I want to create 2 new longitude and 2 new latitudes based on a coordinate and a distance in meters, I want to create a nice bounding box around a certain point. It is for a part of a city and max ±1500 meters. I therefore don't think the curvature of earth has to be taken into account.
So I have 50.0452345 (x) and 4.3242234 (y) and I want to know x + 500 meters, x - 500 meters, y - 500 meters, y + 500 meters
I found many algorithms but almost all seem to deal with the distance between points.
The number of kilometers per degree of longitude is approximately
(pi/180) * r_earth * cos(theta*pi/180)
where theta is the latitude in degrees and r_earth is approximately 6378 km.
The number of kilometers per degree of latitude is approximately the same at all locations, approx
(pi/180) * r_earth = 111 km / degree
So you can do:
new_latitude = latitude + (dy / r_earth) * (180 / pi);
new_longitude = longitude + (dx / r_earth) * (180 / pi) / cos(latitude * pi/180);
As long as dx and dy are small compared to the radius of the earth and you don't get too close to the poles.
The accepted answer is perfectly right and works. I made some tweaks and turned into this:
double meters = 50;
// 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)
//
// 111.32km = 111320.0m (".0" is used to make sure the result of division is
// double even if the "meters" variable can't be explicitly declared as double)
double coef = meters / 111320.0;
double new_lat = my_lat + coef;
// pi / 180 ~= 0.01745
double new_long = my_long + coef / Math.cos(my_lat * 0.01745);
Hope this helps too.
For latitude do:
var earth = 6378.137, //radius of the earth in kilometer
pi = Math.PI,
m = (1 / ((2 * pi / 360) * earth)) / 1000; //1 meter in degree
var new_latitude = latitude + (your_meters * m);
For longitude do:
var earth = 6378.137, //radius of the earth in kilometer
pi = Math.PI,
cos = Math.cos,
m = (1 / ((2 * pi / 360) * earth)) / 1000; //1 meter in degree
var new_longitude = longitude + (your_meters * m) / cos(latitude * (pi / 180));
The variable your_meters can contain a positive or a negative value.
I had to spend about two hours to work out the solution by #nibot , I simply needed a method to create a boundary box given its center point and width/height (or radius) in kilometers:
I don't fully understand the solution mathematically/ geographically.
I tweaked the solution (by trial and error) to get the four coordinates. Distances in km, given the current position and distance we shift to the new position in the four coordinates:
North:
private static Position ToNorthPosition(Position center, double northDistance)
{
double r_earth = 6378;
var pi = Math.PI;
var new_latitude = center.Lat + (northDistance / r_earth) * (180 / pi);
return new Position(new_latitude, center.Long);
}
East:
private static Position ToEastPosition(Position center, double eastDistance)
{
double r_earth = 6378;
var pi = Math.PI;
var new_longitude = center.Long + (eastDistance / r_earth) * (180 / pi) / Math.Cos(center.Lat * pi / 180);
return new Position(center.Lat, new_longitude);
}
South:
private static Position ToSouthPosition(Position center, double southDistance)
{
double r_earth = 6378;
var pi = Math.PI;
var new_latitude = center.Lat - (southDistance / r_earth) * (180 / pi);
return new Position(new_latitude, center.Long);
}
West:
private static Position ToWestPosition(Position center, double westDistance)
{
double r_earth = 6378;
var pi = Math.PI;
var new_longitude = center.Long - (westDistance / r_earth) * (180 / pi) / Math.Cos(center.Lat * pi / 180);
return new Position(center.Lat, new_longitude);
}
Have you checked out: How do I find the lat/long that is x km north of a given lat/long ?
These calculations are annoying at best, I've done many of them. The haversine formula will be your friend.
Some reference: http://www.movable-type.co.uk/scripts/latlong.html
Posting this method for sake of completeness.
Use this method "as it is" to:
Move any (lat,long) point by given meters in either axis.
