SQL select latitude and longitude point within bounding box - sql

I am using google map and I would like to query my SQL to find all point within the bounding box.
In google map, I am using this javascript to get the bounding rectangle
var bounds = map.getBounds();
var ne = bounds.getNorthEast(), sw = bounds.getSouthWest();
var args = {
NW: { lat: ne.lat(), lng: sw.lng() },
NE: { lat: ne.lat(), lng: ne.lng() },
SE: { lat: sw.lat(), lng: ne.lng() },
SW: { lat: sw.lat(), lng: sw.lng() },
}; //NW = North-West, NE = North-East, SE = South-East, SW = South-West
I am, then, using LINQ to select all places from my SQL database:
//nw = North-West, ne = North-East, se = South-East, sw = South-West
double minLat = Math.Min(nw.Lat, Math.Min(ne.Lat, Math.Min(se.Lat, sw.Lat)));
double maxLat = Math.Max(nw.Lat, Math.Max(ne.Lat, Math.Max(se.Lat, sw.Lat)));
double minLng = Math.Min(nw.Lng, Math.Min(ne.Lng, Math.Min(se.Lng, sw.Lng)));
double maxLng = Math.Max(nw.Lng, Math.Max(ne.Lng, Math.Max(se.Lng, sw.Lng)));
return (from rec in tblPlaces.AsNoTracking()
where (rec.Lat >= minLat) && (rec.Lat <= maxLat) && (rec.Lng >= minLng) && (rec.Lng <= maxLng)
select rec).ToList<tblPlace>();
It works well when it is on quite a zoom (google zoom <= 15). But when zooming out to country size (ie. you can see the whole country), it doesn't find the point in my database.
On the debugging, I found the longitude number is way smaller than any point in my database. How is it possible? I zoomed out to see the whole country.
Is the way I select the latitude and longitude is wrong?

I am not an expert in geography but this looks simple. Let us start with longitude. The bounding box could be on one side or across the antimeridian:
-180 0 +180
| |
| +-----+ |
| -10 | x | +10 |
| +-----+ |
| |
| +-----+
| +170 | x | -170
| +-----+
| |
A given longitude exists inside the bounding box if:
lng1 <= lng2 AND (lng1 <= lng AND lng <= lng2) /* both edges on same side */
OR
lng1 > lng2 AND (lng1 <= lng OR lng <= lng2) /* edges on opposite sides */
A given latitude exists inside the bounding box if:
lat1 >= lat2 AND (lat1 >= lat AND lat >= lat2) /* both edges on same side */
OR
lat1 < lat2 AND (lat1 >= lat OR lat >= lat2) /* edges on opposite sides */
If latitudes do not wrap around e.g. in Google Maps API then opposite sides test is not required.
Some tests on db<>fiddle and a jsFiddle showing how LatLngBounds work

I finally found the answer after looking at the tutorial of Latitude and Longitude.
In summary: Latitude is between -90 to 90 and Longitude is between -180 to 180
//Latitude
90 ---------------------- 90
0 ---------------------- 0
-90 ---------------------- -90
//Longitude
-180 0 180
| | |
| | |
| | |
| | |
-180 0 180
Now, the bounding box can overlap. So, the left can be greater than the right or the top can be greater than the bottom in the bounding box (rectangle). Based on How to search (predefined) locations (Latitude/Longitude) within a Rectangular, using Sql? (see the last answer), the solution is simply union all the combination depending on where the bounding box is.
If you can find a better and more efficient solution, I'll award the bounty to you :)
Cheers

Related

Calculate Heading with two GPS Devices

Hi I am planing to use two RTK-GPS devices to get the position and heading of a outdoor robot. The two devices are placed on the left and the right side of the robot with a distance of 2m. The heading should be a value between 0 and 2*PI. 0 if the robot is facing north.
I am trying the following:
double x1 = gps_left->latitude;
double y1 = gps_left->longitude;
double x2 = gps_right->latitude;
double y2 = gps_right->longitude;
double dx = x2-x1;
double dy = y2-y1;
// get the normal of the line
double nx = dy*-1;
double ny = dx;
double angle = atan2(ny ,nx);
somehow the angle always stays the same even if i rotate the robot
double x1 = gps_left->latitude;
double y1 = gps_left->longitude;
double x2 = gps_right->latitude;
double y2 = gps_right->longitude;
double dy = y2-y1;
double nx = cos(x2)*sin(dy);
double ny = cos(x1)*sin(dy) - sin(x1)*cos(x2)*cos(dy);
double angle = atan2(nx,ny);
Can you try this?
0 - heading north,
90 - heading east,
180 - heading south,
270 - heading west.
Like #Mark Booth the resolution might not be enough. We had a similar issue, so we used a magnetometer to fix this. Using this heading information from a magnetometer and GPS we were able to navigate successfully.

