I need a function that maps gps positions to x/y values like this:
getXYpos(GeoPoint relativeNullPoint, GeoPoint p){
deltaLatitude=p.latitude-relativeNullPoint.latitude;
deltaLongitude=p.longitude-relativeNullPoint.longitude;
...
resultX=latitude (or west to east) distance in meters from p to relativeNullPoint
resultY=longitude (or south to north) distance in meters from p to relativeNullPoint
}
i have seen some implementations of "distance of two geoPoints" but they all just calculate the air-line distance. i think the deltaLongitude can be transformed into meters directly but the deltaLatitude depends in the Longitude. does anyone know how this problem can be solved?
Latitude is 0° at the equator and 90° at the pole (and not the opposite). For equator the formula gives 40075 km * cos(0°) / 360 = 111 km. For pole the formula gives 40075 * cos(90°) / 360 = 0 km.
One degree of latitude equals approximately 364,000 feet (69 miles), one minute equals 6,068 feet (1.15 miles), and one-second equals 101 feet. One-degree of longitude equals 288,200 feet (54.6 miles), one minute equals 4,800 feet (0.91 mile), and one second equals 80 feet.
Degrees of latitude are divided into 60 minutes. To be even more precise, those minutes are divided into 60 seconds. One minute of latitude covers about 1.8 kilometers (1.1 miles) and one second of latitude covers about 32 meters (105 feet).
If 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).
To start with, I think you have your latitude and longitude reversed. Longitude measures X, and latitude measures Y.
The latitude is easy to turn into a north-south distance. We know that 360 degrees is a full circle around the earth through the poles, and that distance is 40008000 meters. As long as you don't need to account for the errors due to the earth being not perfectly spherical, the formula is deltaLatitude * 40008000 / 360
.
The tricky part is converting longitude to X, as you suspected. Since it depends on the latitude you need to decide which latitude you're going to use - you could choose the latitude of your origin, the latitude of your destination, or some arbitrary point in between. The circumference at the equator (latitude 0) is 40075160 meters. The circumference of a circle at a given latitude will be proportional to the cosine, so the formula will be deltaLongitude * 40075160 * cos(latitude) / 360
.
Edit: Your comment indicates you had some trouble with the longitude formula; you might have used degrees instead of radians in the call to cos
, that's a common rookie mistake. To make sure there's no ambiguity, here's working code in Python.
def asRadians(degrees):
return degrees * pi / 180
def getXYpos(relativeNullPoint, p):
""" Calculates X and Y distances in meters.
"""
deltaLatitude = p.latitude - relativeNullPoint.latitude
deltaLongitude = p.longitude - relativeNullPoint.longitude
latitudeCircumference = 40075160 * cos(asRadians(relativeNullPoint.latitude))
resultX = deltaLongitude * latitudeCircumference / 360
resultY = deltaLatitude * 40008000 / 360
return resultX, resultY
I chose to use the relativeNullPoint latitude for the X calculation. This has the benefit that if you convert multiple points with the same longitude, they'll have the same X; north-south lines will be vertical.
Edit again: I should have pointed out that this is a very simple formula and you should know its limitations. Obviously the earth is not flat, so any attempt to map it to XY coordinates will involve some compromises. The formula I derived above works best when the area you're converting is small enough to consider flat, and where the slight curvature and non-parallelism of north-south lines can be ignored. There's a whole science to map projections; if you want to see some possibilities a good place to start would be Wikipedia. This specific projection is known as the Equirectangular projection, with some added scaling.
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