I have a method in an API that takes a lat/long coordinate and will find other coordinates within a specified distance. That distance is in radians.
All the math I do these days deals with accounting or maybe x,y coordinates for laying out UI elements, so I appreciate some help validating these numbers.
Lets ignore people in buildings (altitude) and the fact that the planet isn't perfectly spherical. It is my understanding that the supplied method is doing the Haversine formula internally but that detail is isolated from me.
I am looking at the formula for radians:
θ = s /r, where θ is the subtended
angle in radians, s is arc length, and
r is radius
Given the convenient mean radius of the Earth of:
6371 km
(≈3,959 mi)
I have seen other places saying (6378km)
That means 1 radian on Earth is equal to 6371 km in arc length. That would mean the radian for finding coordinates 1 meter apart would be
( 1 / 6371 ) × 10-7
ie - 1.56961231 × 10-7.
Is that right? If not where is the above incorrect?
It's correct. Although I feel a little shame posting this as an answer :)
Related
I have a list of lat/long points which form a path on the surface of the earth.
I have another point on the surface of the earth and I want to find the shortest distance from that point to a point that falls on the path.
While I could approximate the surface of the earth as a plane, accuracy is important.
The distance between points on the path could be anywhere from 1-1000m. The distance to the point not on the path is from 1-50m. The maximum acceptable error is 0.1m.
Any method of calculating this is acceptable, whether assuming a plane, sphere, or the real shape of the earth as long as the error would not exceed 0.1m for any point on land.
This question is marked language agnostic to encourage answers from people not familiar with the language used. The implementation will be in Dart.
It sounds like you need to calculate the Great Circle Distance across the surface of the earth. A Great Circle calculation permits the calculation of distance along the earth's surface between two arbitrary latitudes and longitudes.
A trivial example of a Great Circle calculation would be to migrate along the Earth's surface along the line of equator (zero degrees latitude). Each degree of longitude of migration along that line (for instance from [0 deg N, 90 deg W] to [0 deg N to 91 deg W]) equates to 1/360th of 40,070km, or 111.306km. Moving between two latitudes and two longitudes requires transformation of coordinates and is outside the scope of this quick note.
Summary and equations found here:
https://en.wikipedia.org/wiki/Great-circle_distance
The accuracy you are seeking of 0.1m requires further refinement; using a sphere to approximate the shape of the Earth will limit the accuracy to perhaps 0.5% (see paragraph from Wikipedia, below). Put another way, a 0.5% error of two points 1000 km apart would be 5000 m.
A more formal and precise calculation will use the true shape of the earth, known as a geoid. This is determined by gravitational measurements and is updated from time to time by the geodesic community.
It is possible to determine one's absolute position on Earth to +/- 0.1m (or better) with advanced GPS surveying techniques, such as RTK or satellite correction services (e.g. Omnistar), but determining the path distance to that accuracy is not the same thing. A survey-quality receiver has corrections built into it for current geoids so that it can translate the lat/long/height calculation it makes using GPS signals to the current reference geoid used by the surveying / geodesic community.
You may not actually require 0.1m accuracy for your application; very few applications require 0.1m absolute accuracy over any distance except (for example) geographic determinations of movement in the Earth's surface. Relative accuracy is more important; e.g. measuring the same point at different times. It is more important to know how much a seismic fault moved relative to its position yesterday or last week, or whether a critical point on a pipeline has moved 2 cm to the north in the past year.
I hope this helps.
Cheers
GP
Per Wikipedia:
So long as a spherical Earth is assumed, any single formula for distance on the Earth is only guaranteed correct within 0.5% (though better accuracy is possible if the formula is only intended to apply to a limited area). [7]
I have a SQL database set of places to which I am assigned coordinates (lat, long). I would like to ask those points that lie within a radius of 5km from my point inside. I wonder how to construct a query in a way that does not collect unnecessary records?
Since you are talking about small distances of about 5 km and we are probably not in the direct vicinity of the north or south pole we can work with an approximated grid system of longitude and latitude values. Each degree in latidude is equivalent to a distance of km_per_lat=6371km*2*pi/360degrees = 111.195km. The distance between two longitudinal lines that are 1 degree apart depends on the actual latitude:
km_per_long=km_per_lat * cos(lat)
For areas here in North Germany (51 degrees north) this value would be around 69.98km.
So, assuming we are interested in small distances around lat0 and long0 we can safely assume that the translation factors for longitudinal and latitudinal angles will stay the same and we can simply apply the formula
SELECT 111.195*sqrt(power(lat-#lat0,2)
+power(cos(pi()/180*#lat0)*(long-#long0),2)) dist_in_km FROM tbl
Since you want to use the formula in the WHERE clause of your select you could use the following:
SELECT * FROM tbl
WHERE 111.195*sqrt(power(lat-#lat0,2)
+power(cos(pi()/180*#lat0)*(long-#long0),2)) < 5
The select statement will work for latitude and longitude values given in degree (in a decimal notation). Because of that we have to convert the value inside the cos() function to radians by multiplying it with pi()/180.
