you solve this for lat2 so you could determine the points to the left and right of a given surface point? We already have the solved equation for lon2, so this is the only remaining problem. Please help. Thanks. Date: 06/08/2001 at 12:45:43 From: Doctor Rick Subject: Re: Impossible Equation? Hi, Ryan. I often direct people who inquire about formulas involving latitude and longitude to the following site: Aviation Formulary V1.30 - Ed Williams http://www.best.com/~williams/avform.htm There you will find a formula for "Lat/lon given radial and distance," which is what you seek. Actually, it's more general: given a starting point, ANY initial heading, and a distance, it calculates the latitude and longitude of your destination. I quote: "A point {lat,lon} is a distance d out on the tc radial from point 1 if: lat=asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc)) IF (cos(lat)=0) lon=lon1 // endpoint a pole ELSE lon=mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi ENDIF "This algorithm is limited to distances such that dlon <pi/2, i.e those that extend around less than one quarter of the circumference of the earth in longitude. A completely general, but more complicated algorithm is necessary if greater distances are allowed: lat =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc)) dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat)) lon=mod( lon1-dlon +pi,2*pi )-pi End of quote. Here, "distance d" is a "distance" in radians (an arc, or central angle). You will need to divide your distance (5 miles) by the radius of the earth (3956 miles) to get the arc in radians. You are interested in particular values of the "radial" tc, namely 0, pi/2, pi, and 3*pi/2 radians (N, E, S, W). The first equation (lat) in the general formula above reduces to these four cases: N (tc=0): lat = asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)) = asin(sin(lat1 + d) = lat1 + d S (tc=pi): lat = asin(sin(lat1)*cos(d)-cos(lat1)*sin(d)) = asin(sin(lat1 - d) = lat1 - d E, W (tc=pi/2 or 3*pi/2): lat = asin(sin(lat1)*cos(d)) In the case of N and S, the longitude is just lon = lon1 (the same longitude as the starting point), because north-south lines follow lines of longitude. In the case of E and W, use the equations for dlon and lon to complete the calculation. NOTE: The introduction to the Aviation Formulary page linked above says this: "For the convenience of North Americans I will take North latitudes and West longitudes as positive and South and East negative. The longitude is the opposite of the usual mathematical convention." If you use the convention that east longitudes are positive, you must make the following sign changes. In the first method, lon = mod(lon1+asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi In the second method, lon = mod(lon1+dlon+pi,2*pi)-pi - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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