I am working coordinates system as shown below:
x and y are in meters. I am interested in only positive x,y. I want to convert (x,y) to (lat,lon) and vise versa.
I thought it is simple and I dont see a problem with my solutions below. But I am not getting very correct results.
My Solution:
As we see in the image below, I considered the Latitude and Longitude as angles of 2 circles:
1. (x,y) to (lat,lon)
I applied the Arc Length formula (here), shown below, on both; x and y
Hence, my functions are:
private static double calculateLat(float x) {
int earthRadius = 6371000;
return REF_LOC.getLatitude() + x*360/(2*PI*earthRadius);
}
private static double calculateLong(float y) {
int earthRadius = 6371000;
return REF_LOC.getLongitude() + y*360/(2*PI*earthRadius);
}
REF_LOC is the reference Geo Location for which (x,y) are (0,0). It can be any point on earth.
2. (lat,lon) to (x,y)
For this I am simply using this:
int calculateX(double longitude){
Location.distanceBetween(REF_LOC.getLatitude(), REF_LOC.getLongitude(),
REF_LOC.getLatitude(), lonDeg, results);
return results[0];
}
int calculateY(double latitude){
Location.distanceBetween(REF_LOC.getLatitude(), REF_LOC.getLongitude(),
latDeg, REF_LOC.getLongitude(), results);
return results[0];
}
But I am getting inconsistent results. First I use solution 1 and convert some value (x,y) to (lat,long). But when I use the same (lat,long) back to (x,y) using solution 2, I get about 2 meters difference in x and 10 meters in y. Can anyone help me identify the problem please?
Alluding to my comment on spherical vs elliptical calculations, another way to look at difference in distance calculations for spherical vs elliptical distance is to use the 2 available utilities:
// For REF_LOC = (70, 20)
// Compute a lat/lng due east of a REF_LOC and compute distance using both
// available methods.
LatLng P = SphericalUtil.computeOffsetOrigin(REF_LOC, 20000, 90.0);
// And then compute a distance
d = SphericalUtil.computeDistanceBetween(REF_LOC, P);
Location.distanceBetween(REF_LOC.latitude, REF_LOC.longitude, P.latitude, P.longitude, results);
// d = 20000.000000000036
// results[0] = 20081.818
// and for a REF_LOC = (0, 20)
// d = 20000.000000000127
// results[0] = 20022.377
What's also interesting is the SphericalUtil.computeOffsetOrigin() produces error
in latitude increasing from equator to pole making it non-symmetrical. Yet the resulting distances are essentially exact.
I'd recommend using SphericalUtil.computeOffsetOrigin to compute the X/Y breaking
it into the latitude and longitude offsets as you are doing.
Finally demonstrating the SphericalUtil solution:
// Start with some arbitrary x/y relative to REF_LOC
double Rx = 125.0;
double Ry = 73.0;
// Compute a lat/lon from REF_LOC to the point
LatLng Rll = new LatLng(SphericalUtil.computeOffsetOrigin(REF_LOC, Ry, 180).latitude,
SphericalUtil.computeOffsetOrigin(REF_LOC, Rx, 270).longitude);
// And recompute the x/y components of the lat/lon
double Rxx = SphericalUtil.computeDistanceBetween(REF_LOC, new LatLng(REF_LOC.latitude, Rll.longitude));
double Ryy = SphericalUtil.computeDistanceBetween(REF_LOC, new LatLng(Rll.latitude, REF_LOC.longitude));
Resulting in:
Rx/Ry = (125.0, 73.0)
Rxx/Ryy = (125.00000004545973, 73.00000000137051)
Acceptable error I assume.
So the parting question is - what does the x/y really represent?
Reference the source for both utilities for more information:
Location
SphericalUtil
I'm adding a PNG file as an own floorplan on top of Google Maps with the Google Maps Android API with the following code:
GroundOverlayOptions groundOverlayOptions = new GroundOverlayOptions();
BitmapDescriptor bitmapDescriptor = BitmapDescriptorFactory.fromAsset("building-d.png");
groundOverlayOptions.image(bitmapDescriptor);
groundOverlayOptions.anchor(0, 1);
LatLng buildingSW = new LatLng(47.014815, 8.305098);
LatLng buildingNE = new LatLng(47.015148, 8.305440);
LatLng buildingNW = new LatLng(47.015168, 8.305144);
LatLng buildingSE = new LatLng(47.014792, 8.305385);
Location swLoc = locationFromLatLng(buildingSW);
Location seLoc = locationFromLatLng(buildingSE);
Location nwLoc = locationFromLatLng(buildingNW);
Location neLoc = locationFromLatLng(buildingNE);
float angle = swLoc.bearingTo(nwLoc);
groundOverlayOptions.bearing(angle);
float width = swLoc.distanceTo(seLoc);
groundOverlayOptions.position(buildingSW, width);
mMap.addGroundOverlay(groundOverlayOptions);
Now I know that in the PNG there is a room at pixel coordinates 422/301, 708/301, 422/10 and 708/10 (those are the corners). I'd like to draw a polygon over the GroundOverlay covering that room. How should I do that? Do I need to convert my pixel-coordinates to LatLng and if so, how?