If you have to work with larger distances (>500km) then it is probably better to apply the appropriate distance formula used in navigation like
cos(delta)=cos(lat0)*cos(lat)*cos(long-long0) + sin(lat0)*sin(lat)
After calculating the actual angle delta by applying acos() you simply multiply that value by the earth's radius R = 6371km = 180/pi()*111.195km and you have your desired distance (see here: Wiki: great circle distance)
Update (reply to comment):
Not sure what you intend to do. If there is only one reference position you want to compare against then you can of course precompile your distance calculation a bit like
SELECT #lat0:=51,#long0:=-9; -- assuming a base position of: 51°N 9°E
SELECT #rad:=PI()/180,#fx:=#rad*6371,#fy:=#fx*cos(#rad*#lat0);
Your distance calculation will then simplify to just
SELECT #dist:=sqrt(power(#fx*(lat-#lat0),2)+power(#fy*(long-#long0),2))
with current positions in lat and long (no more cosine functions necessary). It is up to you whether you want to store all incoming positions in the database first or whether you want to do the calculations somewhere outside in Spring, Java or whatever language you are using. The equations are there and easy to use.
I would go with Euklid. dist=sqrt(power(x1-x2,2)+power(y1-y2,2)) . It works everywhere. Maybe you have to add a conversion to the x/y-coordinates, if degrees can't be translated in km that easy.
Than you can go and select everything you like WHERE x IS BETWEEN (x-5) AND (x+5) AND y IS BETWEEN (y-5) AND (y+5) . Now you can check the results with Euklid.
With an optimisation of the result order, you can get better results at first. Maybe there's a way to take Euklid to SQL, too.
I am storing a point as latitude and longitude in a Mysql server. They are both float(10,6). Given a radius, say 100 yards or 100 meters, how can I calculate points around the center. I was thinking of using GIS but I heard it is incomplete or very limited in functionality.
1) Tranform the lat/lon to a cartesian based coordinate system of units meter,
the tranformed point is now at (x,y)
2) Use school mathemathics (polar coordinates) to calculate the points:
2a) create points on circle at (0,0) with polar coordinates (r* sin (phi), r* cos(phi)), r in meters, phi in radians
2b) add (x,y) to all that resulting points, to move the circle points from center (0,0) to (x,y)
3) tranform all points back to lat/lon
I need to calculate the distance between two points, but not in the regular way. I need to know 'the east to west distance' + 'the north to south distance'. I guess this is more simple then the regular 'as the crow flies' calculation but i still can't figure out how to do it.
I want to do this using a MySQL query and preferably have the result returned in km. One of the points will be a constant in the query and the other point is a point from the DB so something like SELECT abs(longitude-39.12345)...+abs(latitude... AS Distance FROM shops WHERE shopname='Bingo'.
Thanks in advance!
The north-to-south distance is proportional to the difference in the latitudes. It's about 1 nautical mile per minute of arc (the circumference of the earth is about 21600 nautical miles).
The east-to-west distance is proportional to the difference in the longitudes, but it also varies with the latitude (e.g. it's zero at the poles): I think it's proportional to the cosine of latitude.
Your answer depends on the accuracy required in your answer. If you don't need an answer more accurate than a spherical earth model, you can use a solution similar to the one given by Captain Tom. If you require more accuracy, you'll need to assume the earth is an oblate spheroid. See http://en.wikipedia.org/wiki/Vincenty%27s_formulae for a couple of solutions.
The east-west difference between two points at different latitudes is a distinct number in degrees of longitude, but converting this to miles is problematic because the miles per degree vary according to the latitude. For example, Los Angles and New York City are 44.3 degrees of longitude apart, but converting this to miles would result in a larger number at LA's latitude than at NYC's latitude, since latitude lines are longest at the equator and shrink to zero at the poles.
A reasonable convention would be to count the E-W distance as the average of the two distances calculated at the two latitudes.
You can determine the distance between any two geocode points using the Great-Circle Distance. Here is another decent article on the subject.
The following is an example in C# which shows how to do this:
var earth_radius = Constants.EarthRadius; // 6377.8 km
var dLat = ToRadians(fromLatitude - toLatitude);
var dLon = ToRadians(fromLongitude - toLongitude);
var a =
Math.Sin(dLat / 2) *
Math.Sin(dLat / 2) +
Math.Cos(ToRadians(fromLatitude)) *
Math.Cos(ToRadians(toLatitude)) *
Math.Sin(dLon / 2) *
Math.Sin(dLon / 2);
var c = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)));
var distanceInKilometers = earth_radius * c;
Use simple trig. The normal "as the crow flies" distance is the hypotenuse of the right triangle formed with the two points you have in mind at either ends of the hypotenuse.
Just look at this http://en.wikipedia.org/wiki/Hypotenuse and the solution should become clear.