What are these GPS data mean, and how to correctly convert them into decimal degrees?

I have a set of household GPS coordinates data, and the format in excel sheet looks like the following (edited for confidential reason):
ID GPSN GPSS GPSE GPSW
1 211234 -9 890123 -9
2 211255 -9 890155 -9
...
My questions are: what kind of GPS coordinates this is (looks like UTM data)? How do I accurately convert them into decimal degrees that only containing a longitude and a latitude (or X, Y data)? Do I need some kind of zoning information to do this correctly? Thanks
I doubt that a GPS receiver would put out UTM coordinates. It looks to me like latitude and longitude in degrees/minutes/seconds (DDMMSS). If so, then one way to do it is the following, in simple Python. The Convert Coordinate Notation tool in ArcGIS might be useful, but you'll have to reformat the data first, probably using Python.
import csv
import sys
# A function that takes 211234, treats it as 21°12'34",
# and returns 21.209444.
def convertToDegrees(DMS):
dms = DMS
dms = int(dms)
seconds = dms % 100
if 60 <= seconds:
print "More than 60 seconds! " + str(DMS) + " is not degrees/minutes/seconds!"
dms /= 100
minutes = dms % 100
if 60 <= minutes:
print "More than 60 minutes! " + str(DMS) + " is not degrees/minutes/seconds!"
dms -= minutes
degrees = dms / 100
degrees += (minutes / 60.0)
degrees += (seconds / (60.0 * 60.0))
if 180 < degrees or -180 > degrees:
print "In " + str(DMS) + ", degrees is outside [-180, 180]: " + str(degrees)
return degrees
# Input and output files from command line parameters
inFilename = sys.argv[1]
outFilename = sys.argv[2]
readFirstRow = False
with open(inFilename, "rb") as inFile:
reader = csv.reader(inFile)
with open(outFilename, "wb") as outFile:
writer = csv.writer(outFile)
# Loop through the rows
for row in reader:
if (not readFirstRow):
# Write the header row only once
writer.writerow(["ID", "latitude", "longitude"])
readFirstRow = True
else:
# Convert this row to latitude and longitude
latitude = 0
longitude = 0
if "-9" != row[1]:
latitude = convertToDegrees(row[1])
if "-9" != row[2]:
latitude = -1 * convertToDegrees(row[2])
if "-9" != row[3]:
longitude = convertToDegrees(row[3])
if "-9" != row[4]:
longitude = -1 * convertToDegrees(row[4])
writer.writerow([row[0], latitude, longitude])
To make sure you get it right, you'll want to confirm that the GPS was putting out latitude and longitude and find out which datum it used (probably WGS 1984).

