I'm interested in working with coordinates, and I was wondering, how is it possible to get the distance between two points (coordinates) in meters. After a long search I found the Haversine Formula, and the Objective-C implementation of it here.
It's the following (I modified it a little bit for myself):
- (CGFloat)directMetersFromCoordinate:(CLLocation *)from toCoordinate:(CLLocation *)to {
static const double DEG_TO_RAD = 0.017453292519943295769236907684886;
static const double EARTH_RADIUS_IN_METERS = 6372797.560856;
double latitudeArc = (from.coordinate.latitude - to.coordinate.latitude) * DEG_TO_RAD;
double longitudeArc = (from.coordinate.longitude - to.coordinate.longitude) * DEG_TO_RAD;
double latitudeH = sin(latitudeArc * 0.5);
latitudeH *= latitudeH;
double lontitudeH = sin(longitudeArc * 0.5);
lontitudeH *= lontitudeH;
double tmp = cos(from.coordinate.latitude*DEG_TO_RAD) * cos(to.coordinate.latitude*DEG_TO_RAD);
return EARTH_RADIUS_IN_METERS * 2.0 * asin(sqrt(latitudeH + tmp*lontitudeH)); }
My question is:
How is it possible to get 100 meters distance (for latitude and for longitude) for the current location?
This formula is too complicated for me, I don't understand it, so I can't "code it back" to get the result what I want.
I need only the actual location (Paris, Tokyo, London, New York, whatever), and a (float) number for latitude and a (float) number for longitude, which (float) number represents 100 meters distance from the actual location.
If you open this page, here you can "calculate out" the 100 meters distance between two points ("actual" and the one 100 meters away).
For example:
Point1: 47.0, 19.0
Point2: 47.0, 19.0013190
---That's 0.1000 km (100 m) distance.
Point1: 47.0, 19.0
Point2: 47.0008995, 19.0
---is also 0.1000 km (100 m) distance.
Here you can see, that at that coordinate (latitude 47.0 and longitude 19.0) 100 meters distance is 0.0008995 (latitude) and 0.0013190 (longitude).
And I want to get these data with the help of the Haversine Formula, just don't know, how.
Could you help me to figure it out?
Thanks!
UPDATE:
Thank you for the answers, right now I don't have time to try them out, but as I understood, I didn't explain exactly what I want.
Maybe this is a better example, how I want to use these "100 meters":
So, right now I'm at the coordinate "lat x" and "lon y". There is another point (let's say a hotel) at another given coordinate, "lat a" and "lon b".
My question is: how can I calculate out, if this hotel is (less than) 100 meters from me? So it doesn't matter, if it's only 5 meters or 99 meters, both of them are less (or equal) than 100 meters far from me.
With the code what I provided, I can calculate this out, that's what that formula is for.
But let's say, I have a million of other coordinates (hotel locations) that I want to work with. And I need only a list of them, what are (less than) 100 meters away from me. So yes, that's a circle around me with a radius of 100 meters, and I need the results within that.
It would take much more time and resource to take all the coordinates of these "million" hotels and calculate the distance one by one, that's why I thought it would be much easier to calculate out, how much 100 meters are in latitude and longitude (changes the value as we are on different locations, that' why I can't use simply the ones what I calculated out in the example above).
So if I would know how much 100 meters are in latitude and longitude for example at London's coordinate (if I'm there), I could simply get the list of the hotels what are (less than) 100 meters far from me, by a simple dividing:
if
((hotelLocation.coordinate.latitude <= (myLocation.coordinate.latitude + "100metersInLatitude")) || (hotelLocation.coordinate.latitude >= (myLocation.coordinate.latitude - "100metersInLatitude")))
&&
((hotelLocation.coordinate.longitude <= (myLocation.coordinate.longitude + "100metersInLongitude")) || (hotelLocation.coordinate.longitude >= (myLocation.coordinate.longitude - "100metersInLongitude")))
{
NSLog(#"Be Happy :-) !");
}
I just need these "100metersInLatitude" and "100metersInLongitude", calculated always from "myLocation".
Whoa, I hope, somebody will understand what I just wrote down, because it's not easy for me, neither... :-)))
Assuming you have a point with latitude and longitude, and you want to find another point that is a distance d on a bearing b, when the distance is small (you said "100 meters" which is very small on the surface of the earth) then you can do a simple approximation - treating the earth's surface locally as "flat". Here is a simple C program that implements this (using the numbers you had above). I updated it to include the "accurate" formulation as well - it's just a few more calculations, but it is accurate at all distances (and not just the short ones). The equation I used came from the link you referenced - subheading "Destination point given distance and bearing from start point"
updated - I moved the accurate calculation into a separate function, and added a loop to compute the new point for all integer bearings from 0 to 359, printing out every 30th. This give you the "circle" I talked about in my initial comment.
