I'm working on Google maps which will show multiple vehicles. I've two points(one point which will be the old position and other will be the present position) this positioning system will help write an equation of a line passing through these two points. Which might intersect with other line and inform me about the probability of collision.
But the problem is that I'm fetching the coordinates from the GPS module and it will give the location in Latitude and Longitude format.
I'll need x, y coordinates for this writing the equation of the line passing through them. I've already explored most of the method in different web pages, but the problem is that they will ask for some screen size(or map bounds) which are kind of not compatible with my type of method.
Questions: What is the method to convert Latitude and Longitude to (x, y) coordinates, just like if we see from the space and earth was flat and taking Gulf of Guinea (Lat: 0°, Lon: 0°) as the origin.
If your task requires high accuracy you have to use Latitude/Longitude and to solve Great Circle intersection task. The Earth is not flat.
If you can accept existence of some error in your calculations AND all of your vehicles are located in small limited area (up to 100km, although it depends on error you can accept) you may assume this confined area as flat.
For instance, if one your vehicle is located at N10.0 E10.0, second one is located at N10.1 E10.2, you may choose N10.0 E10.0 as the origin.
As a result, these two vehicles will have the following (X, Y) coords (it assumes that axis X goes along equator):
1) (0.0km 0.0km)
2) (21.86km 11.1km)
X of second vehicle is (40000km / 360 degrees) * cos(10.0) * (10.2 - 10.0) = 21.86km
Y of second vehicle is (40000km / 360 degrees) * (10.1 - 10.0) = 11.1km
If you will try to apply flat line-line intersection for vehicles located in 10 000 km from each other - your calcutions most probably will be incorrect.
Related
I'm trying to put a point at the top of Mount Everest in Cesium. My most likely candidate as of last night was the code I borrowed to do geodetic to ecef conversion (from PySatel.coord). Upon review this morning, it appears to be correct:
a = 6378.137
b = 6356.7523142
esq = 6.69437999014 * 0.001
e1sq = 6.73949674228 * 0.001
f = 1 / 298.257223563
def geodetic2ecef(lat, lon, alt):
"""Convert geodetic coordinates to ECEF.
Units are degrees and kilometers.
"""
lat, lon = radians(lat), radians(lon)
xi = sqrt(1 - esq * sin(lat))
x = (a / xi + alt) * cos(lat) * cos(lon)
y = (a / xi + alt) * cos(lat) * sin(lon)
z = (a / xi * (1 - esq) + alt) * sin(lat)
return x, y, z
I pulled the lat / lon / alt for the peak of Mt. Everest from Wikipedia. I multiplied the ECF coordinates provided by the above code by 1000(m/km) before positioning the object in my CZML. I get an ECF location of: [302995.41122130124, 5640733.98308375, 2981975.8695256836]. With the default terrain provider (described in the tutorial), this point is significantly higher than the peak of Mt. Everest.
Here's the relevant CZML snippet:
{"position":
{"cartesian": [302995.41122130124, 5640733.98308375, 2981975.8695256836]},
"id": "ellipsoid-1",
"ellipsoid":
{
"radii": {"cartesian": [3545.5375159540376,
164.44985193756034,
164.62702908803794]},
"material": {"solidColor": {"color": {"rgba": [0, 255, 0, 100]}}}
},
"orientation": {"unitQuaternion": [0.00014107125875577922,
-0.011462389405915903,
-0.010254110199791062,
-0.70702315200093502]}
}
There are several factors at work here.
First, the source data that Cesium uses for terrain may have a lower than expected height for the peak of Mount Everest. We use the CGIAR SRTM dataset, so this item in their FAQ is relevant:
Why do some mountain regions have peaks significantly lower than they should be?
As mentioned earlier, many original data voids are concentrated in mountainous areas and in snow-covered regions. Hence, many peaks in high-mountain areas are actually interpolated. Without using a high resolution co-variable for the interpolation, the interpolation fails to identify that the data void is actually a peak, and tends to “flatten” the peak, leading to underestimates in the true elevation for that region. This issue is largely resolved in Version 4.
They say that it is largely resolved in v4, the version Cesium uses, so hopefully this first factor is not the actual problem.
Second, our processing of the source terrain data for use with Cesium may flatten the peak a bit. This problem will be corrected soon, hopefully within the next couple of months.
Third, Wikipedia provides the height as an elevation above mean sea level (MSL). MSL is a complicated surface that is hard to work with mathematically, so your geodetic2ecef is not doing so. Instead, it, like Cesium, is assuming that the altitude is relative to the WGS84 ellipsoid, which is a much nicer surface to work with.
NGA has a web site that can be used to find the height of MSL above the WGS84 ellipsoid, also known as the geoid height:
http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm96/intpt.html
It reports that for the peak of Mount Everest (27° 59′ 17″ N, 86° 55′ 31″ E), MSL is 28.73 meters below WGS84. If you subtract that number from the altitude of the peak reported on Wikipedia, you should get closer, at least.