"Get 100 meters out from" Haversin Formula

I'm interested in working with coordinates, and I was wondering, how is it possible to get the distance between two points (coordinates) in meters. After a long search I found the Haversine Formula, and the Objective-C implementation of it here.
It's the following (I modified it a little bit for myself):
- (CGFloat)directMetersFromCoordinate:(CLLocation *)from toCoordinate:(CLLocation *)to {
static const double DEG_TO_RAD = 0.017453292519943295769236907684886;
static const double EARTH_RADIUS_IN_METERS = 6372797.560856;
double latitudeArc = (from.coordinate.latitude - to.coordinate.latitude) * DEG_TO_RAD;
double longitudeArc = (from.coordinate.longitude - to.coordinate.longitude) * DEG_TO_RAD;
double latitudeH = sin(latitudeArc * 0.5);
latitudeH *= latitudeH;
double lontitudeH = sin(longitudeArc * 0.5);
lontitudeH *= lontitudeH;
double tmp = cos(from.coordinate.latitude*DEG_TO_RAD) * cos(to.coordinate.latitude*DEG_TO_RAD);
return EARTH_RADIUS_IN_METERS * 2.0 * asin(sqrt(latitudeH + tmp*lontitudeH)); }
My question is:
How is it possible to get 100 meters distance (for latitude and for longitude) for the current location?
This formula is too complicated for me, I don't understand it, so I can't "code it back" to get the result what I want.
I need only the actual location (Paris, Tokyo, London, New York, whatever), and a (float) number for latitude and a (float) number for longitude, which (float) number represents 100 meters distance from the actual location.
If you open this page, here you can "calculate out" the 100 meters distance between two points ("actual" and the one 100 meters away).
For example:
Point1: 47.0, 19.0
Point2: 47.0, 19.0013190
---That's 0.1000 km (100 m) distance.
Point1: 47.0, 19.0
Point2: 47.0008995, 19.0
---is also 0.1000 km (100 m) distance.
Here you can see, that at that coordinate (latitude 47.0 and longitude 19.0) 100 meters distance is 0.0008995 (latitude) and 0.0013190 (longitude).
And I want to get these data with the help of the Haversine Formula, just don't know, how.
Could you help me to figure it out?
Thanks!
UPDATE:
Thank you for the answers, right now I don't have time to try them out, but as I understood, I didn't explain exactly what I want.
Maybe this is a better example, how I want to use these "100 meters":
So, right now I'm at the coordinate "lat x" and "lon y". There is another point (let's say a hotel) at another given coordinate, "lat a" and "lon b".
My question is: how can I calculate out, if this hotel is (less than) 100 meters from me? So it doesn't matter, if it's only 5 meters or 99 meters, both of them are less (or equal) than 100 meters far from me.
With the code what I provided, I can calculate this out, that's what that formula is for.
But let's say, I have a million of other coordinates (hotel locations) that I want to work with. And I need only a list of them, what are (less than) 100 meters away from me. So yes, that's a circle around me with a radius of 100 meters, and I need the results within that.
It would take much more time and resource to take all the coordinates of these "million" hotels and calculate the distance one by one, that's why I thought it would be much easier to calculate out, how much 100 meters are in latitude and longitude (changes the value as we are on different locations, that' why I can't use simply the ones what I calculated out in the example above).
So if I would know how much 100 meters are in latitude and longitude for example at London's coordinate (if I'm there), I could simply get the list of the hotels what are (less than) 100 meters far from me, by a simple dividing:
if
((hotelLocation.coordinate.latitude <= (myLocation.coordinate.latitude + "100metersInLatitude")) || (hotelLocation.coordinate.latitude >= (myLocation.coordinate.latitude - "100metersInLatitude")))
&&
((hotelLocation.coordinate.longitude <= (myLocation.coordinate.longitude + "100metersInLongitude")) || (hotelLocation.coordinate.longitude >= (myLocation.coordinate.longitude - "100metersInLongitude")))
{
NSLog(#"Be Happy :-) !");
}
I just need these "100metersInLatitude" and "100metersInLongitude", calculated always from "myLocation".
Whoa, I hope, somebody will understand what I just wrote down, because it's not easy for me, neither... :-)))
Assuming you have a point with latitude and longitude, and you want to find another point that is a distance d on a bearing b, when the distance is small (you said "100 meters" which is very small on the surface of the earth) then you can do a simple approximation - treating the earth's surface locally as "flat". Here is a simple C program that implements this (using the numbers you had above). I updated it to include the "accurate" formulation as well - it's just a few more calculations, but it is accurate at all distances (and not just the short ones). The equation I used came from the link you referenced - subheading "Destination point given distance and bearing from start point"