Python method to move any point by defined meters.
def translate_latlong(lat,long,lat_translation_meters,long_translation_meters):
''' method to move any lat,long point by provided meters in lat and long direction.
params :
lat,long: lattitude and longitude in degrees as decimal values, e.g. 37.43609517497065, -122.17226450150885
lat_translation_meters: movement of point in meters in lattitude direction.
positive value: up move, negative value: down move
long_translation_meters: movement of point in meters in longitude direction.
positive value: left move, negative value: right move
'''
earth_radius = 6378.137
#Calculate top, which is lat_translation_meters above
m_lat = (1 / ((2 * math.pi / 360) * earth_radius)) / 1000;
lat_new = lat + (lat_translation_meters * m_lat)
#Calculate right, which is long_translation_meters right
m_long = (1 / ((2 * math.pi / 360) * earth_radius)) / 1000; # 1 meter in degree
long_new = long + (long_translation_meters * m_long) / math.cos(lat * (math.pi / 180));
return lat_new,long_new
Working Python code to offset coordinates by 10 metres.
def add_blur(lat, long):
meters = 10
blur_factor = meters * 0.000006279
new_lat = lat + blur_factor
new_long = long + blur_factor / math.cos(lat * 0.018)
return new_lat, new_long
if you don't have to be very exact then: each 10000 meters is about 0.1 for latitude and longitude.
for example I want to load locations 3000 meters around point_A from my database:
double newMeter = 3000 * 0.1 / 10000;
double lat1 = point_A.latitude - newMeter;
double lat2 = point_A.latitude + newMeter;
double lon1 = point_A.longitude - newMeter;
double lon1 = point_A.longitude + newMeter;
Cursor c = mDb.rawQuery("select * from TABLE1 where lat >= " + lat1 + " and lat <= " + lat2 + " and lon >= " + lon1 + " and lon <= " + lon2 + " order by id", null);
public double MeterToDegree(double meters, double latitude)
{
return meters / (111.32 * 1000 * Math.Cos(latitude * (Math.PI / 180)));
}
var meters = 50;
var coef = meters * 0.0000089;
var new_lat = map.getCenter().lat.apply() + coef;
var new_long = map.getCenter().lng.apply() + coef / Math.cos(new_lat * 0.018);
map.setCenter({lat:new_lat, lng:new_long});
See from Official Google Maps Documentation (link below) as they solve on easy/simple maps the problems with distance by countries :)
I recommended this solution to easy/simply solve issue with boundaries that you can know which area you're solving the problem with boundaries (not recommended globally)
Note:
Latitude lines run west-east and mark the position south-north of a point. Lines of latitude are called parallels and in total there are 180 degrees of latitude. The distance between each degree of latitude is about 69 miles (110 kilometers).
The distance between longitudes narrows the further away from the equator. The distance between longitudes at the equator is the same as latitude, roughly 69 miles (110 kilometers) . At 45 degrees north or south, the distance between is about 49 miles (79 kilometers). The distance between longitudes reaches zero at the poles as the lines of meridian converge at that point.
Original source 1
Original source 2
Official Google Maps Documentation: Code Example: Autocomplete Restricted to Multiple Countries
See the part of their code how they solve problem with distance center + 10 kilometers by +/- 0.1 degree
function initMap(): void {
const map = new google.maps.Map(
document.getElementById("map") as HTMLElement,
{
center: { lat: 50.064192, lng: -130.605469 },
zoom: 3,
}
);
const card = document.getElementById("pac-card") as HTMLElement;
map.controls[google.maps.ControlPosition.TOP_RIGHT].push(card);
const center = { lat: 50.064192, lng: -130.605469 };
// Create a bounding box with sides ~10km away from the center point
const defaultBounds = {
north: center.lat + 0.1,
south: center.lat - 0.1,
east: center.lng + 0.1,
west: center.lng - 0.1,
};
const input = document.getElementById("pac-input") as HTMLInputElement;
const options = {
bounds: defaultBounds,
componentRestrictions: { country: "us" },
fields: ["address_components", "geometry", "icon", "name"],
origin: center,
strictBounds: false,
types: ["establishment"],
};
This is what I did in VBA that seems to be working for me. Calculation is in feet not meters though
Public Function CalcLong(OrigLong As Double, OrigLat As Double, DirLong As String, DirLat As String, DistLong As Double, DistLat As Double)
Dim FT As Double
Dim NewLong, NewLat As Double
FT = 1 / ((2 * WorksheetFunction.Pi / 360) * 20902230.971129)
If DirLong = "W" Then
NewLat = CalcLat(OrigLong, OrigLat, DirLong, DirLat, DistLong, DistLat)
NewLong = OrigLong - ((FT * DistLong) / Cos(NewLat * (WorksheetFunction.Pi / 180)))
CalcLong = NewLong
Else
NewLong = OrigLong + ((FT * DistLong) / Math.Cos(CalcLat(OrigLong, OrigLat, DirLong, DirLat, DistLong, DistLat) * (WorksheetFunction.Pi / 180)))
CalcLong = NewLong
End If
End Function
Public Function CalcLat(OrigLong As Double, OrigLat As Double, DirLong As String, DirLat As String, DistLong As Double, DistLat As Double) As Double
Dim FT As Double
Dim NewLat As Double
FT = 1 / ((2 * WorksheetFunction.Pi / 360) * 20902230.971129)
If DirLat = "S" Then
NewLat = (OrigLat - (FT * DistLat))
CalcLat = NewLat
Else
NewLat = (OrigLat + (FT * DistLat))
CalcLat = NewLat
End If
End Function
Original poster said:
"So I have 50.0452345 (x) and 4.3242234 (y) and I want to know x + 500 meters..."
