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]
?
So i'm trying to make a simple calculator to calculate a wilks score. I'm not getting the correct answer which i'm sure is because i'm not setting up the formula correctly. If I pass in a weight of 180, squat of 300, bench of 300, and deadlift of 400 in lbs I should be getting a wilks of 305.78 but i'm getting 2.0414858^-5
Heres my calcWilks method
public double calcWilks(double weight, double squat, double bench, double deadLift) {
double a = -216.0475144;
double b = 16.2606339;
double c = -0.002388645;
double d = -0.00113732;
double e = Math.pow(7.01863, -6);
double f = Math.pow(-1.291, -8);
double x = weight;
double coeff;
double score;
coeff = 500 / (a + (b*x) + (c* Math.pow(x, 2))+ (d* Math.pow(x, 3))
+ (e* Math.pow(x, 4)) + (f* Math.pow(x, 5)));
double total = squat + bench + deadLift;
score = coeff* total;
return score;
}
and heres a link to the actual formaula https://en.wikipedia.org/wiki/Wilks_Coefficient
I'm trying to use the Male formula
Thanks for any help!
The e coefficient's value in the wikipedia page is
7.01863E-06
but you've used
Math.pow(7.01863, -6)
That's not the same thing:
7.01863E-06 = 7.01863 * Math.pow(10, -6).
Just use the value 7.01863E-06 (or 7.01863e-6) directly.
(Same problem with f)
Also, note that the inputs to the formula should be in kilograms; not in pounds, as you state in the question.
I need the size of the black part of this image:
I've done some research about how to find it in normal math, and I was pointed to this website: Website
The final answer on getting it was
(from MathWorld - A Wolfram Web Resource: wolfram.com)
where r is the radius of the first circle, R the radius of the second circle, and d the distance between the two centers.
The code I tried to use to get the size of this was the following:
float r = getRadius1();
float R = e.getRadius1();
float deltaX = Math.abs((getX() + getRadius()) - (e.getX() + e.getRadius()));
float deltaY = Math.abs((getY() + getRadius()) - (e.getY() + e.getRadius()));
float d = (float) Math.sqrt(Math.pow(deltaX, 2) + Math.pow(deltaY, 2));
float part, part2, part3;
//Chopping it in parts, because it's easier.
part = (float) (Math.pow(r,2) * Math.acos(
Math.toRadians((Math.pow(d, 2) + Math.pow(r, 2) - Math.pow(R, 2))/(2*d*r))));
part2 = (float) (Math.pow(R,2) * Math.acos(
Math.toRadians((Math.pow(d, 2) + Math.pow(R, 2) - Math.pow(r, 2))/(2*d*R))));
part3 = (float) (0.5 * Math.sqrt((-d + r + R) * (d+r-R) * (d-r+R) * (d+r+R)));
float res = part + part2 - part3;
Main.log(res + " " + part + " " + part2 + " " + part3+ " "
+ r + " " + R + " " + d);
//logs the data and System.out's it
I did some testing, and the output was this:
1345.9663 621.6233 971.1231 246.78008 20.0 25.0 43.528286
So that indicates that the size of the overlapping part was bigger than the circle itself (which is r^2 * PI).
What did I do wrong?
Just a guess (as stated in my comment): try removing the Math.toRadians(...) conversion.
Since there are no degrees involved in the formula but rather radii, I assume the parameter to cos-1(...) is already a value in radians.
If I remove the conversion and run your code, I get the following overlap area size: 11.163887023925781 which seems plausible since the length of the overlap segment on the line between the two centers is 20 + 25 - 43.5 = 1.5 (approximated)
Edit:
If I set the distance to 5 (the smaller circle is completely contained in the bigger one but touches its edge) I get the overlap area size 1256.63 which is exactly the area of the smaller circle (202 * Π). The calculation doesn't seem to work if the distance is smaller than the difference of the radii (i.e. in your case smaller than 5), but that might just be a problem of numerical representation (the normal datatypes might not be able to represent some of the intermediate results).
I'm learning java and have a bit of code I am trying to write that should round 24.9999999 to 25. Instead, it goes to 8.
import java.io.*;
import java.util.*;
public class RadiusOfCircle
{
public static void main(String args[])
{
Scanner kbInput = new Scanner(System.in);
System.out.print("What is the area? _");
double area = kbInput.nextInt();
System.out.println("Radius of your circle is " + Math.sqrt( area / Math.PI));
double radius = Math.sqrt( area / Math.PI);
System.out.println("\nChecking your work now...\n area = pi*(r^2)\n " + area + " = 3.14 * (" + radius + ")^2");
double radiusSqrd = Math.pow(radius, 2);
System.out.println(" " + area + " = 3.14 * " + radiusSqrd);
System.out.println(" " + area + " = " + Math.PI * radiusSqrd);
System.out.println(area + " = " + (Math.round(radiusSqrd)));
System.out.println("Are the two values the same? \nIf yes, your code is correct! \nIf no, try again!");
}
}
Also, when it asks for keyboard input of what the area is, I put in 25.
This is the output:
What is the area? _25
Radius of your circle is 2.8209479177387813
Checking your work now...
area = pi*(r^2)
25.0 = 3.14 * (2.8209479177387813)^2
25.0 = 3.14 * 7.957747154594766
25.0 = 24.999999999999996
25.0 = 8
Are the two values the same?
If yes, your code is correct!
If no, try again!
You're rounding radiusSquared only, rather than Math.PI * radiusSquared. Fixing that should get the result you expect.
System.out.println(" " + area + " = " + Math.PI * radiusSqrd);
System.out.println(area + " = " + (Math.round(radiusSqrd)));
Shouldn't that be:
double value = Math.PI * radiusSqrd;
System.out.println(" " + area + " = " +value );
System.out.println(area + " = " + (Math.round(value )));
You omitted to multiply by PI:
System.out.println(area + " = " + Math.round(Math.PI * radiusSqrd));
Executing this gives the expected result.
float and double were designed for engineering problems, which have large positive powers of 10.. they cannot express negative powers of 10 accurately. In such cases use BigDecimal.
Run this simple code from Joshua Bloch's book Effective Java to get a sense of the extend of inaccuracy when dealing with negative powers and using double to store them. The answer should acutally be zero, but turns out to be something else entirely!
double funds = 1.00;
int itemsBought = 0;
for (double price = .10; funds >= price; price += .10) {
funds -= price;
itemsBought++;
}
System.out.println(itemsBought + ” items bought.”);
System.out.println(“Change: $” + funds);
}
If you want "accurate" math in Java, you should use the BigDecimal class, rather than either of the built-in floating point primitive types. double and float are always going to have issues like this, due to the nature of floating point arithmetic.
Please note that the recommended constructor for BigDecimal uses a String, not any of the numeric types. Since you are getting input from the console, this should be easy to implement.
You can use something like this:
double d = 10.938;
BigDecimal bd = new BigDecimal(d);
bd = bd.setScale(0,BigDecimal.ROUND_HALF_UP);
System.out.println(bd);
bd = new BigDecimal(d);
bd = bd.setScale(1,BigDecimal.ROUND_HALF_UP);
System.out.println(bd);
bd = new BigDecimal(d);
bd = bd.setScale(2,BigDecimal.ROUND_HALF_UP);
System.out.println(bd);
or like this:
double d = 10.938;
DecimalFormat decimalFormat = new DecimalFormat("#");
System.out.println(decimalFormat.format(d));
decimalFormat = new DecimalFormat("#.#");
System.out.println(decimalFormat.format(d));
decimalFormat = new DecimalFormat("#.##");
System.out.println(decimalFormat.format(d));