What is the best way to determine if a point X is in the 100 meter radius of point Y?
Is there a method on CLLocation?
Thanks
see
- (CLLocationDistance) distanceFromLocation:(const CLLocation *)location
documentation
It calculates the distance from another CLLocation object.
you can use this method:
// proximity distance calculation
static const double kDegToRad = 0.017453292519943295769236907684886;
static const double kEarthRadiusM = 6372797.560856;
+ (double)distanceInMetersFromLoc:(CLLocation *)from toLoc:(CLLocation *)to
{
return kEarthRadiusM * [self radianArcFrom:from.coordinate to:to.coordinate];
}
+ (double)radianArcFrom:(CLLocationCoordinate2D)from to:(CLLocationCoordinate2D)to
{
double latitudeArc = (from.latitude - to.latitude) * kDegToRad;
double longitudeArc = (from.longitude - to.longitude) * kDegToRad;
double latitudeHS = sin(latitudeArc * 0.5);
latitudeHS *= latitudeHS;
double lontitudeHS = sin(longitudeArc * 0.5);
lontitudeHS *= lontitudeHS;
double factor = cos(from.latitude * kDegToRad) * cos(to.latitude * kDegToRad);
return 2.0 * asin(sqrt(latitudeHS + factor * lontitudeHS));
}
Compare the distance as
if([distanceInMetersFromLoc:location1 to:location2] < 100)
{
// your condition is satisfied. you can write your code here
}
Related
I would like this code to return a newly constructed geopoint.
I need this,
GeoPoint prjTest=new GeoPoint(vxi+x,vyi+y);
to stick somewhere and return prjTest. I'm new to programming and I don't know well synthax.I tried many things, I can keep guessing for a long time. Please help. Thanks.
public class ProjectileTest
{
public ProjectileTest(float vi, float angle) /** renamed velocity -> vi */
{
//Starting location
double xi = 0, yi = 100;
final double gravity = -9.81;
//timeSlice declares the interval before checking the new location.
double timeSlice = 0.001; /** renamed time -> timeSlice */
double totalTime = 0; /** renamed recordedTime -> totalTime */
double vxi = vi * Math.cos(Math.toRadians(angle)); /** renamed xSpeed -> vxi */
double vyi = vi * Math.sin(Math.toRadians(angle)); /** renamed ySpeed -> vyi */
//This (secondsTillImpact) seems to give a very accurate time whenever the angle is positive.
double secondsTillImpact = Math.sqrt(2 * yi / -(gravity));
/** Not sure I agree. Does this formula take into account the upward motion
* of the projectile along its parabolic arc? My suspicion is that this
* formula only "works" when the initial theta is: 180 <= angle <= 360.
*
* Compare with the result predicted by quadratic(). Discarding the zero
* intercept which can't work for us (i.e. the negative one, because time
* only moves forward) leaves us with an expected result very close to the
* observed result.
*/
double y;
double x;/** Current position along the y-axis */
do {
// x = x + (xSpeed * time);
x = vxi * totalTime; /** Current position along the x-axis */
// y = y + (ySpeed * time);
y = yi + vyi * totalTime + .5 * gravity * (totalTime * totalTime);
// ySpeed = ySpeed + (gravity * time);
double vy = vyi + gravity * totalTime; /** Current velocity of vector y-component */
System.out.println("X: " + round2(x) + " Y: " + round2(y) + " YSpeed: " + round2(vy));
totalTime += timeSlice;
}
while (y > 0);
////////////////////////////++++++++ GeoPoint prjTest=new GeoPoint(vxi+x,vyi+y);
System.out.println("Incorrectly expected seconds: " + secondsTillImpact + "\nResult seconds: " + totalTime);
quadratic((.5 * gravity), vyi, yi);
}
public double round2(double n) {
return (int) (n * 100.0 + 0.5) / 100.0;
}
public void quadratic(double a, double b, double c) {
if (b * b - 4 * a * c < 0) {
System.out.println("No roots in R.");
} else {
double dRoot = Math.sqrt(b * b - 4 * a * c); /** root the discriminant */
double x1 = (-b + dRoot) / (2 * a); /** x-intercept 1 */
double x2 = (-b - dRoot) / (2 * a); /** x-intercept 2 */
System.out.println("x-int one: " + x1 + " x-int two: " + x2);
}
}
}
I'm looking to draw a line on a map from current user position with his bearing/heading.
