Given two gps coordinates the length difference can be calculated by using the haversine formula. But what about the other way around:
Compute the length difference in meter for a given Lat/Long double
Compute the Lat/Long double for a given length in meters
I know this is not exactly possible since it differs from the point on the earth you are, but is it possible to approximate this or something similiar? This does not have to be very precise.
If your displacements aren't too great (less than a few KM), use the quick and dirty estimate that 111,111 meters in the y direction is 1 degree (of latitude) and 111,111 * cos(latitude) meters in the x direction is 1 degree (of longitude).
Alternatively:
//Position, decimal degrees
lat = 51.0
lon = 0.0
//Earth’s radius, sphere
R=6378137
//offsets in meters
distanceNorth = 100
distanceEast = 100
//Coordinate offsets in radians
dLat = distanceNorth/R
dLon = distanceEast/(R*Cos(Pi*lat/180))
//OffsetPosition, decimal degrees
latO = lat + dLat * 180/Pi
lonO = lon + dLon * 180/Pi
This should return:
latO = 51,00089832
lonO = 0,001427437
Related
I have a longitude and latitude stored as a geometry with SRID 4326.
I want to make a line that is exactly 1000 meters long that is 90 degrees (to the right).
I know that the conversion from a longitudal/latitudal degree to a meter varies about where you are on the globe. That is why I will pass a reference long/lat that can be taken into consideration.
I am looking for something "good enough" that assumes that the distance you want will be no greater than 100 miles.
Given a long/lat and a meter distance of 1000 meters, return to me the size of the line long/lat degrees.
This question is specific to one example but I am looking for a general solution, because I have many functions that work upon "SRID Units" and the SRID I work with is long/lat (4326) but I always want to deal with meters and not degrees.
The hope is that I can call scaler function to convert the meters I want to the 4326 units.
Some hacks I have considered for finding X meters is converting the geometry into a geography and using the STBuffer(X) to make a circle with a radius equal to that of X, then create a long line that intersects the buffer and to find the point of the intersection, which will be the long/lat of exactly X meters away. This seems very hacky and inefficient but might be the most accurate
Edit:
To find the deltaX and deltaY the function is doing the the following:
DeltaX = Cos(#AngleRads) * #DistanceInDegrees;
DeltaY = Sin(#AngleRads) * #DistanceInDegrees;
I can supply #AngleRads and #DistanceInDegrees.
The problem is that #DistanceInDegrees has to match 4326 units (degrees). Is it even possible to find a #DistanceInDegrees that will correspond to 1000 meters no matter what angle is given?
When I use the formula
#DistanceInDegrees = (#Meters / 6371008.7714) * (180 / pi()) / cos(#Lat1 * pi()/180) and a angle of 90 degrees, then the length of the line is 1002 meters, close but not identically 1000.. If I use a degree of 45 the length of the line is 1191.67 meters.
If I understand your question, this can be done with a little math.
This also assumes the mean radius of the earth to be 6,371,008.7714 meters.
Example
Declare #Lat1 float = -37.786570
Declare #Lng1 float = 145.178179
Declare #Meters float = -1000
Select Lat1 = #Lat1
,Lng1 = #Lng1
,Lat2 = #Lat1
,Lng2 = #Lng1 + (#Meters / 6371008.7714) * (180 / pi()) / cos(#Lat1 * pi()/180)
Returns
The results can be validated here
After your EDIT, It seems you are looking for DISTANCE and BEARING
Here is a Table-Valued Function which may help
Example
Select * from [dbo].[tvf-Geo-Distance-Bearing](-37.786570,145.178179,1000,-45)
Returns
RetLat RetLng
-37.7802105711301 145.170133170589
Now, when I calculate the distance, I get 999.99999448737 meters
The Table-Valued Function If Interested
CREATE Function [dbo].[tvf-Geo-Distance-Bearing] (#Lat1 float,#Lng1 float,#Dist Float,#Degr float)
Returns #LatLng Table (RetLat float,RetLng Float)
As
Begin
Declare #Lat2 float,#Lng2 float,#eRad float = 6371008.7714
Select #Lat1 = RADIANS(#Lat1)
,#Lng1 = RADIANS(#Lng1)
,#Degr = RADIANS(#Degr)
Set #Lat2 = Asin(Sin(#Lat1) * Cos(#Dist / #eRad ) + Cos(#Lat1) * Sin(#Dist / #eRad ) * Cos(#Degr ))
Set #Lng2 = #Lng1 + Atn2(Sin(#Degr) * Sin(#Dist / #eRad ) * Cos(#Lat1), Cos(#Dist / #eRad ) - Sin(#Lat1) * Sin(#Lat2))
Insert Into #LatLng
Select Degrees(#Lat2)
,Degrees(#Lng2)
Return;
End
I want to overlay great circle arcs between airports on a map using longitudes and latitudes.
I can already get the distance and bearing from the initial and final coordinates but now I need to produce the points on the curve to plot through.
What I would like is a formula which takes the origin, destination, and a distance, and returns the latitude/longitude of the point that lies at that distance on the path between the origin and the destination.
I'm currently approximating the earth by a sphere and using radians -- eventually I'll add in spheroid corrections.
currlat = oldlat + d * sin (angle)/ (radius);
currlon = oldlon + d * cos (angle)/ (radius * cos(oldlat));
where d is distance travelled and angle is in radians. This is assuming circumference of earth at 40000km both at equator and through the poles. You can convert in radians...
