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]
?
I create 60 DataPoint (for every minute in hour) and show them on the graph. Set 7 labels for x axis. Labels with date.
The leftmost label and the rightmost label do not coincide with the beginning and end of the x axis. This screenshot shows mismatches:
Code:
private void updateGraph(){
DataPoint[] dataPoints = new DataPoint[mCurrencyStampList.size()];
int i = 0;
Double minY = null;
Double maxY = 0D;
for(CurrencyStamp stamp : mCurrencyStampList){
dataPoints[i] = new DataPoint(stamp.getDate(), stamp.getClose());
if(maxY < stamp.getClose()){
maxY = stamp.getClose().doubleValue();
}
if(minY == null || minY > stamp.getClose()){
minY = stamp.getClose().doubleValue();
}
i++;
}
LineGraphSeries<DataPoint> points = new LineGraphSeries<>(dataPoints);
mGraphView.addSeries(points);
DateFormat dateFormat = android.text.format.DateFormat.getTimeFormat(getContext());
mGraphView.getGridLabelRenderer().setLabelFormatter(new DateAsXAxisLabelFormatter(getActivity(), dateFormat));
mGraphView.getGridLabelRenderer().setTextSize(32);
mGraphView.getGridLabelRenderer().setNumHorizontalLabels(7);
double minX = mCurrencyStampList.get(0).getDate().getTime();
double maxX = mCurrencyStampList.get(mCurrencyStampList.size()-1).getDate().getTime();
mGraphView.getViewport().setMinimalViewport(minX, maxX, minY == null ? 0 : minY, maxY);
mGraphView.getViewport().setXAxisBoundsManual(true);
mGraphView.getViewport().setYAxisBoundsManual(true);
}
The divisions of the graph do not coincide in time. How to fix this?
upd:
I think I understand why the extreme lines are moving to centre. unix timestamp lose precision when convertion to double. i don't know how fix this. Same problem on Mpandroidchart library. I try trial version of AnyChart, nice worked for me (in constructor of points used own format. not float and double) but this trial
I migrate to MPAndroid and applied coefficient. Just set the minimum and maximum along the x axis:
xAxis.setAxisMinimum(0);
xAxis.setAxisMaximum(100000);
Create coefficient:
long xMin = mCurrencyStampList.get(0).getDate().getTime();
long xMax = mCurrencyStampList.get(mCurrencyStampList.size() - 1).getDate().getTime();
float xCoefficient = (xMax - xMin) / 100000;
Multiple x value to different:
long currentTS = s.getDate().getTime();
Float x = (currentTS - xMin) / xCoefficient;
And in the widget display the return value:
XAxis xAxis = mLineChart.getXAxis();
xAxis.setValueFormatter((value, axis) -> {
Long time = (long) (value * xCoefficient + xMin);
Date date = new Date(time);
DateFormat format;
if(mInterval == Interval.OneHour || mInterval == Interval.SixHours || mInterval == Interval.OneDay){
format = android.text.format.DateFormat.getTimeFormat(getContext());
}
else {
format = android.text.format.DateFormat.getDateFormat(getContext());
}
return format.format(date);
});
not a perfect solution but works for unix timestamp
I currently have a PID algorithm to control my robots turns in an autonomous state. My robot has encoders, on each motor, which there are four of, and also a BNO055IMU. Furthermore each motor is a never rest 40 motor from Andymark, and unfortunately I am stuck with encoders that do 3 pulses. I would like to improve the accuracy of my turns either by using a different algorithm or improving my current one.
My Current Turning Code:
public void turn(int angle, Direction DIRECTION, double timeOut, int sleepTime, double kp, double ki, double kd) {
double targetAngle = imu.adjustAngle(imu.getHeading() + (DIRECTION.value * angle));
double acceptableError = 0.5;
double currentError = 1;
double prevError = 0;
double integral = 0;
double newPower;
double previousTime = 0;
timeoutClock.reset();
while (opModeIsActive() && (imu.adjustAngle(Math.abs(currentError)) > acceptableError)
&& !timeoutClock.elapsedTime(timeOut, MasqClock.Resolution.SECONDS)) {
double tChange = System.nanoTime() - previousTime;
previousTime = System.nanoTime();
tChange = tChange / 1e9;
double imuVAL = imu.getHeading();
currentError = imu.adjustAngle(targetAngle - imuVAL);
integral += currentError * ID;
double errorkp = currentError * kp;
double integralki = integral * ki * tChange;
double dervitive = (currentError - prevError) / tChange;
double dervitivekd = dervitive * kd;
newPower = (errorkp + integralki + dervitivekd);
newPower *= color;
if (Math.abs(newPower) > 1.0) {newPower /= newPower;}
driveTrain.setPower(newPower, -newPower);
prevError = currentError;
DashBoard.getDash().create("TargetAngle", targetAngle);
DashBoard.getDash().create("Heading", imuVAL);
DashBoard.getDash().create("AngleLeftToCover", currentError);
DashBoard.getDash().update();
}
driveTrain.setPower(0,0);
sleep(sleepTime);
}
NOTES:
when driveTrain.setPower(x,y); is called the left parameter is the power set to the left side and the right parameter sets the right side.
Direction is an enum that stores wither -1, or 1 to switch between left and right turns.
Dashboard.getDash.create is solely to keep a log on what is going on.
imu.adjustAngle does the following:
public double adjustAngle(double angle) {
while (angle > 180) angle -= 360;
while (angle <= -180) angle += 360;
return angle;
}
imu.getHeading() is self explanatory it gets the yaw of the robot.
My current values for pid constants. (They work pretty well.)
KP_TURN = 0.005,
KI_TURN = 0.0002,
KD_TURN = 0,
ID = 1;
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 have a data values which vary from +PI to -PI radian.
I need to get minimum rotation (in radians) required to go from old value to new value like:
float rotationRequired(float oldValue, float newValue){
return newValue - oldValue;
}
but simply subtracting won't do, because to go from -179 degree to +179 degree one does not need to rotate full circle around, just 2 degree clockwise. Because -PI = +PI in a circle are technically same rotation. Also the values can be in any range, i.e 740 = 360 + 360 + 20, hence only 20.
I'm thinking about breaking values into sin and cos values, subtract and then atan :
double oldY = Math.sin(oldValue);
double oldX = Math.cos(oldValue);
double newY = Math.sin(newValue);
double newX = Math.cos(newValue);
float delta = (float) Math.atan2( (newY - oldY),(newX - oldX) );
But still its not giving correct results, can anyone suggest another method ?
Just do the subtraction, and then limit the result to +/-180 by adding or subtracting 360 as necessary (the % operator may help here...)
I converted angles to degrees and used this method to suggest what minimum rotation is required and in what direction:
public static int suggestRotation(int o, int n){
//--convert to +0 to +360 range--
o = normalize(o);
n = normalize(n);
//-- required angle change --
int d1 = n - o;
//---other (360 - abs d1 ) angle change in reverse (opp to d1) direction--
int d2 = d1 == 0 ? 0 : Math.abs(360 - Math.abs(d1))*(d1/Math.abs(d1))*-1;
//--give whichever has minimum rotation--
if(Math.abs(d1) < Math.abs(d2)){
return d1;
}else {
return d2;
}
}
private static int normalize(int i){
//--find effective angle--
int d = Math.abs(i) % 360;
if(i < 0){
//--return positive equivalent--
return 360 - d;
}else {
return d;
}
}