Calculating Bearing from UTM (Latitude, Longitude) - geolocation

I want to calculate the relative bearing between two geo-coordinate points. I've gone ahead and converted the coordinates to UTM, but need assistance on figuring out the actual bearing. I'm aware that UTM is only good enough for same-zone calculations, which is fine as the computation will be done across very small distances.
Say point1 is my location 44.4N,-97.7W
and point2 is the location I'd like to get the relative bearing to: 44.4N, -103.3W
Since point 2 is directly to the left of point 1, I'd interpret that as 270 degrees (North being 0 or 360 degrees).
I found this formula: arctan((y1-y2)/(x1-x2))
but it's result don't make sense to me when I plot the points and measure the angles.

FYI- For now, I'm using SQL Server 2008 Spatial Data Types, Spatial Functions, and some T-SQL function I found on the web.
ALTER FUNCTION dbo.GeographyBearing (
#Point1 geography,
#Point2 geography )
RETURNS FLOAT
AS
BEGIN
DECLARE #Bearing DECIMAL(18,15)
DECLARE #Lat1 FLOAT = RADIANS(#Point1.Lat)
DECLARE #Lat2 FLOAT = RADIANS(#Point2.Lat)
DECLARE #dLon FLOAT = RADIANS(#Point2.Long - #Point1.Long)
IF (#Point1.STEquals(#Point2) = 1)
SET #Bearing = NULL
ELSE
SET #Bearing = ATN2(
SIN(#dLon)*COS(#Lat2),
(COS(#Lat1)*SIN(#Lat2)) - (SIN(#Lat1)*COS(#Lat2)*COS(#dLon))
)
SET #Bearing = (DEGREES(#Bearing) + 360) % 360
RETURN ISNULL(#Bearing,0);
END
GO
DECLARE #Vienna geography = geography::Point(16.37, 48.21, 4326)
DECLARE #Moscow geography = geography::Point(37.60, 55.75, 4326)
SELECT dbo.GeographyBearing(#Vienna,#Moscow)

Please find here my answer related to an open source coordinates converter. It might be helpfull for you given your need.

My answer in Javascript http://jsfiddle.net/efwjames/NVhg6/
//stop location - the radii end point
var x1 = 44.9631;
var y1 = -93.2492;
//bus location from the southeast - the circle center
var x2 = 44.95517;
var y2 = -93.2427;
var radians = getAtan2((y1 - y2), (x1 - x2));
function getAtan2(y, x) {
return Math.atan2(y, x);
};
$("#output").text(radians);
var newdeg = radians * (180 / Math.PI);
$("#deg").append(newdeg);
var coordNames = ["N", "NE", "E", "SE", "S", "SW", "W", "NW", "N"];
var directionid = Math.round(newdeg / 45);
if (directionid < 0) {
directionid = directionid + 8
};
$("#dir").append("The vehicle is moving " + coordNames[directionid]);

Related

How to detect object clicked in WebGL?

