I'm drawing a point which reflects my current location. I've got some shapes on the map as well. However, due to changing GPS accuracy I need to show that I've crossed the line on map even if a LatLng point is still before it but the accuracy is ~20(m).
What I have:
- line's start and end points
- circle's center
- circle's radius (in meters)
I'm able to calculate the distance between line and circle's center point but that gives me the value like: 0.00987506668990474 which gives me really nothing because this value does not reflect the distance in meters and I can't really convert that to meters and compare with accuracy radius.
I'm not even sure if I'm on good target to get that information. Maybe there's some another method to get the information if there's an intersection or not.
thanks for any hints
update:
using the distance * 110km I'm getting much better results.
Happy face = intersection not detected - distance < radius, Sad face = intersection detected - distance > radius.
Yellow zone edge is mine line segment. As you can see it works when the intersection is on the top(/bottom), but not on the left(/right) side.
That's my calculation algorithm calculating distance from line segment to point. Forgive the mess... I'm still struggling with that, so it's not optimized yet:
public static double pointLineSegmentDistance(final List<Double> point, final List<List<Double>> line)
{
List<Double> v = line.get(0);
List<Double> w = line.get(1);
double d = pointPointSquaredDistance(v, w);
double t;
List<Double> calculateThis;
if (d > 0)
{
boolean test = (t = ((point.get(0) - v.get(0)) * (w.get(0) - v.get(0)) + (point.get(1) - v.get(1)) * (w.get(1) - v.get(1))) / d) < 0;
if (test)
{
calculateThis = v;
}
else
{
if (t > 1)
{
calculateThis = w;
}
else
{
calculateThis = new ArrayList<Double>();
calculateThis.add(v.get(0) + t * (w.get(0) - v.get(0)));
calculateThis.add(v.get(1) + t * (w.get(1) - v.get(1)));
}
}
}
else
{
calculateThis = v;
}
return Math.sqrt(pointPointSquaredDistance(point, calculateThis));
}
public static double pointPointSquaredDistance(final List<Double> v, final List<Double> w)
{
double dx = v.get(0) - w.get(0);
double dy = v.get(1) - w.get(1);
return dx * dx + dy * dy;
}
final List point - contains 2 elements - (0) latitude (1) longitude
List> line - contains 2 lists - (0) line segment start point (0/0)lat/(0/1)long (1) line segment end point (1/0)lat/(1/1)long
Finally I've solved my problem. It comes the solution is really very simple. Much simpler than I thought. There's maputil library available here:
https://github.com/googlemaps/android-maps-utils
That library privides a function named "isLocationOnPath":
boolean isLocationOnPath(LatLng point, List<LatLng> polyline, boolean geodesic, double tolerance)
Point - is a center or my point.
Polyline - for my needs is an edge of a polygon,
Geodesic - true (of course)
Tolerance - (in meters) is my accuracy taken from GPS accuracy - basicaly is a circle radius.
Details about the method:
http://googlemaps.github.io/android-maps-utils/javadoc/com/google/maps/android/PolyUtil.html#isLocationOnPath-LatLng-java.util.List-boolean-
I hope that helps.
You have two possibilities:
1) The cheap way:
You calcualte the distances between line start/end point and circle center with the android api (distanceBetween()). The result will be in meters, and correctby means of point to point distance calculation. Take the minimum of both start / end point to center circle distances.
If the line segment is very long related to the circle radius, the normal distance of the line to center, using this kind of caluclation is wrong
2) The better way, but more complex:
You transform the line start and end and the circle center to cartesian x,y space with unit = meter and calculate the distance to the line
segment using cartesian math. (see distance to line segment calculation)
First check option1 because it is just one line of code.
Your result of 0.00987506668990474 this looks like a distance in degrees, instead of meters. multiply it with 111km gives a value of about 1km. (at equator)
From the man page for XFillPolygon:
If shape is Complex, the path may self-intersect. Note that contiguous coincident points in the path are not treated as self-intersection.
If shape is Convex, for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined.
If shape is Nonconvex, the path does not self-intersect, but the shape is not wholly convex. If known by the client, specifying Nonconvex instead of Complex may improve performance. If you specify Nonconvex for a self-intersecting path, the graphics results are undefined.
I am having performance problems with fill XFillPolygon and, as the man page suggests, the first step I want to take is to specify the correct shape of the polygon. I am currently using Complex to be on the safe side.
Is there an efficient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non-convex or complex?
You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.
