Say I have a square which consists of four CLLocationCoordinate2D points, which are in lat, lon, and I want to find the area of the square in meters. I convert the CLLocationCoordinate2D points into MKMapPoints, and I find the area in X-Y space. However, the area I find is in the units of MKMapPoint, which don't directly translate to meters. How can I translate this area in MKMapPoint-space back into meters?
The MapKit function MKMetersBetweenMapPoints makes this easier.
For example, if you wanted to get the area of the currently displayed region:
MKMapPoint mpTopLeft = mapView.visibleMapRect.origin;
MKMapPoint mpTopRight = MKMapPointMake(
mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width,
mapView.visibleMapRect.origin.y);
MKMapPoint mpBottomRight = MKMapPointMake(
mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width,
mapView.visibleMapRect.origin.y + mapView.visibleMapRect.size.height);
CLLocationDistance hDist = MKMetersBetweenMapPoints(mpTopLeft, mpTopRight);
CLLocationDistance vDist = MKMetersBetweenMapPoints(mpTopRight, mpBottomRight);
double vmrArea = hDist * vDist;
The documentation states that the function takes "into account the curvature of the Earth."
You can use the Haversine formula to calculate it, assuming that the earth is a perfect sphere.
To understand how lat/lon vs meters works in the context of the earth, you may find it interesting to read about Nautical miles.
You can find some more resources and some sample code by googling objective-c Haversine formula.
Enjoy!
Related
Latitude: 22.744812,
Longitude: 75.892578
The above would be considered my center point.
And now I need to determine the latitude and longitude points from center point 1000 meter outward to each NSWE corners. So I would have a central long/lat, N, S, E and W long/lat..
So I would end up with 4 additional lat/long pairs.
What I am trying to resolve is a formula, preferably that can be done on a standard calculator to determine these 4 NSWE points based on the central point.
You could use MapKit for that:
- (CLLocationCoordinate2D *) calculateSquareCoordinates:(CLLocation*)center withRadius:(float)radius{
MKCoordinateRegion region = MKCoordinateRegionMakeWithDistance(center.coordinate, radius*2, radius*2);
CLLocationCoordinate2D points[4];
points[0] = CLLocationCoordinate2DMake(region.center.latitude - region.span.latitudeDelta/2, region.center.longitude - region.span.longitudeDelta/2);
points[1] = CLLocationCoordinate2DMake(region.center.latitude + region.span.latitudeDelta/2, region.center.longitude - region.span.longitudeDelta/2);
points[2] = CLLocationCoordinate2DMake(region.center.latitude + region.span.latitudeDelta/2, region.center.longitude + region.span.longitudeDelta/2);
points[3] = CLLocationCoordinate2DMake(region.center.latitude - region.span.latitudeDelta/2, region.center.longitude + region.span.longitudeDelta/2);
return points;
}
and just call
CLLocationCoordinate2D *fourPoints = [self calculateSquareCoordinates:center withRadius:1000];
on your code.
you will have to use the Haversine formula to calculate the Lat/Long based on distance from a starting Lat/Long. have a look at this Link
The average radius of the earth is around 6371000 metres. This means that
1 degree of lattitude is equivalent to 6371000 * PI / 180 metres
(NB: PI = 3.14159... etc). However, 1 degree of longitude depends on the lattitude that you are. At the equator, one degree of longitude corresponds to the same distance in metres as 1 degree of lattitude. However, at the north and south poles, all longitude values are the same point (i.e. the pole itself), so 1 degree of longitude at the poles is zero metres. The formula for longitude is
1 degree of longitude is equivalent to 637100 * PI / 180 * COS(Lattitude)
where COS is the trigonometric cosine function. If you make these conversions, then you can do the calculation on a standard calculator. However, be aware that these are approximations that work well over short distances (e.g. less than a few hundred kilometers), but over long distances (e.g. thousands of kilometers) they become more and more inaccurate.
On the iPhone, I get the user's location in decimal degrees, for example: latitude 39.470920 and longitude = -0.373192; That's point A.
I need to create a line with another GPS coordinate, also in decimal degrees, point B. Then, calculate the distance (perpendicular) between the line from A to B and another point C.
The problem is I get confused with the values in degrees. I would like to have the result in meters. What's the conversion needed? How will the final formula to compute this look like?
Why don't you use CLLocations distanceFromLocation: method? It will tell you the precise distance between the receiver and another CLLocation.
CLLocation *locationA = [[CLLocation alloc] initWithLatitude:12.123456 longitude:12.123456];
CLLocation *locationB = [[CLLocation alloc] initWithLatitude:21.654321 longitude:21.654321];
CLLocationDistance distanceInMeters = [locationA distanceFromLocation:locationB];
// CLLocation is aka double
[locationA release];
[locationB release];
It's as easy as that.
(CLLocationDistance)distanceFromLocation:(const CLLocation *)location is the method to get the distance from on CLLocation to another.
Your problem is also of finding the shortest line between a line (A,B) and point C.
I guess if your 3 CLLocations are near ( less than a few kilometers apart), you can do the math "as if" the coordinates are points on a single plane, and use this in C++, or this or this and just use the CLLocations "as if" they were x and y coordinates on a plane.
If your coordinates are far away, or exact accuracy is important then the spherical shape of the earth matters, and you need to do things using great circle distance and other geometry on the face of a sphere.