#include <stdio.h>
#include <math.h>
double radians(double x) {
return acos(0.0) * x / 90.0;
}
void calcNewPosition(double lat, double lon, double bearing, double d, double *newLat, double *newLon) {
double lat1, lon1, br, pi;
double Re = 6371000;
// convert everything to radians first:
lat1 = radians(lat);
lon1 = radians(lon);
br = radians(bearing);
pi = 2 * acos(0.0);
double lat2, lon2;
lat2 = asin( sin(lat1) * cos(d/Re) +
cos( lat1 ) * sin( d / Re ) * cos(br ) );
lon2 = lon1 + atan2(sin(br) * sin( d / Re ) * cos( lat1 ), \
cos( d / Re ) - sin(lat1 ) * sin( lat2 ) );
*newLat = 180. * lat2 / pi;
*newLon = 180. * lon2 / pi;
}
int main(void) {
double lon = 47., lat=19.;
double newLongitude, newLatitude;
double dx, dy, dLong, dLat;
double Re = 6371000, d = 100, bearing = 0.0;
double pi;
double lat1, lon1, br;
// convert everything to radians first:
lat1 = radians(lat);
lon1 = radians(lon);
br = radians(bearing);
pi = 2 * acos(0.0);
// approximate calculation - using equirectangular approximation
// and noting that distance between meridians (lines of longitude)
// get closer at higher latitudes, with cos(latitude).
dx = d * sin(br); // distance in E-W direction
dy = d * cos(br); // distance in N-S direction
dLat = 360 * dy / (2.0 * pi * Re); // convert N-S to degrees latitude
dLong = 360 * dx / (2.0 * pi * Re * cos(lat1)); // convert E-W to degrees longitude
newLatitude = lat + dLat;
newLongitude = lon + dLong;
printf("simple forumula: the new position is %.8lf lon, %.8lf lat\n", newLongitude, newLatitude);
// more accurate formula: based on http://www.movable-type.co.uk/scripts/latlong.html
double lat2, lon2;
calcNewPosition(lat, lon, bearing, d, &lat2, &lon2);
printf("more accurate: the new position is %.8lf lon, %.8lf lat\n", lon2, lat2);
// now loop over all bearings and compute the "circle of points":
int iBearing;
double lonArray[360], latArray[360];
for(iBearing = 0; iBearing < 360; iBearing++) {
calcNewPosition(lat, lon, (double)iBearing, d, &latArray[iBearing], &lonArray[iBearing]);
if (iBearing % 30 == 0) printf("bearing %03d: new lat = %.8lf, new lon = %.8lf\n", iBearing, latArray[iBearing], lonArray[iBearing]);
}
return 0;
}
The output of this is
simple forumula: the new position is 47.00000000 lon, 19.00089932 lat
more accurate: the new position is 47.00000000 lon, 19.00089932 lat
bearing 000: new lat = 19.00089932, new lon = 47.00000000
bearing 030: new lat = 19.00077883, new lon = 47.00047557
bearing 060: new lat = 19.00044966, new lon = 47.00082371
bearing 090: new lat = 19.00000000, new lon = 47.00095114
bearing 120: new lat = 18.99955034, new lon = 47.00082371
bearing 150: new lat = 18.99922116, new lon = 47.00047557
bearing 180: new lat = 18.99910068, new lon = 47.00000000
bearing 210: new lat = 18.99922116, new lon = 46.99952443
bearing 240: new lat = 18.99955034, new lon = 46.99917629
bearing 270: new lat = 19.00000000, new lon = 46.99904886
bearing 300: new lat = 19.00044966, new lon = 46.99917629
bearing 330: new lat = 19.00077883, new lon = 46.99952443
As you can see, it is accurate to within a fraction of a meter (your code gave 19.0008995 - it is actually possible that your result was "wrong" in the last digit as these two methods agree to 8 significant digits even though they use different equations).
The question isn't really answerable if the OP wants a location a distance (100 meters) from the current location without a desired bearing being provided, there being an infinite number of points in a circle around the point.
So, this answer may or may not be what the OP wants, it is a way with CLLocation to calculate the distance between two points.