Also it assumes the angle (direction) is with reference to equator line.
Obviously this needs spheroid corrections.
if you go south sin values will turn negative and north it will go positive. If you go west cos will turn negative and east it will turn positive.
d * sin(angle) and d * cos(angle) gives you the change. and you just calculate the new lat/long on that basis scaling up against circumference of earth.
I have a CLLocationCoordinate2D (c1) and a CLLocation (l1), so I have lat/long values for each point and I can calculate the distance in meters between them using:
[c1 distanceFromLocation:l1]
How can I find the coordinates of a point (c2) 100 meters closer to l1 than c1 (along the same bearing)?
I have calculated it using basic trig using the following:
used the difference in latitude and longitude to calculate the hypotenuse and angle
used the ratio between the distance to cl and the distance to c2 to get the hypotenuse of a triangle ending at c2
used cos and sin to calculate the longitude and latitude of c2
But this seems like a hacky way of doing it as it doesn't take into account of curvature and seems to be using latitude and longitude in a way they are not supposed to be used. It does seem to work over short distances though.
After a bit of research, I found a basic formula for calculating curved distance between two points on the earth's surface:
dlon = lon2 - lon1
dlat = lat2 - lat1
a = { sin(dlat/2) }^2 + [ cos(lat1) * cos(lat2) * { sin(dlon/2) }^2 ]
c = 2 * arcsin(min(1,sqrt(a)))
d = R * c
Although Rickay's answer was helpful, I eventually used the following excellent library which has many useful functions for Core Location calculations: https://github.com/100grams/CoreLocationUtils
I have latitude and longitude of a point.I have to find out the latitude and longitude of another point from a relative distance from the known point.For example point A has some location with latitude and longitude.What is the latitude and longitude after moving 1000m south and 500m west from point A.Is there any direct equation to find this? Thanks in advance
Note the accepted answer is basically the flat earth projection equations:
x = δlon * EarthRadius * cos( lat )
y = δlat * EarthRadius
For better accuracy over larger distances, you should compute the final lat/lon from a typical bearing/range calculation. See the section Destination point given distance and bearing from start point at this website: http://www.movable-type.co.uk/scripts/latlong.html
Instead of looking up an equation you can calculate as follows. Let R be the radius of the Earth. Let a be the current latitude and b be the current longitude. Then if you move δx metres east (negative for west) then δy metres south, calculating the new longitude can be done as follows.
Intersecting a horizontal plane with the Earth at the current latitude will give you a circle of radius R*cos(a). So to convert δx to the change in longitude, you get something like
δlong = δx * 2π / (2π * R * cos(a)) = δx / (R * cos (a))
The change in latitude is easier, since it doesn't depend on the current position. You're always moving around a great circle through the two poles. Then δlat = δy / R. (Of course you need to mod out by 2 π at some point).
I have to see if some GPS coordinates are in a circle that I create.when I say I am creating that circle I' refering to this: I have lat1, long1,
my actual location and I want to test if there is any data surrounding this location but
not far then 1km. I am trying to use circle inequation : (x2-a2)2+(y2-b2)2 <R2 where a=lat1, b=long1 and R=radius. I know that R=1km but how do I transform 1km into data that can be compared to GPS coordinates? and x, y are every value from a collection that are testet to see if they fit.
Lat/Lng coordinates don't work with the Phytagoras Formula, as Earth is a sphere, and not flat...
Take a look at this manual:
http://www.movable-type.co.uk/scripts/latlong.html
You need the distance equation, to see if the distance between the lat/lng you have and the lat/lng of the center of the circle is smaller than R.
It's not that simple. You can't generally use the euclidean distance. This is how i think you could do it:
The correct way
The correct way to do this would be to calculate the great circle distance (the shortest walking distance on the surface), which I never really understood, but it's pretty easy to google (and adamk already gave you a link to it).
The lazy way
Assume that your points are not too close to the poles and that they are close to each other. Then you can use the latitude and longitude as if they were euclidean coordinates. (this is basically some cylindrical projection). 1 degree of latitude will then be (earth_circumference / 360) long and 1 degree of longitude will be (cos(lat) * earth_circumference / 360) long.
the complete code would look something like this:
double distance_lazy(double lat1, double lon1, double lat2, double lon2){
double xDist = (lat2 - lat1) * EARTH_CIRCUMFERENCE / 360.0;
double yDist = cos(lat1) * (lon2 - lon1) * EARTH_CIRCUMFERENCE / 360.0;
return sqrt(xDist^2 + yDist^2);
}
If your distances will be only a few km from each other, this should work pretty ok for the whole europe ... something like that.
The "weird definition" way
Another thing you can say is, that "distance" is the length straight line between the points, even though it goes through the earth. Then this would become calculating the 3D coordinates of a point and then their euclidean distance
double distance_straight_line(double lat1, double lon1, double lat2, double lon2){
double x1 = cos(lat1) * cos(lon1) * EARTH_RADIUS;
double y1 = sin(lat1) * cos(lon1) * EARTH_RADIUS;
double z1 = sin(lon1) * EARTH_RADIUS;
double x2 = cos(lat2) * cos(lon2) * EARTH_RADIUS;
double y2 = sin(lat2) * cos(lon2) * EARTH_RADIUS;
double z2 = sin(lon2) * EARTH_RADIUS;
return sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2);
}
Again this will work as expected for points close to each other, this time they can be anywhere in the world. If you give it points far from each other, the output will be correct, but prety useless (unless you are very good at digging).
Hope it helps.