Firstly, I am not using 3Js in my Orbits app because I encountered a number of limitations including, but not limited to, issues with texture resolution and my requirement for complex lighting equations but I would like to implement something like 3Js' raycaster to allow me to detect the object clicked by the user.
I'm new to WebGL, but an "old hand" in software development so I'm looking for some hints about where to start.
The approach is as follows:
You generate your scene twice, once normally which is displayed and the second, with the objects uniquely coloured but not displayed. Then you use gl.readPixels from the second scene using the position on the first and decode the colour to identify the object.
Now I have to implement it myself.
Picking spheres
When picking spheres, or objects that are separated (not one inside another) you can use a simple distance from ray to very quickly get the closest object.
Example
The function returns a function that does the calculation. As it is only the closest you are interested in the distances can remain as squares. The distance from the camera is held as a unit distance along the ray.
function distanceFromRay() {
var dSqr, ox, oy, oz, vx, vy, vz;
function distanceSqr(px, py, pz) {
const ax = px - ox, ay = py - oy, az = pz - oz;
const u = (ax * vx + ay * vy + az * vz) / dSqr;
distanceSqr.unit = u;
if (u > 0) { // is past origin
const bx = ox + vx * u - px, by = oy + vy * u - py, bz = oz + vz * u - pz;
return bx * bx + by * by + bz * bz; // dist sqr to closest point on ray
}
return Infinity;
}
distanceSqr.unit = 0;
distanceSqr.setRay(x, y, z, xx, yy, zz) { // ray from origin x, y,z,
// infinite length along xx,yy,zz
(ox = x, oy = y, oz = z);
(vx = xx, vy = yy, vz = zz);
dSqr = vx * vx + vy * vy + vz * vz;
}
return distanceSqr;
}
Usage
There is a one time setup call;
// setup
const distToRay = distanceFromRay();
At the start of a frame that requires a pick, calculate the pick ray and set it. Also set the min distance from ray and eye.
// at start of frame set pick ray
distToRay.setRay(eye.x, eye.y, eye.z, pointer.ray.x, pointer.ray.y, pointer.ray.y);
var minDist = maxObjRadius * maxObjRadius;
var nearestObj = undefined;
var eyeDist = Infinity;
Then for each pickable object get the distance by passing the objects center and comparing it to any previous (in frame) found distance, objects radius, and distance from eye.
// per object
const dis = distToRay(obj.pos.x, obj.pos.y, obj.pos.z);
if (dis < obj.radius && dis < minDist && distToRay.unit > 0 && distToRay.unit < eyeDist ) {
minDist = dis;
eyeDist = distToRay.unit;
nearestObj = obj;
}
At the end of the frame if nearestObj is not undefined it will hold a reference to the picked object.
// end of frame
if (nearestObj) {
// you have the closest object
}

Object picking with Ray casting in elm-webgl

Demo almost (?) working example: https://ellie-app.com/4h9F8FNcRPya1/1
For demo: Click to draw ray, and rotate camera with left and right to see ray. (As the origin is from the camera, you can't see it from the position it is created)
Context
I am working on an elm & elm-webgl project where I would like to know if the mouse is over an object when clicked. To do is I tried to implement a simple ray cast. What I need is two things:
1) The coordinate of the camera (This one is easy)
2) The coordinate/direction in 3D space of where was clicked
Problem
The steps to get from 2D view space to 3D world space as I understand are:
a) Make coordinates to be in a range of -1 to 1 relative to view port
b) Invert projection matrix and perspective matrix
c) Multiply projection and perspective matrix
d) Create Vector4 from normalised mouse coordinates
e) Multiply combined matrices with Vector4
f) Normalise result
Try so far
I have made a function to transform a Mouse.Position to a coordinate to draw a line to:
getClickPosition : Model -> Mouse.Position -> Vec3
getClickPosition model pos =
let
x =
toFloat pos.x
y =
toFloat pos.y
normalizedPosition =
( (x * 2) / 1000 - 1, (1 - y / 1000 * 2) )
homogeneousClipCoordinates =
Vec4.vec4
(Tuple.first normalizedPosition)
(Tuple.second normalizedPosition)
-1
1
inversedProjectionMatrix =
Maybe.withDefault Mat4.identity (Mat4.inverse (camera model))
inversedPerspectiveMatrix =
Maybe.withDefault Mat4.identity (Mat4.inverse perspective)
inversedMatrix2 =
Mat4.mul inversedProjectionMatrix inversedPerspectiveMatrix
to =
Vec4.vec4
(Tuple.first normalizedPosition)
(Tuple.second normalizedPosition)
1
1
toInversed =
mulVector inversedMatrix2 to
toNorm =
Vec4.normalize toInversed
toVec3 =
vec3 (Vec4.getX toNorm) (Vec4.getY toNorm) (Vec4.getZ toNorm)
in
toVec3
Result
The result of this function is that the rays are too much to the center to where I click. I added a screenshot where I clicked in all four of the top face of the cube. If I click on the center of the viewport the ray will be correctly positioned.
It feels close, but not quite there yet and I can't figure out what I am doing wrong!
After trying other approaches I found a solution:
getClickPosition : Model -> Mouse.Position -> Vec3
getClickPosition model pos =
let
x =
toFloat pos.x
y =
toFloat pos.y
normalizedPosition =
( (x * 2) / 1000 - 1, (1 - y / 1000 * 2) )
homogeneousClipCoordinates =
Vec4.vec4
(Tuple.first normalizedPosition)
(Tuple.second normalizedPosition)
-1
1
inversedViewMatrix =
Maybe.withDefault Mat4.identity (Mat4.inverse (camera model))
inversedProjectionMatrix =
Maybe.withDefault Mat4.identity (Mat4.inverse perspective)
vec4CameraCoordinates = mulVector inversedProjectionMatrix homogeneousClipCoordinates
direction = Vec4.vec4 (Vec4.getX vec4CameraCoordinates) (Vec4.getY vec4CameraCoordinates) -1 0
vec4WorldCoordinates = mulVector inversedViewMatrix direction
vec3WorldCoordinates = vec3 (Vec4.getX vec4WorldCoordinates) (Vec4.getY vec4WorldCoordinates) (Vec4.getZ vec4WorldCoordinates)
normalizedVec3WorldCoordinates = Vec3.normalize vec3WorldCoordinates
origin = model.cameraPos
scaledDirection = Vec3.scale 20 normalizedVec3WorldCoordinates
destination = Vec3.add origin scaledDirection
in
destination
I left it as verbose as possible, if someone finds I use incorrect terminology please make a comment and I will update the answer.
I am sure there are lots of optimisations possible (Multiplying matrices before inverting or combining some of the steps.)
Updated the ellie app here: https://ellie-app.com/4hZ9s8S92PSa1/0