In contrast, consider the case where the polygon is not self-intersecting, and it consists of a set of points in a list where the consecutive points form the boundary. In this case it is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):
For each consecutive pair of edges of the polygon (each triplet of points), compute the z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross product of these vectors:
given p[k], p[k+1], p[k+2] each with coordinates x, y:
dx1 = x[k+1]-x[k]
dy1 = y[k+1]-y[k]
dx2 = x[k+2]-x[k+1]
dy2 = y[k+2]-y[k+1]
zcrossproduct = dx1*dy2 - dy1*dx2
The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.
If there are N points, make sure you calculate N cross products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).
If the polygon is self-intersecting, then it fails the technical definition of convexity even if its directed angles are all in the same direction, in which case the above approach would not produce the correct result.
This question is now the first item in either Bing or Google when you search for "determine convex polygon." However, none of the answers are good enough.
The (now deleted) answer by #EugeneYokota works by checking whether an unordered set of points can be made into a convex polygon, but that's not what the OP asked for. He asked for a method to check whether a given polygon is convex or not. (A "polygon" in computer science is usually defined [as in the XFillPolygon documentation] as an ordered array of 2D points, with consecutive points joined with a side as well as the last point to the first.) Also, the gift wrapping algorithm in this case would have the time-complexity of O(n^2) for n points - which is much larger than actually needed to solve this problem, while the question asks for an efficient algorithm.
#JasonS's answer, along with the other answers that follow his idea, accepts star polygons such as a pentagram or the one in #zenna's comment, but star polygons are not considered to be convex. As
#plasmacel notes in a comment, this is a good approach to use if you have prior knowledge that the polygon is not self-intersecting, but it can fail if you do not have that knowledge.
#Sekhat's answer is correct but it also has the time-complexity of O(n^2) and thus is inefficient.
#LorenPechtel's added answer after her edit is the best one here but it is vague.
A correct algorithm with optimal complexity
The algorithm I present here has the time-complexity of O(n), correctly tests whether a polygon is convex or not, and passes all the tests I have thrown at it. The idea is to traverse the sides of the polygon, noting the direction of each side and the signed change of direction between consecutive sides. "Signed" here means left-ward is positive and right-ward is negative (or the reverse) and straight-ahead is zero. Those angles are normalized to be between minus-pi (exclusive) and pi (inclusive). Summing all these direction-change angles (a.k.a the deflection angles) together will result in plus-or-minus one turn (i.e. 360 degrees) for a convex polygon, while a star-like polygon (or a self-intersecting loop) will have a different sum ( n * 360 degrees, for n turns overall, for polygons where all the deflection angles are of the same sign). So we must check that the sum of the direction-change angles is plus-or-minus one turn. We also check that the direction-change angles are all positive or all negative and not reverses (pi radians), all points are actual 2D points, and that no consecutive vertices are identical. (That last point is debatable--you may want to allow repeated vertices but I prefer to prohibit them.) The combination of those checks catches all convex and non-convex polygons.
Here is code for Python 3 that implements the algorithm and includes some minor efficiencies. The code looks longer than it really is due to the the comment lines and the bookkeeping involved in avoiding repeated point accesses.
TWO_PI = 2 * pi
def is_convex_polygon(polygon):
"""Return True if the polynomial defined by the sequence of 2D
points is 'strictly convex': points are valid, side lengths non-
zero, interior angles are strictly between zero and a straight
angle, and the polygon does not intersect itself.
NOTES: 1. Algorithm: the signed changes of the direction angles
from one side to the next side must be all positive or
all negative, and their sum must equal plus-or-minus
one full turn (2 pi radians). Also check for too few,
invalid, or repeated points.
2. No check is explicitly done for zero internal angles
(180 degree direction-change angle) as this is covered
in other ways, including the `n < 3` check.
"""
try: # needed for any bad points or direction changes
# Check for too few points
if len(polygon) < 3:
return False
# Get starting information
old_x, old_y = polygon[-2]
new_x, new_y = polygon[-1]
new_direction = atan2(new_y - old_y, new_x - old_x)
angle_sum = 0.0
# Check each point (the side ending there, its angle) and accum. angles
for ndx, newpoint in enumerate(polygon):
# Update point coordinates and side directions, check side length
old_x, old_y, old_direction = new_x, new_y, new_direction
new_x, new_y = newpoint
new_direction = atan2(new_y - old_y, new_x - old_x)
if old_x == new_x and old_y == new_y:
return False # repeated consecutive points
# Calculate & check the normalized direction-change angle
angle = new_direction - old_direction
if angle <= -pi:
angle += TWO_PI # make it in half-open interval (-Pi, Pi]
elif angle > pi:
angle -= TWO_PI
if ndx == 0: # if first time through loop, initialize orientation
if angle == 0.0:
return False
orientation = 1.0 if angle > 0.0 else -1.0
else: # if other time through loop, check orientation is stable
if orientation * angle <= 0.0: # not both pos. or both neg.