updated - I moved the accurate calculation into a separate function, and added a loop to compute the new point for all integer bearings from 0 to 359, printing out every 30th. This give you the "circle" I talked about in my initial comment.
#include <stdio.h>
#include <math.h>
double radians(double x) {
return acos(0.0) * x / 90.0;
}
void calcNewPosition(double lat, double lon, double bearing, double d, double *newLat, double *newLon) {
double lat1, lon1, br, pi;
double Re = 6371000;
// convert everything to radians first:
lat1 = radians(lat);
lon1 = radians(lon);
br = radians(bearing);
pi = 2 * acos(0.0);
double lat2, lon2;
lat2 = asin( sin(lat1) * cos(d/Re) +
cos( lat1 ) * sin( d / Re ) * cos(br ) );
lon2 = lon1 + atan2(sin(br) * sin( d / Re ) * cos( lat1 ), \
cos( d / Re ) - sin(lat1 ) * sin( lat2 ) );
*newLat = 180. * lat2 / pi;
*newLon = 180. * lon2 / pi;
}
int main(void) {
double lon = 47., lat=19.;
double newLongitude, newLatitude;
double dx, dy, dLong, dLat;
double Re = 6371000, d = 100, bearing = 0.0;
double pi;
double lat1, lon1, br;
// convert everything to radians first:
lat1 = radians(lat);
lon1 = radians(lon);
br = radians(bearing);
pi = 2 * acos(0.0);
// approximate calculation - using equirectangular approximation
// and noting that distance between meridians (lines of longitude)
// get closer at higher latitudes, with cos(latitude).
dx = d * sin(br); // distance in E-W direction
dy = d * cos(br); // distance in N-S direction
dLat = 360 * dy / (2.0 * pi * Re); // convert N-S to degrees latitude
dLong = 360 * dx / (2.0 * pi * Re * cos(lat1)); // convert E-W to degrees longitude
newLatitude = lat + dLat;
newLongitude = lon + dLong;
printf("simple forumula: the new position is %.8lf lon, %.8lf lat\n", newLongitude, newLatitude);
// more accurate formula: based on http://www.movable-type.co.uk/scripts/latlong.html
double lat2, lon2;
calcNewPosition(lat, lon, bearing, d, &lat2, &lon2);
printf("more accurate: the new position is %.8lf lon, %.8lf lat\n", lon2, lat2);
// now loop over all bearings and compute the "circle of points":
int iBearing;
double lonArray[360], latArray[360];
for(iBearing = 0; iBearing < 360; iBearing++) {
calcNewPosition(lat, lon, (double)iBearing, d, &latArray[iBearing], &lonArray[iBearing]);
if (iBearing % 30 == 0) printf("bearing %03d: new lat = %.8lf, new lon = %.8lf\n", iBearing, latArray[iBearing], lonArray[iBearing]);
}
return 0;
}
The output of this is
simple forumula: the new position is 47.00000000 lon, 19.00089932 lat
more accurate: the new position is 47.00000000 lon, 19.00089932 lat
bearing 000: new lat = 19.00089932, new lon = 47.00000000
bearing 030: new lat = 19.00077883, new lon = 47.00047557
bearing 060: new lat = 19.00044966, new lon = 47.00082371
bearing 090: new lat = 19.00000000, new lon = 47.00095114
bearing 120: new lat = 18.99955034, new lon = 47.00082371
bearing 150: new lat = 18.99922116, new lon = 47.00047557
bearing 180: new lat = 18.99910068, new lon = 47.00000000
bearing 210: new lat = 18.99922116, new lon = 46.99952443
bearing 240: new lat = 18.99955034, new lon = 46.99917629
bearing 270: new lat = 19.00000000, new lon = 46.99904886
bearing 300: new lat = 19.00044966, new lon = 46.99917629
bearing 330: new lat = 19.00077883, new lon = 46.99952443
As you can see, it is accurate to within a fraction of a meter (your code gave 19.0008995 - it is actually possible that your result was "wrong" in the last digit as these two methods agree to 8 significant digits even though they use different equations).
The question isn't really answerable if the OP wants a location a distance (100 meters) from the current location without a desired bearing being provided, there being an infinite number of points in a circle around the point.
So, this answer may or may not be what the OP wants, it is a way with CLLocation to calculate the distance between two points.
Create two CLLocation point and use the method Have you looked at theCLLocationmethod- (CLLocationDistance)distanceFromLocation:(const CLLocation *)location`.
CLLocation *location1 = [[CLLocation alloc] initWithLatitude:)latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:)latitude2 longitude:longitude2];
double distance = [location1 distanceFromLocation:location2];
The radius of the earth at equator = 6,371 km. The equator is divided into 360 degrees of longitude, so each degree at the equator represents approximately 111.32 km. Moving away from the equator towards a pole this distance decreases to zero at the pole.To calculate the distance at different latitudes multiply it by the cosine of the latitude
3 decimal places,0.001 degrees aproximates to
111.32 meters at equator
96.41meters at 30 degrees N/S
78.71 meters at 45 degrees N/S
55.66 meters at 60 degrees N/S
28.82 meters at 75 degrees N/S
For for small distances (100 meters) Pythagoras’ theorem can be used on an equirectangular projection to calculate distance. This is less complicated than Haversine or Spherical Law of Cosines.
var R = 6371; // km
lat/lng in radians
In pseudo code as I don't know Objective-C
var x = (lng2-lng1) * cos((lat1+lat2)/2);
var y = (lat2-lat1);
var d = sqrt(x*x + y*y) * R;

