I've looked around and seen mention of the haversine formula to determine distance between two coordinates (lat1, lng1) and (lat2, lng2).
I've implemented this code:
function haversineGreatCircleDistance(
$latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
// convert from degrees to radians
$latFrom = deg2rad($latitudeFrom);
$lonFrom = deg2rad($longitudeFrom);
$latTo = deg2rad($latitudeTo);
$lonTo = deg2rad($longitudeTo);
$latDelta = $latTo - $latFrom;
$lonDelta = $lonTo - $lonFrom;
$angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
return $angle * $earthRadius;
}
And am trying to determine:
1) what units this is returning? (goal being in feet)
2) is this equation written the right way?
For example what should be the distance between these two points?
(32.8940695525,-96.7926336453) and (33.0642604502, -96.8064332754)?
I'm getting 18968.0903312 from the formula above.
Thanks!
1) what units this is returning? (goal being in feet)
Whatever units in which you supply the Earth's radius.
2) is this equation written the right way?
Test it. You can compare your results with an existing Haversine formula implementation, like this one.
Related
Athena only allows to calculate the distance of the buffer in decimal degrees but this value varies with respect to the latitude in the globe, tate to obtain a distance according to the following formula but it is not consistent in Mexico.
Athena function like this : ST_Buffer(geometry, double)
Athena geospatial functions
So, is posible obtain the corresponding distance in decimal degrees over a custom point in map , ex : get the decimal degree for point x, y like that distance in meters is 300 mts
Currently I use the following formula to approximate the decimal degrees but some buffers are quite horrible although it meets the minimum required
SELECT
ST_Buffer(ST_GeometryFromText( shape_wkt) ,
abs(5000.0 * 360.0 / (2.0 * pi() * cos( latitud )* 6400000.0) ) ) AS
dinamic_buffer_5000
5000 is buffer in meters
6400000.0 earth radius in meters
Some useffull questions :
gps-coordinates-in-degrees-to-calculate-distances
Calculate distance in meters using results in degrees
calculating-latitude-longitude-x-miles-from-point
A possible alternative is the following
To obtain the decimal degrees relative to a point one could:
Generate a second point at a distance d for this you would have to implement this formula, where the bearing does not matter
With this second point calculate the distance in Athena that will return the distance in decimal degrees, as input for the buffer function.
As an approximate is good alternative
Now how implement the second point ?....Here is the formula
I will try to convert to SQL code if can :
After a test I realize that even with the difference of distance it is not possible to obtain the buffer in an optimal way.
In this case the distance to the lower point was 300 meters, after obtaining the distance in decimal degrees with Athena an oblate shape is obtained, it changes the degree of inclination of the point by 90 degrees but it only generates a slightly larger shape.
Destination point given distance and bearing from start point
Source code (zory im edit for test my sql ):
destinationPoint(distance, bearing, radius=6371e3) {
// sinφ2 = sinφ1⋅cosδ + cosφ1⋅sinδ⋅cosθ
// tanΔλ = sinθ⋅sinδ⋅cosφ1 / cosδ−sinφ1⋅sinφ2
// see mathforum.org/library/drmath/view/52049.html for derivation
const dist_ang = distance / radius; // angular distance in radians
const angulo = Number(bearing).toRadians();
const rad_lat = this.lat.toRadians();
const rad_lon = this.lon.toRadians();
console.log("distance", distance);
console.log("radius", radius);
console.log("angular distance in radians", dist_ang);
console.log("bearing", Number(bearing));
console.log("bearing angulo ", angulo );
console.log("lat.toRadians", rad_lat);
console.log("lon.toRadians", rad_lon);
console.log("lon",this.lon);
console.log("lat",this.lat);
const sinφ2 = Math.sin(rad_lat) * Math.cos(dist_ang) + Math.cos(rad_lat) * Math.sin(dist_ang) * Math.cos(angulo);
const φ2 = Math.asin(sinφ2); //lat
console.log("φ2",φ2); //lat
console.log("sinφ2",sinφ2);
const y = Math.sin(angulo) * Math.sin(dist_ang) * Math.cos(rad_lat);
const x = Math.cos(dist_ang) - Math.sin(rad_lat) * sinφ2;
console.log("y",y);
console.log("x",x);
const λ2 = rad_lon + Math.atan2(y, x); //lon
console.log("λ2",λ2);
const lat = φ2.toDegrees();//lat
const lon = λ2.toDegrees();//lon
console.log("lon2",lon);
console.log("lat2",lat);
return new LatLonSpherical(lat, lon);
}
Given 2 coordinates (point 1 and 2 in red) in WGS84 I need to find the coordinates of the point perpendicular (point 3) to the line at a given distance.