return False
# Accumulate the direction-change angle
angle_sum += angle
# Check that the total number of full turns is plus-or-minus 1
return abs(round(angle_sum / TWO_PI)) == 1
except (ArithmeticError, TypeError, ValueError):
return False # any exception means not a proper convex polygon
The following Java function/method is an implementation of the algorithm described in this answer.
public boolean isConvex()
{
if (_vertices.size() < 4)
return true;
boolean sign = false;
int n = _vertices.size();
for(int i = 0; i < n; i++)
{
double dx1 = _vertices.get((i + 2) % n).X - _vertices.get((i + 1) % n).X;
double dy1 = _vertices.get((i + 2) % n).Y - _vertices.get((i + 1) % n).Y;
double dx2 = _vertices.get(i).X - _vertices.get((i + 1) % n).X;
double dy2 = _vertices.get(i).Y - _vertices.get((i + 1) % n).Y;
double zcrossproduct = dx1 * dy2 - dy1 * dx2;
if (i == 0)
sign = zcrossproduct > 0;
else if (sign != (zcrossproduct > 0))
return false;
}
return true;
}
The algorithm is guaranteed to work as long as the vertices are ordered (either clockwise or counter-clockwise), and you don't have self-intersecting edges (i.e. it only works for simple polygons).
Here's a test to check if a polygon is convex.
Consider each set of three points along the polygon--a vertex, the vertex before, the vertex after. If every angle is 180 degrees or less you have a convex polygon. When you figure out each angle, also keep a running total of (180 - angle). For a convex polygon, this will total 360.
This test runs in O(n) time.
Note, also, that in most cases this calculation is something you can do once and save — most of the time you have a set of polygons to work with that don't go changing all the time.
To test if a polygon is convex, every point of the polygon should be level with or behind each line.
Here's an example picture:
The answer by #RoryDaulton
seems the best to me, but what if one of the angles is exactly 0?
Some may want such an edge case to return True, in which case, change "<=" to "<" in the line :
if orientation * angle < 0.0: # not both pos. or both neg.
Here are my test cases which highlight the issue :
# A square
assert is_convex_polygon( ((0,0), (1,0), (1,1), (0,1)) )
# This LOOKS like a square, but it has an extra point on one of the edges.
assert is_convex_polygon( ((0,0), (0.5,0), (1,0), (1,1), (0,1)) )
The 2nd assert fails in the original answer. Should it?
For my use case, I would prefer it didn't.
This method would work on simple polygons (no self intersecting edges) assuming that the vertices are ordered (either clockwise or counter)
For an array of vertices:
vertices = [(0,0),(1,0),(1,1),(0,1)]
The following python implementation checks whether the z component of all the cross products have the same sign
def zCrossProduct(a,b,c):
return (a[0]-b[0])*(b[1]-c[1])-(a[1]-b[1])*(b[0]-c[0])
def isConvex(vertices):
if len(vertices)<4:
return True
signs= [zCrossProduct(a,b,c)>0 for a,b,c in zip(vertices[2:],vertices[1:],vertices)]
return all(signs) or not any(signs)
I implemented both algorithms: the one posted by #UriGoren (with a small improvement - only integer math) and the one from #RoryDaulton, in Java. I had some problems because my polygon is closed, so both algorithms were considering the second as concave, when it was convex. So i changed it to prevent such situation. My methods also uses a base index (which can be or not 0).
These are my test vertices:
// concave
int []x = {0,100,200,200,100,0,0};
int []y = {50,0,50,200,50,200,50};
// convex
int []x = {0,100,200,100,0,0};
int []y = {50,0,50,200,200,50};
And now the algorithms:
private boolean isConvex1(int[] x, int[] y, int base, int n) // Rory Daulton
{
final double TWO_PI = 2 * Math.PI;
// points is 'strictly convex': points are valid, side lengths non-zero, interior angles are strictly between zero and a straight
// angle, and the polygon does not intersect itself.
// NOTES: 1. Algorithm: the signed changes of the direction angles from one side to the next side must be all positive or
// all negative, and their sum must equal plus-or-minus one full turn (2 pi radians). Also check for too few,
// invalid, or repeated points.