And by the way: Do I really have to use PNGs for GroundOverlays and is there no other supported vector-format like eps, pdf, ...?
Having seen your comment to the other answer, let me complete with some code:
Having set the "origin" in latlng 47.014816, 8.305098, you have to convert those coordinates to mercator and you can do something similar to the below:
public boolean initializeByTwoCouplesOfCooordsAndScale(double[] coordAreal, double[] coordBreal, double[] coordAvirtual, double[] coordBvirtual, double scalingFactor) {
if (coordAreal[0] == coordBreal[0] && coordAvirtual[1] == coordBvirtual[1] && coordAreal[1] == coordBreal[1] && coordAvirtual[0] == coordBvirtual[0]) {
System.err.println("Coordinates must not be the same!");
return false;
}
// aPoint is considered the "origin" point (0,0)
aPoint = coordAreal;
bPoint = coordAvirtual;
// now calculate the angle of the Real world coordinate for the points
double deltaRy = coordBreal[1] - coordAreal[1];
double deltaRx = coordBreal[0] - coordAreal[0];
double aR = Math.atan2(deltaRy, deltaRx);
// Now calculate the angle of the virtual world coordinates
double deltaVy = coordBvirtual[1] - coordAvirtual[1];
double deltaVx = coordBvirtual[0] - coordAvirtual[0];
double aV = Math.atan2(deltaVy, deltaVx);
// Set the transformation angle as the difference between the real and the virtual angles.
mPhi= (aR - aV);
// Set the scaling factor as the provided one
mScale = (scalingFactor);//scaling factor is in function below
// Calculate the scaling factor error correction using the distances of the two systems.
return true;
}
public static double getScalingFactor(double latitude) {
return 1 / (Math.cos(Math.toRadians(latitude)));
}
So you can call the method:
initializeByTwoCouplesOfCooordsAndScale(new double[]{MERCATOR_LNG,MERCATOR_LAT},//real coordinates for point A REMEMBER: LNG,LAT = x,y!
new double[]{0d,0d}, //Virual coordinates for point A
new double[]{MERCATOR_POINT_B_LNG, MERCATOR_POINT_B_LAT},//real point B
new double[]{X_METERS,Y_METERS},//coordinates in meters of point B in virtual map
getScalingFactor(47.014816));
then you can transform with this function:
public double[] transform(double[] coord) {
double[] transCoord = new double[2];
double xscaled = (coord[0] - bPoint[0]) * mScale; // XXX bPoint is the position of origin point in the "VIRTUAL" world. [0] is the x coordinate
double yscaled = (coord[1] - bPoint[1]) * mScale;
transCoord[0] = (xscaled * Math.cos(mPhi)) - (yscaled * Math.sin(mPhi)) + aPoint[0]; //aPoint is the point with real coordinates of origin!
transCoord[1] = (xscaled * Math.sin(mPhi)) + (yscaled * Math.cos(mPhi)) + aPoint[1];
return transCoord;
}
you can find online a way to convert latlng to mercator, it just a bunch of math ;)
You should work in this way:
Your indoor map positions should be relative to a specific point (BOTTOM-LEFT is 0,0 let's say), then all the other positions will be relative to that point in meters, so you will endup in values under 100meters usually.
Having this you have to "move, rotate and scale" the indoor map with respect to the world.
Just take a map on a desktop which is not LAT/LNG and find the coordinates for the same indoor points you have (usually we get real and indoor position for bottom-left and top-right positions) so you can find where it should be in the world.
Take a look also at the scaling factor (depending on the latitude, the map must be scaled)
https://en.wikipedia.org/wiki/Mercator_projection#Scale_factor
We calculate that value by doing something like 1/cos(latitudeINradians)
public static double getScalingFactor(double latitude) {
return 1 / (Math.cos(Math.toRadians(latitude)));
}
Let me know if you can find a way, otherwise i will search and try to strip our code
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 am fairly new to Android programming, but I am getting pretty good at it (I think: ))
What I am doing is building a situated stories app. It is an app that places audio files at certain GPS markers and enables the user to listen to them at specific locations.
The next step is moving audio files. What I want to do is set a marker at a specific position in a city. (done). Next I want to check the location of a second marker that moves in a circle around it.
What I have so far is this:
public void checkCircularPosition(){
/*
* could be a solution?