Minimum distance between a point and a line in latitude, longitude

I have a line with two points in latitude and longitude
A: 3.222895, 101.719751
B: 3.227511, 101.724318
and 1 point
C: 3.224972, 101.722932
How can I calculate minimum distance between point C and a line consists of point A and B?
It will be convenient if you can provide the calculation and objective-c code too. The distance is around 89 meters (using ruler in Google Earth).
Thanks to mimi and this great article http://www.movable-type.co.uk/scripts/latlong.html but they don't give the whole picture. Here is a detail one. All this points are collected using Google Earth using Placemark to mark the locations. Make sure lat/long are set to decimal degrees in Preferences.
lat A = 3.222895
lon A = 101.719751
lat B = 3.222895
lon B = 101.719751
lat C = 3.224972
lon C = 101.722932
Earth radius, R = 6371
1. First you have to find the bearing from A to C and A to B.
Bearing formula
bearingAC = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) )
bearingAB = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) )
φ is latitude, λ is longitude, R is earth radius
2. Find A to C distance using spherical law of cosines
distanceAC = acos( sin(φ₁)*sin(φ₂) + cos(φ₁)*cos(φ₂)*cos(Δλ) )*R
3. Find cross-track distance
distance = asin(sin(distanceAC/ R) * sin(bearingAC − bearing AB)) * R
Objective-C code
double lat1 = 3.227511;
double lon1 = 101.724318;
double lat2 = 3.222895;
double lon2 = 101.719751;
double lat3 = 3.224972;
double lon3 = 101.722932;
double y = sin(lon3 - lon1) * cos(lat3);
double x = cos(lat1) * sin(lat3) - sin(lat1) * cos(lat3) * cos(lat3 - lat1);
double bearing1 = radiansToDegrees(atan2(y, x));
bearing1 = 360 - ((bearing1 + 360) % 360);
double y2 = sin(lon2 - lon1) * cos(lat2);
double x2 = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(lat2 - lat1);
double bearing2 = radiansToDegrees(atan2(y2, x2));
bearing2 = 360 - ((bearing2 + 360) % 360);
double lat1Rads = degreesToRadians(lat1);
double lat3Rads = degreesToRadians(lat3);
double dLon = degreesToRadians(lon3 - lon1);
double distanceAC = acos(sin(lat1Rads) * sin(lat3Rads)+cos(lat1Rads)*cos(lat3Rads)*cos(dLon)) * 6371;
double min_distance = fabs(asin(sin(distanceAC/6371)*sin(degreesToRadians(bearing1)-degreesToRadians(bearing2))) * 6371);
NSLog(#"bearing 1: %g", bearing1);
NSLog(#"bearing 2: %g", bearing2);
NSLog(#"distance AC: %g", distanceAC);
NSLog(#"min distance: %g", min_distance);
Actually there's a library for this. You can find it here https://github.com/100grams/CoreLocationUtils
Calculate bearing for each: C to A , and C to B:
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1)*Math.sin(lat2) -
Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
var brng = Math.atan2(y, x).toDeg();
dLon= lon2-lon1;
Calculate cross-track distance:
var dXt = Math.asin(Math.sin(distance_CB/R)*Math.sin(bearing_CA-bearing_CB)) * R;
R is the radius of earth, dXt is the minimum distance you wanted to calculate.
Code to carry out this calculation is posted at here.
This implements an accurate solution in terms of ellipsoidal geodesics.
For the basic geodesic calculations, you can use
GeographicLib or the port of these algorithms to C which are included in version 4.9.0 of PROJ.4. This C interface is documented here.
Here's the result of compiling and running intercept.cpp:
$ echo 3.222895 101.719751 3.227511 101.724318 3.224972 101.722932 | ./intercept
Initial guess 3.225203 101.7220345
Increment 0.0003349040566247297 0.0003313413822354505
Increment -4.440892098500626e-16 0
Increment 0 0
...
Final result 3.225537904056624 101.7223658413822
Azimuth to A1 -135.1593040635131
Azimuth to A2 44.84069593652217
Azimuth to B1 134.8406959363608
Distance to line is 88.743m:
$ echo 3.224972 101.722932 3.225537904056624 101.7223658413822 | GeodSolve -i
-45.15927221 -45.15930407 88.743
See post here:
https://stackoverflow.com/a/33343505/4083623
For distance up to a few thousands meters I would simplify the issue from sphere to plane.
Then, the issue is pretty simply as a easy triangle calculation can be used:
We have points A and B and look for a distance X to line AB. Then:
Location a;
Location b;
Location x;
double ax = a.distanceTo(x);
double alfa = (Math.abs(a.bearingTo(b) - a.bearingTo(x))) / 180
* Math.PI;
double distance = Math.sin(alfa) * ax;
If you know how to calculate the distance of two points, get the distances between each two points, you get AB, AC, and BC. You want to know the closest distance between point C and line AB.
First get the value of P
P=(AB+BC+AC)/2
Using P, you need to get S
S=SQRT((P(P-AC)(P-AB)(P-AC))
SQRT means square root. Then you get what you want by
2*S/AB