But the line always have the same direction even if I rotate the phone to another direction. I wonder if I miss something in my calculation.
final la1 = userPosition.latitude;
final lo1 = userPosition.longitude;
const r = 6367; // earth radius
const d = 40; // distance
const dist = d / r;
final bearing = vector.radians(direction.value);
final la2 =
asin(sin(la1) * cos(dist) + cos(la1) * sin(dist) * cos(bearing));
final lo2 = lo1 +
atan2(sin(bearing) * sin(dist) * cos(la1),
cos(dist) - sin(la1) * sin(la2));
return LatLng(la2, lo2);
As you can see in the screenshots bellow, I create 2 lines with a different bearing (check map orientation), but they look the same.
import 'dart:math';
import 'package:google_maps_flutter/google_maps_flutter.dart';
import 'package:vector_math/vector_math.dart';
LatLng createCoord(LatLng coord, double bearing, double distance) {
var radius = 6371e3; //meters
var delta = (distance) / radius; // angular distance in radians
var teta = radians(bearing);
var phi1 = radians(coord.longitude);
var lambda1 = radians(coord.latitude);
var phi2 = asin(sin(phi1) * cos(delta) + cos(phi1) * sin(delta) * cos(teta));
var lambda2 = lambda1 +
atan2(sin(teta) * sin(delta) * cos(phi1),
cos(delta) - sin(phi1) * sin(phi2));
lambda2 = (lambda2 + 3 * pi) % (2 * pi) - pi; // normalise to -180..+180°
return LatLng(degrees(lambda2), degrees(phi2)); //[lon, lat]
}
I found a piece of code on web. It calculates the Minimum bounding rectangle by a given lat/lon point and a distance.
private static void GetlatLon(double LAT, double LON, double distance, double angle, out double newLon, out double newLat)
{
double dx = distance * 1000 * Math.Sin(angle * Math.PI / 180.0);
double dy = distance * 1000 * Math.Cos(angle * Math.PI / 180.0);
double ec = 6356725 + 21412 * (90.0 - LAT) / 90.0;
double ed = ec * Math.Cos(LAT * Math.PI / 180);
newLon = (dx / ed + LON * Math.PI / 180.0) * 180.0 / Math.PI;
newLat = (dy / ec + LAT * Math.PI / 180.0) * 180.0 / Math.PI;
}
public static void GetRectRange(double centorlatitude, double centorLogitude, double distance,
out double maxLatitude, out double minLatitude, out double maxLongitude, out double minLongitude)
{
GetlatLon(centorlatitude, centorLogitude, distance, 0, out temp, out maxLatitude);
GetlatLon(centorlatitude, centorLogitude, distance, 180, out temp, out minLatitude);
GetlatLon(centorlatitude, centorLogitude, distance, 90, out minLongitude, out temp);
GetlatLon(centorlatitude, centorLogitude, distance, 270, out maxLongitude, out temp);
}
double ec = 6356725 + 21412 * (90.0 - LAT) / 90.0; //why?
double ed = ec * Math.Cos(LAT * Math.PI / 180); // why?
dx / ed //why?
dy / ec //why?
6378137 is the equator radius, 6356725 is polar radius, 21412 =6378137 -6356725.
from the link, I know a little of the meanings. But these four lines, I don't know why. Could you please help to give more information? Could you please help to let me know the derivation of the formula?
From the link, in the section "Destination point given distance and bearing from start point", it gives another formula to get the result. What is the derivation of the formula?
From this link , I know the derivation of the Haversine Formula, it's very informative. I don't think the formula in the section of "Destination point given distance and bearing from start point" is just a simple reversion of Haversine.
Thanks a lot!
This is a prime example of why commenting your code makes it more readable and maintainable. Mathematically you are looking at the following:
double ec = 6356725 + 21412 * (90.0 - LAT) / 90.0; //why?
This is a measure of eccentricity to account for the equatorial bulge in some fashion. 21412 is, as you know, the difference in earth radius between the equator and pole. 6356725 is the polar radius. (90.0 - LAT) / 90.0 is 1 at the equator, and 0 at the pole. The formula simply estimates how much bulge is present at any given latitude.
double ed = ec * Math.Cos(LAT * Math.PI / 180); // why?
(LAT * Math.PI / 180) is a conversion of latitude from degrees to radians. cos (0) = 1 and cos(1) = 0, so at the equator, you are applying the full amount of the eccentricity while at the pole you are applying none. Similar to the preceding line.
dx / ed //why?
dy / ec //why?