I could manage to make the math to compute this perpendicular point, but when displayed on the map, the point seems to be at a wrong place, probably because of the projection.
What I want on a map:
And what I have instead on the map:
How can I take into account the projection so that the point on the map appears perpendicular to the line? The algorithm below to compute the point comes from here: https://math.stackexchange.com/questions/93424/calculate-rectangle-coordinates-from-line-and-height
public static Coords ComputePerpendicularPoint(Coords first, Coords last, double distance)
{
double slope = -(last.Lon.Value - first.Lon.Value) / (last.Lat.Value - first.Lat.Value);
// number of km per degree = ~111km (111.32 in google maps, but range varies between 110.567km at the equator and 111.699km at the poles)
// 1km in degree = 1 / 111.32km = 0.0089
// 1m in degree = 0.0089 / 1000 = 0.0000089
distance = distance * 0.0000089 / 100; //0.0000089 => represents around 1m in wgs84. /100 because distance is in cm
double t = distance / Math.Sqrt(1 + (slope * slope));
Coords perp_coord = new Coords();
perp_coord.Lon = first.Lon + t;
perp_coord.Lat = first.Lat + (t * slope);
return perp_coord;
}
Thank you in advance!
I have a list of X,Y coordinates that represents a road. For every 5 meters, I need to calculate the angle of the tangent on this road, as I have tried to illustrate in the image.
My problem is that this road is not represented by a mathematical function that I can simply derive, it is represented by a list of coordinates (UTM33N).
In my other similar projects we use ArcGIS/ESRI libraries to perform geographical functions such as this, but in this project I need to be independent of any software that require the end user to have a license, so I need to do the calculations myself (or find a free/open source library that can do it).
I am using a cubic spline function to make the line rounded between the coordinates, since all tangents on a line segment would just be parallell to the segment otherwise.
But now I am stuck. I am considering simply calculating the angle between any three points on the line (given enough points), and using this to find the tangents, but that doesn't sound like a good method. Any suggestions?
In the end, I concluded that the points were plentiful enough to give an accurate angle using simple geometry:
//Calculate delta values
var dx = next.X - curr.X;
var dy = next.Y - curr.Y;
var dz = next.Z - curr.Z;
//Calculate horizontal and 3D length of this segment.
var hLength = Math.Sqrt(dx * dx + dy * dy);
var length = Math.Sqrt(hLength * hLength + dz * dz);
//Calculate horizontal and vertical angles.
hAngle = Math.Atan(dy/dx);
vAngle = Math.Atan(dz/hLength);
I have 2 coordinates and would like to do something seemingly straightforward. I want to figure out, given:
1) Coordinate A
2) Course provided by Core Location
3) Coordinate B
the following:
1) Distance between A and B (can currently be done using distanceFromLocation) so ok on that one.
2) The course that should be taken to get from A to B (different from course currently traveling)
Is there a simple way to accomplish this, any third party or built in API?
Apple doesn't seem to provide this but I could be wrong.
Thanks,
~Arash
EDIT:
Thanks for the fast responses, I believe there may have been some confusion, I am looking to get the course (bearing from point a to point b in degrees so that 0 degrees = north, 90 degrees = east, similar to the course value return by CLLocation. Not trying to compute actual turn by turn directions.
I have some code on github that does that. Take a look at headingInRadians here. It is based on the Spherical Law of Cosines. I derived the code from the algorithm on this page.
/*-------------------------------------------------------------------------
* Given two lat/lon points on earth, calculates the heading
* from lat1/lon1 to lat2/lon2.
*
* lat/lon params in radians
* result in radians
*-------------------------------------------------------------------------*/
double headingInRadians(double lat1, double lon1, double lat2, double lon2)
{
//-------------------------------------------------------------------------
// Algorithm found at http://www.movable-type.co.uk/scripts/latlong.html
//
// Spherical Law of Cosines
//
// Formula: θ = atan2( sin(Δlong) * cos(lat2),
// cos(lat1) * sin(lat2) − sin(lat1) * cos(lat2) * cos(Δlong) )
// JavaScript:
//
// var y = Math.sin(dLon) * Math.cos(lat2);
// var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
// var brng = Math.atan2(y, x).toDeg();
//-------------------------------------------------------------------------
double dLon = lon2 - lon1;
double y = sin(dLon) * cos(lat2);
double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
return atan2(y, x);
}
See How to get angle between two POI?