I will assume the units of the x and y values he gave there were in meters (and not degrees Longitude, Latitude). If so then he is stating measurements to 0.1 micrometer, so I will assume he needs similar accuracy for the translated output. I also will assume by "+500 meters" etc. he meant
the direction to be due North-South and due East-West.
He refers to a reference point:
"2 new latitudes based on a coordinate";
but he did not give the Longitude and Latitude,
so to explain the procedure concretely I will give
the Latitudes and Longitudes for the corners of the
500 meter box he requested around the point
[30 degrees Longitude,30 degrees Latitude].
The exact solution on the surface of the GRS80 Ellipsoid is
given with the following set of functions
(I wrote these for the free-open-source-mac-pc math program called "PARI"
which allows any number of digits precision to be setup):
\\=======Arc lengths along Latitude and Longitude and the respective scales:
dms(u)=[truncate(u),truncate((u-truncate(u))*60),((u-truncate(u))*60-truncate((u-truncate(u))*60))*60];
SpinEarthRadiansPerSec=7.292115e-5;\
GMearth=3986005e8;\
J2earth=108263e-8;\
re=6378137;\
ecc=solve(ecc=.0001,.9999,eccp=ecc/sqrt(1-ecc^2);qecc=(1+3/eccp^2)*atan(eccp)-3/eccp;ecc^2-(3*J2earth+4/15*SpinEarthRadiansPerSec^2*re^3/GMearth*ecc^3/qecc));\
e2=ecc^2;\
b2=1-e2;\
b=sqrt(b2);\
fl=1-b;\
rfl=1/fl;\
U0=GMearth/ecc/re*atan(eccp)+1/3*SpinEarthRadiansPerSec^2*re^2;\
HeightAboveEllipsoid=0;\
reh=re+HeightAboveEllipsoid;\
longscale(lat)=reh*Pi/648000/sqrt(1+b2*(tan(lat))^2);
latscale(lat)=reh*b*Pi/648000/(1-e2*(sin(lat))^2)^(3/2);
longarc(lat,long1,long2)=longscale(lat)*648000/Pi*(long2-long1);
latarc(lat1,lat2)=(intnum(th=lat1,lat2,sqrt(1-e2*(sin(th))^2))+e2/2*sin(2*lat1)/sqrt(1-e2*(sin(lat1))^2)-e2/2*sin(2*lat2)/sqrt(1-e2*(sin(lat2))^2))*reh;
\\=======
I then plugged the reference point [30,30]
into those functions at the PARI command prompt
and had PARI solve for the point +/- 500 meters away
from it, giving the two new Longitudes and
two new Latitudes that the original poster asked for.
Here is the input and output showing that:
? dms(solve(x=29,31,longarc(30*Pi/180,30*Pi/180,x*Pi/180)+500))
cpu time = 1 ms, real time = 1 ms.