Create two CLLocation point and use the method Have you looked at theCLLocationmethod- (CLLocationDistance)distanceFromLocation:(const CLLocation *)location`.
CLLocation *location1 = [[CLLocation alloc] initWithLatitude:)latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:)latitude2 longitude:longitude2];
double distance = [location1 distanceFromLocation:location2];
The radius of the earth at equator = 6,371 km. The equator is divided into 360 degrees of longitude, so each degree at the equator represents approximately 111.32 km. Moving away from the equator towards a pole this distance decreases to zero at the pole.To calculate the distance at different latitudes multiply it by the cosine of the latitude
3 decimal places,0.001 degrees aproximates to
111.32 meters at equator
96.41meters at 30 degrees N/S
78.71 meters at 45 degrees N/S
55.66 meters at 60 degrees N/S
28.82 meters at 75 degrees N/S
For for small distances (100 meters) Pythagoras’ theorem can be used on an equirectangular projection to calculate distance. This is less complicated than Haversine or Spherical Law of Cosines.
var R = 6371; // km
lat/lng in radians
In pseudo code as I don't know Objective-C
var x = (lng2-lng1) * cos((lat1+lat2)/2);
var y = (lat2-lat1);
var d = sqrt(x*x + y*y) * R;
I'm looking for a smooth way to calculate the distance between two GPS Points, so I get the result like: "You have to go x meters up and y meters to the left - so I can work with a 2d-coordinate system, where I have my position as (0,0) and the other positions is showing the distance in (x, y) in meters from my position.
My idea was to calculate the distance between the points using the haversine formula. (This returns my hypotenuse)
In addition to that, I'm calculating the bearing between this two points. This is my alpha.
With this two values, I wanted to use basic trigonometry functions to resolve my problem.
So I tried to calculate:catheti_1 = sin(alpha) * hypotenuse, catheti_2 = cos(alpha) * hypotenuse.
Maybe I'm doing something wrong, but my results are useless at the moment.
So my question is: How can I calculate the distance in x and y direction between two GPS points?
I'm calculating alpha in the following procedure:
public static double bearingTo(GPSBean point1, GPSBean point2) {
double lat1 = Math.toRadians(point1.latitude);
double lat2 = Math.toRadians(point2.latitude);
double lon1 = Math.toRadians(point1.longitude);
double lon2 = Math.toRadians(point2.longitude);
double deltaLong = lon2 - lon1;
double y = Math.sin(deltaLong) * Math.cos(lat2);
double x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1)
* Math.cos(lat2) * Math.cos(deltaLong);
double bearing = Math.atan2(y, x);
return (Math.toDegrees(bearing) + 360) % 360;
}
I just implemented your code, using approximate coordinates of NYC and Boston as reference points, and implementing the Haversine formula as found at http://www.movable-type.co.uk/scripts/latlong.html (which you didn't show):
long1 = -71.02; lat1 = 42.33;
long2 = -73.94; lat2 = 40.66;
lat1 *=pi/180;
lat2 *=pi/180;
long1*=pi/180;
long2*=pi/180;
dlong = (long2 - long1);
dlat = (lat2 - lat1);
// Haversine formula:
R = 6371;
a = sin(dlat/2)*sin(dlat/2) + cos(lat1)*cos(lat2)*sin(dlong/2)*sin(dlong/2)
c = 2 * atan2( sqrt(a), sqrt(1-a) );
d = R * c;
When I run this code, I get d = 306, which agrees with the answer from the above site.
For the bearing I get 52 deg - again, close to what the site gave.
Without seeing the rest of your code it's hard to know why your answer is different.
Note: when the two points are close together, you could make all kinds of approximations, but this code should still work - the formula has good numerical stability because it's using the sin of the difference between longitudes, latitudes (rather than the difference of the sin).
Addendum:
Using your code for x, y (in your question), I get sensible values for the distance - agreeing with the "proper" answer to within 120 m (which isn't bad since one is a straight line approximation and the other follows the curvature of the earth). So I think your code is basically OK now you fixed the typo.
