Latitude and longitude - gps

Actually I have the decimal values of latitude and longitude.
In a 2d referential, are those values the x an y coordenates?
example :
in a position P
latitude = 41.15 longitude = -8.64
So, in a 2d dimension P is defined by (41.15,-8.64) ?
Thanks

No, its exactly opposite:
The coordinate (lat, lon) corresponds to the pair (y,x)
So when passinge lat, lon to mathematical routines, like point in polygon calculations
pass in order (lon, lat).
longitude raises parallell to the aequator, which corresponds to our x achsis direction we normally use in cartesian (x,y) systems.
Unfortunatley for historical reasons, the latitude is often written before longitude. (The cause might be that the latitude was easier to determine than the longitude.)
This all leads for us SW developpers to the bad situation, that sometimes functions use (lat,lon) order, sometimes when working with transformations from (lon,lat) to (x,y) or mathematical routines, the order lon, lat must be used. Be careful, every person I know someday has accidentally exchanged that.
In your example:
P is related to (-8.64), 41.15).
But lat,lon are spherical coordinates, for most application you must convert them to cartesian (x,y) .
But this is another question.

Be careful with units. In systems like Google Maps, those numbers are usually in degrees. Usually.
They might also be in radians, though, so at least make sure of what unit the API is using.

Related

How to calculate Altitude using GPS latitude and longitude

How to calculate Altitude from GPS Latitude and Longitude values.What is the exact mathematical equation to solve this problem.
It is possible for a given lat,lon to determine the height of the ground (above sea level, or above Referenz Elipsoid).
But since the earth surface, mountains, etc, do not follow a mathematic formula,
there are Laser scans, performed by Satelites, that measured such a height for e.g every 30 meters.
So there exist files where you can lookup such a height.
This is called a Digital Elevation Modell, or short (DEM)
Read more here: https://en.wikipedia.org/wiki/Digital_elevation_model
Such files are huge, very few application use that approach.
Many just take the altidude value as delivered by the GPS receiver.
You can find some charts with altitude data, like Maptech's. Each pixel has a corresponding lat, long, alt/depth information.
As #AlexWien said these files are huge and most of them must be bought.
If you are interest of using these files I can help you with a C++ code to read them.
You can calculate the geocentric radius, i.e., the radius of the reference Ellipsoid which is used as basis for the GPS altitude. It can be calculated from the the GPS latitude with this formula:
Read more about this at Wikipedia.

Formula/algorithm to offset GPS coordinations

I have GPS coordinates provided as degrees latitude, longitude and would like to offset them by a distance and an angle.E.g.: What are the new coordinates if I offset 45.12345, 7.34567 by 22km along bearing 104 degrees ?Thanks
For most applications one of these two formulas are sufficient:
"Lat/lon given radial and distance"
The second one is slower, but makes less problems in special situations (see docu on that page).
Read the introduction on that page, and make sure that lat/lon are converted to radians before and back to degrees after having the result.
Make sure that your system uses atan2(y,x) (which is usually the case) and not atan2(x,y) which is the case in Excell.
The link in the previous answer no longer works, here is the link using the way back machine:
https://web.archive.org/web/20161209044600/http://williams.best.vwh.net/avform.htm
The formula is:
A point {lat,lon} is a distance d out on the tc radial from point 1 if:
lat=asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
IF (cos(lat)=0)
lon=lon1 // endpoint a pole
ELSE
lon=mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
ENDIF
This algorithm is limited to distances such that dlon <pi/2, i.e those that extend around less than one quarter of the circumference of the earth in longitude. A completely general, but more complicated algorithm is necessary if greater distances are allowed:
lat =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat))
lon=mod( lon1-dlon +pi,2*pi )-pi

GEOS C API - calculating areas with WGS84 coords (SRID=4326)