// 2. No check is explicitly done for zero internal angles(180 degree direction-change angle) as this is covered
// in other ways, including the `n < 3` check.
// needed for any bad points or direction changes
// Check for too few points
if (n <= 3) return true;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
// Get starting information
int old_x = x[n-2], old_y = y[n-2];
int new_x = x[n-1], new_y = y[n-1];
double new_direction = Math.atan2(new_y - old_y, new_x - old_x), old_direction;
double angle_sum = 0.0, orientation=0;
// Check each point (the side ending there, its angle) and accum. angles for ndx, newpoint in enumerate(polygon):
for (int i = 0; i < n; i++)
{
// Update point coordinates and side directions, check side length
old_x = new_x; old_y = new_y; old_direction = new_direction;
int p = base++;
new_x = x[p]; new_y = y[p];
new_direction = Math.atan2(new_y - old_y, new_x - old_x);
if (old_x == new_x && old_y == new_y)
return false; // repeated consecutive points
// Calculate & check the normalized direction-change angle
double angle = new_direction - old_direction;
if (angle <= -Math.PI)
angle += TWO_PI; // make it in half-open interval (-Pi, Pi]
else if (angle > Math.PI)
angle -= TWO_PI;
if (i == 0) // if first time through loop, initialize orientation
{
if (angle == 0.0) return false;
orientation = angle > 0 ? 1 : -1;
}
else // if other time through loop, check orientation is stable
if (orientation * angle <= 0) // not both pos. or both neg.
return false;
// Accumulate the direction-change angle
angle_sum += angle;
// Check that the total number of full turns is plus-or-minus 1
}
return Math.abs(Math.round(angle_sum / TWO_PI)) == 1;
}
And now from Uri Goren
private boolean isConvex2(int[] x, int[] y, int base, int n)
{
if (n < 4)
return true;
boolean sign = false;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
for(int p=0; p < n; p++)
{
int i = base++;
int i1 = i+1; if (i1 >= n) i1 = base + i1-n;
int i2 = i+2; if (i2 >= n) i2 = base + i2-n;
int dx1 = x[i1] - x[i];
int dy1 = y[i1] - y[i];
int dx2 = x[i2] - x[i1];
int dy2 = y[i2] - y[i1];
int crossproduct = dx1*dy2 - dy1*dx2;
if (i == base)
sign = crossproduct > 0;
else
if (sign != (crossproduct > 0))
return false;
}
return true;
}
For a non complex (intersecting) polygon to be convex, vector frames obtained from any two connected linearly independent lines a,b must be point-convex otherwise the polygon is concave.
For example the lines a,b are convex to the point p and concave to it below for each case i.e. above: p exists inside a,b and below: p exists outside a,b
Similarly for each polygon below, if each line pair making up a sharp edge is point-convex to the centroid c then the polygon is convex otherwise it’s concave.
blunt edges (wronged green) are to be ignored
N.B
This approach would require you compute the centroid of your polygon beforehand since it doesn’t employ angles but vector algebra/transformations
Adapted Uri's code into matlab. Hope this may help.
Be aware that Uri's algorithm only works for simple polygons! So, be sure to test if the polygon is simple first!
% M [ x1 x2 x3 ...
% y1 y2 y3 ...]
% test if a polygon is convex
function ret = isConvex(M)
N = size(M,2);
if (N<4)
ret = 1;
return;
end
x0 = M(1, 1:end);
x1 = [x0(2:end), x0(1)];
x2 = [x0(3:end), x0(1:2)];
y0 = M(2, 1:end);
y1 = [y0(2:end), y0(1)];
y2 = [y0(3:end), y0(1:2)];
dx1 = x2 - x1;
dy1 = y2 - y1;
dx2 = x0 - x1;
dy2 = y0 - y1;
zcrossproduct = dx1 .* dy2 - dy1 .* dx2;
% equality allows two consecutive edges to be parallel
t1 = sum(zcrossproduct >= 0);
t2 = sum(zcrossproduct <= 0);
ret = t1 == N || t2 == N;
end
I have a map with overlays which i want to cache -
on each place the user visited on the map (which is a rectangle area) - i check if i have a cache of the overlays that reside in this rectangle .
In order to improve caching (so if the user was previously on the same rectangle,except that now he is a few meters away from the previous rectangle) - i want to "round" the coordinates.
This way, each time the user is in a rectange - i check if this rectangle is similar to previously cached rectangles and if so i bring the cached result .