How do I implement the area calculator using latitude and longitueds

Can any one help me , How do I implement the area calculator using in group of latitude and longitudes .
For Example One person walk around the building , I am getting the latitude and longitudes while walking time . I want calculate the how much square feet that building was constructed.
Well, I´m going to try to help you, but I´m not going to give the complete answer.
I think, the first step is convert lats and longs in a cartesian coordinate system. You should calculate the center of all points. (A simple median).
Second step, convert all points to ENU coordinates centered in its center:
This step, I did, and here you are:
Constants:
#define DEGREES_TO_RADIANS (M_PI/180.0)
#define WGS84_A (6378137.0) // WGS 84 semi-major axis constant in meters
#define WGS84_E (8.1819190842622e-2) // WGS 84 eccentricity
Structs:
//To change to ECEF
typedef struct{
double x;
double y;
double z;
} ECEFCoordinate;
typedef struct{
double east;
double north;
double up;
} ENUCoordinate;
Methods, (you need past through ECEF):
#pragma mark Geodetic utilities definition
-(ECEFCoordinate) ecefFromLatitude:(double)lat longitude:(double)lon andAltitude:(double)alt
{
double clat = cos(lat * DEGREES_TO_RADIANS);
double slat = sin(lat * DEGREES_TO_RADIANS);
double clon = cos(lon * DEGREES_TO_RADIANS);
double slon = sin(lon * DEGREES_TO_RADIANS);
double N = WGS84_A / sqrt(1.0 - WGS84_E * WGS84_E * slat * slat);
ECEFCoordinate ecef;
ecef.x = (N + alt) * clat * clon;
ecef.y = (N + alt) * clat * slon;
ecef.z = (N * (1.0 - WGS84_E * WGS84_E) + alt) * slat;
return ecef;
}
// Converts ECEF to ENU coordinates centered at given lat , lon (with ECEFCenter)
-(ENUCoordinate)enuFromECEFCenter:(ECEFCoordinate)ecefCenter withLat:(double)lat andLon:(double)lon fromEcef:(ECEFCoordinate)ecef
{
double clat = cos(lat * DEGREES_TO_RADIANS);
double slat = sin(lat * DEGREES_TO_RADIANS);
double clon = cos(lon * DEGREES_TO_RADIANS);
double slon = sin(lon * DEGREES_TO_RADIANS);
double dx = ecefCenter.x - ecef.x;
double dy = ecefCenter.y - ecef.y;
double dz = ecefCenter.z - ecef.z;
ENUCoordinate enu;
enu.east = -slon*dx + clon*dy;
enu.north = -slat*clon*dx - slat*slon*dy + clat*dz;
enu.up = clat*clon*dx + clat*slon*dy + slat*dz;
return enu;
}
Last step: (I think the easy way is use triangles, from center to two consecutive points), calculate the area of a bunch of points in a cartesian coordinate system (east,north). Same than (x,y).
Good luck.
Last help two calculate the Area, I think you can find more help (and maybe best way) trough internet.

how to calculate angle between two path lines on google map using lat-long coordinates