I create a polygon where each x/y point is WGS84 format
lat/long values.
The polygons are good approximations to circles and sectors of
radius R (each circumference/arc point is a projected lat/long
value of distance R from a centre/apex coordinate - which I have
verified is correct by computing the Haversine distance between
the edge and reference points and getting a value of R back) .
I use GEOSSetSRID(4326) to indicate the coords are WGS84 format.
GEOSGetSRID() confirms the SRID is set.
Use of GEOSArea then gives a value not even remotely close to
the expected value.
I do not see what else I can programmatically do.
If I set the points in cartesian format, and then set the SRID to
4326, will GEOS implicitly convert the polygon points to WGS84 ??
Is the basic GEOS C API incapable of doing the above ??
Dos SRID have no meaning to the API at all ??
Any info/pointers to correct usage/solutions would be much appreciated.
TIA.
The distance that is given is something like degrees between the two points. In actuality, the GEOS API (at least the C++ interface) is units agnostic; the units it gives the distance in is based on whatever you passed in.
In general, multiplying the result you get by 111000 gives you a fairly accurate measurement in meters. For area, you have to do 111000^2.

How to group latitude/longitude points that are 'close' to each other?

I have a database of user submitted latitude/longitude points and am trying to group 'close' points together. 'Close' is relative, but for now it seems to ~500 feet.
At first it seemed I could just group by rows that have the same latitude/longitude for the first 3 decimal places (roughly a 300x300 box, understanding that it changes as you move away from the equator).
However, that method seems to be quite lacking. 'Closeness' can't be significantly different than the distance each decimal place represents. It doesn't take into account that two locations may have different digits in the 3rd (or any) decimal place, but still be within the distance that place represents (33.1239 and 33.1240).
I've also mulled over the situation where Point A, and Point C are both 'close' to Point B (but not each other) - should they be grouped together? If so, what happens when Point D is 'close' to point C (and no other points) - should it be grouped as well. Certainly I have to determine the desired behavior, but how would either be implemented?
Can anyone point me in the right direction as to how this can be done and what different methods/approaches can be used?
I feel a bit like I'm missing something obvious.
Currently the data is an a MySQL database, use by a PHP application; however, I'm open to other storage methods if they're a key part in accomplishing this. here.
There are a number of ways of determining the distance between two points, but for plotting points on a 2-D graph you probably want the Euclidean distance. If (x1, y1) represents your first point and (x2, y2) represents your second, the distance is
d = sqrt( (x2-x1)^2 + (y2-y1)^2 )
Regarding grouping, you may want to use some sort of 2-D mean to determine how "close" things are to each other. For example, if you have three points, (x1, y1), (x2, y2), (x3, y3), you can find the center of these three points by simple averaging:
x(mean) = (x1+x2+x3)/3
y(mean) = (y1+y2+y3)/3
You can then see how close each is to the center to determine whether it should be part of the "cluster".
There are a number of ways one can define clusters, all of which use some variant of a clustering algorithm. I'm in a rush now and don't have time to summarize, but check out the link and the algorithms, and hopefully other people will be able to provide more detail. Good luck!
Use something similar to the method you outlined in your question to get an approximate set of results, then whittle that approximate set down by doing proper calculations. If you pick your grid size (i.e. how much you round off your co-ordinates) correctly, you can at least hope to reduce the amount of work to be done to an acceptable level, although you have to manage what that grid size is.
For example, the earthdistance extension to PostgreSQL works by converting lat/long pairs to (x,y,z) cartesian co-ordinates, modelling the Earth as a uniform sphere. PostgreSQL has a sophisticated indexing system that allows these co-ordinates, or boxes around them, to be indexed into R-trees, but you can whack something together that is still useful without that.
If you take your (x,y,z) triple and round off- i.e. multiply by some factor and truncate to integer- you then have three integers that you can concatenate to produce a "box name", which identifies a box in your "grid" that the point is in.
If you want to search for all points within X km of some target point, you generate all the "box names" around that point (once you've converted your target point to an (x,y,z) triple as well, that's easy) and eliminate all the boxes that don't intersect the Earth's surface (tricker, but use of the x^2+y^2+z^2=R^2 formula at each corner will tell you) you end up with a list of boxes target points can be in- so just search for all points matching one of those boxes, which will also return you some extra points. So as a final stage you need to calculate the actual distance to your target point and eliminate some (again, this can be sped up by working in Cartesian co-ordinates and converting your target great-circle distance radius to secant distance).