Also, if the user is zoomed out and his rectangle is contained within a bigger (previously cached) rectangle - then I also can use the cached rectangle.
Any suggestions ?
If you're just looking at how to group the coordinates, decide on the maximum difference between coordinates in x and y or latitude and longtitude you want. Then there are two ways you can go about grouping them. The first is easier, but it will be slow if you have many points.
Say we have a data structure called cachedPoints, a max distance between related points called maxdistance and a new point we're trying to check to see if it's close to the other called point.
for each cachedPoint in cachedPoints
{
if (point.x - cachedPoint.x < maxdistance)
{
if (point.y - cachedPoint.y < maxdistance)
{
cachedPoint.incrementvisits();
}
}
}
the other way is to use a data structure sorted by x or latitude, and then search to see if there is a cachedpoint with an x or latitude within maxdistance of point, then check the y or longtitude. It'd be a bit faster, but it would take some kind of a hash to implement and adds a bunch of complexity you may not need.
Hopefully that's what you're asking.
If you set up a data structure like:
var a = { 'scales' : [50, 100, 200, 400, 1000],
'cachedRects': [{'location': 'rect-large-1234-5678.png', x: 1234, y: 5678, scale: 3}
{'location': 'rect-small-1240-5685.png', x: 1240, y: 5685, scale: 1} ]
}
you can use the modulo function to do this:
var currentx = GetCurrentX();
var currenty = GetCurrentY();
var currentScale = GetCurrentScale();
var rectFound = false;
foreach(rect in a.cachedRects) {
if (rect.scale === currentScale
&& currentx % a.scales[currentScale] === rect.x
&& currenty % a.scales[currentScale] === rect.y) {
rectFound = true;
useOverlay(rect);
break;
}
}
if(!rectFound) {
//could loop again for a larger rectangle of a lower scale.
}
the above may or may not turn out to be valid JS - I've not tried to run it. I hope you get the gist, anyway.
Hey You can Add Marker in Google map v2 in android.
Here i give code for add marker
MarkerOptions mOpt = new MarkerOptions();
mOpt.position(new LatLng(userHstry.getMyLatlng().latitude, userHstry.getMyLatlng().longitude)); // map.clear();
mOpt.title("Address : " + userHstry.getAddress()).snippet("Date : " + userHstry.getDate() + " , Time : " + userHstry.getTime());
map.addMarker(mOpt);
In my web/MySQL application I have something like this to get distance between two points:
6371 * acos(cos(radians(-19.83996)) * cos(radians(lat)) * cos(radians(-43.94910) - radians(lng)) + sin(radians(-19.83996)) * sin(radians(lat)))
But I tested in SQLite and these mathematical functions (acos, cos, radians, sin) do not exist. Is there something equivalent for me to calculate the distance directly in the database?
However, I have an iPhone application that uses this method to calculate. Works perfectly, but now I need to perform this same search in the database in an Android application.
Thanks in advance.
UPDATE
I have 9000 points to calculate distance and obtain the 5 nearby locations of a given point.
Here is what I would do:
Take your given point. Measure a (that is an arbitrary value to be refined) ~2km wide square around it and take the values for the east/west/north/south bounds.
Make a query for elements inside this square. This one is easy, you just have to
select * from points where lat between ? and ? and lon between ? and ?
Count your result. Not enough result (less than 5, obviously, but I would say twice that to be sure), retry with a larger radius. Too much (say, more than 100), try again with a smaller radius.
Once you have enough, load them, make sure all 5 elements you need are not only in the Xkm wide square, but also in the Xkm radius circle (to avoid having a potential closer element not detected by the previous approximation).
Another approach
Valid only if your given point is relatively close to those you are searching.
Measure a local approximation of a flat earth. Close to your point, you can consider a linear relation between lat, lon, and distance. That allows you to make a request sorted by a simple calculus. (multiplication and addition). Again, select a little more points in order to make the proper calculation after the SQLite request.