Use Haversine formula to Calculate distance (in km) between two points specified by latitude/longitude (in numeric degrees)
from: Haversine formula - R. W. Sinnott, "Virtues of the Haversine"
Sky and Telescope, vol 68, no 2, 1984
http://www.census.gov/cgi-bin/geo/gisfaq?Q5.1
Example usage from form:
result.value = LatLon.distHaversine(lat1.value.parseDeg(), long1.value.parseDeg(), * lat2.value.parseDeg(), long2.value.parseDeg());
Javascript :
LatLon.distHaversine = function(lat1, lon1, lat2, lon2) {
var R = 6371; // earth's mean radius in km
var dLat = (lat2-lat1).toRad();
var dLon = (lon2-lon1).toRad();
lat1 = lat1.toRad(), lat2 = lat2.toRad();
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1) * Math.cos(lat2) * Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return d;
}
If anybody is interested to have a simpler formula that anyone can understand. Here is mine, it works for Sweden, but you can adapt it to work anywhere by making a more general formula for calculation of longfactor.
Hope you can understand even if it is written in an odd language.
<gpsDist lat1,long1,lat2,long2> all parameters in 1/100000 degree.
Example: <getDist 5950928,1327120,5958505,1302241> => 16303
Same at https://gps-coordinates.org/distance-between-coordinates.php => 16.35 KM.
<var $latFactor,1.112>
<function getDist,
-<var $longFactor,<calc 0.638 - ($1/100000-55)*0.0171,3>>
-<var $latDist,<calc ($3-$1)*$latFactor>>
-<var $longDist,<calc ($4-$2)*$longFactor>>
-<sqrt $latDist*$latDist + $longDist*$longDist>
->
/Bertil Friman
Good day, I'm fairly new to Objective C Dev and am enquiring to an implementation of the midpoint formula located here http://www.movable-type.co.uk/scripts/latlong.html.
Formula:
Bx = cos(lat2).cos(Δlong)
By = cos(lat2).sin(Δlong)
latm = atan2(sin(lat1) + sin(lat2), √((cos(lat1)+Bx)² + By²))
lonm = lon1 + atan2(By, cos(lat1)+Bx)
My Implementation of this formula in Objective C is.
- (CLLocationCoordinate2D) getMidPointCoords
{
double dLon = (self.toCoordinate.longitude - self.fromCoordinate.longitude) * (M_PI/180);
double Bx = cos(self.toCoordinate.latitude)*cos(dLon);
double By = cos(self.toCoordinate.latitude)*sin(dLon);
double latM = atan2(sin(self.fromCoordinate.latitude)+sin(self.toCoordinate.latitude), sqrt( (cos(self.fromCoordinate.latitude)+Bx)*(cos(self.fromCoordinate.latitude)+Bx) + By*By) );
double lonM = self.fromCoordinate.longitude + atan2(By, cos(self.fromCoordinate.latitude) + Bx);
CLLocationCoordinate2D midPoint;
midPoint.latitude = latM;
midPoint.longitude = lonM;
}
When I debug this code though it is clearly returning the incorrect coordinates.
So essentially my Question is "Is this because of the fact that I am using doubles?" or is my implementation of this formula simply flauwed?
Thank you in advance to any assistance, or insight provided.
The trigonometric functions you are using (sin, cos, atan2) require their parameter to be in radians, not degrees.
For example, in this line:
double Bx = cos(self.toCoordinate.latitude)*cos(dLon);
the cos(self.toCoordinate.latitude) is wrong because self.toCoordinate.latitude is in degrees but cos requires radians.
Everywhere that you call a trigonometric function, first convert the parameter from degrees to radians (multiply degrees by (M_PI/180.0)).
In addition, in this line:
double lonM = self.fromCoordinate.longitude + atan2(...
the self.fromCoordinate.longitude also needs to be converted to radians because the rest of the expression is also in radians.
Finally, at the end latM and lonM will have the midpoint but in radians (not degrees).
So when setting midPoint.latitude and midPoint.longitude, you have to convert latM and lonM from radians back to degrees by multiplying them by (180.0/M_PI).
Let me just start with the code.
- (NSPoint*) pointFromPoint:(NSPoint*)point withDistance:(float)distance towardAngle:(float)angle; {
float newX = distance * cos(angle);
float newY = distance * sin(angle);
NSPoint * anNSPoint;
anNSPoint.x = newX;
anNSPoint.y = newY;
return thePoint;
}
This should, based on my knowledge, be perfect. It should return and x value of 0 and a y value of 2 if I call this code.
somePoint = [NSPoint pointFromPoint:somePoint withDistance:2 towardAngle:90];
Instead, I get and x value of 1.05 and a y of 1.70. How can I find the x and y coordinates based on an angle and a distance?
Additional note: I have looked on math.stackexchange.com, but the formulas there led me to this. I need the code, not the normal math because I know I will probably screw this up.