%1172 = [29, 59, 41.3444979398934670450280297216509190843055]
? dms(solve(x=29,31,longarc(30*Pi/180,30*Pi/180,x*Pi/180)-500))
cpu time = 1 ms, real time = 1 ms.
%1173 = [30, 0, 18.6555020601065329549719702783490809156945]
? dms(solve(x=29,31,latarc(30*Pi/180,x*Pi/180)+500))
cpu time = 1,357 ms, real time = 1,358 ms.
%1174 = [29, 59, 43.7621925447500548285775757329518579545513]
? dms(solve(x=29,31,latarc(30*Pi/180,x*Pi/180)-500))
cpu time = 1,365 ms, real time = 1,368 ms.
%1175 = [30, 0, 16.2377963202802863245716034907838199823349]
?
Is there a Java operator that would create a value equal to count but in fewer lines?
double theta = //Some value;
int count = 0;
while (theta > 0) {
theta -= pi * (1.0 / 8.0);
count += 1;
}
You've just implemented division by repeated subtraction.
So, if you had actual real numbers in Java, you could do ...
int result = (int) (theta / (Math.PI / 8)) + 1;
... to get the same result. However due to repeated rounding errors in your code, that has many more steps than a simple division, the answers will diverge a bit. The results will be the same for thetas up to around 55 million.
The rounding errors are there because float and double are not accurate representations of real numbers.
See also: Is floating point math broken?
Assuming theta > 0, something like this should work:
int count = (int)Math.ceil (theta/(Math.PI * (1.0 / 8.0)));
or
int count = (int)Math.ceil (theta*8/Math.PI);
How about this: Math.ceil( theta/(pi * (1.0 / 8.0)))?
I have the following code:
String price="9000.89";
double basePrice =10000;
String dis= String.valueOf(((basePrice - Double
.parseDouble(price)) * 100 / basePrice));
System.out.println(dis);
It is giving me the output as "9.991100000000007", but i need it in 2 decimal points. Any idea how to do that?
Expected output : 9.99
String price="9000.89";
double basePrice =10000;
String dis = String.format("%.2f",((basePrice - Double.parseDouble(price)) * 100 / basePrice));
System.out.println(dis);
I would recommend
double dis = ((basePrice - Double.parseDouble(price)) * 100 / basePrice);
System.out.format("%1.2f", dis);
You can round using:
Math.round(number * 100.0) / 100.0;
I have two Lines: L1 and L2. I want to calculate the angle between the two lines. L1 has points: {(x1, y1), (x2, y2)} and L2 has points: {(x3, y3), (x4, y4)}.
How can I calculate the angle formed between these two lines, without having to calculate the slopes? The problem I am currently having is that sometimes I have horizontal lines (lines along the x-axis) and the following formula fails (divide by zero exception):
arctan((m1 - m2) / (1 - (m1 * m2)))
where m1 and m2 are the slopes of line 1 and line 2 respectively. Is there a formula/algorithm that can calculate the angles between the two lines without ever getting divide-by-zero exceptions? Any help would be highly appreciated.
This is my code snippet:
// Calculates the angle formed between two lines
public static double angleBetween2Lines(Line2D line1, Line2D line2)
{
double slope1 = line1.getY1() - line1.getY2() / line1.getX1() - line1.getX2();
double slope2 = line2.getY1() - line2.getY2() / line2.getX1() - line2.getX2();
double angle = Math.atan((slope1 - slope2) / (1 - (slope1 * slope2)));
return angle;
}
Thanks.
The atan2 function eases the pain of dealing with atan.
It is declared as double atan2(double y, double x) and converts rectangular coordinates (x,y) to the angle theta from the polar coordinates (r,theta)
So I'd rewrite your code as
public static double angleBetween2Lines(Line2D line1, Line2D line2)
{
double angle1 = Math.atan2(line1.getY1() - line1.getY2(),
line1.getX1() - line1.getX2());
double angle2 = Math.atan2(line2.getY1() - line2.getY2(),
line2.getX1() - line2.getX2());
return angle1-angle2;
}
Dot product is probably more useful in this case. Here you can find a geometry package for Java which provides some useful helpers. Below is their calculation for determining the angle between two 3-d points. Hopefully it will get you started:
public static double computeAngle (double[] p0, double[] p1, double[] p2)
{
double[] v0 = Geometry.createVector (p0, p1);
double[] v1 = Geometry.createVector (p0, p2);
double dotProduct = Geometry.computeDotProduct (v0, v1);
double length1 = Geometry.length (v0);
double length2 = Geometry.length (v1);
double denominator = length1 * length2;
double product = denominator != 0.0 ? dotProduct / denominator : 0.0;
double angle = Math.acos (product);
return angle;
}
Good luck!