Insert cos_lat_rad,sin_lat_rad,cos_lon_rad,sin_lon_rad in to your table
contentValues.put("cos_lat_rad", Math.cos(deg2rad(latitude)));
contentValues.put("sin_lat_rad", Math.sin(deg2rad(latitude)));
contentValues.put("cos_lon_rad", Math.cos(deg2rad(longitude)));
contentValues.put("sin_lon_rad", Math.sin(deg2rad(longitude)));
degree to radian
public static double deg2rad(double deg) {
return (deg * Math.PI / 180.0);
}
query , distance in km
Cursor c=database.dis(String.valueOf(Math.cos((double) distance / (double) 6380)), Math.cos(deg2rad(latitude)), Math.sin(deg2rad(latitude)), Math.cos(deg2rad(longitude)), Math.sin(deg2rad(longitude)));
query
public Cursor dis(String dis, double cos_lat_rad, double sin_lat_rad, double cos_lon_rad, double sin_lon_rad) {
Cursor cursor = sqLiteDatabase.rawQuery("SELECT * ,(" + sin_lat_rad + "*\"sin_lat_rad\"+" + cos_lat_rad + "*\"cos_lat_rad\"*(" + sin_lon_rad + "*\"sin_lon_rad\"+" + cos_lon_rad + "*\"cos_lon_rad\")) AS \"distance_acos\" FROM parish WHERE ("+sin_lat_rad+" * \"sin_lat_rad\" +"+ cos_lat_rad +"* \"cos_lat_rad\" * (+"+sin_lon_rad +"* \"sin_lon_rad\" + "+cos_lon_rad +"* \"cos_lon_rad\")) >"+dis+ " ORDER BY \"distance_acos\" DESC ", null);
return cursor;
}
convert distance_acos to km
if(c.moveToFirst())
do {
double distance_acos= c.getDouble(c.getColumnIndex("distance_acos"));
String Distance=String.valueOf(Math.acos(distance_acos) * 6380);
}while (c.moveToNext());
Reply of Angel does not solve the problem for SQLite since it still includes ACOS function which does not exist in SQLite
What I have concluded after a thorough research is that one should pick a rough estimation to use in SQLite with the formula below:
Distance estimation for km:
SQRT(SQUARE((TO_LAT-FROM_LAT)*110)+
SQUARE((TO_LONG-FROM_LONG)*COS(TO_LAT)*111))
For windows:
Install minGw full options. modify environment variable: system variable path, including
c:\mingw\bin
test functionality command:
g++ --version
copy files: extension-functions.c, sqlite3.h, sqlite3ext.h in sqlite3 program directory. Go to sqlite3 directory and compile:
gcc -shared -I "path" -o libsqlitefunctions.so extension-functions.c
(path = path of sqlite3ext.h; i.e. C:\sqlite3)
If the program is built so that loading extensions is permitted, the following will work:
sqlite> SELECT load_extension('./libsqlitefunctions.so');
sqlite> select cos(radians(45));
0.707106781186548
SQLite Distance implementation:
From: https://www.movable-type.co.uk/scripts/latlong.html https://en.wikipedia.org/wiki/Haversine_formula
Distance
This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points (ignoring any hills they fly over, of course!).
Haversine
Formula:
a = sin²(Δφ/2) + cos φ1 * cos φ2 * sin²(Δλ/2)
c = 2 * atan2( √a, √(1−a) )
c = 2 *
d = R * c
where φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!
JavaScript:
const R = 6378136.6 ; // meters equatorial radius
const φ1 = lat1 * Math.PI/180; // φ, λ in radians
const φ2 = lat2 * Math.PI/180;
const Δφ = (lat2-lat1) * Math.PI/180;
const Δλ = (lon2-lon1) * Math.PI/180;
const a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
const c = 2 * Math.asen(MIN (1, Math.sqrt(a))); //sqlite implementation
const d = R * c; // in metres
Distance = 2 * R * ASIN( MIN (1, SQRT( SIN( (RADIANS(lat1)-RADIANS(lat2))/2 )^2 + COS( RADIANS(lat1) )*COS( RADIANS(lat2) )*SIN( (RADIANS(long1)-RADIANS(long2))/2 )^2 )))
Physical Properties of Earth https://en.wikipedia.org/wiki/Earth_ellipsoid :Ecuatorial radius: 6378.1366 Kms. Average radius: 6367 Kms
Constant = 2 * 6378136.6 = 12756273.2
SQLite query command with coordinates taken from table PAR:
ROUND (
12756273.2 * ASIN(
MIN (1 ,
SQRT(
POWER( SIN(RADIANS(PAR.Lat1 - PAR.Lat2)/2) , 2) +
COS(RADIANS(PAR.Lat1)) * COS(RADIANS(PAR.Lat2)) * POWER ( SIN(RADIANS(PAR.Long1 - PAR.Long2)/2) , 2)
)
)
)
, 0) AS Distance
In android we have a Location class, we need initialize the location class and in it we have a method called distanceBetween() which gives the distance between two geopoints.
Refer this LINK