This page has information on computing geoid heights programmatically:
http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm96/egm96.html
I recommend interpolating over the 15-minute geoid height file instead of computing heights from coefficients.
A couple of other notes not directly related to the question:
Cesium has code to convert LLA (we call it Cartographic) to Cartesian. See Ellipsoid.cartographicToCartesian.
You can specify coordinates in CZML in cartographicDegrees or cartographicRadians instead of cartesian, and then Cesium will do the conversion for you automatically. You still have to adjust for the geoid when specifying the height, however. Also, don't forget that longitude is first.
Proj4js - a port of the famous proj4 library - might do the job for you.
My question is somewhat related to this similar one, which links to a pretty complex solution - but what I want to understand is the result of this:
Using a Mysql Geometry field to store a small polygon I duly ran
select AREA(myPolygon) where id =1
over it, and got an value like 2.345. So can anyone tell me, just what does that number represent seeing as the stored values were long/lat sets describing the polygon?
FYI, the areas I am working on are relatively small (car parks and the like) and the area does not have to be exact - I will not be concerned about the curvature of the earth.
2.345 of what? Thanks, this is bugging me.
The short answer is that the units for your area calculation are basically meaningless ([deg lat diff] * [deg lon diff]). Even though the curvature of the earth wouldn't come into play for the area calculation (since your areas are "small"), it does come into play for the calculation of distance between the lat/lon polygon coordinates.
Since a degree of longitude is different based on the distance from the equator (http://en.wikipedia.org/wiki/Longitude#Degree_length), there really is no direct conversion of your area into m^2 or km^2. It is dependent on the distance north/south of the equator.
If you always have rectangular polygons, you could just store the opposite corner coordinates and calculate area using something like this: PHP Library: Calculate a bounding box for a given lat/lng location
The most "correct" thing to do would be to store your polygons using X-Y (meters) coordinates (perhaps UTM using the WGS-84 ellipsoid), which can be calculated from lat/lon using various libraries like the following for Java: Java, convert lat/lon to UTM. You could then continue to use the MySQL AREA() function.
I have a database full of rows if coordinate pairs like this:
ux: 6643641
uy: 264274
uz: NULL
I have been tasked to make all these coordinates appear on google maps as points of interest, but nobody could tell me what the hell those coordinates were.
What I need for Google Maps is longitude and lengtitude coordinates. I know the one can be converted to the other, but nothing more.
I realize this might not be the correct place to ask about coordinate systems, but I honestly couldn't think of any other place to state the question.
Thanks for any help!
That's my bad, I now see that there is more data for each row:
CoordSystemNumber: 23
CoordSystemName: EUREF89 UTM Sone 33
I think that format is called UTM. You need to know the Zone and Hemisphere to complete the conversion. Is there other data associated with this?
Tell me if this seems helpful :
x = 882880 meters
y = -4924482 meters
z = 3944130 meters
Geocentric latitude and longitude are not commonly used, but they are defined by
latitude = arctan( z / sqrt( x^2 + y^2 ) )
longitude = arctan( y / x )
Taken from here :
http://www.cv.nrao.edu/~rfisher/Ephemerides/earth_rot.html
see this too :
http://en.wikipedia.org/wiki/Geographic_coordinate_system
This wikipedia article might offer some help.
The coordinates are often chosen such that one of the numbers represent vertical position, and two or three of the numbers represent horizontal position. A common choice of coordinates is latitude, longitude and elevation.
I have a map of the US, and I've got a lot of pixel coordinates on that map which correspond to places of interest. These are dynamically drawn on the map at runtime according to various settings and statistics.
I now need to switch over to a different map of the US which is differently sized, and the top left corner of the map is in a slightly place in the ocean.
So I've manually collected a small set of coordinates for each map that correspond to one another. For example, the point (244,312) on the first map corresponds to the point (598,624) on the second map, and (1323,374) on the first map corresponds to (2793,545) on the second map, etc.
So I'm trying to decide the translation for the X and Y dimensions. So given a set of points for the old and new maps, how do I find the x' = A*x + C and y' = B*x + D equations to automatically translate any point from the old map to the new one?
You have the coordinates of two points on both maps, (x1,y1), (x'1, y'1), (x2, y2) and (x'2, y'2).
A = (x'1 - x'2)/(x1 - x2)
B = (y'1 - y'2)/(y1 - y2)
C = x'1 - A x1
D = y'1 - B y1
P.S. Your equations imply a simple scaling/translation from one map to another. If you're worried about using different projections from globe to plane, the equations will be more complicated.
To get result more robust against inaccuracies more that two points may help.
In this case if you assume only shift and scaling Least squares fit may help: Wikipedia
Basically you minimize sum( (Axi+B-xi')^2 + (Cyi+D-yi')^2 ) by selecting optimal A,B,C,D.
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.