Depending on how much work you want to put in this one, I would suggest looking at Tree Traversal Algorithms (check the column on the right), things like A* alpha star, that you can use to find your find from one point to another, even if obstacles are in-between.
If I understand you correctly, you have the current location and you have some other location. You want to find the distance (as the crow flies) between the two points, and to find a walking path between the points.
To answer your first question, distanceFromLocation will find the distance across the earth's surface between 2 points, that is it follows the curvature of the earth, but it will give you the distance as the crow flies. So I think you're right about that.
The second question is a much harder. What you want to do is something called path-finding. Path finding, require's not only a search algorithm that will decide on the path, but you also need data about the possible paths. That is to say, if you want to find a path through the streets, the computer has to know how the streets are connected to each other. Furthermore, if you're trying to make a pathfinder that takes account for traffic and the time differences between taking two different possible paths, you will need a whole lot more data. It is for this reason that we usually leave these kinds of tasks up to big companies, with lots of resources, like Google, and Yahoo.
However, If you're still interested in doing it, check this out
http://www.youtube.com/watch?v=DoamZwkEDK0
has anybody already programmed a iphone compass heading tilt compensation?
i have got some approaches, but some help or a better solution would be cool!
FIRST
i define a vector Ev, calculated out of the cross product of Gv and Hv. Gv is a gravity vector i build out of the accelerometer values and Hv is an heading vector built out the magnetometer values.
Ev stands perpendicular on Gv and Hv, so it is heading to horizonatl East.
SECOND
i define a vector Rv, calculated out of the cross product Bv and Gv. Bv is my looking vector and it is defined as [0,0,-1]. Rv is perpendicular to Gv and Bv and shows always to the right.
THIRD
the angle between these two vectors, Ev and Rv, should be my corrected heading. to calculate the angle i build the dot product and thereof the arcos.
phi = arcos ( Ev * Rv / |Ev| * |Rv| )
Theoretically it should work, but maybe i have to normalize the vectors?!
Has anybody got a solution for this?
Thanks, m01d
Yep. You DEFINITELY have to normalize.
This is from my code that I use to extract the orientation of the device.
Gravity is obtained as the x,y,z of the accelerometer
and compass is obtained from the x,y,z of the heading function
gravity.normalize();
compass.normalize();
compassEast=gravity.cross(compass);
compassEast.normalize();
compassNorth=compassEast.cross(gravity);
compassNorth.normalize();
Let me know if you need the full code.
Also, for those who havnt yet seen the iphone 4s gyroscope in action: its amazing! I swapped the above input to gravity and compass for the equivalents from the gyro and the result is stable and smooth and awesome :) Go Apple.
I didn't receive the source code but I set up my own example. You can see the project and code here: http://www.sundh.com/blog/2011/09/stabalize-compass-of-iphone-with-gyroscope/
yes, i did it like described above. but the result is not very accurate. i think with smoother accelerometer values it should work that way. because of this i have choosen to do the tilt compensation by adding/subtracting the accelermoter values of the corresponding axis to/from the compass values.
Here iss my code for the solution above, but its not a final working solution:
- (void)locationManager:(CLLocationManager *)manager didUpdateHeading:(CLHeading *)newHeading{
if (newHeading != nil) {
float Ax = accelerationValueX;
float Ay = accelerationValueY;
float Az = accelerationValueZ;
float filterFactor = 0.2;
Mx = [newHeading x] * filterFactor + (Mx * (1.0 - filterFactor));
My = [newHeading y] * filterFactor + (My * (1.0 - filterFactor));
Mz = [newHeading z] * filterFactor + (Mz * (1.0 - filterFactor));
float counter = ( -pow(Ax, 2)*Mz + Ax*Az*Mx - pow(Ay, 2)*Mz + Ay*Az*My );
float denominator = ( sqrt( pow((My*Az-Mz*Ay), 2) + pow((Mz*Ax-Mx*Az), 2) + pow((Mx*Ay-My*Ax), 2) ) * sqrt(pow(Ay, 2)+pow(-Ax, 2)) );
headingCorrected = (acos(counter/denominator)* (180.0 / M_PI)) * filterFactor + (headingCorrected * (1.0 - filterFactor));
}
...
}