arcgis-10.0coordinate systemcoordinatesdistancekriging
I need to calculate distance between two stations in a map in order to calculate empirical variogram in ordinary kriging to interpolate temperature data.
Calculating distance in kriging using Euclidean distance formula between two points on XY axes, and I found that not sufficient because I specified Lag_units and Lag_distance in meters when I programmed kriging.
Can I use the haversine formula to calculate distance between real coordinates instead of Euclidean distance?
Using the Pythagorean formula on positions given in latitude and longitude makes as little sense as, say, computing the area of a circle using the formula for a square: although it produces a number, there is no reason to suppose it ought to work.
Although at small scales any smooth surface looks like a plane, the accuracy of the Pythagorean formula depends on the coordinates used. When those coordinates are latitude and longitude on a sphere (or ellipsoid), we can expect that
Distances along lines of longitude will be reasonably accurate.
Distances along the Equator will be reasonably accurate.
All other distances will be erroneous, in rough proportion to the differences in latitude and longitude.
The error depends on the start and end point of the distance calculations. However, because both the sphere and ellipsoid have a circular symmetry around the axis, the error depends only on the difference of the longitudes, so to study this error we might as well take the point of origin to be at the Prime Meridian. Because both the sphere and ellipsoid are symmetric under a north-south reflection, we only need to study points of origin in the southern hemisphere. For any such point we may draw a contour map of the relative error, equal to [Pythagorean calculation] / [True distance].
The Pythagorean formula, using the mean radius of the earth, is
Pythagorean distance = 6371000. * Sqrt[dx^2 + dy^2]] * pi / 180 meters
where dx is the difference in longitudes and dy is the difference in latitudes, both in degrees. (The difference in longitude values is reduced modulo 360 to give the correct value of dx when crossing the antimeridian; not doing so would introduce artificially large errors that tell us nothing about the Pythagorean formula itself.)
The following plots show the relative error compared to the correct distance on the WGS 84 ellipsoid for latitudes from -70 to 0 in increments of 10 degrees. The horizontal coordinate is the difference in longitudes and the vertical coordinate is the latitude of the destination. Light regions have relatively small error: the contour lines are at 1, 1.01, 1.02, 1.05, 1.1, 1.2, 1.5, 2, etc. (The pure white areas in the corners are places where the error goes beyond the range of these contours.) The red dots show the point of origin.
The vertical white bands testify to the correctness of expectation (1): Pythagorean distances are accurate when there is a small difference in longitudes. The horizontal white bands at low latitudes confirm expectation (2): near the Equator, horizontal distances are reasonably accurate. Otherwise, as witnessed by the extensive darker regions, at all other distances the Pythagorean formula is bad.
We can make quantitative estimates of the maximum error attained for pairs of nearby points (within, say, a few hundred kilometers of each other). Scale--using an appropriate value for the radius--is true along the meridian but along a circle of latitude it errs approximately by the secant of the latitude. For example, at a latitude of 40 degrees the secant is 1.31, implying the Pythagorean formula will give distances about 31% too large in the east-west direction. (This is evident in the upper right contour plot, for a point of origin at -40 degrees latitude, where the region immediately east-west of the red dot lies between the 1.2 and 1.5 contours.) Short distances in all other directions will be too large by some amount between 0% and 31%; longer distances may err by even more (as the contour plots show).
Haversine is used for the great circle distance (shortest distance following the sphere curvature). So it does not take the height into account.
For straight line distance, I would first convert your lat/long/radius+alt triplets to XYZ in a 3D cartesian system (see here). Then you can apply your equation.
Because your points are very close from each other, you could also change you coordinates to a projected coordinate system, then you can use your equation for 3D distance with all units in meters. For an exact solution, you should select an custom azimuthal equidistant projection centered on one of your points.
Note : because you are near the equator, you could also use an approximation of the size of a degree (0.5% error if you assume that they are the same in lat and long) in order to compute your distances in meter. 2*pi*R=360degree, so the mean value of a degree along the meridian = 111132 m
Best Answer
What you're looking into is measuring a distance on google maps. Unfortunatly, this is not a question to which an easy answer can be given, but you might consider one of these ways :
Let me now focus on this last bullet : google maps uses a coordinate system called "Pseudo Mercator" or "Web Mercator" which have the code EPSG:3857, I live in Morocco, so the coordinate system I should use is EPSG:32630, depending on where you live, you should pick a convenient coordinate system for accurate results. Once you pick your coordinate system, your should look for a the mathematical tranformation between EPSG:3857 and EPSG:XXXXX (your local CS) or leave it to a third party library to carry such transformation (depends on which tools you're using).
Edit : just forgot to mention that if you want a third party library that can handle tranformations, look up Proj4 or PROJ4NET for C# programming.