The fiddling around comes down to making sure you don't have to search too many boxes, but at the same time don't bring in too many extra points. I've found it useful to index each point on several different grids (e.g. resolutions of 1Km, 5Km, 25Km, 125Km etc). Ideally you want to be searching just one box, remember it expands to at least 27 as soon as your target radius exceeds your grid size.
I've used this technique to construct a spatial index using Lucene rather than doing calculations in a SQL databases. It does work, although there is some fiddling to set it up, and the indices take a while to generate and are quite big. Using an R-tree to hold all the co-ordinates is a much nicer approach, but would take more custom coding- this technique basically just requires a fast hash-table lookup (so would probably work well with all the NoSQL databases that are the rage these days, and should be usable in a SQL database too).
Maybe overkill, but it seems to me a clustering problem: distance measure will determine how the similarity of two elements is calculated. If you need a less naive solution try Data Mining: Practical Machine Learning Tools and Techniques, and use Weka or Orange
If I were tackling it, I'd start with a grid. Put each point into a square on the grid. Look for grids that are densely populated. If the adjacent grids aren't populated, then you have a decent group.
If you have adjacent densely populated grids, you can always drop a circle at the center of each grid and optimize for circle area vs (number of points in the circle * some tunable weight). Not perfect, but easy. Better groupings are much more complicated optimization problems.
Facing a similar issue, I've just floor the Longitude and Latitude until I got the required 'closeness' in meters. In my case, floor to 4 digits got me locations grouped when they are approx. 13 meters apart.
If the Long or Lat are negatives - replace floor with ceil
First FLOOR (or CEIL) to required precision and then GROUP on the rounded long and lat.
The code to measure distance between two geo locations was borrowed from Getting distance between two points based on latitude/longitude
from math import sin, cos, sqrt, atan2, radians
R = 6373.0
lat1 = radians(48.71953)
lon1 = radians(-73.72882)
lat2 = radians(48.719)
lon2 = radians(-73.728)
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
distance = (R * c)*1000
print("Distance in meters:", round(distance))
Distance in meters: 84
As expected, the distance is larger in the south, and smaller in the north - for the same angle.
For the same coordinates, but on the equator, the distance is 109 meters (modify the latitudes to 0.71953 and 0.719).
I modified the number of digits in the following and always kept one-click on Long and one on Lats, and measured the resulting distances:
lat1 = radians(48.71953)
lon1 = radians(-73.72882)
lat2 = radians(48.71954)
lon2 = radians(-73.72883)
Distance in meters 1
lat1 = radians(48.7195)
lon1 = radians(-73.7288)
lat2 = radians(48.7196)
lon2 = radians(-73.7289)
Distance in meters 13
lat1 = radians(48.719)
lon1 = radians(-73.728)
lat2 = radians(48.720)
lon2 = radians(-73.729)
Distance in meters 133
lat1 = radians(48.71)
lon1 = radians(-73.72)
lat2 = radians(48.72)
lon2 = radians(-73.73)
Distance in meters 1333
Summary: Floor / Ceil the longitude and latitude to 4 digits, will help you group on locations that are approximately 13 meters apart.
This number changes depending on the above equation: larger near the equator and smaller in the north.
If you are considering latitude and longitude there are several factors to be considered in real time data: obstructions, such as rivers and lakes, and facilities, such as bridges and tunnels. You cannot group them simply; if you use the simple algorithm as k means you will not be able to group them. I think you should go for the spatial clustering methods as partitioning CLARANS method.

How to calculate area which was compose with mulit- Coordinates?

as topic, the Coordinates value (Latitude and Longitude) is known , these Coordinates will compose as polygonal area , my question is how to calculate the area of the polygonal that is base the geography ?
thanks for your help .
First you would need to know whether the curvature of the surface would be significant. If it is a relatively small then you can get a good approximation by projecting the coordinates onto a plane.
Determine units of measure per degree of latitude (eg. meters per degree)
Determine units of meature per degree of longitude at a given latitude (the conversion factor varies as you go North or South)
Convert latitude and longitude pairs to (x,y) pairs in the plane
Use an algorithm to compute area of a polygon. See StackOverflow 451425 or Paul Bourke
If you are calculating a large area then spherical techniques must be used.
If I understand your question correctly - triangulation should help you. Basically you break the polygonal to triangles in such a way that they don't overlap and sum their areas.