Swift 3.0+
Only calculate distance between two coordinates:
let distance = source.distance(from: destination)
When you have array of locations:
To get distance from array of points use below reduce method.
Here locations is array of type CLLocation.
let calculatedDistance = locations.reduce((0, locations[0])) { ($0.0 + $0.1.distance(from: $1), $1)}.0
Here you will get distance in meters.
I am looking for a way to calculate the distance between 2 points on the globe. We've been told to use Haversine, which works fine to calculate the shortest distance between the 2 points.
Now, I'd like to calculate the "long distance" between to points. So suppose you have 2 cities, A in the west and B in the east. I want to know the distance from B to A if I would travel eastwards around the globe and then reach A coming from the west.
I've tried changing a couple of things in the haversine function, but doesn't seem to work.
Anyone know how I can simply do this using small adjustments to the haversine function?
This is what I'm using now:
lat1, lat2, lng1, lng2 are in radians
part1 = sin(lat2) * sin(lat1);
part2 = cos(lat2) * cos(lat1) * cos(lng1 - lng2);
distance = 6378.8 * acos(part1 + part2);
The way I see it is that you can draw a circle around the globe between the 2 cities. The long distance the the circumference of that circle minus the short distance. But in contrary of what was replied, the circle's length is not equal to the earth's circumference. This is only the case for 2 points on the equator.
Tnx
Jeroen
The circumference of the earth is approx 40,075KM, work out the short distance and subtract it from that.
I wanted to test the mapKit and wanted to make my own overlay to display the accuracy of my position.
If i have a zoom factor of for example .005 which radius does my circle around me has to have(If my accuracy is for example 500m)?
Would be great to get some help :)
Thanks a lot.
Look at the documentation for MKCoordinateSpan, which is part of the map's region property. One degree of latitude is always approx. 111 km, so converting the latitudeDelta to meters and then getting to the meters per pixel should be easy. For longitudinal values it is not quite so easy as the distance covered by one degree of longitude varies between 111 km (at the equator) and 0 km (at the poles).
My way to get meters per pixel:
MKMapView *mapView = ...;
CLLocationCoordinate2D coordinate = ...;
MKMapRect mapRect = mapView.visibleMapRect;
CLLocationDistance metersPerMapPoint = MKMetersPerMapPointAtLatitude(coordinate.latitude);
CGFloat metersPerPixel = metersPerMapPoint * mapRect.size.width / mapView.bounds.size.width;
To add to another answer, a difference of one minute of latitude corresponds to one nautical mile: that's how the nautical mile was defined. So, converting to statute miles, 1 nautical mile = 1.1508 statue miles, or 6076.1 ft. or 1852 meters.
When you go to longitude, the size of the longitude circles around the Earth shrink as latitude increases, as was noted on the previous answer. The correct factor is that
1 minute of longitude = (1852 meters)*cos(theta),
where theta is the latitude.
Of course, the Earth is not a perfect sphere, but the simple calculation above would never be off by more than 1%.
as title how to? i have tried the code from google earth, but seem like the result is different with the google map calculation result. below provided the code i did
-(double)GetDistance:(double)lat1 long1:(double)lng1 la2:(double)lat2 long2:(double)lng2 {
//NSLog(#"latitude 1:%.7f,longitude1:%.7f,latitude2:%.7f,longtitude2:%.7f",lat1,lng1,lat2,lng2);
double radLat1 = [self rad:lat1];
double radLat2 = [self rad:lat2];
double a = radLat1 - radLat2;
double b = [self rad:lng1] -[self rad:lng2];
double s = 2 * asin(sqrt(pow(sin(a/2),2) + cos(radLat1)*cos(radLat2)*pow(sin(b/2),2)));
s = s * EARTH_RADIUS;
s = round(s * 10000) / 10000;
return s;
}
-(double)rad:(double)d
{
return d *3.14159265 / 180.0;
}
the EARTH_RADIUS value is 6378.138
by using this function by provided two coordinates the result come out is 4.5kM
but when i use google map get direction between two same coordinates, it show me the distance is about 8km
can anyone help to point out the problem of my code?
Since this is tagged iPhone, why not use the built-in distance function rather than rolling your own? location1 and location2 are CLLocation objects.
CLLocationDistance distance = [location1 getDistanceFrom:location2];
Here is a simple code (supposing you just have latitude and longitude of the two points)
CLLocation *startLocation = [[CLLocation alloc] initWithLatitude:startLatitude longitude:startLongitude];
CLLocation *endLocation = [[CLLocation alloc] initWithLatitude:endLatitude longitude:endLongitude];
CLLocationDistance distance = [startLocation distanceFromLocation:endLocation]; // aka double
Don't forget to add MapKit Framework to your project, and import MapKit in your file :
#import <MapKit/MapKit.h>
Google Maps is likely to be giving you the driving distance, whereas the great circle equation you have listed is going to be the straight line surface distance. If there was a straight line surface road directly from point A to point B, Google Maps would likely give you the same distance as the equation you have there.
Since
getDistanceFrom:
isDeprecated
Try use the
[newLocation distanceFromLocation:oldLocation
You should be able to use the google API directly to calculate either great circle distance or driving distance depending on your application needs.
See GLatLong::distanceFrom and GDirections::getDistance.