The above seems to be the fractional additions to distance in both the x and y directions attributable to the bulge at any given lat/lon used in the newLon newLat computation to arrive at the new location.
I haven't done any research into the code snippet you found, but mathematically, this is what is taking place. Hopefully that will steer you in the right direction.
Haversine Example in C
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double m2ft (double l) { /* convert meters to feet */
return l/(1200.0/3937.0);
}
double ft2smi (double l) { /* convert feet to statute miles*/
return l/5280.0;
}
double km2smi (double l) { /* convert km to statute mi. */
return ft2smi(m2ft( l * 1000.0 ));
}
static const double deg2rad = 0.017453292519943295769236907684886;
static const double earth_rad_m = 6372797.560856;
typedef struct pointd {
double lat;
double lon;
} pointd;
/* Computes the arc, in radian, between two WGS-84 positions.
The result is equal to Distance(from,to)/earth_rad_m
= 2*asin(sqrt(h(d/earth_rad_m )))
where:
d is the distance in meters between 'from' and 'to' positions.
h is the haversine function: h(x)=sin²(x/2)
The haversine formula gives:
h(d/R) = h(from.lat-to.lat)+h(from.lon-to.lon)+cos(from.lat)*cos(to.lat)
http://en.wikipedia.org/wiki/Law_of_haversines
*/
double arcradians (const pointd *from, const pointd *to)
{
double latitudeArc = (from-> lat - to-> lat) * deg2rad;
double longitudeArc = (from-> lon - to-> lon) * deg2rad;
double latitudeH = sin (latitudeArc * 0.5);
latitudeH *= latitudeH;
double lontitudeH = sin (longitudeArc * 0.5);
lontitudeH *= lontitudeH;
double tmp = cos (from-> lat * deg2rad) * cos (to-> lat * deg2rad);
return 2.0 * asin (sqrt (latitudeH + tmp*lontitudeH));
}
/* Computes the distance, in meters, between two WGS-84 positions.
The result is equal to earth_rad_m*ArcInRadians(from,to)
*/
double dist_m (const pointd *from, const pointd *to) {
return earth_rad_m * arcradians (from, to);
}
int main (int argc, char **argv) {
if (argc < 5 ) {
fprintf (stderr, "Error: insufficient input, usage: %s (lat,lon) (lat,lon)\n", argv[0]);
return 1;
}
pointd points[2];
points[0].lat = strtod (argv[1], NULL);
points[0].lon = strtod (argv[2], NULL);
points[1].lat = strtod (argv[3], NULL);
points[1].lon = strtod (argv[4], NULL);
printf ("\nThe distance in meters from 1 to 2 (smi): %lf\n\n", km2smi (dist_m (&points[0], &points[1])/1000.0) );
return 0;
}
/* Results/Example.
./bin/gce 31.77 -94.61 31.44 -94.698
The distance in miles from Nacogdoches to Lufkin, Texas (smi): 23.387997 miles
*/
I assume 6356725 has something to do with the radius of the earth. Check out this answer, and also take a look at the Haversine Formula.
I am following the quaternion tutorial: http://www.raywenderlich.com/12667/how-to-rotate-a-3d-object-using-touches-with-opengl and am trying to rotate a globe to some XYZ location. I have an initial quaternion and generate a random XYZ location on the surface of the globe. I pass that XYZ location into the following function. The idea was to generate a lookAt vector with GLKMatrix4MakeLookAt and define the end Quaternion for the slerp step from the lookAt matrix.