*
radius = 250; //offset in meters
gpsLatCenter = 5.1164; //?how can i make this accurate in meters?
gpsLonCenter = 52.0963; //??how can i make this accurate in meters?
degree = 0; //should be variable over time (full circle in 1Hr, 3600sec --> 360/3600 = 0,1 deg/s)
radian;
radian = (degree/180)*Math.PI;
gpsCircleLat = gpsLatCenter+Math.cos(radian)*radius;
gpsCircleLon = gpsLonCenter-Math.sin(radian)*radius;
*/
}
Now, I have checked this code in adobe flash, which made a movie clip move around in a circle. So I know the calculations are somewhat right. But, I have no way of calculating the latitude and longitude of the resulting coordinates.
EDIT!!
i found the solution with the help posted below. still a lot of work to figure out how to use the results. anyway, i posted the resulting function below.
to make this work, you need _radius wich is 6371 (earth's radius), a bearing, a distance, and a start location.
thanks a lot guys!
public static void destinationPoint(double brng, double dist) {
dist = dist/_radius; // convert dist to angular distance in radians
brng = Math.toRadians(brng); //
double lat1 = Math.toRadians(_lat);
double lon1 = Math.toRadians(_lon);
double lat2 = Math.asin( Math.sin(lat1)*Math.cos(dist) + Math.cos(lat1)*Math.sin(dist)*Math.cos(brng) );
double lon2 = lon1 + Math.atan2(Math.sin(brng)*Math.sin(dist)*Math.cos(lat1), Math.cos(dist)-Math.sin(lat1)*Math.sin(lat2));
lon2 = (lon2+3*Math.PI) % (2*Math.PI) - Math.PI; // normalise to -180..+180º
Log.i(APPTAG, ""+Math.toDegrees(lat2));
Log.i(APPTAG, ""+Math.toDegrees(lon2));
Location movLoc = new Location("");
movLoc.setLatitude(Math.toDegrees(lat2));
movLoc.setLongitude(Math.toDegrees(lon2));
Log.i(APPTAG, ""+movLoc);
}
You should check the section Destination point given distance and bearing from start point at this website: http://www.movable-type.co.uk/scripts/latlong.html
That website has the proper formula for using your start point (gpsLatCenter/gpsLonCenter) and bearing (degree in you code) to compute the final lat/lon (gpsCircleLat/gpsCircleLon).
How do you determine if LocationX is between LocationA and LocationB? Is there an Android way to determine it?
UPDATE
I am asking here if LocationX is on the BuildingA or on the BuildingB. I have the upper left and lower right coordinates of the two buildings.
UPDATE2
Let say LocationX is a person. Main goal is to determine if that person is on the BuildingA or is on the BuildingB.
UPDATE3
Let say that BuildingA and BuildingB are rectangle/square. so we just need upper left and lower right latitude and longitude.
UPDATE4
Let me to add code snippet from #ligi 's response. But the code below is failed.
Location locationA = new Location("A");
locationA.setLatitude(35.70217034224572);
locationA.setLongitude(139.74359443782043);
Location locationB = new Location("B");
locationB.setLatitude(35.69994863045598);
locationB.setLongitude(139.74633029101562);
Location locationX = new Location("X");
// TODO result: X:169.661 Y:347.359
locationX.setLatitude(35.70066307001967);
locationX.setLongitude(139.74450102446747);
//distanceAB > (distanceXA + distanceXB - threshold)
float threshold = 0f;
float distanceAB = locationA.distanceTo(locationB);
float distanceXA = locationX.distanceTo(locationA);
float distanceXB = locationX.distanceTo(locationB);
boolean result = distanceAB > (distanceXA + distanceXB - threshold);
assertTrue(result);
calculate 3 distances ( e.g. via Location.distanceTo ) : distanceAB , distanceXA , distanceXB
then check the following: (distanceAB == (distanceXA + distanceXB))
and you might introduce a bit of fuzzynes with some threshold bigger than 0 as in the example check - depending on the use case - as in reality things will not be optimal strung
I see that the problem has changed with the updates - the new problem could be addressed e.g. with Rect.contains - just build a Rect for each building and check with Rect.contains if your person is in the building.
A point (a,b) is contained in the rectangle defined by the two points (x0,y0) and (x1,y1) if minimum(x0,x1) < a < maximum(x0,x1) and minimum(y0,y1) < b < maximum(y0,y1).
A point (a,b) is on the line connecting (x0,y0) to (x1,y1) if a - y0 = ((y1-y0)/(x1-x0))*(b - x0), assuming x1 != x0. If x0 does equal x1, (a,b) is on the line if and only if a also equals x0.
The distance between the point (a,b) and the point (x0,y0) is the square root of (a-x0)^2+(b-y0)^2.