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

I'm looking for an algorithm which when given a latitude and longitude pair and a vector translation in meters in Cartesian coordinates (x,y) would give me a new coordinate. Sort of like a reverse Haversine. I could also work with a distance and a heading transformation, but this would probably be slower and not as accurate. Ideally, the algorithm should be fast as I'm working on an embedded system. Accuracy is not critical, within 10 meters would be good.

Yeah, that would be fine as I'm expecting <1km offsets.

whuber Correct answer

10 years agoIf your displacements aren't too great (less than a few kilometers) and you're not right at the poles, use the quick and dirty estimate that 111,111 meters (111.111 km) in the y direction is 1 degree (of latitude) and 111,111 * cos(latitude) meters in the x direction is 1 degree (of longitude).

I suppose this could work; what would the range be - within 10 degrees of the poles?

@Thomas: Actually, you can be very close to the poles. I checked against a UTM calculation using equal x- and y-displacements of 1400 m (so the total displacement is 2 km). The results are good to 8.6 meters or better. The worst latitude (for this direction and amount of displacement) is 81 degrees: the approximation actually gets more accurate as you move north and its error stays below 10 meters until you get beyond 89.6 degrees!

Incidentally, these magic numbers of 111,111 are easy to remember by knowing some history: the French originally *defined* the meter so that 10^7 meters would be the distance along the Paris meridian from the equator to the north pole. Thus, 10^7 / 90 = 111,111.1 meters equals one degree of latitude to within the capabilities of French surveyors two centuries ago.

Thanks! Within 0.4 degrees is much better than I could have hoped for.

So with the formula if I wanted to move +100m in the y direction from say 10.0 N, 10.0 E, would I just add 100 / 111111? If moving in the x direction +100m, would it be 100÷(111,111×(cos 10))? Just making sure I've got this right.

@Thomas Yes, that's right. Note how the second formula *expands* the apparent x-displacement (by virtue of dividing by a number less than 1) as it should, because a degree of longitude gets smaller as you move towards the poles from the equator. The only potential hitch is to make sure you and your software platform agree on what "cos" means: it had better interpret cos(10) as the cosine of 10 *degrees*, not 10 radians! (If not, 10 degrees = 10 * pi / 180 radians illustrates the simple conversion.) At this point, the code offered by @haakon_d should make complete sense to you.

Good point - must remember to convert to radians!

Somebody attempted to edit this answer to replace "meters" by "km." They probably were reading the comma "," in the European sense of a decimal point. I follow the American convention (which I believe is the convention of international publications, too) of using a comma to separate long digit strings into groups of three and a decimal point "." instead of the comma. (This usage is clearly shown in previous comments.) To avoid any ambiguity I have edited the answer to show clearly what the comma and point mean.

Somebody else attempted to edit this answer to replace the cosine of the latitude by the cosine of the longitude. That's definitely wrong; I rolled back the edit.

@whuber -- that's an interesting story for sure but I'm not sure that numbers get much _"easier to remember"_ than 111111... (but maybe things were different "back in 2010". :-) +

@whuber A million years later, but just figured I would point out that the 'safe' option which everybody tends to understand is using spaces to indicate the thousand separator. 111 111 meters or 111 111,111 meters or 111 111.111 meters will be understood by virtually everyone without difficulty.

Note: In Java, cos() takes radians by default. You can use Math.toRadians to convert.

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`

I just want to point out that this is identical to the answer I provided except you have replaced my value of 111,111 meters per degree by 111,319.5. Your value is slightly better at high latitudes but slightly worse in the lower latitudes (from 0 to about 40 degrees). Either value meets the stated requirements of accuracy.

+1 for providing code. Note that it's more accurate than you suspect (the error is typically less than 5 m over 2000 m).

I wondered if I should add a remark in my answer that this is an identical solution to yours except for the value of R, but left it out due to brevity. When it comes to precision you’re right as long as you don’t add any rotational errors to the system. Using offsets measured in a local projected coordinate system the rotational errors may grow quite large.

That's an excellent point: we have implicitly assumed the x-displacement is at least close to *true* east-west and the y-displacement is close to north-south. If not, they have to be converted into equivalent E-W and N-S displacements (not just "eastings" and "northings") before computing their lat-lon equivalents.

The d distance parameter of the Aviation Formulary equations is in radians, e.g. (distance/radius-of-earth).

@haakon_d can you tell me if i want to turn 90 degrees how can i apply this to your formula. I am new to this and i am interested in your off sett algorithm. I want to make an object move arbitrarily on the map however i want to include turning as well. Can you give me some ideas on this.

I made a Java implementation. Not the most advanced, but ready for use. `double lat; double lon; //Earth’s radius, choose the one that fits your needs //final int r=6378137; final int r=6372814; //offsets in meters double dn = j; double de = i; //Coordinate offsets in radians double dLat = dn/r; double dLon = de/(r*Math.cos(Math.PI*lat/180)); //OffsetPosition, decimal degrees double latO = lat + dLat * 180/Math.PI; double lonO = lon + dLon * 180/Math.PI;` *Edit: Fixed the markdown.

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.

Thanks for this - looks ideal. Though I can't figure out if the distance is in meters or some other measurement.

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

License under CC-BY-SA with attribution

Content dated before 6/26/2020 9:53 AM

underdark 10 years ago

So you'd be fine modeling the earth as a sphere?