- (void)rotateToLocationX:(float)x andY:(float)y andZ:(float)z {
// Turn on the interpolation for smooth rotation
_slerping = YES; // Begin auto rotating to this location
_slerpCur = 0;
_slerpMax = 1.0;
_slerpStart = _quat;
// The eye location is defined by the look at location multiplied by this modifier
float modifier = 1.0;
// Create a look at vector for which we will create a GLK4Matrix from
float xEye = x;
float yEye = y;
float zEye = z;
//NSLog(#"%f %f %f %f %f %f",xEye, yEye, zEye, x, y, z);
_currentSatelliteLocation = GLKMatrix4MakeLookAt(xEye, yEye, zEye, 0, 0, 0, 0, 1, 0);
_currentSatelliteLocation = GLKMatrix4Multiply(_currentSatelliteLocation,self.effect.transform.modelviewMatrix);
// Turn our 4x4 matrix into a quat and use it to mark the end point of our interpolation
//_currentSatelliteLocation = GLKMatrix4Translate(_currentSatelliteLocation, 0.0f, 0.0f, GLOBAL_EARTH_Z_LOCATION);
_slerpEnd = GLKQuaternionMakeWithMatrix4(_currentSatelliteLocation);
// Print info on the quat
GLKVector3 vec = GLKQuaternionAxis(_slerpEnd);
float angle = GLKQuaternionAngle(_slerpEnd);
//NSLog(#"%f %f %f %f",vec.x,vec.y,vec.z,angle);
NSLog(#"Quat end:");
[self printMatrix:_currentSatelliteLocation];
//[self printMatrix:self.effect.transform.modelviewMatrix];
}
The interpolation works, I get a smooth rotation, however the ending location is never the XYZ I input - I know this because my globe is a sphere and I am calculating XYZ from Lat Lon. I want to look directly down the 'lookAt' vector toward the center of the earth from that lat/lon location on the surface of the globe after the rotation. I think it may have something to do with the up vector but I've tried everything that made sense.
What am I doing wrong - How can I define a final quaternion that when I finish rotating, looks down a vector to the XYZ on the surface of the globe? Thanks!
Is the following your meaning:
Your globe center is (0, 0, 0), radius is R, the start position is (0, 0, R), your final position is (0, R, 0), so rotate the globe 90 degrees around X-asix?
If so, just set lookat function eye position to your final position, the look at parameters to the globe center.
m_target.x = 0.0f;
m_target.y = 0.0f;
m_target.z = 1.0f;
m_right.x = 1.0f;
m_right.y = 0.0f;
m_right.z = 0.0f;
m_up.x = 0.0f;
m_up.y = 1.0f;
m_up.z = 0.0f;
void CCamera::RotateX( float amount )
{
Point3D target = m_target;
Point3D up = m_up;
amount = amount / 180 * PI;
m_target.x = (cos(PI / 2 - amount) * up.x) + (cos(amount) * target.x);
m_target.y = (cos(PI / 2 - amount) * up.y) + (cos(amount) * target.y);
m_target.z = (cos(PI / 2 - amount) * up.z) + (cos(amount) * target.z);
m_up.x = (cos(amount) * up.x) + (cos(PI / 2 + amount) * target.x);
m_up.y = (cos(amount) * up.y) + (cos(PI / 2 + amount) * target.y);
m_up.z = (cos(amount) * up.z) + (cos(PI / 2 + amount) * target.z);
Normalize(m_target);
Normalize(m_up);
}
void CCamera::RotateY( float amount )
{
Point3D target = m_target;
Point3D right = m_right;
amount = amount / 180 * PI;
m_target.x = (cos(PI / 2 + amount) * right.x) + (cos(amount) * target.x);
m_target.y = (cos(PI / 2 + amount) * right.y) + (cos(amount) * target.y);
m_target.z = (cos(PI / 2 + amount) * right.z) + (cos(amount) * target.z);
m_right.x = (cos(amount) * right.x) + (cos(PI / 2 - amount) * target.x);
m_right.y = (cos(amount) * right.y) + (cos(PI / 2 - amount) * target.y);
m_right.z = (cos(amount) * right.z) + (cos(PI / 2 - amount) * target.z);
Normalize(m_target);
Normalize(m_right);
}
void CCamera::RotateZ( float amount )
{
Point3D right = m_right;
Point3D up = m_up;
amount = amount / 180 * PI;
m_up.x = (cos(amount) * up.x) + (cos(PI / 2 - amount) * right.x);
m_up.y = (cos(amount) * up.y) + (cos(PI / 2 - amount) * right.y);
m_up.z = (cos(amount) * up.z) + (cos(PI / 2 - amount) * right.z);
m_right.x = (cos(PI / 2 + amount) * up.x) + (cos(amount) * right.x);
m_right.y = (cos(PI / 2 + amount) * up.y) + (cos(amount) * right.y);
m_right.z = (cos(PI / 2 + amount) * up.z) + (cos(amount) * right.z);
Normalize(m_right);
Normalize(m_up);
}
void CCamera::Normalize( Point3D &p )
{
float length = sqrt(p.x * p.x + p.y * p.y + p.z * p.z);
if (1 == length || 0 == length)
{
return;
}
float scaleFactor = 1.0 / length;
p.x *= scaleFactor;
p.y *= scaleFactor;
p.z *= scaleFactor;
}
The answer to this question is a combination of the following rotateTo function and a change to the code from Ray's tutorial at ( http://www.raywenderlich.com/12667/how-to-rotate-a-3d-object-using-touches-with-opengl ). As one of the comments on that article says there is an arbitrary factor of 2.0 being multiplied in GLKQuaternion Q_rot = GLKQuaternionMakeWithAngleAndVector3Axis(angle * 2.0, axis);. Remove that "2" and use the following function to create the _slerpEnd - after that the globe will rotate smoothly to XYZ specified.