How do I display a full 360 degrees using Actionscript 2 and Trigonometry?

I'm creating a game that uses trigonometry to calculate and display distance and degrees in dynamic text boxes. I'm calculating the distance of my cursor from center of a movie clip. And using that center of the movie clip, I'm trying to calculate and display a full 360º as my cursor moves around the swf. I have the distance part of the game working but the part that displays degrees is not working properly. The dynamic text box only display from 90º thru 270º. Instead of going past 270º to 360º/0º to 90º, it just counts back down from 270º to 90º. Below is my actionscript. I'd greatly appreciate any help or suggestions. Thanks!
//Mouse and Dynamic Text Boxes-------------------------
Mouse.hide();
onMouseMove = function () {
feedback.text = "You are moving your mouse";
cursor._x = _xmouse;
cursor._y = _ymouse;
updateAfterEvent();
xmouse_value.text = Math.atan2((a), (b));
ymouse_value.text = Math.round(radians*180/Math.PI)
updateAfterEvent();
};
Mouse.addListener(myListener);
//distance (RANGE)
_root.onEnterFrame = function () {
xmid = Stage.width/2;
ymid = Stage.height/2;
a = _root._ymouse-ymid;
b = _root._xmouse-xmid;
c = Math.sqrt(Math.pow(a, 2)+Math.pow(b, 2));
feedbacka.text = Math.round(a);
feedbackb.text = Math.round(b);
feedbackc.text = Math.round(c/30.4);
updateAfterEvent();
var radians:Number;
var degrees:Number;
//Calculcate Radians
//Radians specify an angle by measuring the length around the path of the circle.
radians = Math.atan2((c), (b))
//calculate degrees
//the angle the circle is in relation to the center point
//update text box inside circle
radians_txt = Math.round(radians*360/Math.PI);
degrees_txt = Math.round(radians*180/Math.PI);
updateAfterEvent();
//getting past 270 degrees
radians2_txt = Math.round(radians/Math.PI);
radians2_txt = Math.floor(radians + -270);
}
The parameters to atan2 should be the delta-y and delta-x between the two points, but you are passing the distance between the two points and the delta-x. Try this instead:
radians = Math.atan2(a, b);
The next problem is to convert the radians into degrees. To convert radians to degrees, you can do this:
degrees_txt = radians * 180 / Math.PI;
Note that atan2 returns from between -Math.PI / 2 to Math.PI / 2. When converted to degrees, this range becomes -180 to 180. To convert to 0 to 360, you can add 360 to the result if it is negative:
if(degrees_txt < 0) degrees_txt += 360;