I want to calculate angle between two path lines on Google map. I have the lat-long coordinates of the end points of the lines.
Please suggest If there is other information I can use to do this that is available from Google maps.
If the distances are small you can use this method.
The accuracy reduces with large distances and the further you move from the equator.
First find bearings with computeHeading()
computeHeading(from:LatLng, to:LatLng) .Returns the heading from one LatLng to another LatLng. Headings are expressed in degrees clockwise from North within the range [-180,180).
function getBearings(){
var spherical = google.maps.geometry.spherical;
var point1 = markers[0].getPosition();// latlng of point1
var point2 = markers[1].getPosition();
var point3 = markers[2].getPosition();
var bearing1 = google.maps.geometry.spherical.computeHeading(point1,point2);
var bearing2 = google.maps.geometry.spherical.computeHeading(point2,point3);
var angle =getDifference(bearing1, bearing2);
return angle;
}
You can then use this function to calculate angle between the bearings.
function getDifference(a1, a2) {
al = (a1>0) ? a1 : 360+a1;
a2 = (a2>0) ? a2 : 360+a2;
var angle = Math.abs(a1-a2)+180;
if (angle > 180){
angle = 360 - angle;
}
return Math.abs(angle);
}
Calculate both bearings using an API you have.
Then write a function angleDiff(angle1, angle2).
There are two possible angleDiff types: One that retunrs negative angles, too.
And one that only delivers positive angles.
Test that function with these test cases:
angleDiff(350, 10): diff = 20
10, 350: diff = 20 (or -20)
170, 190 :diff = 20
190, 170: diff 20 (or -20)
0,360 and 360,0: diff = 0

How to move 2 xna sprites away from eachother dynamically?

I have two items, lets call them Obj1 and Obj2... Both have a current position pos1 and pos2.. Moreover they have current velocity vectors speed1 and speed2 ... How can I make sure that if their distances are getting closer (with checking current and NEXT distance), they will move farther away from eachother ?
I have a signed angle function that gives me the signed angle between 2 vectors.. How can I utilize it to check how much should I rotate the speed1 and speed2 to move those sprites from eachother ?
public float signedAngle(Vector2 v1, Vector2 v2)
{
float perpDot = v1.X * v2.Y - v1.Y * v2.X;
return (float)Math.Atan2(perpDot, Vector2.Dot(v1, v2));
}
I check the NEXT and CURRENT distances like that :
float currentDistance = Vector2.Distance(s1.position, s2.position);
Vector2 obj2_nextpos = s2.position + s2.speed + s2.drag;
Vector2 obj1_nextpos = s1.position + s1.speed + s1.drag;
Vector2 s2us = s2.speed;
s2us.Normalize();
Vector2 s1us = s1.speed;
s1us.Normalize();
float nextDistance = Vector2.Distance(obj1_nextpos , obj2_nextpos );
Then depending whether they are getting bigger or smaller I want to move them away (either by increasing their current speed at the same direction or MAKING THEM FURTHER WHICH I FAIL)...
if (nextDistance < currentDistance )
{
float angle = MathHelper.ToRadians(180)- signedAngle(s1us, s2us);
s1.speed += Vector2.Transform(s1us, Matrix.CreateRotationZ(angle)) * esc;
s2.speed += Vector2.Transform(s2us, Matrix.CreateRotationZ(angle)) * esc;
}
Any ideas ?
if objects A and B are getting closer, one of the object components (X or Y) is opposite.
in this case Bx is opposite to Ax, so only have to add Ax to the velocity vector of object B, and Bx to velocity vector of object A
If I understood correctly, this is the situation and you want to obtain the two green vectors.
The red vector is easy to get: redVect = pos1 - pos2. redVect and greenVect2 will point to the same direction, so the only step you have is to scale it so its length will match speed2's one: finalGreenVect2 = greenvect2.Normalize() * speed2.Length (although I'm not actually sure about this formula). greenVect1 = -redVect so finalGreenVect1 = greenVect1.Normalize() * speed1.Length. Then speed1 = finalGreenVect1 and speed2 = finalGreenVect2. This approach will give you instant turn, if you prefer a smooth turn you want to rotate the speed vector by:
angle = signedAngle(speed) + (signedAngle(greenVect) - signedAngle(speed)) * 0.5f;
The o.5f is the rotation speed, adjust it to any value you need. I'm afraid that you have to create a rotation matrix then Transform() the speed vector with this matrix.
Hope this helps ;)

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