// Rotate the globe using Slerp interpolation to an XYZ coordinate
- (void)rotateToLocationX:(float)x andY:(float)y andZ:(float)z {
// Turn on the interpolation for smooth rotation
_slerping = YES; // Begin auto rotating to this location
_slerpCur = 0;
_slerpMax = 1.0;
_slerpStart = _quat;
// Create a look at vector for which we will create a GLK4Matrix from
float xEye = x;
float yEye = y;
float zEye = z;
_currentSatelliteLocation = GLKMatrix4MakeLookAt(xEye, yEye, zEye, 0, 0, 0, 0, 1, 0);
// Turn our 4x4 matrix into a quat and use it to mark the end point of our interpolation
_slerpEnd = GLKQuaternionMakeWithMatrix4(_currentSatelliteLocation);
}
How do I calculate the angle in degrees between the coordinates of two POIs (points of interest) on an iPhone map application?
I'm guessing you try to calculate the degrees between the coordinates of two points of interest (POI).
Calculating the arc of a great circle:
+(float) greatCircleFrom:(CLLocation*)first
to:(CLLocation*)second {
int radius = 6371; // 6371km is the radius of the earth
float dLat = second.coordinate.latitude-first.coordinate.latitude;
float dLon = second.coordinate.longitude-first.coordinate.longitude;
float a = pow(sin(dLat/2),2) + cos(first.coordinate.latitude)*cos(second.coordinate.latitude) * pow(sin(dLon/2),2);
float c = 2 * atan2(sqrt(a),sqrt(1-a));
float d = radius * c;
return d;
}
Another option is to pretend you are on cartesian coordinates (faster but not without error on long distances):
+(float)angleFromCoordinate:(CLLocationCoordinate2D)first
toCoordinate:(CLLocationCoordinate2D)second {
float deltaLongitude = second.longitude - first.longitude;
float deltaLatitude = second.latitude - first.latitude;
float angle = (M_PI * .5f) - atan(deltaLatitude / deltaLongitude);
if (deltaLongitude > 0) return angle;
else if (deltaLongitude < 0) return angle + M_PI;
else if (deltaLatitude < 0) return M_PI;
return 0.0f;
}
If you want the result in degrees instead radians, you have to apply the following conversion:
#define RADIANS_TO_DEGREES(radians) ((radians) * 180.0 / M_PI)
You are calculating the 'Bearing' from one point to another here. There's a whole bunch of formula for that, and lots of other geographic quantities like distance and cross-track error, on this web page:
http://www.movable-type.co.uk/scripts/latlong.html
the formulae are in several formats so you can easily convert to whatever language you need for your iPhone. There's also javascript calculators so you can test your code gets the same answers as theirs.
If the other solutions dont work for you try this:
- (int)getInitialBearingFrom:(CLLocation *)first
to:(CLLocation *)second
{
float lat1 = [self degreesToRad:first.coordinate.latitude];
float lat2 = [self degreesToRad:second.coordinate.latitude];
float lon1 = [self degreesToRad:first.coordinate.longitude];
float lon2 = [self degreesToRad:second.coordinate.longitude];
float dLon = lon2 - lon1;
float y = sin (dLon) * cos (lat2);
float x1 = cos (lat1) * sin (lat2);
float x2 = sin (lat1) * cos (lat2) * cos (dLon);
float x = x1 - x2;
float bearingRadRaw = atan2f (y, x);
float bearingDegRaw = bearingRadRaw * 180 / M_PI;
int bearing = ((int) bearingDegRaw + 360) % 360; // +- 180 deg to 360 deg
return bearing;
}
For final bearing, simply take the initial bearing from the end point to the start point and reverse it (using θ = (θ+180) % 360).
You need these 2 helpers:
-(float)radToDegrees:(float)radians
{
return radians * 180 / M_PI;
}
-(float)degreesToRad:(float)degrees
{
return degrees * M_PI /180;
}