Algorithm for offsetting a latitude/longitude by some amount of meters

I find that Aviation Formulary, here is great for these types of formulas and algorithms. For your problem, check out the "lat/long given radial and distance":here

Note that this algorithm might be a bit too complex for your use, if you want to keep use of trigonometry functions low, etc.

As Liedman says in his answer Williams’s aviation formulas are an invaluable source, and to keep the accuracy within 10 meters for displacements up to 1 km you’ll probably need to use the more complex of these.

But if you’re willing to accept errors above 10m for points offset more than approx 200m you may use a simplified flat earth calculation. I think the errors still will be less than 50m for offsets up to 1km.

 //Position, decimal degrees
 lat = 51.0
 lon = 0.0

 //Earth’s radius, sphere
 R=6378137

 //offsets in meters
 dn = 100
 de = 100

 //Coordinate offsets in radians
 dLat = dn/R
 dLon = de/(R*Cos(Pi*lat/180))

 //OffsetPosition, decimal degrees
 latO = lat + dLat * 180/Pi
 lonO = lon + dLon * 180/Pi 

This should return:

 latO = 51,00089832
 lonO = 0,001427437

It might make sense to project the point first. You could make something like this pseudo-code:

falt_coordinate = latlon_to_utm(original_koordinate)
new_flat_coordinate = flat_coordinate + (x,y)
result_coordinate = utm_to_latlon(new_flat_coordinate)

where (x,y) is the desired offset.

You don't need to use utm, any flat coordinate system, that makes sense in your area will do.

What software are you working with?

I created a simple custom map on Google Maps that illustrates the estimation algorithm mentioned by the accepted answer (1/111111 == one meter). Feel free to see and play with it here:

https://drive.google.com/open?id=1XWlZ8BM00PIZ4qk43DieoJjcXjK4z7xe&usp=sharing

Vincenty's direct formula should do the job.

Here is python code for whuber's answer

from math import cos, radians

def meters_to_lat_lon_displacement(m, origin_latitude):
    lat = m/111111
    lon = m/(111111 * cos(radians(origin_latitude)))
    return lat, lon

Here is Swift version used on kodisha answer on SO:

extension CLLocationCoordinate2D {
  
  func earthRadius() -> CLLocationDistance {
    let earthRadiusInMetersAtSeaLevel = 6378137.0
    let earthRadiusInMetersAtPole = 6356752.314
    
    let r1 = earthRadiusInMetersAtSeaLevel
    let r2 = earthRadiusInMetersAtPole
    let beta = latitude
    
    let earthRadiuseAtGivenLatitude = (
      ( pow(pow(r1, 2) * cos(beta), 2) + pow(pow(r2, 2) * sin(beta), 2) ) /
        ( pow(r1 * cos(beta), 2) + pow(r2 * sin(beta), 2) )
    )
    .squareRoot()
    
    return earthRadiuseAtGivenLatitude
  }
  
  func locationByAdding(
    distance: CLLocationDistance,
    bearing: CLLocationDegrees
  ) -> CLLocationCoordinate2D {
    let latitude = self.latitude
    let longitude = self.longitude
    
    let earthRadiusInMeters = self.earthRadius()
    let brng = bearing.degreesToRadians
    var lat = latitude.degreesToRadians
    var lon = longitude.degreesToRadians
    
    lat = asin(
      sin(lat) * cos(distance / earthRadiusInMeters) +
        cos(lat) * sin(distance / earthRadiusInMeters) * cos(brng)
    )
    lon += atan2(
      sin(brng) * sin(distance / earthRadiusInMeters) * cos(lat),
      cos(distance / earthRadiusInMeters) - sin(lat) * sin(lat)
    )
    
    let newCoordinate = CLLocationCoordinate2D(
      latitude: lat.radiansToDegrees,
      longitude: lon.radiansToDegrees
    )
    
    return newCoordinate
  }
}

extension FloatingPoint {
  var degreesToRadians: Self { self * .pi / 180 }
  var radiansToDegrees: Self { self * 180 / .pi }
}