dx1 = x2-x1;
dy1 = y2-y1;
dx2 = x4-x3;
dy2 = y4-y3;
d = dx1*dx2 + dy1*dy2; // dot product of the 2 vectors
l2 = (dx1*dx1+dy1*dy1)*(dx2*dx2+dy2*dy2) // product of the squared lengths
angle = acos(d/sqrt(l2));
The dot product of 2 vectors is equal to the cosine of the angle time the length of both vectors. This computes the dot product, divides by the length of the vectors and uses the inverse cosine function to recover the angle.
Maybe my approach for Android coordinates system will be useful for someone (used Android PointF class to store points)
/**
* Calculate angle between two lines with two given points
*
* #param A1 First point first line
* #param A2 Second point first line
* #param B1 First point second line
* #param B2 Second point second line
* #return Angle between two lines in degrees
*/
public static float angleBetween2Lines(PointF A1, PointF A2, PointF B1, PointF B2) {
float angle1 = (float) Math.atan2(A2.y - A1.y, A1.x - A2.x);
float angle2 = (float) Math.atan2(B2.y - B1.y, B1.x - B2.x);
float calculatedAngle = (float) Math.toDegrees(angle1 - angle2);
if (calculatedAngle < 0) calculatedAngle += 360;
return calculatedAngle;
}
It return positive value in degrees for any quadrant: 0 <= x < 360
You can checkout my utility class here
The formula for getting the angle is tan a = (slope1-slope2)/(1+slope1*slope2)
You are using:
tan a = (slope1 - slope2) / (1 - slope1 * slope2)
So it should be:
double angle = Math.atan((slope1 - slope2) / (1 + slope1 * slope2));
First, are you sure the brackets are in the right order? I think (could be wrong) it should be this:
double slope1 = (line1.getY1() - line1.getY2()) / (line1.getX1() - line1.getX2());
double slope2 = (line2.getY1() - line2.getY2()) / (line2.getX1() - line2.getX2());
Second, there are two things you could do for the div by zero: you could catch the exception and handle it
double angle;
try
{
angle = Math.atan((slope1 - slope2) / (1 - (slope1 * slope2)));
catch (DivideByZeroException dbze)
{
//Do something about it!
}
...or you could check that your divisors are never zero before you attempt the operation.
if ((1 - (slope1 * slope2))==0)
{
return /*something meaningful to avoid the div by zero*/
}
else
{
double angle = Math.atan((slope1 - slope2) / (1 - (slope1 * slope2)));
return angle;
}
Check this Python code:
import math
def angle(x1,y1,x2,y2,x3,y3):
if (x1==x2==x3 or y1==y2==y3):
return 180
else:
dx1 = x2-x1
dy1 = y2-y1
dx2 = x3-x2
dy2 = y3-y2
if x1==x2:
a1=90
else:
m1=dy1/dx1
a1=math.degrees(math.atan(m1))
if x2==x3:
a2=90
else:
m2=dy2/dx2
a2=math.degrees(math.atan(m2))
angle = abs(a2-a1)
return angle
print angle(0,4,0,0,9,-6)
dx1=x2-x1 ; dy1=y2-y1 ; dx2=x4-x3 ;dy2=y4-y3.
Angle(L1,L2)=pi()/2*((1+sign(dx1))* (1-sign(dy1^2))-(1+sign(dx2))*(1-sign(dy2^2)))
+pi()/4*((2+sign(dx1))*sign(dy1)-(2+sign(dx2))*sign(dy2))
+sign(dx1*dy1)*atan((abs(dx1)-abs(dy1))/(abs(dx1)+abs(dy1)))
-sign(dx2*dy2)*atan((abs(dx2)-abs(dy2))/(abs(dx2)+abs(dy2)))