A working version of your function, which accepts values in degrees instead of radians, would look like this:
- (NSPoint)pointFromPoint:(NSPoint)origin withDistance:(float)distance towardAngle:(float)angle
{
double radAngle = angle * M_PI / 180.0;
return NSMakePoint(origin.x + distance * cos(radAngle), point.y + distance * sin(radAngle));
}
Your problem is you're giving the angle in degrees (e.g. 90), but the math is expecting it in radians. Try replacing the 90 with M_PI_2
I have some problems figuring out where my error is. I got the following:
Have an image and corresponding GPS coordinates of its top-left and bottom-right vertices.
E.g:
topLeft.longitude = 8.235128;
topLeft.latitude = 49.632383;
bottomRight.longitude = 8.240547;
bottomRight.latitude = 49.629808;
Now a have an Point that lies in that map:
p.longitude = 8.238567;
p.latitude = 49.630664;
I draw my image in landscape fullscreen (1024*748).
Now I want to calculate the exact Pixel position (x,y) of my point.
For doing that I am trying to use the great circle distance approach from here: Link.
CGFloat DegreesToRadians(CGFloat degrees)
{
return degrees * M_PI / 180;
};
- (float) calculateDistanceP1:(CLLocationCoordinate2D)p1 andP2:(CLLocationCoordinate2D)p2 {
double circumference = 40000.0; // Erdumfang in km am Äquator
double distance = 0.0;
double latitude1Rad = DegreesToRadians(p1.latitude);
double longitude1Rad = DegreesToRadians(p1.longitude);
double latititude2Rad = DegreesToRadians(p2.latitude);
double longitude2Rad = DegreesToRadians(p2.longitude);
double logitudeDiff = fabs(longitude1Rad - longitude2Rad);
if (logitudeDiff > M_PI)
{
logitudeDiff = 2.0 * M_PI - logitudeDiff;
}
double angleCalculation =
acos(sin(latititude2Rad) * sin(latitude1Rad) + cos(latititude2Rad) * cos(latitude1Rad) * cos(logitudeDiff));
distance = circumference * angleCalculation / (2.0 * M_PI);
NSLog(#"%f",distance);
return distance;
}
Here is my code for getting the Pixel position:
- (CGPoint) calculatePoint:(CLLocationCoordinate2D)point {
float x_coord;
float y_coord;
CLLocationCoordinate2D x1;
CLLocationCoordinate2D x2;
x1.longitude = p.longitude;
x1.latitude = topLeft.latitude;
x2.longitude = p.longitude;
x2.latitude = bottomRight.latitude;
CLLocationCoordinate2D y1;
CLLocationCoordinate2D y2;
y1.longitude = topLeft.longitude;
y1.latitude = p.latitude;
y2.longitude = bottomRight.longitude;
y2.latitude = p.latitude;
float distanceX = [self calculateDistanceP1:x1 andP2:x2];
float distanceY = [self calculateDistanceP1:y1 andP2:y2];
float distancePX = [self calculateDistanceP1:x1 andP2:p];
float distancePY = [self calculateDistanceP1:y1 andP2:p];
x_coord = fabs(distancePX * (1024 / distanceX))-1;
y_coord = fabs(distancePY * (748 / distanceY))-1;
return CGPointMake(x_coord,y_coord);
}
x1 and x2 are the points on the longitude of p and with latitude of topLeft and bottomRight.
y1 and y2 are the points on the latitude of p and with longitude of topLeft and bottomRight.
So I got the distance between left and right on longitude of p and distance between top and bottom on latitude of p. (Needed for calculate the pixel position)
Now I calculate the distance between x1 and p (my distance between x_0 and x_p) after that I calculate the distance between y1 and p (distance between y_0 and y_p)
Last but not least the Pixel position is calculated and returned.
The Result is, that my point is on the red and NOT on the blue position:
Maybe you find any mistakes or have any suggestions for improving the accuracy.
Maybe I didn't understand your question, but shouldn't you be using the Converting Map Coordinates methods of MKMapView?
See this image
I used your co-ordinates, and simply did the following:
x_coord = 1024 * (p.longitude - topLeft.longitude)/(bottomRight.longitude - topLeft.longitude);
y_coord = 748 - (748 * (p.latitude - bottomRight.latitude)/(topLeft.latitude - bottomRight.latitude));
The red dot markes this point. For such small distances you don't really need to use great circles, and your rounding errors will be making things much more inaccurate