Calculating Lat/Lng X miles from point

Question

I am wanting to find a latitude and longitude point given a bearing, a distance, and a starting latitude and longitude.

This appears to be the opposite of this question (Distance between lat/long points).

I have already looked into the haversine formula and think it's approximation of the world is probably close enough.

I am assuming that I need to solve the haversine formula for my unknown lat/long, is this correct? Are there any good websites that talk about this sort of thing? It seems like it would be common, but my googling has only turned up questions similar to the one above.

What I am really looking for is just a formula for this. I'd like to give it a starting lat/lng, a bearing, and a distance (miles or kilometers) and I would like to get out of it a lat/lng pair that represent where one would have ended up had they traveled along that route.

Answer 1

I'd be curious how results from this formula compare with Esri's pe.dll.

(citation).

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

Here's an html page for testing.

Answer 2

Bunch of worked out math samples are here Calculate distance, bearing and more between Latitude/Longitude points which includes the solution to "Destination point given distance and bearing from start point".

Answer 3

If you're interested in better accuracy, there's Vincenty. (Link is to the 'direct' form, which is exactly what you're after).

There are quite a few existing implementations, if you're after code.

Also, a question: You're not assuming the traveller maintains the same bearing for the entire trip, are you? If so, then these methods aren't answering the right question. (You'd be better off projecting to mercator, drawing a straight line, then un-projecting the result.)

Answer 4

If you were in a plane, then the point that is r meters away at a bearing of a degrees east of north is displaced by r * cos(a) in the north direction and r * sin(a) in the east direction. (These statements more or less define the sine and cosine.)

Although you are not in a plane--you're working on the surface of a curved ellipsoid that models the Earth's surface--any distance less than a few hundred kilometers covers such a small part of the surface that for most practical purposes it can be considered flat. The only remaining complication is that one degree of longitude does not cover the same distance as a degree of latitude. In a spherical Earth model, one degree of longitude is only cos(latitude) as long as a degree of latitude. (In an ellipsoidal model, this is still an excellent approximation, good to about 2.5 significant figures.)

Finally, one degree of latitude is approximately 10^7 / 90 = 111,111 meters. We now have all the information needed to convert meters to degrees:

The northwards displacement is r * cos(a) / 111111 degrees;

The eastwards displacement is r * sin(a) / cos(latitude) / 111111 degrees.

For example, at a latitude of -0.31399 degrees and a bearing of a = 30 degrees east of north, we can compute

cos(a) = cos(30 degrees) = cos(pi/6 radians) = Sqrt(3)/2 = 0.866025.
sin(a) = sin(30 degrees) = sin(pi/6 radians) = 1/2 = 0.5.
cos(latitude) = cos(-0.31399 degrees) = cos(-0.00548016 radian) = 0.999984984.
r = 100 meters.
east displacement = 100 * 0.5 / 0.999984984 / 111111 = 0.000450007 degree.
north displacement = 100 * 0.866025 / 111111 = 0.000779423 degree.

Whence, starting at (-78.4437, -0.31399), the new location is at (-78.4437 + 0.00045, -0.31399 + 0.0007794) = (-78.4432, -0.313211).

A more accurate answer, in the modern ITRF00 reference system, is (-78.4433, -0.313207): this is 0.43 meters away from the approximate answer, indicating the approximation errs by 0.43% in this case. To achieve higher accuracy you must use either ellipsoidal distance formulas (which are far more complicated) or a high-fidelity conformal projection with zero divergence (so that the bearing is correct).