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I have a set of lat/long coordinates and need to offset the value to the "left" by < 10 meters. I calculate left by getting the direction my instrument is facing and use basic trigonometry to get the displacement in terms of lat/long. There is another similar answer on this forum but that does not talk about such small offsets. Algorithm for offsetting a latitude/longitude by some amount of meters

d_lat = shift * math.cos(math.radians(90 - theta))
d_long = shift * math.sin(math.radians(90 - theta))

How do I accurately convert this small shift from meters to latitude and longitude values? Since my offset is by less than 10 meters I need it to be fairly accurate.

user
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1 Answers1

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Use GeographicLib:

from geographiclib.geodesic import Geodesic

geod = Geodesic.WGS84


lat1 = 0
lon1 = 0
theta = 0 #direction from North, clockwise 
azi1 = theta - 90 #(90 degrees to the left)
shift = 10 #meters


g = geod.Direct(lat1, lon1, azi1, shift)


lat2 = g['lat2']
lon2 = g['lon2']


print("lat2 = {:.8f}, lon2 = {:.8f}.".format(lat2, lon2))


lat2 = 0.00000000, lon2 = -0.00008983.
Gabriel De Luca
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