I am reverse engineering a transportation visualization app. I need to find out the latitude for the origin of their data feed. Specifically what XY 0,0 is. The only formulas I have found calculate distance between two points, or location of a bearing/distance.
They use the XY to display a map in a very legacy application. The XY is in FEET.
I have these coordinates:
47.70446615506108, -122.34469839507263: x=1268314, y=260622
47.774182540800616,-122.3412994737105: x=1269649, y=286031
47.60024792289405, -122.32767331735774: x=1271767, y=222532
47.57012494413499, -122.29129609983679: x=1280532, y=211374
I need to find out what the latitude and longitude of x=0, y=0 is and what the formula would be to find this out.
They have two data feeds, one is more current than the other. The feed with the most current data does NOT include latitude, longitude, but only XY. I am trying to extrapolate based on their less current, yet more informative (includes lat, lon) data feed what 0,0 is so I can simply convert their (more current) data feed's XY coordinates to latitude and longitude.
If you look at the first 2 lines of data, and subtract the latitude
47.7044 - 47.7741 = -0.06972 degrees
There are 60 nautical miles per degree of latitude, and 6076 feet per nautical mile.
-.06972 * 60 * 6076 = 25,415 ft
Subtracting the two 'Y' values:
260662 - 286031 = 25,409 ft
So indeed that seems to prove the X and Y values are in feet.
If you take any of the Y values, and convert back to degrees, for example
260622 ft / ( 6076 ft/nm ) / ( 60 nm/degree ) = .71
286031 ft / 6076 / 60 = .78
So subtracting those values from the latitudes of (47.70 and 47.77) gives you very close to exactly 47 degrees, which should be your y=0 point.
For longitude, a degree is 60 nautical miles at the equator and 0 miles at the poles. So the number of miles per degree has to be multiplied by the cosine of the latitude, so approx cos(47 degrees), or .68. So instead of 6076 nm per degree, it's about 4145 nm.
So for the X values,
1268314 ft / ( 4145 ft/nm ) / ( 60 nm/degree ) = 5.10 degrees
1269649 ft / 4145 / 60 = 5.10 degrees
These X numbers increase as the latitude increases (less negative), so I believe you should add 5.1 degrees, which means the X base point is about
-122.3 + 5.1 = 117.2 West longitude for your x=0 point.
This is roughly the position of Spokane WA.
So given X=1280532, Y=211374
Lat = 47 + ( 211374 / 6096 / 60 ) = 47.58
Lon = -117.2 - ( 1280532 / ( 6096 * cos(47.58)) / 60 ) = -122.35
Which is roughly equivalent to the given data 47.57 and -122.29
The variance may be due to different projections - the X,Y system may be a "flattened" projection as opposed to lat/long which apply to a spherical projection? So to be accurate you may yet need more advanced math or that open source library :)
This question may also be helpful, it contains code for calculating great circle distances:
Calculate distance between two latitude-longitude points? (Haversine formula)
There are many different coordinate systems. You need to find out the what the coordinate systems are for both the lat/lon's (e.g. WGS84 etc) and x/y's first (e.g. some sort of projected system probably).
Once you have that information there are several tools you can use to do conversions and manipulations. One example (of a free open source coding library) is proj4.
Ask them what coordinate system they're using! (or if you got the dataset from some database, look at the metadata for the dataset and it should tell you. Otherwise I'd be skeptical of its value)
Most likely this is one of the state plane coordinate systems. They're for localized areas of the earth (kind of like UTM), and are frequently used for surveying.
You can use CORPSCON (or other GIS programs; ExpertGPS will do this if you have the GIS Option Pack but it's not free. I forget whether GPSBabel does conversion) to convert between lat/long and any of the state plane coordinate systems. You'll also need to know which datum the coordinates are in. WGS84 and NAD83 are very close but NAD27 is different.
You've got good advice on coordinate systems already, so I'll just chime in with the library I've used with great success in the past.
Geotrans is approved for use by the US Department of Defence, so you can be sure that it is well tested. You can grab it from here:
http://earth-info.nga.mil/GandG/geotrans/index.html
That might not be the right link as that page talks about the application, not the library. I expect the library is in the Developers package. Licensing terms were very liberal from memory, but make sure you review the terms before using it commercially.
Edit:
An interesting discussion on Geotrans licensing can be found here:
http://www.mail-archive.com/debian-legal#lists.debian.org/msg39263.html
Over here, I said this:
In Java, I would use the OpenMap converter from a point's expression in UTM to one using Latitude and Longitude (assuming a WGS-84 ellipsoid which is most commonly used in GPS).
OpenMap is open source and I would post a link to their download page but they have a short license script in the way. So, to avoid being rude, I won't deep link. Instead, head to their homepage and click Downloads.
That should either solve your problem directly or at least point you towards a useful algorithm.
I've used Brenor Brophey's gPoint PHP class to do this on a couple of occasions. Solid results, GPL code, and easily deployed. Recommended.